1. Concepts in quantum mechanics - Bartholomew Andrews · 2020-01-09 · 1. Concepts in quantum...

1. Concepts in quantum mechanics

This is mainly a review of ideas that you’ve already encountered in the 2nd year Quantum Mechanics and Applications of Quantum Mechan-

ics courses. These ideas are important for the rest of the course.

1.1. Eigenvalues and eigenfunctions

We wish to find the quantum state, Ψ, of an atom or a molecule. The internal interactions of the atom or molecule (i.e. the electric and

magnetic interactions between all the particles that make up the system) are described by a Hamiltonian H`

, and the state of the system is

given by the Schrödinger equation

(1.1)ä Ñ¶ Ψ

¶ t

= H`

Ψ.

There will be a set of states, Ψn = Φn ã-i Ωn t, that are special. They are special because their entire time-dependence is contained in the factor

ã-ä Ωn t. The functions Φn do not depend on time and they are called the eigenfunctions of the system. We often refer to them as stationary

states to emphasize the fact that there is no time-dependence other than the ã-ä Ωn t factor - the probability distribution is unchanging in time.

If we substitute one of our special states - Ψn = Φn ã-i Ωn t - into the Schrödinger equation we obtain

ä Ñ H-ä ΩnL Φn ã-ä Ωn t

= H`

Φn ã-ä Ωn t.

The ã-ä Ωn t factor cancels out and we are left with an equation which is independent of time:

(1.2)H`

Φn = En Φn

or in Dirac notation

(1.3)H`

n] = En n]This is the time-independent Schrödinger equation. We have introduced the quantity En = Ñ Ωn, which is the energy eigenvalue correspond-

ing to the eigenfunction Φn. Our first aim is to find these eigenvalues and eigenfunctions.

In the case of the hydrogen atom, we can solve the time-independent Schrödinger equation directly to find the eigenvalues and eigenfunctions.

This will be our starting point, and it forms the foundation for the whole of atomic and molecular physics. For atoms or molecules with more

than one electron, the Schrödinger equation cannot be solved exactly. Nevertheless, it is possible to calculate the energy levels to a high

degree of precision using approximate methods.

1.2. Degeneracy, commuting operators and simultaneous eigenfunctions

When two or more eigenfunctions have the same eigenvalue, they are degenerate. We encounter degeneracy all of the time in atomic and

molecular physics. Suppose we have two states - we’ll label them 1\ and 2\ for now - that are both eigenstates of our Hamiltonian with

the same energy: H`

1] = En 1], H` 2] = En 2]. The labelling (1 and 2) is arbitrary and not helpful in telling us about the properties of the

states. What’s more, when 1\ and 2\ are both eigenstates with the same eigenvalue, so too is any linear combination of them:

H` HΑ 1] + Β 2]M = EnHΑ 1\ + Β 2]M, for any arbitrary Α and Β. So we have an infinity of possible eigenfunctions, none apparently any more

special than any other.

What we like to do when we have such a degeneracy is to find a physically meaningful operator which commutes with the Hamiltonian - let’s

just call it X`

for now - AX`

, H` E = 0. Because X

` commutes with the Hamiltonian we can find states that are simultaneous eigenfunctions of

both H`

and X`. This procedure identifies special linear combinations of 1\ and 2\ - special because these particular combinations are

eigenfuntions of X`

as well as being eigenfunctions of H`

. These states are obviously still degenerate with respect to H`

, but are not necessarily

degenerate with respect to X`

. Let’s assume that they’re not for the moment. Then, we can label these states according to their eigenvalues of

H`

and X` - we’ll call them n, a\ and n, b\. They have the following properties:

H`

n, a] = En n, a],H`

n, b] = En n, b],X`

n, a] = a n, a],X`

n, b] = b n, b].You see that we have distinguished the states according to their eigenvalues - the states are just lists that specify the eigenvalues of the two

commuting operators (in an order that has to be understood). This is a much more helpful and physically meaningful way of labelling states -

we instantly know the properties of the states we are talking about.

Now it could be that our two states are also degenerate with respect to X`

(i.e. a = b). Then, we can’t distinguish the states according to the

eigenvalues of H`

and X`

alone. When this happens, we look for a third operator, Y`, that commutes with the other two, and we find the states

that are simultaneous eigenfunctions of H`

, X`

and Y`, labelling them (say) n, a, p\ and n, a, q\ where p and q are the eigenvalues of Y

`. If it

happens that p = q, then we continue with the procedure, finding a fourth commuting observable, and so on until we’ve found unique states.

It’s always possible to do this. Though we’ve illustrated the procedure using just two states, it clearly generalizes to an arbitrary number of

degenerate states.

Atomic & Molecular Physics Mike Tarbutt, 2014

1

Now it could be that our two states are also degenerate with respect to X`

(i.e. a = b). Then, we can’t distinguish the states according to the

eigenvalues of H`

and X`

alone. When this happens, we look for a third operator, Y`, that commutes with the other two, and we find the states

that are simultaneous eigenfunctions of H`

, X`

and Y`, labelling them (say) n, a, p\ and n, a, q\ where p and q are the eigenvalues of Y

`. If it

happens that p = q, then we continue with the procedure, finding a fourth commuting observable, and so on until we’ve found unique states.

It’s always possible to do this. Though we’ve illustrated the procedure using just two states, it clearly generalizes to an arbitrary number of

degenerate states.

1.3. Constants of the motion

We often describe dynamical systems with the help of quantities that stay constant. If you throw a ball, for example, it travels in a parabola,

which you can resolve into a horizontal and vertical motion. The horizontal momentum is a constant of the motion (if there’s no resistance). If

the ball is spinning, this angular momentum is also a constant of the motion. The same ideas are important in quantum mechanics. Here, we

have operators corresponding to observables. An important result in quantum mechanics is that when an operator commutes with the

Hamiltonian the corresponding observable does not change in time. This means that if we make a measurement of the observable, the possible

outcomes and their probabilities do not change in time. So when AX`

, H` E = 0 we say that X

` is a constant of the motion. We also know that

when AX`, H

` E = 0 we can find states that are simultaneous eigenfunctions of both H`

and X`

, and that we label these states with their eigenval-

ues. All we are doing when we specify the eigenstate in this way is to list the quantum numbers that correspond to the constants of the

motion, the properties that do not change in time. We often call these the “good quantum numbers”.

To take an example, consider the energy eigenstates of the hydrogen atom, which we can write as n, l, s, ml, ms\. These quantum numbers

label (respectively) the energy, the orbital angular momentum, the spin angular momentum, the projection of the oribtal angular momentum

onto the z-axis, and the projection of the spin angular momentum onto the z-axis. All these quantities are constants of the motion (because of

the conservation of energy and the conservation of angular momentum).

1.4. Perturbation theory

The energy levels of atoms and molecules can be determined to very high precision, both experimentally and theoretically. Very precise

theoretical calculations are possible even though there are no exact solutions for atoms with more than one electron. The solution proceeds

through a series of approximations. An important approximation method is perturbation theory. We want to find the eigenvalues of the

Hamiltonian H`

but we cannot do it exactly. So we divide H`

up into a part that we can solve, H`

0, plus a much smaller additional part H`

':

H`

= H`

0 + H`

'. We solve for H`

0 and find the eigenfunctions and eigenvalues

(1.4)H`

0 Φn = En

0Φn

or, in Dirac notation,

(1.5)H`

0 n] = En

0n].

The perturbation H`

' shifts the energy eigenvalues, so that the new eigenvalues are En1 = En

0 + DEn. The result of first-order perturbation

theory is that the energy shift is the expectation value of the perturbation, evaluated using the zeroth-order eigenstates:

(1.6)DEn = YH`

'] = Yn H`

' n]. = à Φn

*H`

' Φn â V

Here, as a reminder, I’ve written down the result using both Dirac notation and the explicit integral form. We will use Dirac notation much of

the time, and you should be familiar with what it means. In the second form above (YH`

']), the expectation value is simply indicated by the

angle brackets, and we have to remember to use the zeroth-order eigenstates.

First-order perturbation theory also tells us that the perturbation changes the eigenstates. The corrected eigenstates are given by

(1.7)n '\ = n\ + âk¹n

Yk H`

' n]En

0 - Ek

0n_ BEquivalently, Φn ' = Φn + â

k¹n

Ù Φk

*H`

' Φn â V

En0 - E

k

0ΦnF.

1.5. Perturbation theory with degeneracy

The above results apply to non-degenerate states. In atomic physics, however, we are almost always dealing with states that are degenerate,

so we need to know how to handle this. States that have the same energy in the absence of the perturbation usually have slightly different

energies in the presence of the perturbation. We say that the perturbation removes or ‘lifts’ the degeneracy. To give a simple example, think

of a single electron in an atom with an orbital angular momentum l = 1. The projection of this angular momentum onto the z-axis can take

three possible values with quantum numbers m = -1, 0, +1. If there are no external fields applied there is nothing special about the z-axis,

and so the energy cannot depend on the value of m. We say that the state is 3-fold degenerate. Let’s apply, as a small perturbation, a magnetic

field in the z-direction. The electron in its orbit is a little current loop with an associated magnetic moment, the value of m tells us which way

this magnetic moment is pointing, and the energy depends on the relative orientation of the magnetic moment and the applied magnetic field.

So now the energy does depend on m - the perturbation has lifted the degeneracy.


2

Without perturbation HH=H0LStates are degenerate

With perturbation HH=H0+H'LDegeneracy is lifted

When there’s degeneracy, we can still use the perturbation results given above, but we have to be careful. You can immediately see this from

equation (1.7): when there's degeneracy, En0 = E

k

0 for some n, k - then there are terms in the sum where the denominator is zero, and the

result seems to be infinite. The problem arises because, as discussed above, all linear combinations Α 1\ + Β 2\ are eigenfunctions when 1\and 2\ are degenerate. Suppose we try to calculate the energy shift using equation (1.6) when two of the zeroth-order eigenfunctions are

degenerate. We have a choice about which linear combination of these eigenfunctions to use (i.e. the values of Α and Β), and the result of the

calculation depends on this choice (the choice of "basis" or "representation"). So how do we know which choice is the correct one?

Answer: choose the linear combinations such that Yk H`

' n] = 0 for all cases where En0 = E

k

0. When Yk H`

' n] = 0 we say that the

perturbation H`

' does not connect states n\ and k\.

Provided we do this, we can just use standard perturbation theory (i.e. the above results) in the usual way. How do we find the linear

combinations of degenerate states that satisfy the above requirement? The formal method is as follows: calculate all the matrix elements

Yk H`

' n] in whatever representation you like, arrange these matrix elements in a matrix, then diagonalize it. The diagonalization procedure

picks out the correct linear combinations and gives us the perturbed set of eigenstates and eigenvalues. We won’t use this method. Instead,

we’ll use a method that is easier and brings a lot more physical insight. It is based on identifying the constants of the motion. Here are the

steps:

1. Identify an observable, X`, that is a constant of the motion for both the unperturbed and perturbed system. This means that X

` commutes

both with H`

0 and H`

0 + H`

'. There is usually a physical argument that helps to identify the things that remain constant - for example, if the

pertubation doesn’t apply any torque, the angular momentum is still a constant of the motion.

2. Find the set of states that are simultaneous eigenfunctions of H`

0 and X`

. It can be shown that the perturbation does not connect these states,

i.e. Yn, a H`

' n, b] = 0 if a and b are two different eigenvalues of X`

, both with the same zeroth-order energy En.

3. Using these states, apply the standard results of perturbation theory (as above).

Don't worry if this all seems a bit abstract at this point - the examples we'll meet along the way will make it clearer.

Proof

You don’t need to know the proof of the above result, but here it is for interest. Since X`

commutes with both H`

0 and H`

0 + H`

', it must also

commute with H`

', i.e. X`

H`

' = H`

' X`. So

Yn, a X`

H`

' n, b] - Yn, a H`

' X`

n, b] = 0

\ Ha - bL Yn, a H`

' n, b] = 0.

In the second step, we’ve used the fact that the states are eigenstates of X`

with eigenvalues a and b, and also the fact that X`

is a Hermitian

operator (it corresponds to an observable). From the above, we immediately see that Yn, a H`

' n, b] = 0 when a ¹ b.

1.6. Energy level structure through a hierarchy of approximations

We find the energy level structure of complicated atoms and molecules by making a series of approximations. In fact, even for the simplest

atom - hydrogen - we need to make this series of approximations to fully describe the energy level structure. A consideration of the hydrogen

atom is a nice introduction to the various interactions and splittings that we’ll encounter.

First we solve the Schrödinger equation for a single electron moving in the Coulomb potential of the nucleus. This gives us a set of energy

levels which we call the gross energy level structure. The spacing between the low-lying energy levels is in the 1-10eV range. Expressed as a

wavelength it is in the range from 100-1000nm. So transitions between these levels produce spectral lines in the uv, visible or near-ir.

This is not the end of the story for hydrogen, because there are interactions that we haven’t yet included. Most important are effects due to

relativity. The electron moves at about 1% the speed of light, so relativistic effects are small, but not completely negligible. Because the

effects are small, we can treat them using perturbation theory. The electron is moving through the electric field of the nucleus, but motion

through an electric field produces a magnetic field, and so the electron sees this motional magnetic field. The electron also possesses an

intrinsic magnetic dipole moment - the magnetic moment associated with its intrinsic spin. There is an interaction between this magnetic

moment and the motional magnetic field which causes the energy levels to shift and to split up into components, according to the value of the

angular momentum. This splitting is called the fine structure. In hydrogen, the fine-structure splitting is about 10-4 - 10-5eV. The effects of

relativity are stronger in heavier atoms, and the fine-structure splitting in heavy atoms can be as large as 10-1 eV.

This is still not the end of the story, because we still have to consider interactions of the electron with properties of the nucleus other than its

charge. In particular, the nucleus has a spin, and a magnetic moment associated with this spin. There is an interaction between the nuclear and

electron magnetic moments which splits the energy levels up even further. This is called the hyperfine structure, and it too can be calculated

to high accuracy using perturbation theory. The splittings are typically about 10-5eV.


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This is still not the end of the story, because we still have to consider interactions of the electron with properties of the nucleus other than its

charge. In particular, the nucleus has a spin, and a magnetic moment associated with this spin. There is an interaction between the nuclear and

electron magnetic moments which splits the energy levels up even further. This is called the hyperfine structure, and it too can be calculated

to high accuracy using perturbation theory. The splittings are typically about 10-5eV.

1.7. Summary of the main points

è For a one-electron atom, we can solve the Schrödinger equation to find the eigenvalues and eigenfunctions.

è For many electron atoms, we use approximate methods (in this course: first-order perturbation theory).

è Energy levels of atoms and molecules are often degenerate. Use additional quantum numbers (e.g. angular momentum quantum numbers)

to distinguish degenerate states.

è We find all the details of the energy-level structure through a hierarchy of approximations.

1.8. To do

Do question 1 from problem set 1. If necessary, use the notes from the 2nd year Quantum Mechanics lecture course to help you, or a

book on quantum mechanics.

Make sure you are comfortable with Dirac notation, and the concepts of degeneracy, commuting operators and simultaneous

eigenfunctions. Useful reading are the notes from the 2nd year quantum mechanics course.


4

2. Gross structure of one electron atoms

This is revision of material you have already covered in the 2nd year Quantum Mechanics and Applications of Quantum Mechanics courses,

so we will just bring out the important points here.

2.1. Bohr model and orders of magnitude

You will have encountered the Bohr model of the hydrogen atom before. Although the Bohr model is an ad hoc model, it gives many of the

correct results for the hydrogen atom (as we’ll see later), and it is a useful starting point for understanding the orders of magnitudes of various

atomic properties. In the Bohr model, an electron of mass me and charge -e moves in a circular orbit around a much heavier nucleus of

charge +e, and the angular momentum is quantized in integer units of Ñ. To keep the electron in orbit with radius r and speed v, the centrifu-

gal and electrostatic forces must balance:

(2.1)me v

2

r

=e

2

4 Π Ε0 r2

.

The quantization of angular momentum (L) is expressed simply as

(2.2)L = me v r = n Ñ,

where n is an integer.

The size of an atom and the Bohr radius

From these two equations we find the orbital radius as:

(2.3)r = n2

4 Π Ε0 Ñ2

me e2

= n2

a0,

where we’ve defined the quantity

(2.4)a0 =4 Π Ε0 Ñ2

me e2

,

which is known as the Bohr radius. It depends only on fundamental constants and has the value 5.292 ´ 10-11m. This is (roughly speaking)

the size of hydrogen in its ground state. For many-electron atoms the greater Coulomb attraction to the nucleus is largely offset by the

electron-electron repulsion, and so other atoms are roughly the same as this.

The speed of the electron and the fine-structure constant

From the same equations, we get the speed of the electron as

(2.5)v =1

n

e2

4 Π Ε0 Ñ c

c =1

n

Α c,

where we’ve defined the dimensionless quantity

(2.6)Α =e

2

4 Π Ε0 Ñ c

,

which is known as the fine structure constant. It is a fundamental constant and has the approximate value 1/137. We see that the speed of the

electron in the ground state is Α c » c 137. Because the speed is a small fraction of the speed of light, we can use non-relativistic quantum

mechanics to a fairly good approximation. However, to understand the finer details of atomic structure we will need to consider relativistic

effects.

The energy levels and the Rydberg constant

Adding together the kinetic energy, 1

2me v

2, and the potential energy, e2 H4 Π Ε0 r), and using the above formulae for v and r, we get the total

energy to be

(2.7)En = -1

2 n2

Α2

me c2.

The factor 1 2 Α2me c

2 is known as the Rydberg energy and, in electron-volts, has the value 13.6eV. After dividing the Rydberg energy by

h c, we get the Rydberg constant, which is usually given the symbol R¥. It is the most precisely measured quantity in physics. Also note that

the kinetic energy T, and the potential energy V are related by T = -1 2 V , which is an example of the virial theorem.

The magnetic moment and the Bohr magneton

The orbiting electron is a current loop, and associated with this current loop is a magnetic dipole moment. Recall from electromagnetism that

the magnetic moment of a current loop where the current I encloses an area A is simply Μ = I A. The current is I =âq

ât. In the ground state,

the time taken for the electron to complete an orbit is t = 2 Π a0 v, so the current in the loop is I = e Α c H2 Π a0L. The area of the current

loop is Π a02, and so the magnetic moment is


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(2.8)Μ =1

2e a0 Α c =

e Ñ

2 me

= ΜB,

where we’ve defined

(2.9)ΜB =e Ñ

2 me

,

which is known as the Bohr magneton. It sets the scale for atomic magnetic moments and has the value 9.274 ´ 10-24J T

-1.

2.2. Schrödinger equation for a one-electron atom

Our task is to calculate the energy levels and wavefunctions of a single electron of charge -e and mass me moving in the field of a nucleus of

charge Z e and mass M . We will usually be talking about the hydrogen atom, in which case the nucleus is a proton and Z = 1, but our

treatment will also apply to other one-electron systems such as the He+ ion (Z = 2) or more exotic one-electron atoms (see later). For now,

we approximate the nucleus as a point particle.

The time-independent Schrödinger equation for the relative motion is

(2.10)-Ñ2

2 m

Ñ2

+V HrL Ψ = E Ψ,

where the potential V HrL is the Coulomb potential,

(2.11)V HrL = -Z e

2

4 Π Ε0 r

,

and m is the reduced mass of the electron-nucleus system

(2.12)m =me M

me + M

.

The electron is much lighter than the nucleus, and so it does most of the moving. But we are interested in obtaining precise results and so we

have not assumed that the nucleus is stationary. Our equation describes the relative motion of the two particles, which is why it is the

reduced mass that appears and not the electron mass. Since M >> me the reduced mass is close to the mass of an electron, m » me.

2.3. Angular part

We will solve equation (2.10) in spherical polar coordinates, Hr, Θ, ΦL, but the potential depends only on the distance between the electron and

the nucleus, i.e. only on the coordinate r and not on the angles Θ and Φ. We call it a central potential. This is a crucial point. We can discover

a great deal about the atom without needing to know the actual form of the potential (equation (2.11)) - we just have to know that it depends

only on r. This idea becomes particularly important when, later, we come to many-electron atoms. In spherical polar coordinates equation

(2.10) is

(2.13)-Ñ2

2 m

1

r2

¶

¶ r

r2

¶

¶ r

+Ñ2

L` 2

2 m r2

+ V HrL ΨHr, Θ, ΦL = E ΨHr, Θ, ΦL,

where Ñ L is the angular momentum operator, defined by Ñ L = r´ p. Note carefully the way we have defined L - it is a dimensionless

operator equal to the angular momentum operator divided by Ñ (some text books use L itself as the angular momentum operator, so be aware

of this possible difference). In spherical polar coordinates the squared angular momentum operator is

(2.14)Ñ2

L

` 2

= -Ñ2

1

sin Θ

¶

¶ Θsin Θ

¶

¶ Θ+

1

sin2 Θ

¶2

¶ Φ2

,

and the z-component of the angular momentum operator is

(2.15)Ñ L`

z = -ä Ñ¶

¶ Φ.

Because V HrL depends only on r, the wavefunction separates into a product of a radial part and an angular part and there are a complete set of

solutions of the form

(2.16)ΨHr, Θ, ΦL = RHrL YHΘ, ΦL.The angular part, YHΘ, ΦL, should be familiar to you. It is an eigenfunction of the operator L

` 2

with eigenvalues lHl + 1L and l having non-zero

integer values. It is also an eigenfunction of L`

z with eigenvalues m , an integer in the range -l £ m £ l. The l and m appear as formal constants

in the separation of variables and the constraints on their values are a consequence of requiring the wavefunction to be single-valued and finite

everywhere. More important is their physical meaning: l is the quantum number that specifies the value of the angular momentum, while m is

the quantum number that specifies the projection of the angular momentum onto the z-axis. Note that we have a set of functions YHΘ, ΦL, one

function for each eigenstate specified by the quantum numbers l and m, and so we label them using subscripts - Yl,mHΘ, ΦL. To summarize


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(2.17)L` 2

Yl,mHΘ, ΦL = lHl + 1L Yl,mHΘ, ΦL,(2.18)L

`z Yl,mHΘ, ΦL = m Yl,mHΘ, ΦL.

You can look up the mathematical forms for the Yl,m functions in a text book. The Φ dependence is simply eä m Φ, while the Θ dependence is

given by the associated Legendre polynomials, Pl,mHΘL. Below are polar plots showing the angular probability distributions Yl,mHΘ, ΦL 2 for a

range of values of Hl, mL. Note that (because we have taken the modulus squared) these distributions do not depend on Φ. In the polar plots, Θ

is measured from the vertical axis and the distance from the origin of a point on the curve is proportional to the value of the function in that

direction.

H0,0L

H1,0L H1,±1L

H2,0L H2,±1L H2,±2L

H3,0L H3,±1L H3,±2L H3,±3L

For any given value of l there are 2 l + 1 possible values of m. When we sum the Yl,mHΘ, ΦL 2 over all the values of m for a given l, the

result is independent of Θ and Φ, and so is spherically symmetric:

âm=-l

l

Yl,mHΘ, ΦL 2 is independent of Θ and Φ.

Notation

We use a letter to indicate the electron’s state of orbital angular momentum. For historical reasons the letters are

l=0 l=1 l=2 l=3 l=4 l=5

s p d f g h

with higher values of l then following alphabetically.

2.4. Radial part

The radial part of the problem is found by solving the radial equation which is

(2.19)-Ñ2

2 m

1

r2

¶

¶ r

r2

¶

¶ r

+Ñ2

lHl + 1L2 m r

2+ V HrL RHrL = E RHrL.


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It is usual to cast this in a slightly simpler form using the substitution PHrL = r RHrL, which gives us

(2.20)-Ñ2

2 m

â2P

â r2

+Ñ2

lHl + 1L2 m r

2+ V HrL P = E P.

The formal solution can be found in most quantum mechanics text books, but we can learn a lot just by looking at the above equation. The first

term on the left hand side is the radial kinetic energy. We can consider the bracketed term as an effective potential for the motion:

(2.21)Veff HrL =Ñ2

lHl + 1L2 m r

2+ V HrL.

In this effective potential, the first term is a centrifugal barrier proportional to lHl + 1L while the second term is the Coulomb potential,

equation (2.11). The effective potential for hydrogen is plotted below for both l = 0 (solid line) and l = 1 (dashed line). The two terms in the

effective potential have opposite signs - the Coulomb potential pulls the electron towards the nucleus while the centrifugal barrier keeps the

electron away from the nucleus. At small enough distances, the centrifugal barrier (scaling as 1 r2) overwhelms the Coulomb attraction

(scaling only as 1 r) and the electron cannot reach the nucleus. States with zero angular momentum, l = 0, are the exception to this - for them

the centrifugal barrier vanishes and they can reach the nucleus. So for states with l ¹ 0, we can expect the wavefunction to go to zero at the

origin, whereas for states with l = 0 we expect the wavefunction to have a maximum at the origin.

l=0

l=1

0 2 4 6 8 10-15

-10

-5

0

5

10

r H10-10

mL

Veff

HeVL

Effective potential , VeffHrL, for hydrogen

At large distances, as r ® ¥, the potential energy terms fall off and the radial kinetic energy dominates. For the bound states that we are

interested in (negative values of E) the wavefunctions must then fall off exponentially as expH-Κ rL, where Κ2 = -2 m E Ñ2.

The solution of the radial equation introduces a new integer from the requirement that the wavefunction behaves correctly as r ® 0 and as

r ® ¥ - this is the principal quantum number n which can take all integer values greater than the value of l. The solutions are labelled by the

two quantum numbers n and l and we write Rn,lHrL. The first six radial functions for hydrogen (Z = 1) are plotted in the figure below as a

function of position r. For reference (don’t try to remember them!), the equations for these radial functions are also given.

Important things to note about these radial functions:

ì The radial coordinate r is scaled by the quantity a0. This is the “Bohr radius”. We see from the plots (or the formulae) that it tells us the

approximate size of the radial wavefunctions. It has the value 5.292 ´ 10-11m, and is given by

(2.22)a0 =4 Π Ε0 Ñ2

me e2

.

ì At large distances, all the wavefunctions fall off exponentially, with the scale for the exponential decay being n a0. The wavefunction

extends further for larger values of the principal quantum number n. In fact, the expectation value of r scales as n2 (you can check this

yourself).

ì The plots and formulae above are for Z = 1. For other values of Z, simply replace a0 by a0 Z everywhere. This makes sense: increasing

the nuclear charge pulls the electron in closer to the nucleus.

ì The wavefunctions for l = 0 are largest at the the nucleus (i.e. at the origin), whereas those for l ¹ 0 are zero at the nucleus. This will be

important later when we come to consider the subtle effects due to the finite size of the nucleus and its magnetic moment.

ì The number of nodes in the radial wavefunction is equal to n - l - 1.


8

n=1

R1,0=2K 1

a0

O32expK-

r

a0

O

H1,0L

0 5 10 15ra0

0.5

1.0

1.5

2.0

a032

Rn,l

n=2

R2,0 = 2K 1

2a0

O32K1-r

2a0

OexpK-r

2a0

O

R2,1 =2

3

K 1

2a0

O32K r

2a0

OexpK-r

2a0

O

H2,0L

H2,1L

5 10

-0.2

0.2

0.4

0.6

a032

Rn,l

n=3

R3,0=2K 1

3a0

O32B1-2r

3a0

+2

3K r

3a0

O2FexpK-r

3a0

O

R3,1=4 2

3K 1

3a0

O32K r

3a0

OB1-1

2

r

3a0

FexpK-r

3a0

O

R3,2=2 2

3 5

K 1

3a0

O32K r

3a0

O2

expK-r

3a0

O

H3,0L

H3,1LH3,2L

5 10 15ra0

-0.1

0.1

0.2

0.3

0.4

a032

Rn,l

2.5. Energy levels

The same procedure that gives us the radial functions also gives us the energy eigenvalues. These are

(2.23)En = -Z

2e

4m

2 H4 Π Ε0L2 Ñ2n

2

This is an important result, but it is not easy to remember in this form. We can rewrite it by introducing an extremely important quantity called

the fine structure constant:

(2.24)Α =e

2

4 Π Ε0 Ñ c

.

This is a dimensionless fundamental constant which compares electromagentic quantites (e and c) with Planck’s constant. It has been

measured to extremely high precision and its value (I’ll just write down the first 3 digits, though many more are known) is 1 137. In terms of

Α, the energy levels are

(2.25)En = -1

2

Z2 Α2

n2

m c2.

Consider what this means for the ground-state of the hydrogen atom (Z = 1, n = 1, m > me). me c2 is the rest mass of the electron and the

equation tells us that (in the ground state) the binding energy to the proton is simply 1 2 Α2 times this rest mass. The energy 1

2me Α2

c2 is

called the Rydberg energy and its value is 13.6eV. Note that it is defined in terms of the full electron mass, not the reduced mass. This means

that the ionisation energy of hydrogen is slightly smaller than the Rydberg energy, smaller in fact by the ratio m me. After dividing the

Rydberg energy by h c, we get the Rydberg constant, which is usually given the symbol R¥. It is the most precisely measured constant in

physics, with a relative uncertainty of 5 parts in 1012: R¥ = 10, 973, 731.568539 ± 0.000055 m-1.

Note that, although the eigenfunctions of a one-electron atom are specified by three quantum numbers, n, l, m, the energy levels depend only

on the quantum number n. This is a special case which has its origin in the 1 r form of the Coulomb potential. For all other atoms we will

find that the energies depend on both n and l.

The energy levels of hydrogen are plotted in the figure below, labelled by the quantum numbers n and l. It is common to show the energy

levels of atoms on a diagram similar to this one. Note once again that for hydrogen, and only for hydrogen, states of the same n but different

l are degenerate (have the same energy). This is not the case for other atoms.


9

1

2

3

45

¥

nl=0 l=1 l=2 l=3

1s

2s

3s

2p

3p 3d

Energy HeVL

Notation

We specify the state of the electron using the quantum numbers n and l, but for l we use the letter (see above) instead of the number.

Remember that l can only take integer values smaller than n. So the allowed states of hydrogen are 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f etc.

These are indicated in the energy level diagram.

2.6. Hydrogen-like systems

As already noted above, our treatment is also valid for other one-electron atoms. Important examples are He+, Li2+ etc. We see from equation

(2.25) that the energies scale as Z2 - so (for example) the ionization energy of He+ is 4 times that of hydrogen, the ionization energy of Li2+

is 9 times that of hydrogen etc. Another important one-electron atom is deuterium. It has Z = 1, but the nuclear mass is twice that of hydrogen

so the reduced mass is slightly different. This shifts all the energy levels by a small amount - you can work out the size of this shift yourself.

Also interesting and relevant is positronium - a bound state of an electron and a positron. This atom is unstable because the electron and

positron annihilate, but it lives for long enough (~1Μs) to study it. Its reduced mass is simply half the electron mass and so all the energies are

halved.

A topical example of a hydrogen-like atom is antihydrogen - a positron bound to an antiproton. Through an amazing research effort, it is now

possible to produce and store many thousands of antihydrogen atoms. Researchers are now turning their attention to the study of the energy

level structure of antihydrogen to find out whether it is exactly the same as that of hydrogen. They will also test whether it behaves the same

under gravity as normal matter. This research tests some of our most basic laws of physics.


è Orders of magnitude: size of atom is order a0, speed of electron is order Α c, energy levels are order Α2 times the electron rest-mass

energy, magnetic moments are order ΜB.

è Because the Coulomb potential is a central potential the solution for the hydrogen atom separates into an angular part and a radial part.

è The angular solutions are the spherical harmonics. They are the eigenfunctions of L` 2

and L`

z with eigenvalues lHl + 1L and m respectively.

è The characteristic size of the ground-state radial function is a0, the Bohr radius. The wavefunction extends to larger distances for higher

values of n.

è The eigenfunctions are zero at the origin, except for s-states.

è The energies are proportional to 1 n2 and are degenerate with respect to l.

è The energies are proportional to Z2 and to the reduced mass m.

2.8. To do

Check that you can derive the results for the Bohr atom written down in this handout.

Make sure that you’ve understood all the material in this handout. Revise any parts that are unfamiliar. Useful reading are the lecture

notes for 2nd year Quantum Mechanics, chapter 2 of Softley, chapter 2 of Woodgate, or chapter 3 of Rae (‘Quantum Mechanics’).

Do questions 2-5 from problem set 1.


10

3. Angular momentum

3.1. Properties of angular momentum

You have already met the two types of angular momentum - orbital angular momentum and spin angular momentum. The eigenstates of

orbital angular momentum can be represented by spherical harmonics. The spin eigenstates cannot be represented using spatial functions - spin

is just different. Nevertheless, both kinds of angular momentum can be represented and handled using exactly the same mathematics - this is

the quantum mechanics of angular momentum in general.

An angular momentum operator Ñ J` is a vector operator that satisfies the following commutation relations:

(3.1)AJx

`, Jy

` E = ä Jz

`, AJy

`, Jz

` E = ä Jx

`, AJz

`, J

`xE = ä Jy

`.

Since the components of J` do not commute, we cannot have states that are simultaneous eigenstates of more than one component. However,

each of the components commutes with the operator J` 2

:

(3.2)AJx, J2E = AJy, J

2E = AJz, J2E = 0.

This means that we can form states that are simultaneous eigenstates of J` 2

and one of the components of J`, chosen by convention to be Jz

`.

These correspond to quantum mechanical constants of the motion, which we denote using the quantum numbers j and m. We label the

eigenstates using these quantum numbers, so they are written as j, m\ and have the following properties:

(3.3)J` 2

j, m^ = jH j + 1L j, m^(3.4)Jz

`j, m] = m j, m].

The quantum number j is called the angular momentum quantum number, but you should be careful to remember that j is not itself the

eigenvalue - the important operator is J` 2

whose eigenvalue is jH j + 1L. The quantum number m is often called the magnetic quantum number

because of the role it plays when a magnetic field is applied to an atom (see later). It can only have the values

(3.5)m = - j, - j + 1, .... j - 1, j.

So for any value of j there are always 2 j + 1 possible values of m. For example, when j = 1 2, there are only two possible values of m,

namely -1 2 and 1 2. When j = 1, m can have three possible values m = -1, 0, 1.

Electrons have spin-1/2. Electrons in atoms can have orbital angular momentum - integer values from zero upwards. Photons have spin-1.

Notation

Note that I have defined the angular momentum operator as Ñ J`. This means that J

` is dimensionless, and has dimensionless eigenvalues.

These multiply Ñ to give the actual angular momentum in SI units. Be careful about this when reading the text books - some authors use this

same notation, but some have J` itself be the angular momentum operator. This can cause confusion over the Ñ’s, so be aware of that.

3.2. The vector model for angular momentum

As emphasized above, the components of an angular momentum operator do not commute with one another. We can still calculate the

expectation values of each component however. It should be obvious that

Y j, m Jz

`j, m] = m.

The expectation values of the other two components can also be calculated, and the result is

(3.6)Y j, m Jx

`j, m] = Y j, m Jy

`j, m] = 0.

[The easiest way to see this is to use the raising and lowering operators J+

`= Jx

`+ ä Jy

`, J-

`= Jx

`- ä Jy

`. Recall (from your quantum mechanics

lectures) that these operators increase/decrease the value of m by one unit without changing the value of j. States of different m are orthogo-

nal, so we immediately get the above results for the expectation values of Jx

` and Jy

`].

These expectation values motivate a semiclassical model of angular momentum known as the vector model. The model is drawn below. We

picture the angular momentum as a vector precessing around the z-axis, in much the same way as the spin of a spinning top or gyroscope

precesses about a vertical axis. The vector itself is labelled by j. This is just a label so that, when we are dealing with more than one angular

momentum, we know which one we are talking about - we have to always remember that the expectation value of J` 2

is not j2 but jH j + 1L.

Because the precession is a precession about the z-axis the z-component of the angular momentum has a constant value - we say that it is a

constant of the motion and we label it as m to remind ourselves that the expectation value of Jz

` is m. The x and y components of the angular

momentum are not constants - they change in time as the angular momentum precesses. This expresses the quantum mechanical idea that,

once the z-component is fixed, the x and y components are uncertain. The time-average of the x-component is zero, as is the time-average of

the y-component. So we see that the classical time-averages behave just like the quantum mechanical expectation values.


11

jm

z

x

y

It is important to appreciate that this vector model is a picture that helps us to understand angular momentum in quantum mechanics. It is a

very helpful picture which provides a lot of insight, especially when we have more than one angular momentum to consider. It can also be

used as an aid to calculations in quantum mechanics, but this has to be done carefully, since its applicability is limited.

3.3. The vector model for coupled angular momenta

In atomic physics, we have to know how to deal with more than one angular momentum. In particular, we have to know the proper way to

add them up. We already need this for a one-electron atom because the electron can have both orbital and spin angular momentum, and we

might need to add these up to make a total angular momentum. For many-electron atoms we’ll need to add up the angular momenta of the

individual electrons to make a total.

Suppose we have two angular momentum operators (again, spin or orbital, it doesn’t matter), J1 and J2. We can add them up to give the total

angular momentum

(3.7)J`

= J1

`+ J2

`.

All three operators, J1

`, J2

` and J

` have the properties listed in the section above. Of course they do - they are all angular momenta.

How should we represent these two angular momenta using the vector model? One obvious way is just to draw two separate vector models

for J1

` and J2

`, which we can combine into a single picture like this:

j1

j2

m1

m2

z

In the picture m1 is the projection of j1 onto the z-axis, and m2 is the projection of j2 onto the z-axis. j1, j2, m1 and m2 are all constants of

the motion. This is a picture where there is no coupling between the two angular momenta - each angular momentum just does its own thing

and we can treat the two completely independently. You can think of this like two independent gyroscopes.

In real physical systems, like atoms and molecules, there is often some kind of interaction that couples together two angular momenta. We

will encounter several interactions of this kind. For example, in hydrogen there is an interaction called the spin-orbit interaction which

introduces a coupling between the orbital angular momentum of the electron and its spin. Before introducing this interaction the two angular

momenta are independent, but after introducing the interaction they are not independent - they are coupled together. You can think of the two

gyroscopes again, but now there is an interaction between them. For example, the gyroscopes might be connected by a weak spring, or there

could be a lump of charge on the flywheel of each gyroscope. Because of the interaction, each gyroscope exerts a torque on the other, and

each starts precessing. In this complicated situation, there is one thing that we can be sure of - there is no external torque acting on the system

as a whole, and so the total angular momentum of the whole system is a constant. What’s more, the precession of the individual angular

momenta is a precession about their resultant (the total angular momentum). All of this translates into a vector model for coupled angular

momenta in quantum mechanics. Here is the picture:


12

j

j1

j2m

z

The important things to understand from this picture are as follows:

ì j1 and j2 precess together about their resultant j.

ì j1 and j2 are constants of the motion, but their projections onto the z-axis are not constants. The picture makes this clear - the projection

of j1 onto the z-axis changes as j1 precesses around j (and the same is true for j2).

ì j is a constant of the motion, and so too is its projection onto the z-axis, which is m.

ì If we make the interaction that is coupling j1 and j2 stronger, the precession about the resultant is faster. The rate of precession is

proportional to the strength of the interaction.

ì There is a one-to-one correspondence between time-averaged quantities in the vector model and the equivalent quantum-mechanical

expectation value.

Athough we’ve put together this model in an ad hoc way, it does have a formal basis (see Woodgate, for example).

3.4. Adding angular momenta

Consider again the addition of two angular momenta, J1

` and J2

`, to make a total J

`= J1

`+ J2

`.

J` 2

has eigenvalues jH j + 1L where the possible values of j are

(3.8)j = j1 - j2 , j1 - j2 +1 ... j1 + j2 - 1, j1 + j2.

This makes sense when you think about adding together the two vectors. The largest total angular momentum you can make is when the two

are lined up (going around in the same direction) and then you just get the sum of the two. The smallest you can make is when the two are

going around in opposite directions - then you get the larger minus the smaller. For example, if we add together j1 = 2 with j1 = 1 2 the

possible values of j are 3/2 and 5/2. Or if we add j1 = 2 with j1 = 1, the possible values of j are 1, 2 and 3.

How shall we write down the total state of the system? Let’s consider an important example, namely that of adding two spin-1/2 angular

momenta, j1 = 1 2, j2 = 1 2. Since each spin-1/2 can have two possible values of m (±1/2), there must be 4 possible states of the combined

system. We could write these out explicitly using the notation j1, m1; j2; m2\. In fact, since we know that j1 and j2 are both 1/2, we can use

the simpler shorthand m1; m2\. Then, the four possible states are

1

2;

1

2_,

1

2; -

1

2_, -

1

2;

1

2_, -

1

2; -

1

2_.

An even nicer shorthand than this is to use an up arrow for 1/2 and a down arrow for -1/2, always with m1 written first and m2 second. This

nicely represents whether each spin is up or down relative to the z-axis. In this notation, the states are

(3.9)m1 m2\ = \, ¯\, ¯ \, ¯ ¯\.

This representation is like the first vector model picture in section 3.3, where the angular momenta are not coupled and we specify the

quantum numbers j1, m1, j2, m2.

There is a second, equally valid, way of writing down these states, and this follows the second vector model picture in section 3.3. Here, we

specify the states by the quantum numbers j1, j2, j, m, using the notation j1; j2; j, m\. Once again, we know that j1 and j2 are both 1/2, so

we can drop them from the notation to keep it a bit simpler - so we just specify j, m\. Now, when j1 = j2 = 1 2, the possible values of j

are 0 and 1. So the allowed states are

(3.10)j, m\ = 1, 1\, 1, 0\, 1, -1\, 0, 0\.

Notice that, whichever way we choose to write them, we always get 4 states. That’s good - the number of states is a physical thing and

shouldn’t depend on how they are represented. Now we come to an important point. There has to be a way of linking the two representations,

i.e. of writing the state in one representation in terms of the states in the other representation. We often do this in quantum mechanics - writing

down a quantum state as a linear superposition of some basis states. For our example of two spin-1/2 particles, the relationship is:

(3.11)1, 1\ = \,


13

(3.12)1, 0\ = ¯\ + ¯ \

2

,

(3.13)1, -1\ = ¯ ¯\,

(3.14)0, 0\ = ¯\ - ¯ \

2

.

3.5. Advanced topic: Proof and generalization of the above

This part is beyond the scope of the course - but is here for information in case you are interested.

Proof

It should be obvious that the only way to make the state 1, 1\ (total angular momentum 1 and projection 1) is to have the two spin-1/2’s

both pointing upwards. Formally this is obtained using the relation Jz

`= J

`1 z + J

`2 z (which follows from equation (3.7)), which tells us that

m = m1 + m2. Similarly, to get the state 1, -1\ we must have the two spin-1/2’s pointing downwards. This explains (3.11) and (3.13). But

what about the other two? We need to introduce the angular momentum raising and lowering operators, defined as J+

`= Jx

`+ ä Jy

`,

J-

`= Jx

`- ä Jy

`. Their effect is to raise or lower the quantum number m by one unit, leaving j unchanged:

(3.15)J+

`j, m] = H j - mL H j + m + 1L j, m + 1_,

(3.16)J-

`j, m] = H j + mL H j - m + 1L j, m - 1_.

Applying J-

` to equation (3.11) we immediately obtain the result (3.12). A second application of J-

` will give us (3.13). Finally, we find the

expression for the state 0, 0\ by requiring it to be orthogonal to the others.

Generalization

When we add together two angular momenta, we can always represent the state in two different, equally valid ways.

Option 1. Because J`

1 and J`

2 are different angular momenta, all components of J`

1 commute with all components of J`

2. So we can write the

state as a simultaneous eigenstate of J1

` 2

, J2

` 2

, J`

1 z and J`

2 z, with eigenvalues j1H j1 + 1L, j2H j2 + 1L, m1 and m2. The state is written

j1, m1; j2 m2\. This is called the uncoupled representation.

Option 2. Since J1

` 2

, J2

` 2

, J` 2

and Jz

` all commute we can write the state in terms of their eigenvalues, j1H j1 + 1L, j2H j2 + 1L, jH j + 1L and m.

The state is written j1; j2; j, m\. This is called the coupled representation.

It is always possible to transform from one representation to another. The coefficients that define this transformation are found using the

raising and lowering operators. You don’t need to know how to do this, but it is helpful for you to know that it can be done.

3.6. Usefulness of the vector model

Let's take a relevant example. Suppose we have an electron with orbital angular momentum l` and spin angular momentum s

`. Later on, we will

see that there is an interaction that couples these two angular momenta - it is called spin-orbit coupling and is very important in describing

atomic structure. The interaction energy is proportional to l`.s` and we will need to find the expectation value of this operator using angular

momentum eigenstates.

We can use the vector model to see how to calculate this expectation value. First we form the the total angular momentum j`

= l`

+ s`

.

Because of the interaction that couples l` and s

`, each precesses around the resultant j

`. The projection quantum numbers ml and ms are not

constants of the motion, but j and m j are constants. So it makes good sense to use a coupled representation where our states are written

l; s; j, m j]. Now, since j`

= l`

+ s`

, j`2

= l`2

+ s`2

+ 2 l`.s`, and so we can write the operator that we care about as

l`.s`

=1

2J j

`2

- l`2

- s`2N,

which is brilliant because our states are eigenstates of everything on the right hand side. We can evaluate the expectation value without any

trouble:

Yl`.s`] =1

2Z j

`2

- l`2

- s`2^ =

1

2@ jH j + 1L - lHl + 1L - sHs + 1LD.

Now we’ll look at another important example. Suppose we want to find the expectation value of lz

` in the case where l

` and s

` are coupled and

so are precessing about their resultant j`. We already know that the answer is not simply ml since this is not a constant of the motion. In the

vector model, we see that l` is precessing around j

` so that the component of l

` perpendicular to j

` will time-average to zero. So to find the time

average of lz

`, we first project l

` onto j

`, and then project j

` onto the z-axis:


14

Xlz

` \ = [ Il.`

j`

M j`

j`2

.z`_

Here, the quantity in the brackets is a vector in the direction of j`, whose magnitude is equal to the projection of l

` onto j

`. That vector is then

projected onto the z-axis. We can write the above as

Xlz

` \ =

Yl`. j`]

X j`2\

X jz

` \.

So we see that, because of the coupling which causes l` to precess around j

`, the expectation value of lz

` is proportional to the expectation

value of jz

`. The expectation values are evaluated using our coupled basis states l; s; j, m\, and we are able to calculate everything. Obvi-

ously, X j`2\ = jH j + 1L and X jz

` \ = m, and we use exactly the same trick as above to calculate Yl`. j`].

The above is an example of a more general result. Suppose we have some system with total angular momentum operator J`, such that

J` 2

j, m^ = jH j + 1L j, m^. Then it is possible to show that for any vector operator V`,

(3.17)Y j, m Vk

`j, m] =

Y j, m J`.V

`j, m]

jH j + 1L Y j, m Jk

`j, m].

Here, Vk

` is any of the three components of the vector. This important result is called the projection theorem. It tells us that the expectation

value of a component of V`

is proportional to the expectation value of the same component of the total angular momentum J`. While V

` can be

any vector at all, in atomic and molecular physics it is almost always an angular momentum. We will need to use this result in this course. The

formal proof is beyond the scope of the course, but the vector model shows us how it comes about.


è There are two types of angular momentum - orbital and spin. Mathematically, they are handled in exactly the same way.

è Every component of J` commutes with J

` 2

.

è The components of J` do not commute with one another.

è j, m\ is a simultaneous eigenfunction of J` 2

[with eigenvalue jH j + 1L ] and of Jz

` [with eigenvalue m]. It is not an eigenstate of either Jx

`

or Jy

`. [N.B. There’s nothing special about the z-axis, just that we have to pick one component, and this is the one chosen by convention].

è For a given j there are 2 j + 1 allowed values for m: They are m = - j, - j + 1, .... j - 1, j.

è Angular momenta can be added together: J`

= J1

`+ J2

`. Then, we have two possible ways of representing the total state of the system. We

can use the “uncoupled representation” j1, m1; j2, m2\, or we can use the “coupled representation” j1; j2; j, m\. In the coupled

representation J` 2

j1; j2; j, m^ = jH j + 1L j1; j2; j, m^ where j can take values between j1 + j2 and j1 - j2 .

è When we add together two spin-1/2’s there are 4 possible states - three with total angular momentum j = 1 (m = -1, 0, +1), and one

with total angular momentum j = 0.

è We often use the vector model as a picture of how angular momenta are coupled together, and to help us calculate expectation values.

3.8. To do

Revise the material in this handout. Useful reading: lecture notes from 2nd year Quantum Mechanics, chapter 4 of Rae (‘Quantum

Mechanics’), section 1.5 of Softley, or appendix C of Woodgate.

Do questions 6 and 7 from problem set 1.


15

4. Fine structure in one-electron atoms

We found the energy levels of hydrogen by solving the Schrödinger equation. This is a non-relativistic equation, and therefore approximate.

Since the equation is approximate, the energy levels are also approximate. As we will see, the approximation is a good one, but we would

like to go further and include the effects of relativity. There are two ways we could do that. One way is to solve the correct relativistic

equation, which is the Dirac equation. This is beyond the scope of this course, so we will use the second way: we introduce relativistic

corrections as perturbations and find the energy shift using first-order perturbation theory. Because the perturbations due to relativity are

small, this procedure works well, and it also gives us a great deal of physical insight.

For our present case of the fine structure in hydrogen, the zeroth-order Hamiltonian is

(4.1)H`

0 =p` 2

2 m

+ V HrLand the zeroth-order eigenfunctions and eigenvalues are the ones already worked out (see lecture 2). When we include relativistic corrections

we find that there are three separate perturbations.

4.1. Relativistic correction to the kinetic energy

The relativistic kinetic energy of the electron is

(4.2)T`

= I p` 2

c2

+ me2

c4M12

- me c2

= me c2 1 +

p` 2

me2

c2

12- me c

2>

p` 2

2 me

-p` 4

8 me3

c2

,

where in the last step we have used a Taylor expansion of the square root to second order in the small quantity p` 2 me

2c

2. The first term is

just the kinetic energy that already appears in the non-relativistic Hamiltonian (i.e. in the Schrödinger equation). The second term is our

perturbation, which we’ll call H`

1 ':

(4.3)H`

1 ' = -p` 4

8 me3

c2

= -1

2 me c2

p` 2

2 me

2

= -1

2 me c2

IH`

0 - V M2.

The energy shift is

(4.4)DE1 = -1

2 me c2

ZIH`

0 - V M2^ = -1

2 me c2

AEn2

- 2 En XV \ + YV 2]E,

where we have used the fact that the expectation value of H`

0 is simply the zeroth order energy En = -1

2

Z2 Α2

n2

me c2, as given by equation

(2.25). Inserting the Coulomb potential for V , we get

(4.5)XV \ = -Z e

2

4 Π Ε0

[ 1

r

_ = -Z e

2

4 Π Ε0

Z

n2

a0

= -Z

2

n2

Α2

me c2,

(4.6)YV 2] =Z

2e

4

H4 Π Ε0L2[ 1

r2

_ =Z

2e

4

H4 Π Ε0L2

1

Hl + 1 2L n3

Z

a0

2

=Z

4

n3

1

Hl + 1 2L Α4

me2

c4.

Here, we have used standard results for the expectation values of 1 r and 1 r2 for a one-electron atom (you can look them up, or work them

out yourself), and we’ve used the expressions for a0 and Α given by equations (2.22) and (2.24). Adding up the terms and simplifying we

reach the result

(4.7)DE1 = -En

Z2 Α2

n2

3

4-

n

Hl + 1 2L .

4.2. Spin-orbit interaction

The electron is moving through the electric field of the nucleus. Motion at velocity v in an electric field E produces a magnetic field

(4.8)B = -1

c2

v ´ E.

This magnetic field interacts with the intrinsic magnetic dipole moment, Μs

`, of the electron and we get an interaction term

(4.9)H`

2 ' = -Μs

`.B

For any radial electric field, we can write the electric field in terms of the potential energy V as

(4.10)E =1

e

â V

â r

r

r

.

Thus, our perturbation is


16

(4.11)H`

2 ' =1

me e c2

1

r

â V

â r

Μs

`.Hp

`´ r

`L,

where we have introduced the momentum p = me v. Next we use the relationship between the magnetic moment and the spin,

Μs

`= -gsJ e Ñ

2 me

N s`, and the definition of the orbital angular momentum, Ñ l

`= r

`´ p

`. We’ll also use the fact that gs = 2 (to a very good approxima-

tion). Finally, we have to introduce an extra factor of 1/2 due to Thomas precession which is a further relativistic effect that comes from a

careful consideration of the Lorentz transformation in a rotating frame. The derivation of this extra factor is beyond the scope of this course.

Using these substitutions, we obtain

(4.12)H`

2 ' =Ñ2

2 me2

c2

â V

â r

1

r

s`.l`

For the case of a one-electron atom, we have V = -Z e2 H4 Π Ε0 rL, and so

1

r

â V

â r

=Z e

2

4 Π Ε0 r3

.

Thus, for a one-electron atom, the perturbation is

(4.13)H`

2 ' =Ñ2

2 me2

c2

Z e2

4 Π Ε0

1

r3

s`.l`

To calculate the energy shift we need to calculate the expectation value of 1 r3 and of s

`.l`. The first is a standard result for a one-electron

atom:

(4.14)[ 1

r3

_ =1

lHl + 1 2L Hl + 1L n3

Z

a0

3

.

To calculate the expectation value of s`.l`

we introduce the total angular momentum of the electron j`

= l`

+ s`. Squaring, we get

j`2

= l`2

+ s`2

+ 2 s`.l` and so

(4.15)s`.l`

=1

2J j

`2

- l`2

- s`2N.

Our zero-order eigenstates are eigenstates of l`2

, s`2

and j`2

with eigenvalues lHl + 1L, sHs + 1L and jH j + 1L respectively. So the expectation

value we need is

(4.16)Ys`.l`] =1

2@ jH j + 1L - lHl + 1L - sHs + 1LD.

Using these expectation values we obtain the expression for the energy shift:

(4.17)DE2 = YH`

2 '] = -En

Z2 Α2

n

@ jH j + 1L - lHl + 1L - sHs + 1LD2 lHl + 1 2L Hl + 1L

Note that this shift does not apply when l = 0.

4.3. Darwin term for l = 0

There is an additional relativistic term that applies to s-states only. It can be obtained from the Dirac equation and it accounts for the break-

down of the non-relativistic approximation as r ® 0 and the potential energy becomes very large. It only applies to s-states because only s-

states have a non-vanishing wavefunction at the origin. The energy shift due to this term is found to be

(4.18)DE3 = -En

Z2 Α2

n

Hl = 0 onlyL.

4.4. Putting it all together

The total energy shift is found by adding up the three individual shifts, DEn j = DE1 + DE2 + DE3, remebering that DE2 only applies when

l ¹ 0 and DE3 only applies when l = 0. The result, valid for all l is

(4.19)DEn, j = -En

Z2 Α2

n2

3

4-

n

j + 1 2.

Important points to note about this result:

ì The fine structure shifts are of order Z2 Α2 smaller than the gross energy. For hydrogen (Z=1), this makes them about 1 1372 = 5 ´ 10-4

times smaller.


17

ì The fine structure is largest for small n.

ì The energy shift depends on n and j, but not on l. This means that states with the same n and j, but different l, remain degenerate when

relativistic effects are included. Consider n = 2 for example. The 2s state can only have j = 1 2, whereas the 2p state has two compo-

nents j = 1 2 and j = 3 2. These latter two have different energies, but the j = 1 2 components of the 2s and 2p states are degenerate.

ì DEn, j is always negative HEn is negative and n ³ j + 1 2). So all states are shifted down in energy.

ì It is the spin-orbit interaction that is responsible for the splitting of levels that have different j. The other two terms only result in a shift,

not a splitting. For this reason, it is the spin-orbit interaction that is most important in many-electron atoms.

The diagram below shows the fine structure splittings for the n = 2 and n = 3 states of hydrogen. The scale of the fine-structure splitting is

indicated. The other fine structure splittings are approximately to scale. The n = 2 to n = 3 energy difference is not on the same scale (it is

vastly larger than the fine structure splittings).

n=2

n=3

l=0 l=1 l=2

j=12 j=12

j=12 j=12

j=32

j=32 j=32j=52

Non-relativistic limit

Non-relativistic limit

4.52 ´ 10-5

eV

4.5. The Dirac result

Although it is outside the syllabus, it is worth writing down the result obtained by solving the Dirac equation directly:

(4.20)En, j = me c2 1 + Z

2Α

21

n - I j +1

2M + BI j +

1

2M2

- Z2 Α2F12

2 -12

.

This remarkably simple formula is an exact result for the energy levels of a one-electron atom (apart from the effects of QED and hyperfine

structure). Again, the energies depend on n and j but not on l. If we expand this formula in powers of Z Α, we find the following:

ì The first term is simply me c2, the rest mass of the electron.

ì The second term is of size Z2 Α2 and is equal to the gross energy En (equation (2.25)) that we find from the Schrödinger equation.

ì The third term is of size Z4 Α4 and is exactly equal to the total fine-structure energy shift that we found above using perturbation theory,

i.e. equation (4.19).

4.6. The Lamb shift

[Outside the scope of this course, but here for interest.] Even this Dirac result is not exactly correct. Even smaller corrections due to Quantum

Electrodynamics (QED) need to be applied. These corrections have their origin in the fluctuations of the vacuum. The fluctuations of the

vacuum cause tiny fluctuations in the position of the electron. This smears out the charge a little bit and the mean potential it sees due to the

field of the nucleus is slightly different to that seen by a point particle. The QED effect can be calculated to extremely high precision. It is

greatest for s states of low n. In n = 2 it results in a small splitting between the 2 s12 and 2 p12 states (remember that these are degenerate

before the QED effects are included). This splitting is called the Lamb shift, named after W.E. Lamb who, along with R.C. Retherford, was

the first to measure this splitting. The splitting is about 10 times smaller than the fine structure splitting. It has now been measured to

extremely high precision and is one of the most exact tests that we have of QED.


è Fine structure is due to relativistic effects.


18

è The most important of these is the spin-orbit interaction which is due to the interaction of the electron’s magnetic moment with the

magnetic field it sees as it moves through the electric field of the nucleus.

è The splitting is into levels labelled by the total angular momentum quantum number j. States of the same n and j, but different l are

degenerate.

è The fine structure splitting is of order Z2 Α2 times smaller than the gross energy.

è The fine structure splitting is largest for states of low n.

4.8. To do

Revise the material in this handout. Useful reading: section 2.10 and 2.11 of Softley, chapter 4 of Woodgate or section 2.3 of Foot.

Do questions 1, 2 and 3 from problem set 2.


19

5. Helium, part I

We wish to find the energy levels of helium. Let’s first start with a qualitative discussion. Start out with the ground-state of He+ - this will set

our zero of energy. Now introduce the second electron. When the electron is far away from the nucleus it sees a distant ball of charge e made

up of the nucleus (charge +2 e) and the first electron (charge -e). This looks just like hydrogen, so the highly excited states of helium should

just be the same as those of hydrogen. We say that, for these highly excited states, the nuclear charge (Z = 2) is screened by the inner

electron so that the effective Z is just 1. This concept of screening is an important one. It’s only when the second electron gets close to the

nucleus (i.e. in the lower states) that it starts to see the nucleus and the first electron separately. Then the screening of the nuclear charge is

incomplete and the energy levels are shifted to lower energies relative to hydrogen.

5.1. A perturbation approach to the energy levels of helium

For two-electrons in the field of an infinitely heavy nucleus, the Schrödinger equation is

(5.1)-Ñ2

2 me

Ñ12

-Ñ2

2 me

Ñ22

-Z e

2

4 Π Ε0 r1

-Z e

2

4 Π Ε0 r2

+e

2

4 Π Ε0 r12

ΨHr1, r2L = E ΨHr1, r2L.Here, Ñ1

2 is the Laplacian operator with respect to the coordinates of electron 1, and similarly for electron 2. The wavefunction ΨHr1, r2L is a

function of the coordinates of both electrons. The first and second terms in the equation are the kinetic energies of the two electrons. The

third and fourth terms are the electrostatic attraction of the electrons to the nucleus. The last term is the electrostatic repulsion between the

two electrons, with r12 = r1 - r2 being the distance between the electrons. We have not included the kinetic energy of the nucleus which is

a very small contribution to the total energy.

There is no exact solution to this equation, so we have to use an approximate method. Another approximate method to find the energy levels

is to treat the electrostatic repulsion as a perturbation, and this is the method we will follow.

Let’s write equation (5.1) as

(5.2)IH`

1 + H`

2 + H`

'M ΨHr1, r2L = E ΨHr1, r2L,where

(5.3)H`

1 = -Ñ2

2 me

Ñ12

-Z e

2

4 Π Ε0 r1

,

and similarly for H`

2, and

(5.4)H`

' =e

2

4 Π Ε0 r12

.

For our first approximation we neglect H`

' and solve IH`

1 + H`

2M Ψ0 = E0 Ψ0. This is easy to do because the electron’s don’t interact and so the

wavefunction separates into a product of two single-particle functions. We write

(5.5)Ψ0 = uaH1L ubH2L.Here, we are using some shorthand notation - label ‘1’ to mean the coordinates of electron 1, and label ‘2’ for the coordinates of electron 2.

Using the usual separation of variables procedure, it’s easy to show that the functions uaH1L and ubH2L satisfy

(5.6)H`

1 ua = Ea ua, H`

2 ub = Eb ub

and

(5.7)E0 = Ea + Eb.

Since equations (5.6) are hydrogenic Schrödinger equations, the functions ua and ub are the eigenfunctions of hydrogen (with Z = 2), and the

energies Ea and Eb are the energies of hydrogen, i.e. equation (2.25) with Z = 2. The subscripts a and b stand for the quantum numbers

n, l, m and n ', l ', m ' that define the hydrogenic states. Recalling that the energy eigenvalues of a one-electron atom are -13.6 Z2 n

2, we see

that (neglecting the electron-electron repulsion) the ground-state energy of the helium atom is -108.8eV. However, we need to be careful

what we mean here - this is the energy that would be needed to remove both electrons from the helium atom. Normally, when we refer to the

energy levels of an atom, we put the zero of energy at the first ionization limit - for helium this is the ground-state of He+, a one-electron

atom with energy -54.4eV. All of this is just to say that, in this approximation where we neglect the electron-electron repulsion, the energy

required to remove one electron from helium is half that required to remove both. So the ground state energy of helium, in this approxima-

tion, is -54.4eV.

It turns out that states with both electrons excited lie above the ionization energy, and the atom rapidly ionizes if both electrons are excited,

leaving He+ and a free electron. For this reason, we’ll confine ourselves to those states where at least one electron is in the ground state - so

we’re interested in the states 1s1s, 1s2s, 1s2p etc.

5.2. The ground state of helium by perturbation theory

For the ground state, both electrons are in the 1s state. The labels a and b are the same, and our unperturbed wavefunction is

Ψ0 = u1 sH1L u1 sH2L. Now we apply first-order perturbation theory to find the energy shift due to the electron-electron repulsion. The answer is

(no need for you to be able to calculate this):


20

For the ground state, both electrons are in the 1s state. The labels a and b are the same, and our unperturbed wavefunction is

Ψ0 = u1 sH1L u1 sH2L. Now we apply first-order perturbation theory to find the energy shift due to the electron-electron repulsion. The answer is

(no need for you to be able to calculate this):

(5.8)DEI1 s2M = [Ψ0

e2

4 Π Ε0 r12

Ψ0_ =1

4 Π Ε0

5

8

Z e2

a0

=5

8Z Α

2me c

2= 34 eV.

So the ground state energy is shifted up by 34eV from its unperturbed value, making the ground-state energy of helium -20.4eV. Notice how

large the energy shift due to the perturbation is! It’s so large that just applying the first order shift is unlikely to be very accurate. The true

ground-state is -24.58eV.

5.3. The variational method

Our perturbation approach was not very accurate for finding the ground state energy of helium. Instead, we can use a different approximate

method which we find to be much more accurate. It is called the variational method, and it is based on the following statement:

è Consider a system with Hamiltonian H`

. The expectation value of H`

evaluated using any arbitrary wavefunction (let’s call it v) is always

larger than or equal to the true ground-state energy, E0:

(5.9)à v*

H`

v â V ³ E0.

The equality itself holds when v is the true ground state wavefunction.

This is an amazing result. It tells us that we can discover the ground state energy of any system just by guessing. We guess a wavefunction,

then evaluate the expectation value of the Hamiltonian using then guess. Whatever guess we make, that expectation value will be larger the

true ground-state energy. The better our guess, the closer we’ll get to the true energy. Usually, we leave some free parameters in the guess

wavefunction and then vary them to find the parameters that gives us the minimum expectation value. This is an excellent way of getting close

to the true energy.

Proof

The proof of this result is outside the syllabus, but the result is amazing and the proof is simple, so let’s look at it.

We have some Hamiltonian H`

, and we'd like to find the energy eigenvalues En and the corresponding eigenfunctions Φn. They satisfy

H`

Φn = En Φn. Let's take any arbitrary function (our guess), call it v. We'll assume it's normalized (if it wasn't already, normalize it!). We don't

yet know what the Φn are, but we know that they must form a complete orthonormal set. So we can expand our function v on the basis of the

complete set of functions Φn:

v = ân

an Φn.

Calculate the expectation value of H using the functions v:

XH` \ = à v

*H`

v â V = à Jân

an

*Φn

*N H` Jâ

mam ΦmN â V = â

nâ

man

*am à Φn

*H Φm â V .

Using the fact that H`

Φm = Em Φm and that the Φn are orthonormal, we get

XH` \ = â

nan

2En.

The ground state is the one that has the lowest energy, E0. So En ³ E0 for all n. Using this, and the fact that Ún an2 = 1, we arrive at the

remarkable result

XH` \ ³ E0.

5.4. The ground state of helium by the variational method

Let’s apply this method to helium. We first need to come up with a guess wavefunction. When we neglected the electron-electron repulsion,

we found a ground state wavefunction Ψ0 = u1 sHr1L u1 sHr2L, where u1 sHrL are the ground-state wavefunctions for a one-electron atom,

u1 sHrL =Z

3

Π a03

12expH-Z r a0L.

The electron-electron repulsion tends to counteract the central pull of the nucleus. The nucleus pulls both electrons towards the centre, but

they cannot both go to the centre because then their mutual repulsion pushes them apart more strongly. This is the same idea that we

introduced above - that the nuclear charge is partly screened by the electron-electron repulsion. We will return to this idea in more detail when

we study many-electron atoms. For now, let this idea inspire our guess wavefunction which will be vHr1, r2L = u1 s

Zeff Hr1L u1 s

Zeff Hr2L, where

u1 s

Zeff HrL =Zeff

3

Π a03

12expH-Zeff r a0L.

This is just the same as the above wavefunction, except with Z replaced by Zeff - an effective nuclear charge. We will not specify what value

it has - instead, we leave it as a free parameter in the variational procedure. Now all we have to do is evaluate Yv H`

v] using this v and the

helium Hamiltonian, H`

= H`

1 + H`

2 + H '`

, where H`

1, H`

2 are given in (5.3) and H`

' in equation (5.4). Here, we have to remember that while

u1 sHr1L is an eigenfunction of H`

1, u1 s

Zeff Hr1L is not (because Zeff ¹ Z). To calculate the expectation value we could tediously evaluate all the

integrals. Instead of doing that, let’s find the results by inspection. Here are the logical steps:


21

This is just the same as the above wavefunction, except with Z replaced by Zeff - an effective nuclear charge. We will not specify what value

it has - instead, we leave it as a free parameter in the variational procedure. Now all we have to do is evaluate Yv H`

v] using this v and the

helium Hamiltonian, H`

= H`

1 + H`

2 + H '`

, where H`

1, H`

2 are given in (5.3) and H`

' in equation (5.4). Here, we have to remember that while

u1 sHr1L is an eigenfunction of H`

1, u1 s

Zeff Hr1L is not (because Zeff ¹ Z). To calculate the expectation value we could tediously evaluate all the

integrals. Instead of doing that, let’s find the results by inspection. Here are the logical steps:

è Remember that Yu1 s H`

1 u1 s] is just the ground-state energy of the one-electron atom, i.e. Yu1 s H`

1 u1 s] = -1

2Z

2 Α2me c

2.

è Recall the virial theorem, which tells us that XT`

1\ = -1 2 XV1\, where T`

1and V1 are the kinetic and potential energies that make up H1.

è It follows that Yu1 s T`

1 u1 s] = -Yu1 s H`

1 u1 s] =1

2Z

2 Α2me c

2.

è T`

1 doesn’t depend on Z, so the Z-dependence in this last result must come only from the wavefunction. Thus, when we replace u1 s by u1 s

Zeff ,

we must get the same result but with Z replaced by Zeff . i.e. Yu1 s

ZeffT`

1 u1 s

Zeff ] =1

2Zeff

2 Α2me c

2.

è Using the virial theorem once again, we see that Xu1 s V1 u1 s\ = -2 Yu1 s T`

1 u1 s] = Z2 Α2

me c2.

è V1 depends linearly on Z, so in the above one power of Z comes from V1 while the other comes from the wavefunction. Thus, when we

replace u1 s by u1 s

Zeff , we must get the same result but with one power of Z replaced by Zeff (and the other left alone). i.e.

Yu1 s

ZeffV1 u1 s

Zeff ] = Z Zeff Α2me c

2.

è Equation (5.8) tells us that Zu1 sH1L u1 sH2L e2

4 Π Ε0 r12

u1 sH1L u1 sH2L^ =5

8Z Α2

me c2. The electron-electron repulsion doesn’t depend on Z, so

the Z dependence must come from the wavefuction. It follows that Zu1 s

Zeff H1L u1 s

Zeff H2L e2

4 Π Ε0 r12

u1 s

Zeff H1L u1 s

Zeff H2L^ =5

8Zeff Α2

me c2.

Putting all this together, we arrive at the required expectation value:

(5.10)XH` \ = Yu1 s

Zeff H1L u1 s

Zeff H2L H`

1 + H`

2 + H`

' u1 s

Zeff H1L u1 s

Zeff H2L] = Zeff2

- 2 Z Zeff +5

8Zeff Α

2me c

2.

Now we find the value of Zeff that minimizes XH` \, by setting

âXH` \

âZeff

= 0. This gives a minimum Zeff = Z - 5 16, which for helium (Z = 2) is

Zeff = 27 16 = 1.69. Putting this back into the above equation gives us a minimum expectation value:

XH` \min = -2.848 Α

2me c

2= -77.5 eV.

As before, this is the energy required to remove both electrons from the ground state, but we really want the energy relative to the ionization

energy. That’s fine because He+ is a one-electron atom whose ground state we can calculate exactly- it’s the -54.4eV we already found

above. So we reach the result that the ground state energy of helium must be less than -23.1eV. This is much closer to the true energy

-24.58eV) than we found by perturbation theory.

Note that we used an extremely simple guess wavefunction, and yet the variational method got us close to the correct result. Using more

sophisticated trial wavefunctions, with additonal parameters to vary, it’s possible to get a result that is exceedingly close to the true energy;

so although we can’t write down an exact formula for the energy levels, it is possible to calculate them numerically to as great a precision as

we could ever want. Note also that the method gives us some insight - using an effective Z smaller than 2 minimized the energy, which is

telling us that our notion of a screened nucleus is a useful one.


è We can find the ground state of helium using perturbation theory, first neglecting the electron-electron repulsion and then including it as a

perturbation.

è When the electron-electron repulsion is neglected the electrons move independently and the wavefunctions are simply products of

hydrogenic wavefunctions.

è The variational method tells us that for any trial wavefunction v, Yv H`

v] is always greater than or equal to the true ground state energy.

è A simple trial wavefunction for helium is to use the hydrogenic wavefunctions with Z replaced by an effective nuclear charge to account

for screening of the nucleus. This gives a good estimate of the ground state energy.

è Using more sophisticated trial wavefunctions the helium ground state can be found to extremely high accuracy.

5.6. To do

Revise the material in this handout, particularly the application of the variational method. Useful reading: chapter 4 of Softley, chapter 5

of Woodgate, or chapter 3 of Foot.


22

6. Helium, part II

6.1. The excited states of helium and exchange symmetry

In the last lecture we looked at the ground state of helium. Now we look at the excited states. Our unperturbed wavefunction is again

(6.1)Ψ0 = uaH1L ubH2L,but now the labels a and b are different. Consider the alternative wavefunction

(6.2)Ψ0 ' = ubH1L uaH2L.It is exactly the same as wavefunction (6.1), but with the coordinates 1 and 2 swapped. Swapping the two electrons cannot possibly change

anything - this is both physically obvious and obvious from inspection of equation (5.1) where swapping coordinates 1 and 2 leaves the

equation unchanged. Both Ψ0 and Ψ0 ' are eigenfunctions of our zeroth order Hamiltonian, both with the same energy eigenvalue - there is a

degeneracy with respect to the exchange of electrons. What’s more, any arbitrary linear combination of these two eigenfunctions,

Α uaH1L ubH2L + Β ubH1L uaH2L is also an eigenfunction. How are we to know what Α and Β should be?

It turns out that the eigenfunctions we actually need are the ones that are either symmetric or antisymmetric under exchange:

(6.3)Ψ+

=1

2

@uaH1L ubH2L + ubH1L uaH2LD =1

2

Huab + ubaL,

(6.4)Ψ-

=1

2

@uaH1L ubH2L - ubH1L uaH2LD =1

2

Huab - ubaL.

In the final expressions, we’ve used a further shorthand, simply writing uaH1L ubH2L as uab and writing ubH1L uaH2L = uba .

Understanding exchange symmetry

In this subsection, we argue why it is these particular eigenfunctions that we need. This is a slightly more advanced topic which can be

skipped if you prefer.

Let's review where we're up to. We'd like to apply perturbation theory to calculate the effect of the electron-electron repulsion. To apply

perurbation theory, all we need to know is the perturabation and the zeroth-order eigenfunctions. We know what the perturbation is, but we're

not sure what the zeroth-order eigenfunctions should be because there's a degeneracy. Α uaH1L ubH2L + Β ubH1L uaH2L is an eigenfunction of the

Hamiltonian for any values of Α and Β. So we need to find the proper values of Α and Β. This is a very common situation, and it is always

resolved by finding a new operator that commutes with the Hamiltonian. The proper eigenfunctions to use are the ones that are simultaneous

eigenfunctions of the Hamiltonian and of this new operator - this new requirement will resolve the arbitrariness of Α and Β.

In the present case, the degeneracy we’ve identified is a degneracy with respect to exchange of labels. So let’s simply introduce an operator

that performs this exchange operation:

(6.5)O` @uaH1L ubH2LD = ubH1L uaH2L

The operator obviously commutes with the Hamiltonian because swapping the two electrons doesn't change the Hamiltonian. It should also be

obvious from the above equation that uaH1L ubH2L is not an eigenfunction of the exchange operator (the state on the right-hand state is just

different to the one on the left-hand side). If we operate with O`

twice, we get back to the state we started with - so O` 2

has eigenvalue 1. It

follows that the eigenvalues of O`

are ±1. The corresponding eigenfunctions are the symmetric and antisymmetric eigenfunctions given by

equations (6.3) and (6.4) above.

6.2. The effect of the electron-electron repulsion in the excited states

There’s something very interesting about these two states, Ψ+ and Ψ-. Think about the distance between the two particles (the expectation

value of r12). Notice that if we reduce this distance to zero (r1 = r2) Ψ- goes to zero - for the Ψ- state the probability of the electrons being

very close together goes to zero. This is not the case for Ψ+ which remains non-zero when the electrons are close together. So the electrons

are much more likely to be close together when they are in state Ψ+ than when they are in the state Ψ-. You can think of this as the overlap-

ping parts of the wavefunctions of the two electrons interfering constructively in Ψ+ but destructively in Ψ-.

Now we turn to the perturbation, H`

'. We can see qualitatively what the perturbation will do. It is an electrostatic repulsion between the

electrons, so its effect must be to raise the energy. The repulsion is obviously largest when the distance between the electrons is small. But

we’ve already argued that the probability of being at small distances apart is greater for Ψ+ than for Ψ-. So the upward energy shift produced

by the perturbation is greater for Ψ+ than for Ψ- - the energy level corresponding to Ψ+ will be higher than the one corresponding to Ψ-.

To calculate the energy shift, we apply first order perturbation theory.

(6.6)DE+= YΨ

+H`

' Ψ+] =

1

2Yuab + uba H

`' uab + uba] = Yuab H

`' uab] + Yuab H

`' uba],

(6.7)DE-= YΨ

-H`

' Ψ-] =

1

2Yuab - uba H

`' uab - uba] = Yuab H

`' uab] - Yuab H

`' uba].

In the last step, we’ve used the fact that Yuba H`

' uba] = Yuab H`

' uab] and Yuba H`

' uab] = Yuab H`

' uba] which follows from the fact that

nothing changes if we simply swap the electrons. It is usual to give names and symbols to these matrix elements. We define


23

In the last step, we’ve used the fact that Yuba H`

' uba] = Yuab H`

' uab] and Yuba H`

' uab] = Yuab H`

' uba] which follows from the fact that

nothing changes if we simply swap the electrons. It is usual to give names and symbols to these matrix elements. We define

(6.8)J = Yuab H`

' uab],and call this the direct integral, and define

(6.9)K = Yuab H`

' uba].which we call the exchange integral. Remember that ua and ub are just the well known eigenfunctions of hydrogen, so the integrals can be

calculated for any state. But for now we’ll just call them J and K and remember that their values depend on which state we are talking about.

It is worth noting that YΨ+H`

' Ψ-] = YΨ-H`

' Ψ+] = 0. This means that the perturbation does not connect the two states of opposite

exchange symmetry. This is an important point - without this the perturbation theory we have used would not be valid.

6.3. Spin

So far, we have understood that for each configuration of the two electrons (apart from the ground state 1 s2) there are two allowed states,

Ψ+ and Ψ-, that have different energies and different properties under the exchange of the electrons - one is symmetric and the other is

antisymmetric. We have not said anything at all about the spins of the two electrons. Our Hamiltonian does not couple the spin degree of

freedom to the spatial degree of freedom, so the total wavefunction is just a product of a spatial part and a spin part. For two electrons there

are 4 possible spin states - as given by equations (3.11) - (3.14). These spin states are reproduced here. Recall that on the left hand side of

each equality the spin state is represented using the total spin S, and its projection M , S, M\ , while on the right hand side the state is

represented by the individual projection quantum numbers m1, m2\1, 1\ = \,

1, 0\ = ¯\ + ¯ \

2

,

1, -1\ = ¯ ¯\,

0, 0\ = ¯\ - ¯ \

2

.

Just as for the spatial wavefunctions, let’s consider the symmetry of these states under exchange of particles. It’s obvious from the right hand

sides that the first three states, the ones with S = 1, are symmetric under exchange, while the fourth, with S = 0, is antisymmetric.

Since there are four spin states and the two spatial states HΨ+, Ψ-) we might imagine that there are now 8 allowed states in total. This is not

the case. To understand why, we have to introduce a new concept - the fact that the total wavefunction for a system of fermions has to be

antisymmetric under the exchange of particles. This means that we are constrained in the way we put together the spatial states and the spin

states - the only ones we are allowed are Ψ+\ 0, 0\ and Ψ-\ 1, M\ (with M = 0, ± 1). So the symmetric spatial state HΨ+L which is the

one of higher energy always goes together with the antisymmetric spin state which is the state of total spin zero. Since there is only one of

these states, we call it a singlet. Similarly, the antisymmetric spatial state HΨ-) which has the lower energy always goes together with the

symmetric spin states which are the three states with total spin one, S = 1. Since there are three of these states, we call it a triplet.

6.4. Energy level diagram

The diagram below shows the effect of the electrostatic repulsion and the antisymmetry requirement, for the state with n = 1, n ' = 2. First,

the direct integral (J) increases the energy and the increase is greater for the 1s2p state than it is for the 1s2s state. So the configurations 1s2s

and 1s2p are not degenerate any more. That the 1s2s state is lower in energy can be understood from the fact that the p-electron cannot get

close to the nucleus and so the attraction to the nucleus is better shielded by the inner s-electron. The exchange integral (K) splits each

configuration into two levels (they are known as terms). The splitting is entirely due to the electrostatic repulsion, but the antisymmetry

requirement means that each term is identified with a given spin-state (either singlet or triplet).


24

n=1,n'=2

unperturbed

1s2p

1s2s

1S

3S

1P

3P

J2 s

J2 p

2K2 p

2K2 s


è The structure of helium can be understood using perturbation theory, the concept of exchange symmetry, and the requirement that the

complete wavefunction HspaceL ´ HspinL has to be antisymmetric.

è For each electron configuration (apart from the ground state) there are two states of different energy due to the effect of electrostatic

repulsion between electrons.

è The state of higher energy is the one whose spatial part is symmetric, and spin part is antisymmetric. It has S = 0 and is called a spin

singlet.

è The state of lower energy is the one whose spatial part is antisymmetric and spin part is symmetric. It has S = 1 and is called a spin triplet.

è For the ground state, only the singlet state exists.

è The low-lying energy levels of helium lie below the corresponding levels of hydrogen, because the screening of the nucleus is incomplete.

è For the high-lying excited states the screening is almost perfect, and the energies are the same as for hydrogen.

6.6. To do

Revise the material in this handout. Useful reading: chapter 4 of Softley, chapter 5 of Woodgate, or chapter 3 of Foot.

Do question 4 from problem set 2.


25

7. The central field approximation, the alkalis, and other many-electron atoms

We now move on to study atoms with many electrons.

7.1. The central field approximation

The Hamiltonian that describes the electrostatic interactions for N electrons and a nucleus of charge Z e is

(7.1)H`

= âi=1

N

-Ñ2

2 me

Ñi2

-Z e

2

4 Π Ε0 ri

+ âi> j

e2

4 Π Ε0 rij

Here, Ñi2 is the Laplacian operator with derivatives taken with respect to the coordinates of the ith electron, ri is the distance of the ith

electron from the nucleus, and rij is the distance between electrons i and j. In the last sum, we have to sum over all electron pairs: so we sum

over all j from 1 to N, and for each j we sum over all i from j + 1 to N (so that we don’t do any double-counting). The Schrödinger equation

is H`

ΨHr1, r2 ... rN L = E ΨHr1, r2 ... rN L, where the wavefunction is a function of the coordinates of every electron.

For an atom with many electrons, the mutual repulsion term is large and we cannot treat it as a perturbation. Instead, we make progress by

guessing that the repulsive terms contain a large component which is spherically symmetric. This is the central field approximation. In this

approximation, each electron moves independently in an effective central potential, V HriL, which is made up of the attraction to the nucleus

plus the central part of the electron-electron repulsion which we write as Úi SHriL. Now we can split up our Hamiltonian into a central part and

a residual part:

(7.2)H`

= H`

0 + H`

1,

where our zeroth-order Hamiltonian is

(7.3)H`

0 = âi=1

N

-Ñ2

2 me

Ñi2

+V HriL ,

and

(7.4)V HriL = -Z e

2

4 Π Ε0 ri

+ SHriL.The residual part, H1, is the part of the electron-repulsion that is not contained in the central potential, i.e.

(7.5)H`

1 = âi> j

e2

4 Π Ε0 rij

- âi=1

N

SHriL.

Because the majority (the central part) of the electrostatic repulsion is now included in the zeroth-order Hamiltonian, H`

1 can now successfully

be treated as a small perturbation.

We’ll come to the question of how we find the correct V HriL later. First, let’s examine the effect of applying this central field approximation.

Notice that H`

0 in equation (7.3) is a sum over single-electron Hamiltonians, H`

0 = Úi H`

0 i. As a result, the wavefunction separates into a

product of single particle wavefunctions, and the Schrödinger equation separates into N equations, one for each electron. Writing

ΨHr1, r2 ... rN L = Ψ1Hr1L Ψ2Hr2L ... ΨN HrN L, we have for each electron the equation

(7.6)-Ñ2

2 me

Ñi2

+V HriL Ψn,l,ml= En,l Ψn,l,ml

.

This is the equation for a single-electron moving in a central potential. We already know that we can separate this into an angular part and a

radial part, with the angular part being the spherical harmonics, Yl,mlHΘ, ΦL, labelled by the orbital angular momentum quantum numbers l and

ml. The solution to the radial part depends on the exact form of V HriL, but we know that there will be a set of functions labelled by n and l.

The energy eigenvalues Ei will also, in general, depend on n and l. This is why (anticipating the result) we have labelled the single-particle

wavefunctions with the indices n, l, ml and the energies with the indices n, l. The total energy is just the sum of the individual energies

E = âi=1

N

Eni ,li.

When we include the spin of each electron, we see that there’s one extra quantum number we need - ms - to completely define the state of

every electron.

Now we can write the complete state of the atom just by writing down the quantum numbers n, l, ml, ms for every electron. In doing so, we

have to take care to obey the Pauli exclusion principle which tells us that no two electrons can have the same set of quantum numbers. Since

the energy does not depend on ml or ms, we normally do not write these down, but we remember that for any state n, l there are two possible

values of ms (±1/2) and 2 l + 1 possible values of ml - giving 2 H2 l + 1L states in total. So, 2 electrons for an s state, 6 electrons for a p state,

10 electrons for a d state, 14 electrons for an f state, and so on.


26

Notation

We write the state of the atom by listing the occupied energy levels, for example 1 s2 2 s

2 2 p6 for the ground state of neon, and

1 s2 2 s

2 2 p6 3 s for the ground state of sodium. This is called the configuration. The electrons that have the same n are said to belong to a

shell. Electrons that have the same values of n and l belong to the same sub-shell. So, for sodium, there are 2 electrons in shell 1, 8 electrons

in shell 2 and 1 electron in shell 3. If we talk about sub-shells there are 2 electrons in 1s, 2 in 2s, 6 in 2p and 1 in 3s. We often abbreviate the

configuration by replacing the configuration of all the inner shells by the corresponding element symbol - so we can write sodium as @NeD 3 s.

The filled shells, sometimes called closed shells, form the core, and the ones outside the closed shells are the valence electrons.

7.2. The self-consistent field

To calculate the energy levels and wavefunctions we still have to solve the radial equation, and to do that we need to know V HrL. The method

that is used is the self-consistent field method, sometimes called the Hartree method. Here are the steps that make up the method:

1. Make some reasonable guess for V HrL.2. Solve the single-electron Schrödinger equation (equation (7.6)) using this V HrL to obtain all the eigenvalues and eigenstates, Φn,l. This is

straightforward to do numerically on a computer.

3. Take the many-electron state to be the product of the N single electron eigenstates of lowest energy, filling up the states according to the

exclusion principle.

4. Calculate the charge distribution due to the electrons, i.e. Ρe = -e Φn,l2. Then calculate the electrostatic potential due to the charge of

the nucleus and the charge distribution of the electrons. This is a new potential V ' HrL and is probably not the same as the original potential

V HrL. We take V ' HrL to be a better approximation to the true potential than our original V HrL.5. Iterate steps 2-4, until the potential obtained converges. Then we will have a self-consistent field, meaning that the field we assume in

solving the Schrödinger equation is exactly the field produced by the nucleus and the electron distribution of our solution.

This procedure proves to be successful in predicting the energy-level structure of many-electron atoms.

7.3. The alkalis and the quantum defect

Of all many-electron atoms, the alkalis are the simplest to understand. They all have a single electron outside spherically symmetric closed

shells. Here are the electron configurations:

Li 1s22s

Na 1s22s22p63s

K 1s22s22p63s23p64s

Rb 1s22s22p63s23p63d104s24p65s

Cs 1s22s22p63s23p63d104s24p64d105s25p66s

For Li and Na, the shells fill up in the order that you would expect - first the n = 1 shell fills, then the n = 2 shell and so on. But for K, there is

a deviation from the obvious ordering. The 4s sub-shell fills before the 3d sub-shell. It’s the same story with Rb, where the 5s sub-shell fills

before either the 4d or the 4f sub-shells. We can understand this ordering using the idea of screening.

For a 1 r potential, as in hydrogen, we know that the energy levels are ordered according to the principal quantum number n only; there is no

dependence on l. For many-electrons, the central potential is not simply 1 r and the energies depend on both n and l. We can get a feel for

what the central potential must look like from some simple arguments. If we take an electron out to large r, the nuclear charge Z e is almost

fully screened by the inner electrons which together have a charge HZ - 1L e. So, at large r, the potential just looks like the e2 4 Π Ε0 r

potential of hydrogen. If we take the electron to very small r, the potential will be dominated entirely by the nucleus and will be Z e2 4 Π Ε0 r.

At intermediate distances the form of the potential will be intermediate between these two forms.

Consider the Na atoms. The states 3s, 3p and 3d are not degenerate (as they are in hydrogen). The large orbital angular momentum of the 3d

electron prevents it from reaching small r, so it spends almost all its time outside the closed shell. The inner electrons are very effective at

screening the nucleus and so the energy of the 3d state is very close to the n = 3 level of hydrogen. For the 3s electron there is no angular

momentum barrier and it is able to penetrate the inner core of electrons and see a stronger potential. So its energy is lowered compared to the

n = 3 level of hydrogen. The 3p level lies somewhere in between. Moving now to K, exactly the same principles apply. We find that the 4s

state is shifted down in energy so far that it is below the 3d state, and this is the reason why the 4s state fills up first.

There is a very simple formula that quite accurately describes the energy levels of the alkalis. It is

(7.7)En,l = -h c R¥

Hn - ∆lL2

Here, R¥ is the Rydberg constant (h c R¥ > 13.6 eV). The formula is almost the same as the energy level formula for hydrogen, but instead

of n2 in the denominator we have Hn - ∆lL2. The quantity ∆l is called the quantum defect. Its value depends on l, and varies from one alkali to

another, but it has almost no dependence on n. Moreover, for l > 2, ∆l is zero to a good approximation. This is as we would expect from the

above arguments which show that for high l the energy levels are hydrogenic. The use of the quantum defect is extremely useful - we can

describe the entire energy level structure of an alkali once we know just three numbers, ∆0, ∆1 and ∆2 (sometimes written ∆s, ∆p and ∆d).

The quantum defects are different from one alkali to another. For sodium, the values are ∆0 = 1.36, ∆1 = 0.86, ∆2 = 0.01. Using these

numbers, and equation (7.7), we can draw the complete energy level diagram for sodium.


27

2

3

4

56

¥n

l=0 l=1 l=2 l=3 Hydrogen

3s

3p

3d

4s

4p

4d 4f

Energy HeVL


è A large part of the electron-electron repulsion can be included into a central potential

è In this central potential, the total wavefunction is a product of single-electron wavefunctions, each labelled by the set of quantum numbers

Hn, l, ml, msL.è For a given n, the s state has the lowest energy because an s electron penetrates quite deeply into the electron core and so the shielding of

the nucleus is incomplete. Next in energy is the p state, then the d state and so on. The levels with high angular momentum are

hydrogenic.

è Knowing the ordering of the energy levels, and applying the exclusion principle, the configurations of all the atoms in the periodic table

can be understood.

è The energy levels and eigenfunctions can be calculated using the self-consistent field method.

è The energy levels of the alkalis are described by a simple formula involving the quantum defect.

7.5. To do

Revise the material in this handout. Useful reading: chapter 4 of Softley, chapter 6 of Woodgate, chapter 4 of Foot.



28

8. Finer details of multi-electron atoms

8.1. LS coupling

So far, we have looked at the energy level structure in the central field approximation and understood how this leads to electron configura-

tions. We still need to account for the effect of the residual electrostatic interaction, given by equation (7.7). This is treated as a perturba-

tion. We will not attempt to calculate the energy shifts and splittings due to this perturbation, but instead understand its effect qualitatively.

Once again, we take the approach of identifying quantities that remain constants of the motion in the presence of the perturbation (in other

words, we look for operators that commute with the zeroth-order Hamiltonian and with the perturbation).

The residual electrostatic interaction is an interaction internal to the atom, and so it cannot change the total orbital angular momentum of the

atom. So it’s a good idea to define the total orbital angular momentum

(8.1)L`

= âili

`.

This is a constant of the motion, and we can be sure that, even in the presence of the perturbation, the energy eigenstates must be eigenstates

of L` 2

(with eigenvalue L HL + 1LL and of Lz

` (with eigenvalue ML). Electrons that form closed sub-shells do not contribute to the sum because

their total angular momentum is zero. Suppose we have two electrons outside closed sub-shells for example, with orbital angular momenta l1`

and l2`. In the vector model, the perturbation due to the residual electrostatic interaction causes l1

` and l2

` to precess about the total orbital

angular momentum L`. This is shown in the picture below. We see that l1, l2, L and ML are the constants of the motion, but ml1

and ml2 are

not.

L

l1

l2

The residual electrostatic interaction has no influence on the spin, but the spin enters into our discussion because of the requirement that the

total wavefunction be antisymmetric under exchange of electrons. This is the same as the discussion for helium. So we also introduce the total

spin angular momentum which is the vector sum of all the spins of the individual electrons:

(8.2)S`

= âisi

`.

This is also a constant of the motion. The energy eigenstates are also eigenstates of S` 2

(with eigenvalue SHS + 1L) and of Sz

` (with eigenvalue

MS). Once again, filled shells do not contribute anything to the total spin, so we only have to sum over the electrons that are outside closed

sub-shells. The vector model for the spins is similar to the one drawn above for the orbital angular momenta.

The perturbation causes a given configuration to split into a set of terms, labelled by the total orbital angular momentum L. When we combine

the spatial and spin parts we have to satisfy the exchange-antisymmetry requirement and this requires us to combine states of definite L with

states of definite total spin S. Just as in helium, states with the same L but different S can have different energies due to the exchange

symmetry requirement, even though the interaction that is causing the energy splitting has nothing to do with the spin.

Notation

This coupling scheme, where we add up the individual li`

to make a total L`, and add up the individual si

` to make a total S

` is called LS-

coupling. In some textbooks, it is referred to as “Russell-Saunders coupling”. A configuration splits up into levels labelled by L and S and

called terms. These terms are degenerate with respect to ML and Ms. The notation used to denote a term is

2 S+1L

Just as we use s, p, d ... to stand for l = 0, 1, 2 ..., we use S, P, D ... to stand for L = 0, 1, 2 ... . We write the pre-superscript as 2 S + 1 to

denote the spin multiplicity. There is also a naming convention that follows the multiplicity as shown in the table below:

S multiplicity name

0 1 singlet

12 2 doublet

1 3 triplet

32 4 quadruplet

So the term 2S is a doublet-S, while 3P is a triplet-P.


29

Working out terms

Let’s consider some configurations, and work out what terms are allowed.

The configuration 1s22s The full 1s2 shell does not contribute to the total angular momenta. So L and S are just given by the 2s electron. So

the only possible term is 2S.

The configuration 1s2s. We have l1 = l2 = 0 and s1 = s2 = 1 2. So we must have L = 0, but we can have S = 0 or S = 1. So there are two

terms - 1S and 3S.

The configuration 2p3p. We have l1 = l2 = 1 and s1 = s2 = 1 2. So we can have L = 0, 1, 2, and S = 0, 1. So there are 6 terms -1S, 3

S, 1P, 3

P, 1D, 3

D.

This is all straightforward when the electrons are not equivalent (i.e. not in the same subshell). When the electrons are equivalent, it’s more

complicated because we have to make sure that the total angular momentum state is antisymmetric. This reduces the number of possible

terms. We already saw an example of this for the ground state of helium (1s2) where the only term is 1S. In the case of the configuration 2p2,

it turns out that the only allowed terms are 1S, 3P and 1D.

8.2. Fine structure in LS coupling

Do not confuse LS coupling with the spin-orbit interaction. They are completely different things. LS coupling is the angular momentum

coupling scheme discussed above. It is useful when we introduce the residual electrostatic interaction which leads to terms described by the

values of L and S. We have not yet said anything about the spin-orbit interaction for these many-electron atoms - that’s what we’re going to

do next.

In lecture 4 we looked at the spin-orbit interaction for a one-electron atom and saw how this interaction led to fine-structure splitting. The

same occurs for many-electron atoms. Let’s write down again the perturbation that we derived in lecture 4 for the spin-orbit interaction - it’s

equation (4.12):

Hso

=Ñ2

2 me2

c2

â V

â r

1

r

l`.s`

In deriving this perturbation, we did not assume a Coulomb potential. We only assumed that the electric field was central (depending only on

r). For many-electron atoms in the central-field approximation, each electron moves independently in a central potential V HrL, and we can use

exactly the same perturbation as above, except now we get a term like this for each electron. For electrons in filled sub-shells the total orbital

angular momentum and spin angular momentum are both zero and they make no contribution to the spin-orbit interaction. So it’s only the

valence electrons we have to consider.

One valence electron

In the case of an alkali, which has only one valence electron, the above perturbation is all we have. For an alkali the total orbital angular

momentum L`

is just the orbital angular momentum of the valence electron L`

= l`. Similarly, S

`= s

`. So we can immediately write the perturba-

tion for an alkali as

(8.3)H`

so

= Β L

`.S

`

where

(8.4)Β =Ñ2

2 me2

c2

â V

â r

1

r

.

To evaluate the energy shift due to the perturbation, we proceed in exactly the same way as before. We introduce the total angular momentum

J`

= L`

+ S`. Then, just as before, we can easily find the expectation value of L

`.S

`:

(8.5)YL` .S` ] =

1

2@JHJ + 1L - LHL + 1L - SHS + 1LD.

So, for all terms that have L ¹ 0, the spin-orbit interaction produces a splitting of the term into two components labelled by the value of J -

one has J = L + 1 2 and the other J = L - 1 2. The energy shifts are 1

2X Β\ @JHJ + 1L - LHL + 1L - SHS + 1LD. The slitting between the two

components with J = L ± 1 2 is

(8.6)DE = EJ=L+12 - EJ=L-12 =1

2X Β\ @HL + 1 2L HL + 3 2L - HL - 1 2L HL + 1 2LD = X Β\ HL + 1 2L = X Β\ Jmax

where Jmax = L + 1 2 is the larger of the two allowed values of J . To calculate the size of the splitting, we have to calculate the expectation

value of 1

r

âV

âr in the central field approximation. This can be done once the central potential V and the wavefunction for the valence electron

have been determined using the self-consistent field method described in lecture 7. Without doing any of that, we can infer the important

properties of fine-structure splitting by considering the behaviour of Z 1

r

âV

âr^:

ì For large n and l, the valence electron remains always outside the closed shells, and the potential is like the Coulomb potential in hydro-

gen. So the fine-structure splitting of these high n, l configurations will be nearly the same as in hydrogen.


30

ì For low n and l (but not l = 0 where there is no splitting), the valence electron penetrates into the core electrons and the shielding of the

nucleus is incomplete. As these electrons approach the nucleus the gradient of the potential becomes very large and so the fine-structure

splitting is larger for these core-penetrating states.

ì For the heavier alkalis the un-shielded nuclear charge is greater and so the gradient of the potential for the core-penetrating states is larger

for the heavier alkalis. This means that we expect fine-structure splitting to be larger for heavier alkalis than for light ones.

To illustrate this last point, the table below gives the fine-structure splitting (in GHz) for the lowest energy 2P term of each alkali.

Element Levels Splitting HGHzLLi 2p 2P32-

2P12 10

Na 3p 2P32-2P12 516

K 4p 2P32-2P12 1730

Rb 5p 2P32-2P12 7123

Cs 6p 2P32-2P12 16 612

Below are plotted the lowest-lying energy levels of Cs, with the fine-structure included. The diagram is to scale (which makes the fine-

structure splitting of the d states invisible).

6s

7s

6p

7p

5d

6d

2S12

2S12

2P12

2P32

2P12

2P32

2D32

2D52

2D32

2D52

-4

-3

-2

-1

0

EnergyHeVL

Two (or more) valence electrons

When there are two valence electrons, the perturbation is the sum of the spin-orbit interaction for each electron individually:

(8.7)H`

so

= Β1 l1

`.s1

`+ Β2 l2

`.s2

`

We can use the vector model to help us calculate the expectation value of this perturbation. The picture we need is below. First of all, l1`

and

l2

` precess rapidly around their resultant L

`= l1

`+ l2

` and s1

` and s2

` precess rapidly around their resultant S

`= s1

`+ s2

`. These precessions are due

to the residual electrostatic interaction (plus exchange symmetry), and the precession is rapid because this interaction is strong (compared to

the spin-orbit interaction). L`

and S`

then precess slowly around their resultant J`

= L`

+ S`. This is due to the spin-orbit interaction, and the

precession is slow because the spin-orbit interaction is much weaker than the residual electrostatic interaction.


31

L

l1

l2

S

s1

s2

J

We can replace quantum mechanical expectation values with time-averages in the vector model. In the time it takes L` and S

` to precess around

J`, the individual angular momenta l1

`, l2

` have precessed many times around their resultants, and so only the components parallel to L

`

contribute in the time-average - the perpendicular components time-average to zero. The same is true for the individual spins. The upshot is

that the expectation values of l1`.s1

` and l2

`.s2

` are both proportional to the expectation value of L

`.S

`. So, we end up being able to write the

perturbation as

(8.8)H`

so

= ΒLS L

`.S

`

where ΒLS depends on the configuration and the term. The same applies when there are more than two valence electrons - we again end up

with the same form for the perturbation. So we have the same result as before - the term splits up into levels specified by the values of J , and

the energy shift is given by

(8.9)DEJ =Β

2@JHJ + 1L - LHL + 1L - SHS + 1LD.

where Β is the expectation value of ΒLS and depends on the configuration and the term. The size of the fine structure splitting will follow the

same trends as outlined above for the alkalis.

The energy interval between adjacent J levels is easily found:

(8.10)DEFS = DEJ - DEJ-1 =Β

2@JHJ + 1L - HJ - 1L JD = Β J .

This result is called the interval rule for fine structure in LS coupling. It tells us that the energy difference between adjacent levels is

proportional to the larger value of J .

As an example, consider a 3P term. This could be in helium, or in another atom with two valence electrons. The term has S = 1 and L = 1, and

the spin-orbit interaction splits this into three levels with total angular momentum J = 0, 1, 2. According to the interval rule, the energy

splitting between J = 2 and J = 1 is double the energy splitting between J = 1 and J = 0. The splitting of the term into levels is shown below:

Β

Β

2Β

Β

2Β3P

3P2

3P1

3P0

Below is a schematic energy level diagram for the 3s4s and 3s4p levels of magnesium. The left hand side shows the configurations. The

residual electrostatic interaction splits each of these into two terms. The spin-orbit interaction further splits the 3P term into 3 levels, with

J = 0, 1, 2. The other terms are not split by the spin-orbit interaction because they have L = 0 or S = 0 (or both).


32

3s4p

3s4s

1S

3S

1P

3P

3P2

3P1

3P0

1P1

1S0

3S1

Configurations Terms Levels


è The central field approximation gives us configurations labelled as n1, l1, n2, l2 ... ...

è The residual electrostatic interaction, plus the exchange anti-symmetry requirement, splits configurations into terms labelled by L and S

as 2 S+1L.

è The spin-orbit interaction splits terms into levels labelled by J . The notation is now 2 S+1LJ .

è The spin-orbit interaction, considered in the context of LS coupling, is described by a perturbation that is proportional to L`.S

`.

è There is no fine-structure splitting if either L or S are zero.

è There is an interval rule for the fine structure levels - the energy difference between adjacent levels is proportional to the larger value of

J .

è Fine-structure splitting is largest for the lowest-lying energy levels because for these the valence electrons penetrate the core electrons

most strongly.

è Fine-structure splitting is larger in heavier atoms than in light ones.

8.4. To do

Revise the material in this handout. Useful reading: chapter 7 of Woodgate, chapter 5 of Foot.



33

9. Hyperfine structure

There is one last level of structure that we need to consider, the hyperfine structure. It is due to properties of the nucleus other than its

charge, most notably the magnetic moment of the nucleus and the electric quadrupole moment of the nucleus. Both are important, but we will

discuss only the effect of the magnetic moment (the effect of the electric quadrupole moment is beyond the scope of this course). Although

the hyperfine structure is very small, it plays an extremely important role in atomic physics.

9.1. General treatment

Let the nucleus have a spin angular momentum Ñ I`. Associated with this spin, and parallel to it, the nucleus will have a magnetic moment ΜI

`.

We usually write

(9.1)ΜI

`= gI ΜN I

`.

Here, ΜN =e Ñ

2 Mp

, where Mp is the proton mass. The g factor, gI , is a constant of proportionality. We might think it would be equal to 2, just

like the g factor for an electron, but it isn’t because the nucleus has structure - the protons and neutrons - and even they have structure - the

quarks. So each nucleus has a different value for gI (but they are all of order 1).

There is an interaction between this nuclear magnetic moment and the magnetic field produced at the nucleus by the electrons, Bel. The

interaction Hamiltonian is

(9.2)H`

hfs = -ΜI

`.Bel.

The magnetic field produced by the electrons is due both to their orbital angular momentum around the nucleus and their intrinsic magnetic

moments proportional to their spins. The total magnetic field is parallel to the total electronic angular momentum J`. The reasoning that leads

to this result should be familiar by now: the magnetic field due to the individual li`

is along L`

because the li`

precess rapidly around L`;

similarly, the magnetic field due to the individual si

` is along S

`; since L

` and S

` precess around J

`, the resultant magnetic field is along J

`. So we

can write Bel µ J`, and our interaction Hamiltonian has the form

(9.3)Hhfs = Ahfs I`.J

`.

We find the energy splitting by applying first-order perturbation theory. We proceed in exactly the same way as we did for fine structure.

First, we introduce the total angular momentum F`. It is the sum of the total electronic angular momentum J

`, and the nuclear spin I

`, and is

now the true total angular momentum of the atom (there is nothing more we need to include!):

(9.4)F`

= I`

+J`.

We can form eigenstates of this total angular momentum, which we write as I; J ; F , MF \, meaning that I` and J

` are coupled to form a

resultant F`. This state is an eigenstate of I

`2

, J` 2

, F` 2

and Fz

`, with eigenvalues IHI + 1L, JHJ + 1L, FHF + 1L and MF respectively. It is not an

eigenstate of Iz

` or of Jz

`. Since F

` 2

= I`2

+ J` 2

+ 2 I`.J

`, we can write

(9.5)Hhfs =Ahfs

2J F

` 2

- I`2

- J` 2N.

Using first-order perturbation theory, and our I; J ; F , MF \ eigenstates, the energy shift is

(9.6)DEF = YI; J ; F , MF H`

hfs I; J ; F , MF ] =A

2@FHF + 1L - IHI + 1L - JHJ + 1LD,

where, A = XAhfs\. The hyperfine interaction causes a splitting of a level into hyperfine components labelled by the total angular momentum

F , which can take values between J - I and J + I. Once again, there is an interval rule which says that the energy difference of adjacent

levels is proportional to the larger value of F :

(9.7)DEhfs = DEF - DEF-1 = A F

The constant A may be positive or negative. When it is positive, the component with largest F has the highest energy, and when it is negative

it is the component of lowest F that is highest in energy.

We can see that the hyperfine interaction will be small - the form of the interaction is similar to the spin-orbit interaction, except with the

nuclear magnetic moment ΜI

` replacing the electron magnetic moment Μs

`. Since ΜI Μs = me Mp >

1

2000, the hyperfine structure is expected

to be about 3 orders of magnitude smaller than the fine-structure. Note that many nuclei have zero spin, in which case the atom has no

hyperfine structure.

This deals with hyperfine structure in general. We gain more insight by looking at some specific and important examples, which is what we

do next.

9.2. Hyperfine structure in the ground state of hydrogen

Consider the ground state of hydrogen, 1s. The electron has no orbital angular momentum, and so the only contribution to Bel comes from the

intrinsic magnetic moment Μs

`= -gs ΜB s

`. To find the magnetic field this produces at the nucleus we can use the following result from

electromagnetism: the magnetic field inside a uniformly magnetized sphere of magnetization M is 2

3Μ0 M . The magnetization is the magnetic

moment per unit volume, and so the magnetization at position r due to the electron’s magnetic moment is the total magnetic moment multi-

plied by the probability density of finding the electron at position r, which is just ΨHrL 2 . This is not a constant everywhere, but it is very

nearly constant over the volume occupied by the nucleus which is what matters to us. So the magnetization at the nucleus is


34

Consider the ground state of hydrogen, 1s. The electron has no orbital angular momentum, and so the only contribution to Bel comes from the

intrinsic magnetic moment Μs

`= -gs ΜB s

`. To find the magnetic field this produces at the nucleus we can use the following result from

electromagnetism: the magnetic field inside a uniformly magnetized sphere of magnetization M is 2

3Μ0 M . The magnetization is the magnetic

moment per unit volume, and so the magnetization at position r due to the electron’s magnetic moment is the total magnetic moment multi-

plied by the probability density of finding the electron at position r, which is just ΨHrL 2 . This is not a constant everywhere, but it is very

nearly constant over the volume occupied by the nucleus which is what matters to us. So the magnetization at the nucleus is

(9.8)MH0L = -gs ΜB s Ψ1 sH0L 2

and the magnetic field this produces is

(9.9)Bel = -2

3gs Μ0 ΜB Ψ1 sH0L 2

s`.

Using equation (9.1) for ΜI

`, and equation (9.2) for the hyperfine Hamiltonian, and the fact that for our single s electron in hydrogen J

`= s

`, we

obtain

(9.10)H`

hfs =2

3gs gI Μ0 ΜB ΜN Ψ1 sH0L 2

I`.J

`,

which has the same form as equation (9.3), with the quantity A now specified: A =2

3gs gI Μ0 ΜB ΜN Ψ1 sH0L 2. We found the hydrogenic

wavefunctions in lecture 2. For the s state with principal quantum number n:

ΨnsH0L 2=

1

Π a03

n3

.

So, for the specific case of the ground state of hydrogen, we get

(9.11)A =2

3gs gI Μ0 ΜB ΜN

1

Π a03

.

The proton spin is I = 1 2, and the electron spin is J = 1 2. So the ground state is split into two hyperfine components, F = 1 and F = 0,

separated by the amount A (see equation (9.7)). The splitting is shown in the diagram below. All of the quantities in the above formula for A

are know (for a proton, the gI factor is 5.585694713). You can look up the values in a table and thus calculate the expected value of the

hyperfine splitting in the ground state of hydrogen. Calculate it as a frequency - you should get 1.42 GHz, close to the (extremely precisely!)

measured value - DEhfs h = 1 420 405 751.7667 ± 0.0009 Hz.

A

4

-3A

4

J=12, I=12F=1

F=0

If a hydrogen atom is in the upper of the two hyperfine states HF = 1), it can decay to the ground state (F = 0) by emitting radiation at this

frequency. The wavelength of this radiation is 21cm. This radiation is emitted by the vast quantities of hydrogen in our own galaxy and all the

galaxies in the universe. It can be detected and mapped using radio telescopes and is extremely important in astronomy and cosmology

The hydrogen maser

In a hydrogen maser, the frequency of the ground-state hyperfine interval of hydrogen is used as an extremely precise and stable frequency

reference. Look it up (e.g. on wikipedia).

9.3. Hyperfine structure in the ground state of caesium

The hyperfine structure in the ground state of Cs is enormously important because the frequency of this hyperfine interval defines the second.

Cs has just one stable isotope, Cs-133, and its nuclear spin is I = 7 2. The ground electronic state of Cs is 6s 2 S12. The J = 1 2 couples to

the I = 7 2 to give two hyperfine components with total angular momentum F = 3, 4, with F = 3 having the lower energy. The splitting is

shown below.


35

7A

4

-9A

4

J=12, I=72

F=4

F=3

The calculation of the size of the hyperfine splitting is similar to that for hydrogen given above. The only difference is in the wavefunction at

the origin, which of course has a different value in Cs than it does in hydrogen. For the ground state of the alkalis, a good approximation for

the wavefunction at the origin is

ΨH0L 2=

Z

Π a03Hn - ∆0L3

.

The only change relative to hydrogen is the appearance of the nuclear charge, Z, in the numerator, and the replacement of n by n - ∆0 in the

denominator, where ∆0 is the quantum defect for s-states that we met in lecture 7. For Cs, Z = 55, n - ∆0 = 1.87, gI = 0.74. So, scaling from

the hydrogen result we would expect the hyperfine interval for Cs to be given approximately as

DECs

DEH

»ZCs

Hn - ∆0LCs3

gI,Cs

gI,H

Fmax,Cs

Fmax,H

=55

1.873

0.74

5.59

4

1= 4.5.

This gives an estimate for the Cs hyperfine frequency of 6.4 GHz. In fact, the frequency is higher - 9.2GHz - but our estimate is not too bad

given that it’s non-relativistic. The important point to notice is that, unlike the fine-structure splitting which increases enormously as we go to

heavier atoms, the size of the hyperfine splitting does not change very much between light and heavy atoms.

The caesium clock and the definition of the second

The hyperfine interval in the ground state of Cs defines the SI unit of time, the second. In a caesium atomic clock, the frequency of this

interval is used as an extremely precise and stable frequency reference. Look this up (e.g. on wikipedia, or the website of the UK National

Physical Laboratory (NPL)).


è Hyperfine structure is due to properties of the nucleus other than its charge, most importantly its magnetic dipole moment and its electric

quadrupole moment.

è The hyperfine hamiltonian is of the form A I`.J

`.

è It leads to the splitting of energy levels into hyperfine components labelled by the total angular momentum F , which takes values from

J - I to J + I.

è Hyperfine structure is much smaller than fine structure.

è The ground state hyperfine splitting of hydrogen can be calculated quite accurately, and has also been measured to very high precision. It

is the origin of the ubiquitous 21cm radiation and is the basis of the hydrogen maser.

è The hyperfine splitting of the ground states of all the alkalis lie in the 1-10 GHz range.

è The hyperfine frequency of the ground state of Cs defines the second.

9.5. To do

Revise the material in this handout. Useful reading: chapter 9 of Woodgate and sections 6.1 & 6.4 of Foot.

Read about the hydrogen maser and caesium atomic clock, e.g. on Wikipedia or the website of the National Physical Laboratory (NPL).



36

10. The Zeeman effect

The Zeeman effect describes the interaction of atoms with a static magnetic field. The total magnetic moment of the atom, Μ`, interacts with

the applied field, B, and the energy depends on the relative orientation of the magnetic moment and the magnetic field. The term that we have

to add to the total Hamiltonian is

(10.1)H`

Z = -Μ`.B,

which we treat as a small perturbation.

To find the total magnetic moment of the atom, we have to add up the contributions from all the electrons. For each electron there is a

magnetic moment due the orbital motion,

Μl

`= -gl ΜB l

`,

and an intrinsic magnetic moment associated with the spin,

Μs

`= -gs ΜB s

`.

Here, the g-factors (along with the constant ΜB) give the constant of proportionality between the magnetic moment and the associated angular

momentum: gl = 1 (exactly) and gs > 2 (it's slightly greater than 2 because of small QED corrections). Note the sign as well - the magnetic

moments point in the opposite direction to the angular momenta. The magnetic moments of the core electrons all cancel out, because their

angular momenta add up to zero, so it's only the valence electrons that matter.

We will see that the energy shifts produced by the perturbation are small, even compared to the fine structure. So we suppose that we have

already found the energy level structure of the atom down to the level of detail of the fine structure, and our job now is to calculate what

happens to these levels when the magnetic field is applied. Remember that the fine structure levels are labelled by the quantum numbers

L, S, J and MJ , so our job is to calculate the energy shift of the eigenstates L; S; J , MJ \ due to the perturbation - i.e. to calculate

XL; S; J , MJ HM L; S; J , MJ \. We'll do a complete calculation in a moment, but first we'll make some simple arguments which will give us

the correct energy shifts up to a factor of order 1.

10.1. Simple calculation

Since each valence electron has a magnetic moment of order ΜB, the total magnetic moment will also be of order ΜB. This total magnetic

moment must be in the same direction as the total angular momentum of the atom, J`. So, by analogy with the above, we can write

(10.2)Μ`

= -gJ ΜB J`,

where gJ is a proportionality constant which we expect to be of order 1 (we’ll calculate it in a moment). It is customary to set up the

coordinate system so that the z-axis is along the magnetic field direction, so that we have B = B z`, where z

` is a unit vector along the z-axis.

Using this, along with equations (10.1) and (10.2) we have

(10.3)H`

Z = gJ ΜB B Jz

`.

The energy shift due to the perturbation is

(10.4)DEZ = YL; S; J , MJ H`

Z L; S; J , MJ ] = gJ ΜB B YL; S; J , MJ Jz

`L; S; J , MJ ] = gJ ΜB B MJ .

In the last step, we’ve used the fact that our states are eigenstates of Jz with eigenvalue MJ , i.e. Jz

`L; S; J , MJ ] = MJ L; S; J , MJ ]. This

result contains all the important information. First, we see that the energy levels of the atom split up into sub-levels with the energy being

proportional to MJ (i.e. the projection of the total electronic angular momentum onto the z-axis). This is what we would expect - the energy

depends on the relative orientation of Μ and B, and Μ is parallel to J while B is parallel to the z-axis. We see that the energy splitting is

proportional to the magnitude of the magnetic field, which again makes good sense. The approximate size of the energy shift is ΜB B. The

value of ΜB is is 9.27 ´ 10-24J T. Converting this to a frequency, we get ΜB h > 14 GHz T. Now, 1T is a very large magnetic field. The

earth’s field, for example, is about 5 ´ 10-5T. So in most situations, we expect the field to be less than 1T and the energy level shifts

produced by the field to be less than about 10 GHz (this is 4 ´ 10-5 eV). This is much smaller than the gross energy structure, the splitting of

configurations into terms by the residual electrostatic interaction, and (for most atoms), the fine-structure splitting. This explains why we have

applied this perturbation last - it’s the smallest of all the effects.

Historical note

Some text books make a distinction between the “normal Zeeman effect”, and the “anomalous Zeeman effect”. This is historical, and has to

do with the discovery of spin. You shouldn’t bother yourself with any of this (unless you’re interested in the history) - there’s just one effect,

it’s called the Zeeman effect, and it produces the energy shift calculated above.

10.2. Calculating gJ

Adding up all the contributions to Μ`, our perturbation is

(10.5)H`

Z = Iâigl ΜB li

`+ gs ΜB si

` M.B = Igl ΜB L`

+ gs ΜB S` M.B = ΜB B IgL Lz

`+ gs Sz

` M.Here, we’re using the total orbital and spin angular momenta once again (L

`= Úi li

`, S

`= Úi si

`, by definition). In the last step we’ve used the

fact that the magnetic field is in the z-direction.

Note that if gs were equal to gl (i.e. 1), we would simply have H`

Z = ΜB BILz

`+ Sz

` M = ΜB B Jz

`, and thus DEZ = ΜB B MJ - i.e. our gJ factor

would just be 1. But gs is not equal to gl, and that makes the calculation slightly more complicated.


37

Note that if gs were equal to gl (i.e. 1), we would simply have H`

Z = ΜB BILz

`+ Sz

` M = ΜB B Jz

`, and thus DEZ = ΜB B MJ - i.e. our gJ factor

would just be 1. But gs is not equal to gl, and that makes the calculation slightly more complicated.

We have to calculate the expectation values of Lz

` and Sz

`, but our states L; S; J , MJ \ are not eigenstates of Lz

` and Sz

`. To do the calculation,

we turn once again to the vector model. The picture is shown below.

J

L

SMJ

B

In this picture L` and S

` are precessing rapidly around their resultant J

`= L

`+ S

`, because of the spin-orbit interaction. The magnetic field exerts

a torque which causes the total angular momentum J` to precess around the z-axis, but this precession is much slower because the interaction

with the magnetic field is much weaker than the spin-orbit interaction. So to calculate the expectation value of Lz

` we first project L

` onto J

`,

and then we project J` onto the z-axis:

(10.6)XLz

` \ = [ IL` .J`

M Jz

`

J` 2

_

Similarly

(10.7)XSz

` \ = [ IS` .J`

M Jz

`

J` 2

_

We can calculate these expectation values using methods we’ve used before. Since S`

= J`

-L`, we have L

`.J

`=

1

2JJ` 2

+ L` 2

- S` 2N. So

(10.8)XLz

` \ = [JJ` 2

+ L` 2

- S` 2N Jz

2 J` 2

_ =JHJ + 1L + LHL + 1L - SHS + 1L

2 JHJ + 1L MJ

Similarly,

(10.9)XSz

` \ = [JJ` 2

+ S` 2

- L` 2N Jz

`

J` 2

_ =JHJ + 1L - LHL + 1L + SHS + 1L

2 JHJ + 1L MJ

Using equations (10.5), (10.8) and (10.9), and using gl = 1, gs = 2, we find the energy shift to be

(10.10)DEZ =3 JHJ + 1L - LHL + 1L + SHS + 1L

2 JHJ + 1L ΜB B MJ

This has exactly the same form as equation (10.4) - DEZ = gJ ΜB MJ - but now we also find the value of gJ (which is known as the Landé g-

factor):

(10.11)gJ =3 JHJ + 1L - LHL + 1L + SHS + 1L

2 JHJ + 1L .

10.3. An example - the Zeeman effect in helium

Let’s apply these results to helium.

In the ground state 1 s2 1S0, there is no angular momentum at all, and so there is no Zeeman effect.

The level 1s2s 1 S0 also has no Zeeman effect, for the same reason.

For the level 1s2s 3 S1, we have L = 0, J = S = 1, and so gJ = 2. The level splits into three sub-levels, with MJ = -1, 0, +1, separated by

2 ΜB B.

For the 1s2p 1 P1 level, we have S = 0, J = L = 1, and so gJ = 1. The level splits into three sub-levels, with MJ = -1, 0, +1, separated by

ΜB B.


38

For the 1s2p 3 P0 level, the total angular momentum is zero, so there is no Zeeman effect.

For the 1s2p 3 P1 level, we have S = 1, L = 1, J = 1, and so gJ = 3 2. The level splits into three sub-levels, with MJ = -1, 0, +1, sepa-

rated by 3 2 ΜB B.

For the 1s2p 3 P2 level, we have S = 1, L = 1, J = 2, and so gJ = 3 2. The level splits into five sub-levels, with MJ = -2, -1, 0, +1, +2,

separated by 3 2 ΜB B.

Now let’s consider a specific transition in helium, say the transition 1s2s 3 S1- 1s2p 3 P2. The spectral line corresponding to this transition

will be split into components in the presence of a magnetic field. Here’s the energy level structure in the field B, with the levels labelled

according to their values of MJ . To work out how many components there will be in the spectral line of this transition, we have to apply the

selection rule DMJ = 0, ± 1. The allowed transitions are shown by the dotted lines labelled a-i in the figure.

1s2s 3S1

1s2p 3P2

-1

0

1

-2

-1

0

1

2

MJ

a b c d e f g h i

1.5 ΜBB

2ΜBB

The figure below shows how the spectral line is divided into its components.The labelling of the components is the same as in the figure

above. We see that the magnetic field splits the spectral line into 9 components, with the frequency separation between each component being

0.5 ΜB B h. For example, if the applied magnetic field is 1mT, the components will be separated by 7MHz. Satisfy yourself that all this is

correct.

0.5

Frequency HΜBBhL

ab cd e fg hi


è The Zeeman effect splits fine-structure energy levels into sub-levels labelled by the MJ quantum number.

è For weak fields the energy shift is linear in B.

è The energy shift is DEZ = gJ ΜB B MJ , where gJ is given by equation (10.11). There are 2 J + 1 sub-levels. The weak-field

approximation means that the Zeeman splitting is small compared to the fine-structure splitting, which is almost always the case.

è The Zeeman splitting results in a spectral line being split into components whose spacing varies linearly with B (in the weak-field

aproximation). The spectrum can be worked out from the energy level diagram and the selection rule DMJ = 0, ± 1.

10.5. To do

Revise the material in this handout. Useful reading: section 8.1 of Woodgate or section 5.5 of Foot.


10.6. Advanced topic - Zeeman effect including hyperfine structure

This is beyond the syllabus for this course, so this is here for your interest.

So far we neglected the hyperfine structure in our discussion of the Zeeman effect. Now let's include it. Taking the Zeeman interaction and the

hyperfine interaction together we have the perturbation


39

(10.12)H`

' = H`

hfs + H`

Z = A I`.J

`+ΜB B Igl Lz

`+ gs Sz

` M,where, as before, we have taken the magnetic field to be in the z direction, and we have gl = 1, gs = 2. The approach to finding the total

energy shifts depends on whether the Zeeman shifts are small or large compared to the hyperfine structure.

Weak magnetic field

We'll first consider the case of small magnetic fields. Here, small means that the Zeeman splitting is much smaller than the hyperfine splitting

(i.e. ΜB B << A.) In this case, we deal with the hyperfine structure first. For a first approximation, we neglect the Zeeman terms HZ in

equation (10.12) and calculate the energy level shifts due to H`

hfs alone. This is the calculation we already did in lecture 9. Now we have

energy levels split by the hyperfine interaction and labelled by the total angular momentum quantum number F (remember that F`

= I`

+ J`).

We write the eigenstates using the labels I; J ; F , MF \ to remind us that they are eigenstates of J` 2

, I`2

, F` 2

and Fz

`.

Next we calculate the energy shift due to the magnetic field using perturbation theory, with these states as our basis states. The perturbation is

H`

Z = ΜB B Igl Lz

`+ gs Sz

` M. Following exactly the same arguments presented above, this can be re-written as H`

Z = gJ ΜB B Jz

` where gJ is

given by equation (10.11). The only remaining task is to calculate XH`

Z\ using our I; J ; F , MF \ eigenstates:

(10.13)DEZ = XH`

Z\ = gJ ΜB B YI; J ; F , MF Jz

`I; J ; F , MF ].

Note that I; J ; F , MF \ is not an eigenstate of Jz

`, but we can again use the vector model to sort this out. In fact, the calculation is almost

exactly the same as in lecture 10. We project J` onto F

`, and then project F

` onto the z-axis, obtaining

(10.14)XJz

` \ = [ IJ` .F` M Fz

`

F` 2

_ =1

2[

JF` 2

+ J` 2

- I`2N Fz

F` 2

_ =FHF + 1L + JHJ + 1L - IHI + 1L

2 FHF + 1L MF

So the energy shift is now given by

(10.15)DEZ = gF ΜB B MF ,

where the new g-factor that has appeared is

(10.16)gF = gJ

FHF + 1L + JHJ + 1L - IHI + 1L2 FHF + 1L .

Notice that, just as before, the energy shift is linear in B and the levels are split according to the quantum number MF - the projection of the

total angular momentum onto the z-axis. The only thing that has changed is the g-factor - from gJ given by equation (10.11) to gF given by

equation (10.16).

Example: the Zeeman effect in the ground state of Cs

We looked at the ground state hyperfine structure of Cs in lecture 9. Now we can calculate the Zeeman effect of this hyperfine structure. In

the ground state of Cs, we have L = 0, S = J = 1 2 and I = 7 2. So we get two hyperfine levels, F = 3 and F = 4, separated by the 9.2GHz

hyperfine interval. The value of gJ for the ground state is gJ = 2. This should be obvious from the fact that the only contribution to the

magnetic moment comes from the spin of the single valence electron, so we must have gJ = gs = 2. Alternatively, you can see it from

equation (10.11) with L = 0 and J = S. From equation (10.16) we get the values of gF : gF = -1 4 for F = 3, and gF = +1 4 for F = 4.

The graph below shows the Zeeman splitting of the two hyperfine energy levels as a function of magnetic field. The zero of energy is placed

(arbitrarily) at the F = 3 energy in zero field. Note that the Zeeman splitting is indeed small compared to the hyperfine splitting, at least for

fields up to about 0.1T. In most applications of interest the magnetic fields tend to be quite a bit smaller than this and so our picture is good. If

we go to higher fields the approximation that we’ve made starts to break down.


40

F

4

3

MF

+4

+3

+2

+10

-1

-2

-3

-4

+3+2

+10

-1

-2

-3

0.00 0.05 0.10 0.15-2

0

2

4

6

8

10

12

Magnetic field HTL

Energ

yh

HHzL

Linear Zeeman effect in the ground state of Cs

Strong magnetic field

Now let’s consider the opposite extreme, where the Zeeman splitting is much larger than the hyperfine structure HΜB B >> A). This is also

easy to calculate. We simply reverse the order in which we apply the perturbations. First we neglect the hyperfine interaction, and calculate

the Zeeman splitting of the fine structure levels. We already did the first step - calculating the Zeeman splitting with the hyperfine structure

neglected - in lecture 10. The Zeeman splitting is simply DEZ = gJ ΜB B MJ . The eigenstates are eigenstates of J` 2

and JZ

`, and also of I

`2

and

IZ

` (since the hyperfine interaction hasn’t yet been introduced). So our basis states are I, MI ; J , MJ \.

Now we treat the hyperfine interaction as a perturbation using these basis states. The energy shift is given by

(10.17)DEhfs = YA I`.J

` ] = YI, MI ; J , MJ A I`.J

`I, MI ; J , MJ ].

In the vector model, J` is precessing rapidly around the z-axis because of the strong Zeeman interaction and so the Jx

` and Jy

` components time-

average to zero. So the only term in I`.J

` that contributes to the expectation value is Iz

`Jz

`. So we obtain

(10.18)DEhfs = YA Iz

`Jz

` ] = A MI MJ .

The total energy shift in a strong field is thus given by

(10.19)DE = gJ ΜB B MJ + A MI MJ .

The picture below shows the energy levels for the ground state of Cs in a large magnetic field. Here the large interval is 2 ΜB B and the small

interval is A 2.

MI MJ

+12

-12+72+52+32+12,

-12-32-52-72

-72-52-32-12+12+32+52+72

A2

2ΜBB

For intermediate strengths of magnetic field, we can’t arrange the terms as a hierarchy of perturbations, and we have no choice but to set up

the matrix representation of H ' and diagonalize this to find the eigenvalues. However, without doing any calculations at all, we can easily link

the energy levels obtained in the weak-field limit to those obtained in the strong field limit using the following two rules: (1) The total

magnetic quantum number is MF = MI + MJ and can be used to label the states at all values of the field, (2) States with the same values of

MF cannot cross one another. I’ll leave it to you to make the diagram showing how the low-field levels connect to the high field ones.


41

For intermediate strengths of magnetic field, we can’t arrange the terms as a hierarchy of perturbations, and we have no choice but to set up

the matrix representation of H ' and diagonalize this to find the eigenvalues. However, without doing any calculations at all, we can easily link

the energy levels obtained in the weak-field limit to those obtained in the strong field limit using the following two rules: (1) The total

magnetic quantum number is MF = MI + MJ and can be used to label the states at all values of the field, (2) States with the same values of

MF cannot cross one another. I’ll leave it to you to make the diagram showing how the low-field levels connect to the high field ones.

Summary of this advanced topic

è The Zeeman effect splits a hyperfine level into sub-levels labelled by the MF quantum number.

è For weak fields the energy shift is linear in B.

è For weak fields, the energy shift is DEZ = gF ΜB B MJ , where gF is given by equation (10.16). There are 2 F + 1 sub-levels. The weak-

field approximation means that the Zeeman splitting is small compared to the hyperfine-structure splitting, which is often the case, but

not always.

è For strong fields, the perturbations are taken in the opposite order - first the Zeeman effect and then the hyperfine interaction. Energy

levels are labelled by the quantum numbers MJ and MI . There is a large splitting of gJ ΜB B MJ due to the magnetic field, and then a

smaller splitting of A MI MJ due to the hyperfine interaction.

è There is no remaining degeneracy in a magnetic field.

è The spectrum can be worked out from the energy level diagram and the selection rule DMF = 0, ± 1.


42

11. Interaction of atoms with an oscillating field

So far, we have studied in some detail the energy level structure of atoms. We have done this for free atoms, and for atoms in the presence of

a static magnetic field. Next, we want to understand what happens when the atom is subjected to an oscillating field - i.e. to an electromag-

netic wave. This is a complicated subject, but we will specialize to a particular case which is fairly simple and is of enormous practical

importance. We consider an atom that has just two energy levels. We suppose that we have already solved the time-independent Schrödinger

equation for the free atom - the eigenfunctions and eigenvalues are Φ1, Φ2 and E1, E2. So that we have a specific example in mind, let’s

consider two levels that are separated by the hyperfine interaction, the frequency difference being Ω0 = HE2 - E1L Ñ. This could, for

example, be the 9.2GHz hyperfine interval in Cs. Now we introduce a magnetic field that is oscillating at frequency Ω, and we want to find

out what happens. We are most interested in the case where the oscillating field is close to being in resonance with the atom, i.e. Ω » Ω0, for

it is at this frequency that we can expect something interesting to happen. A schematic representation of the situation we have is drawn below:

E1

E2

Ñ Ω0 Ñ Ω

Because the magnetic field depends on time, we need to solve the time-dependent Schrödinger equation.

11.1. Solving the time-dependent Schrödinger equation for a two level atom interacting with an oscillating

field

We write the magnetic field as B = B0 cosHΩ tL, where B0 contains all the spatial dependence of the wave. The time-dependent Schrödinger

equation is

(11.1)ä Ñ¶ Y

¶ t

= H`

Y,

where H`

is the total Hamiltonian for the system. This we write as a sum of two parts, H`

= H`

0 + H`

I . H`

0 is the Hamiltonian for the free atom,

whose eigenfunctions and eignevalues we’ve already found: H`

0 Φ1 = E1 Φ1 and H`

0 Φ2 = E2 Φ2. H`

I is the term that describes the interaction

between the atom and the radiation - it is the same interaction that we had before when we studied the Zeeman effect, except that now the

magnetic field is time-varying:

(11.2)H`

I = -Μ`.B0 cosHΩ tL

where Μ` is the atom’s magnetic moment. For definiteness, let’s suppose that the magnetic field is pointing along the z-axis. Then

(11.3)H`

I = -Μ`

zB0 cosHΩtL

Note that, in the absence of the radiation, Φ1 and Φ2 are stationary states, their only time-dependence being the usual ã-ä E1 tÑ and ã-ä E2 tÑ. In

the presence of the radiation, we can, quite generally, expand the full wavefunction in terms of these eigenstates:

(11.4)YHtL = c1HtL Φ1 ã-ä E1 tÑ

+ c2HtL Φ2 ã-ä E2 tÑ.

Here, c1 and c2 are coefficients that depend on time. Our whole task is to calculate these coefficients - once we know them we know the

complete wavefunction at all times and so the problem is solved. We need to substitute this YHtL into the time-dependent Schrödinger

equation. The left hand side is

ä Ñ¶ Y

¶ t

= c1 Φ1 E1 ã-ä E1 tÑ

+ c2 Φ2 E2 ã-ä E2 tÑ

+ ä Ñâ c1

â t

Φ1 ã-ä E1 tÑ

+ ä Ñâ c2

â t

Φ2 ã-ä E2 tÑ

and the right hand side is

H`

Y = H`

0 Y + H`

I Y = c1 Φ1 E1 ã-ä E1 tÑ

+ c2 Φ2 E2 ã-ä E2 tÑ

+ c1HtL H`

I Φ1 ã-ä E1 tÑ

+ c2HtL H`

I Φ2 ã-ä E2 tÑ

So, the Schrödinger equation gives us

(11.5)ä Ñâ c1

â t

Φ1 ã-ä E1 tÑ

+ ä Ñâ c2

â t

Φ2 ã-ä E2 tÑ

= c1HtL H`

I Φ1 ã-ä E1 tÑ

+ c2HtL H`

I Φ2 ã-ä E2 tÑ.

Multiply by Φ1* ãä E1 tÑ and integrate over all space, using the orthonormality of the Φ’s:

ä Ñâ c1

â t

= c1 à Φ1*

H`

I Φ1 â V + c2 ã-ä HE2-E1L tÑ à Φ1

*H`

I Φ2 â V

Because the atom is so much smaller than the wavelength of light, the amplitude B0 is very nearly a constant over the whole region where the

eigenfunctions are non-zero and we can take it outside the integral. Then, using Ω0 = HE2 - E1L Ñ and cosHΩ tL =1

2Hãä Ω t + ã-ä Ω tL, the above

equation becomes


43

Because the atom is so much smaller than the wavelength of light, the amplitude B0 is very nearly a constant over the whole region where the

eigenfunctions are non-zero and we can take it outside the integral. Then, using Ω0 = HE2 - E1L Ñ and cosHΩ tL =1

2Hãä Ω t + ã-ä Ω tL, the above

equation becomes

ä Ñâ c1

â t

= -c1

B0

2Iã

ä Ω t+ ã

-ä Ω tM à Φ1*

Μ`

zΦ1 â V - c2

B0

2Iã

-ä HΩ+Ω0L t+ ã

ä HΩ-Ω0L tM à Φ1*

Μ`

zΦ2 â V

Of the four terms on the right hand side of the above equation, the first two oscillate at frequency Ω and the third at frequency Ω + Ω0. All

three terms oscillate very rapidly, and their effects time-average to zero on any timescale of interest to us. The fourth term oscillates at

frequency Ω - Ω0 which is very much slower then the other three (when Ω » Ω0). For this reason, only this last term has any significant

effect on the atom and we can neglect the others. This approximation is known as the rotating-wave approximation. Applying it, we end up

with the following differential equation:

(11.6)äâ c1

â t

=W

2ã

ä HΩ-Ω0L tc2,

where we have defined the quantity

(11.7)W = -B0

Ñà Φ1

*Μz Φ2 â V = -

B0

ÑX1 Μz 2\.

Similarly, if we return to equation (11.5) and this time multiply by Φ2* ãä E2 tÑ and then integrate and apply the same rotating wave approxima-

tion we get

(11.8)äâ c2

â t

=W

2ã

-ä HΩ-Ω0L tc1.

We can combine (11.6) and (11.8) to obtain an equation for just c2:

(11.9)â2

c2

â t2

+ äHΩ - Ω0L â c2

â t

+W2

4c2 = 0.

This differential equation tells us how the amplitude of state 2 changes with time. If at t = 0 the atom is in level 1 (i.e. c1H0L = 1, c2H0L = 0),

the solution for the probability of being in level 2 (i.e. for c2HtL 2) is

(11.10)c2HtL 2=

W2

W2 + ∆2sin2

1

2W

2+ ∆

2t ,

where we have defined ∆ = Ω - Ω0, which is the detuning of the light from the atom’s resonance frequency.

11.2. Rabi oscillations and Π-pulses

We see that the probability oscillates sinusoidally between the two levels. When the light is on resonance with the atom (∆ = 0), we have

c2HtL 2 = sin2I 1

2W tM - the probability of being in the upper state oscillates between 0 and 1 with the characteristic angular frequency W - the

oscillations are called Rabi oscillations and W is called the Rabi frequency (after I. I. Rabi, whio first demonstrated the effect). As we

increase the frequency detuning of the light from the atomic resonance frequency (increase ∆ ), the amplitude of the oscillations gets

smaller, and the frequency gets larger.

The plot below shows the probability of finding the atom in level 2 as a function of time (expressed as the product W t), for three different

values of the detuning - ∆ = 0 (solid line), ∆ = W (dotted line) and ∆ = 3 W (dashed line).

0 5 10 15 20

0.0

0.2

0.4

0.6

0.8

1.0

W t

Pro

babil

ity

of

bein

gin

level

2

Suppose the atom in level 1 and we apply a pulse of radiation which is on resonance. We can control the Rabi frequency W by controlling the

intensity of the radiation, and we can choose the duration of the pulse, T. If we make W T = Π, the atom will be transferred from level 1 to

level 2 with 100% probability. A pulse that satisfies this condition is called a Π-pulse.


44

Suppose the atom in level 1 and we apply a pulse of radiation which is on resonance. We can control the Rabi frequency W by controlling the

intensity of the radiation, and we can choose the duration of the pulse, T. If we make W T = Π, the atom will be transferred from level 1 to

level 2 with 100% probability. A pulse that satisfies this condition is called a Π-pulse.

To get a feel for the numbers, let’s calculate an approximate size for the Rabi frequency. It is given by equation (11.7) - W =B0

ÑX1 Μ

`z

2\.

For a typical hyperfine transition in any atom, the value of the matrix element is of order the Bohr magneton, ΜB = e Ñ H2 meL. So

W ~1

2

e

me

B0. The magnetic field amplitude of an electromagnetic wave is related to its intensity via I =1

2

c

Μ0

B02. So, we can write

W ~1

2

e

me

2 Μ0 I

c. For a 1W beam of microwaves, of uniform intensity inside a circle of radius 5cm, this value is 91000 rad/s.

Note that although we have treated the case of an atom with just two energy levels, the same theory works extremely well for multi-level

atoms where the radiation is tuned close to resonance to one of the transition frequencies - the atom undergoes Rabi oscillations between

these two particular levels, and nothing happens to any of the other levels.

11.3. Resonance lineshape

We might also wonder how the transition probability depends on the detuning from resonance. To see this, we set W T = Π, so that the

transition probability is 1 on resonance, and then plot equation H11.10) as a function of the detuning ∆. The result is plotted below - this is the

characteristic resonance lineshape for an atom interacting with monochromatic radiation.

-6 -4 -2 0 2 4 6

0.0

0.2

0.4

0.6

0.8

1.0

Detuning , ∆W

Pro

babil

ity

of

bein

gin

level

2

11.4. Where’s the spontaneous emission?

We have found that, when an atom interacts with a near-resonant oscillating field, the probability of being in the excited state oscillates back

and forth. When the resonance condition is met, the amplitude of these oscillations is maximal (from 0 to 1), and the angular frequency of the

oscillations occurs at the Rabi frequency. According to our theory, these oscillations continue for as long as the oscillating field is turned on.

The atom starts out in state 1, absorbs energy from the oscillating field as it gradually transitions to state 2, and then returns the energy to the

field as it returns to state 1. It’s like a continual process of absorption from the field followed by stimulated emission back into the field.

Spontaneous emission seems to play no role in our theory.

If we include spontaneous emission in our theory, we find that the Rabi oscillations are damped out. The oscillations do not continue forever

but gradually decay, and the populations in the two levels eventually reach a steady state. The characteristic damping time is simply the

excited state lifetime.

In the case that we considered above - two hyperfine levels being driven by a near-resonant microwave field - it is an extremely good

approximation to neglect spontaneous emission. The natural lifetime of the upper level turns out to be around a million years - so in any

reasonable experiment, all you ever see are the Rabi oscillations, with no damping, exactly as described by the above theory.

But what if the two levels are two different electronic states of an atom, for example the 6s and 6p states of Cs, separated by a visible

transition. In this case, the lifetime of the upper state is of order 10ns, and the Rabi oscillations are damped out on this timescale. In fact, it is

very difficult indeed to observe Rabi oscillations in this case, because the damping time is so fast. We’ll consider a steady-state picture of the

atom-light interaction in the next lecture.

These arguments about spontaneous emission and damping of the Rabi oscillations are correct, but we still haven’t answered the question of

why spontaneous emission did not just appear naturally in our theory. What is it that we left out of the above theory? The answer is that

we’ve used a quantized atom interacting with a classical electromagnetic field. For a complete and proper description, we should quantize the

electromagnetic field as well. When you quantize the electromagnetic field, you find that there are zero-point fluctuations (just like in a

harmonic oscillator where the ground state energy is not zero but 1 2 Ñ Ω) - they are the fluctuations of the vacuum. These vacuum fluctua-

tions drive the electron, in much the same way as real radiation drives it. For an atom in an excited state, resonant radiation causes stimulated

emission, whereas the vacuum fluctuations cause spontaneous emission. It is only when you quantize the electromagnetic field that an excited

atom is able to decay. Vacuum really does have these zero-point fluctuations - if it did not there would be no light! All of this is outside the

scope of this course - you would learn about it in a course on quantum optics.


è When a two-level atom interacts with radiation near resonance the probability oscillates between the two states.


45

è The behaviour is the same for a multi-level atom - only the two levels that are close to resonance with the radiation matter.

è On resonance, the probability oscillates fully (i.e. between 0 and 1). The angular frequency of this oscillation is the Rabi frequency.

è The Rabi frequency is proportional to the amplitude of the oscillating field, and to the matrix element of the relevant operator between

the two states (in the case of two hyperfine levels, the magnetic dipole operator).

è As the radiation is detuned from the resonance frequency, the oscillation goes faster and the amplitude gets smaller.

è The oscillations are damped by spontaneous emission, with the damping time being the upper state lifetime. This is extremely long for

microwave transitions between hyperfine levels and the damping can be neglected. It is extremely short for the higher frequency

transitions between different electronic states of an atom, and then the Rabi oscillations can be neglected.

11.6. To do

Revise the material in this handout. Useful reading: sections 7.1 & 7.3 of Foot.



46

12. Transitions: the A and B coefficients

12.1. Types of transition

In the last lecture, we considered the specific case of two energy levels separated by the hyperfine interaction, and we saw that a resonant

oscillating magnetic field can drive transitions between these two levels. The characteristic transition rate is set by the Rabi frequency:

W =B0

ÑX1 Μ

`z

2\.

The Rabi frequency is a product of two factors - the amplitude of the oscillating magnetic field, and a transition matrix element. The operator

that appears in the matrix element is the magnetic dipole moment operator. Notice how everything here is magnetic in nature - the transition

is driven by the oscillating magnetic field interacting with the atom’s magnetic moment. Why is that? Why not the oscillating electric field,

and the atom’s electric dipole moment? The answer lies in the nature of the transition that we were considering - a hyperfine transition. Recall

that the hyperfine interaction is an interaction between the magnetic moments of the electron and nucleus. The difference in energy between

hyperfine levels arises due to the relative orientations of the magnetic moments - the energy depends on whether these magnetic moments are

parallel, antiparallel, or somewhere in-between. A transition between hyperfine levels requires the re-orientation of the magnetic moments,

and so it is no surprise that it’s the oscillating magnetic field that drives this transition. The transition between hyperfine levels is an example

of a magnetic dipole transition.

What about transitions where the electronic configuration changes. In this case, the electron has to be driven from one state to another, and it

is the oscillating electric field that is primarily responsible for the transition. This is an example of an electric dipole transition. Once again,

the atom can undergo Rabi oscillations between the two states (though for an electric dipole transition these are likely to be rapidly damped

by spontaneous emission). The new Rabi frequency is similar to the one written above, but we replace B0 by E0, the amplitude of the

oscillating electric field, and we replace the magnetic dipole moment operator HΜ`

zL with the electric dipole moment operator, -e z

`:

W =E0

ÑX1 e z

`2\.

So we see that we have two different kinds of transitions - electric dipole and magnetic dipole transitions. The rate for an allowed electric

dipole transition is much higher than for an allowed magnetic dipole transition, so electric dipole transitions dominate as long as they are

allowed by the selection rules. When they are forbidden by the selection rules, magnetic dipole transitions are important. We return to

selection rules in lecture 13.

There are other higher-order transitions too, for example electric quadrupole transitions. All of these have much slower rates than electric

dipole transitions and are only important when the electric dipole transition is forbidden. In this course, you should just be aware that such

higher-order transitions can occur and that the selection rules governing them are different to the ones for electric dipole transitions.

12.2. The Einstein A and B coefficients

Consider a population of atoms with two energy levels, a ground state of energy E1 and an excited state of energy E2. We know that atomic

energy levels are often degenerate, so let our energy levels have degeneracies g1 and g2. The population in levels 1 and 2 are N1 and N2

respectively. If we put an atom in the excited state we know that it will not remain there forever - it decays back to the ground state, emitting

a photon in the process. This is called spontaneous emission. The excited state decays with a certain rate, let’s call it A21. This rate is a

property of the atom. The decay of level 2 is described by the rate equation

(12.1)â N2

â t

= -A21 N2.

This equation states that the rate of change of the population in level 2 is proportional to the decay rate and to the level 2 population itself.

The solution to this equation is

(12.2)N2HtL = N2H0L expH-t ΤL,where

(12.3)Τ =1

A21

is the lifetime of the excited state. The excited state decays with a characteristic time Τ, which is the inverse of the rate A21.

Now let’s shine some light on the atoms. This introduces two new processes - absorption and stimulated emission.

Consider absorption first. Atoms in level 1 absorb light and are excited into level 2, so in our population of atoms the absorption rate must be

proportional to the population in level 1. What else does the absorption rate depend on? It should depend on some property of the atomic

transition - how strongly this particular transition in this particular atom absorbs light - we call this property the absorption coefficient B12.

The rate of absorption should also depend on how much light there is - we expect the rate of absorption to be proportional to the intensity of

the light. However, it's no good if the light is at the wrong frequency - the absorption only happens if the angular frequency of the light is

close enough to the angular frequency of the transition, Ω21 = HE2 - E1L Ñ. So the quantity that matters is the energy density per unit angular

frequency interval ΡHΩL (sometimes called the spectral energy density). Stimulated emission is the inverse process to absorption, so it’s rate is

also proportional to ΡHΩL and to the atom’s coefficient for stimulated emission B21.


47

Below is the diagram that shows the fundamental processes that can happen, and their rates.

B12ΡHΩ12LN1 B21ΡHΩ12LN2 A21N2

E1, g1, N1

E2, g2, N2

The rate equation that describes the population of level 2 is

(12.4)â N2

â t

= B12 ΡHΩ12L N1 - B21 ΡHΩ12L N2 - A21 N2,

We can also write the equivalent equation for â N1 â t (simply change all the signs!).

Einstein found the relationships between the coefficients A21, B21, B12 using the following clever argument. Suppose we put the the atoms in

a box whose walls are at temperature T, and allow the atoms to come into thermal equilbrium with the blackbody radiation in the box. The

energy density per unit angular frequency of blackbody radiation inside the box is given by the Planck distribution law:

(12.5)ΡHΩL =Ñ Ω3

Π2c

3

1

expHÑ Ω kB TL - 1.

In thermal equilibrium the populations are not changing, so we have â N2 â t = 0. From equation (12.4) we then have

(12.6)B12 ΡHΩ12L N1 - B21 ΡHΩ12L N2 - A21 N2 = 0,

which we can rearrange to give

(12.7)ΡHΩ12L =A21

B21

1

HB12 B21L HN1 N2L - 1.

But in thermal equilibrium, the way population is distributed between two energy levels is given by the Boltzmann distribution:

(12.8)

N2

N1

=g2 expH-E2 kB TLg1 expH-E1 kB TL =

g2

g1

exp -Ñ Ω12

kB T

.

Substituting equation (12.8) into equation (12.7) we get

(12.9)ΡHΩ12L =A21

B21

1

HB12 B21L Hg1 g2L expHÑ Ω12 kB TL - 1.

In order to make this equation consistent with equation (12.5), we must have

(12.10)A21 =Ñ Ω12

3

Π2c

3B21,

and

(12.11)B12 =g2

g1

B21.

Now remember - the A and B coefficients are properties of the atom. They do not depend on the light, or on the atom’s environment. So,

although we have derived the relationships between the A and B coefficients by considering the special case of thermal equilbrium with

blackbody radiation, they must always be true.

In the case where there is no degeneracy, equation (12.11) tells us that the coefficients for absorption and stimulated emission are equal. This

makes good sense, since one process is the inverse of the other. When there’s degeneracy, we just have to take account of the different

number of sub-levels, which is why the factor g1 g2 appears. Equation (12.10) tells us that the A and B coefficients are proportional to one

another. So if an atom absorbs strongly (large B12), its decay rate will be high (large A21), meaning that the upper state lifetime is short.


48

12.3. Steady-state population for two-level atoms

Once the populations have reached a steady-state, we have â N2 â t = 0. From equation (12.4) we have

B12 ΡHΩ12L N1 - B21 ΡHΩ12L N2 - A21 N2 = 0.

If we have just two levels, so that there is no degeneracy (g1 = g2 = 1), then B12 = B21, and the ratio of the populations in the steady state is

(12.12)

N2

N1

=B Ρ

A + B Ρ.

Here, we’ve used the shorter notation B = B21, A = A21.

Remember that Ρ is proportional to the intensity of the light at the resonance frequency. When the intensity is low, the rate of absorption (and

stimulated emission) is small compared to the spontaneous emission rate (i.e. B Ρ << A). In this case, equation (12.12) tells us that N2 << N1

- most of the population stays in the ground state when the radiation is weak, as we would expect. To get a significant fraction of the

population into the upper state, we have to make the absorption rate comparable to the spontaneous emission rate. When these two rates are

equal (B Ρ = A), we have N2 N1 = 1 2 (meaning that 1/3 of the total population is in the upper state). The maximum possible upper state

population occurs when the absorption (and stimulated emission) rate is much larger than the spontaneous emission rate, B Ρ >> A. Then

equation (12.12) tells us that N2 = N1 - i.e. half of all the population is in the upper state. In the steady state, you cannot get more than half

into the upper state, no matter how intense the radiation is. This makes sense - in the limit where the radiation is very intense, the absorption

and stimulated emission rates dominate (spontaneous emission is not important), and since their rates are equal we must end up with half the

population in each state.

12.4. Expressions for the A and B coefficients

It can be shown (see below if you’re interested) that for an electric dipole transition the Einstein B-coefficient is given by

(12.13)B12 =Π

3 Ε0 Ñ2X2 e r

`1\ 2.

This equation relates the absorption rate to a property of the atom - the square of the matrix element of the electric dipole operator.

Now, using equation (12.10) we obtain an expression for the A-coefficient:

(12.14)A21 =g1

g2

Ω213

3 Ε0 Π Ñ c3

X2 e r`

1\ 2

This is an important result - it gives the rate of spontaneous emission, and hence the excited state lifetime (Τ = 1 A) in terms of atomic

properties. The spontaneous emission rate is proportional to the square of the matrix element of the electric dipole operator and to Ω03. The

lifetime of an atomic state scales inversely as the transition frequency cubed. Consider hydrogen for example. The 2p state decays to the

ground state, emitting ultra-violet light (around 2.5 ´ 1015 Hz), and the lifetime is just a few nanoseconds. By contrast, the lifetime of the

upper hyperfine component of the 1s state, which is only separated from the lower hyperfine component by 1.4 ´ 109 Hz, is about 10 million

years.

Advanced topic: derivation of the above

The derivation of the above lies outside the scope of the course, but is given here for completeness.

In the last lecture, we considered the interaction of a two-level atom with monochromatic radiation, and we used the time-dependent

Schrödinger equation to find the probability of being in the excited state - equation (11.10). We can use this result to obtain an expression for

the B-coefficient. To get there, we assume that the atom is interacting with weak broadband radiation, so that Ω - Ω0 >> W for almost all

frequencies. At any one frequency where this holds, equation (11.10) reduces to

c2HtL 2=

W2

HΩ - Ω0L2sin2HHΩ - Ω0L t 2L.

For an electric dipole transition, W2 =e

2E0

2

Ñ2Z12

2 where Z12 = X1 z`

2\. The electric field amplitude is related to the energy density in the

frequency interval from Ω to Ω + â Ω by ΡHΩL â Ω = 1 2 Ε0 E02HΩL. So, for the narrow slice of radiation centred at Ω we have

W2

=2 e

2

Ε0 Ñ2Z12

2ΡHΩL â Ω.

Now integrating over all frequencies in the broadband radiation we find the upper state probability to be

c2HtL 2=

2 e2

Ε0 Ñ2Z12

2 à sin2HHΩ - Ω0L t 2LHΩ - Ω0L2

ΡHΩL â Ω.

Because of the denominator, the integral is dominated by the frequencies Ω that are reasonably close to Ω0. Since the radiation is broadband

and slowly varying over the range where the integrand is significant, we can replace ΡHΩL by its value at Ω0, and then take it outside the

integral. Then, with x = HΩ - Ω0L t 2, we obtain


49

c2HtL 2=

2 e2

Ε0 Ñ2Z12

2ΡHΩ0L t

2à sin2

x

x2

â x.

If we take the limits of integration to ±¥ the integral evaluates to Π. In fact, all that matters is that the limits be large enough, which here

means that they are much larger than ±Π. Since x is linear in t, we can always make the limits large by making t large - i.e. this essentially

means that the radiation is applied for long enough that the system reaches its steady state. So, the integral is replaced by Π. The transition

rate is the transition probability per unit time, R = c2HtL 2 t. So we reach the result

R =Π e

2

3 Ε0 Ñ2D12

2ΡHΩ0L,

where D122 = X2 r

`1\ 2 and we have used the fact that Z12

2 = D122 3 (because all three directions are equivalent). This is the absorption

rate for a two-level atom, which in the Einstein treatment is given by B12 ΡHΩ0L. So we get the expression for the absorption coefficient:

B12 =Π e

2

3 Ε0 Ñ2D12

2.


è There are three fundamental interactions between light and atoms - absorption, stimulated emission, and spontaneous emission.

è The Einstein A and B coefficients relate the rate coefficients for these processes - equations (12.10) and (12.11).

è For a two-level atom, in the steady-state, the fraction of the population in the upper state approaches 1/2 when the absorption rate is

much larger than the spontaneous emission rate.

è In the interaction of atoms with light, the dominant operator is the electric dipole operator. The transition rate is proportional to the

square of the matrix element of this operator between the two states of interest.

12.6. To do

Revise the material in this handout. Useful reading: chapter 3 of Woodgate, section 2.4 and chapter 7 of Thorne, section 1.7 of Foot.



50

13. Selection rules. Atomic spectra

13.1. Atomic spectra and the selection rules

As we’ve seen, an atom can undergo transitions between one state and another. These transitions can be driven by an electromagnetic field

oscillating at one of the atom’s resonant frequencies (leading to absorption or stimulated emission), or the transition can be spontaneous. It it

these transitions between the energy levels that produce the characteristic spectrum of the atom. When white light shines through an atomic

sample, the atoms absorb light at (and only at) the specific frequencies that correspond to the difference in energy between a pair of energy

levels (i.e. at the frequencies fij where h fij = E j - Ei). When the frequency spectrum of the light is then studied (using a spectrometer), one

sees dark lines at these specific frequencies - this is the absorption spectrum of the atom. Similarly, if the atoms are excited, e.g. in a lamp,

they emit light at these same characteristic frequencies. Again, when the spectrum of the light is studied, we see that it contains a specific set

of frequencies - this is the emission spectrum of the atom. It is through the careful study of these spectra that the energy level structure of

atoms has been determined.

The main interaction between an electromagnetic field and an atom is the electric dipole interaction - the electric field of the light drives the

electron in the atom. The electron’s displacement from the nucleus is r, and it has a charge -e, so it produces an electric dipole moment

d`

= -e r`. [If there are many electrons, we just sum over them: d = -e Úi ri.] The electric field of the light, E, interacts with this electric

dipole moment, and the interaction Hamiltonian is -d`.E. Let’s write the electric field of the light as E = Ε E0 cosHΩ tL. As pointed out in the

last lecture, the transition rate, R21, between two states 1\ and 2\ is proportional to the squared matrix element of the electric dipole

moment operator:

(13.1)R21 µ X2 -e r`

1\.Ε 2

If we want to calculate the transition rate, we have to calculate the above matrix element, which for a one-electron atom we can do using the

hydrogenic eigenfunctions and evaluating the integrals. However, very often the transition rate is equal to zero, meaning that the transition is

forbidden. This is important in understanding the spectra of atoms - only certain transitions are allowed. Whether a transition is allowed or

forbidden depends on the quantum numbers of the two states involved in the transition - so we get selection rules that govern how these

quantum numbers are allowed to change in a transition.

Similarly, the rate for a magnetic dipole transition is proportional to the square of X2 Μ`

1\ where Μ`

is the atom’s magnetic dipole moment.

Again, a consideration of when this matrix element is zero gives us the selection rules for magnetic dipole transitions.

13.2. Parity and the parity selection rule

We often refer to the parity of the eigenstates. The parity operation P is the inversion of the coordinates through the origin, r ® -r. All the

eigenstates of an atom are states of definite parity, either +1 or -1: P`

Ψ = ± Ψ. This is a consequence of the fact that electromagnetic interac-

tions conserve parity. The parity of a state is usually referred to as even or odd depending on whether the sign is plus or minus. All the radial

functions are even, and so the parity is specified entirely by the spherical harmonics. Their parity depends on l and is simply H-1Ll. So, states

with even l have even parity, and states with odd l have odd parity.

The parity is a crucial property when we consider transitions between states. Suppose the two states 1\ and 2\ have the same parity. As

outlined above, the rate is proportional to the square of

X2 r`

1\ = à Ψ2*

r Ψ1 â V ,

where the integral is over all space. Clearly r` is an odd operator (it changes sign through the origin). So if the two states have the same parity

(either both odd or both even), the integrand is odd and the integral evaluates to zero. So we have an important selection rule for an electric

dipole transition: the parity must change in the transition. This is a strict selection rule that applies to many-electron atoms just as it does to a

one-electron atom.

The magnetic dipole moment operator is an even operator. This is because the magnetic moment is proportional to the angular momentum,

and the angular momentum does not change sign on reflection through the origin (think of orbital angular momentum, r`

´ p`

- both r` and p

`

change sign and so their product does not). So - the parity must not change in a magnetic dipole transition.

We see that, in this sense, the two types of transition are complementary. If one type is forbidden by the parity selection rule, the other type

will be allowed. Note that all the fine-structure levels of a particular term (levels labelled by J) have the same parity and so transitions

between them are magnetic dipole transitions. The same is true for hyperfine levels.

13.3. Selection rules for the quantum numbers n, l and m

Now let’s find the electric dipole selection rules for the quantum numbers n, l and m. We’ll do it for a one-electron atom, and then generalize.

First suppose that the light is linearly polarized along the z-axis (which could be defined by an applied magnetic field for example). This

means that Ε is a unit vector in the z direction and so r.Ε = z = r cos Θ. So the transition rate is proportional to the square of the following

matrix element:

Xn ', l ', m ' r cos Θ n, l, m\ = normalization ´ à0

¥

Rn',l'*

r Rn,l r2

â r à0

Π

Pl',ml '* cos Θ Pl,ml

sin Θ â Θ à0

2 Π

ãä m' Φ

ã-ä m Φ

â Φ.

Here, Rn,l is the radial part of the hydrogenic wavefunction and depends only on the radial coordinate, and Pl,mlãä m Φ is the angular part of

the wavefunction - the spherical harmonics. The Pl,ml are the associated Legendre polynomials and depend only on Θ. The dependence on Φ is

contained entirely in the ãä m Φ factor. The integral over Φ is zero unless m ' = m.


51

Here, Rn,l is the radial part of the hydrogenic wavefunction and depends only on the radial coordinate, and Pl,mlãä m Φ is the angular part of

the wavefunction - the spherical harmonics. The Pl,ml are the associated Legendre polynomials and depend only on Θ. The dependence on Φ is

contained entirely in the ãä m Φ factor. The integral over Φ is zero unless m ' = m.

Next suppose that the light is linearly polarized along x. Then we have r.Ε = x = r sin Θ cos Φ. Writing out just the Φ integral this time:

Xn ', l ', m ' r cos Θ n, l, m\ µ à0

2 Π

ãä ml ' Φ cos Φ ã

-ä ml Φâ Φ =

1

2à

0

2 Π

ãä Hm'-m+1L Φ

â Φ +1

2à

0

2 Π

ãä Hm'-m-1L Φ

â Φ.

The integral is zero unless m ' = m ± 1. A moment’s consideration shows that the same result applies for light linearly polarized along y. So

we reach the following selection rules for m:

(13.2)D m = 0 Hlight linearly polarized parallel to zL,(13.3)D m = ± 1 Hlight linearly polarized perpendicular to zL.

Turning now to the l quantum number, we see that its change is governed by the integral over Θ. To evaluate this integral, we need to know

something about the associated Legendre polynomials, in particular the following recursion relations (don’t try to remember these!):

sin Θ Pl,m =Pl+1,m - Pl-1,m

2 l + 1,

cos Θ Pl,m =Hl - m + 1L Pl+1,m + Hl + mL Pl-1,m

H2 l + 1L .

Using these, and the fact that the Pl,m are orthogonal to one another, we see that (irrespective of the polarization) we must have l ' = l ± 1. So

the selection rule is:

(13.4)D l = ± 1 Helectric dipoleL.Finally, consider the change in the n quantum number, which is governed by the radial integral. The integral is an overlap integral between the

radial wavefunctions weighted by r (i.e. overlap in the outer regions contributes more to the integral than in the inner regions of the atom).

The integral is always non-zero so there is no selection rule on the change in n - anything is allowed; to find the transition rate the integral has

to be evaluated.

D n = anything Helectric dipoleL.For a many-electron atom in the central field approximation, the many electron wavefunction is just a product of one-electron wavefunctions

with angular parts given by the spherical harmonics just as in hydrogen. So the selection rules are the same (applied now to the particular

electron that’s making the transition). Note though that, for a many electron atom, these selection rules hold within the central field approxi-

mation. They hold to the extent that the central field approximation is a good approximation.

13.4. Selection rules for the quantum numbers L, S, J and MJ

The selection rule on J follows from the conservation of angular momentum, the fact that a photon has a spin of 1, and the addition of angular

momentum in quantum mechanics (recall: adding J1 and J2 the allowed values go, in integer steps, from J1 - J2 to J1 + J2). So we have

the selection rule D J = 0, ± 1, plus the additional rule that a transition between J ' = 0 and J = 0 is forbidden (because you can’t add 0 and 1

and get 0, not even in quantum mechanics!). Applying the same reasoning to the component of angular momentum in the z-direction, we get

the selection rule DMJ = 0, ± 1, plus the additional rule that MJ ' = 0 to MJ = 0 is forbidden when D J = 0. These selection rules are strict

ones, as they are based on a conservation law.

The selection rule for S is simple: DS = 0. This follows from the fact that the electric dipole operator acts in the coordinate space of the

electrons, and has nothing to do with the spin - so the spin can’t change in the transition.

The selection rule for L now follows directly from the other two and the fact that J`

= L`

+ S`. Since S doesn’t change in the transition, the

change in L has to be the same as the change in J . So the selection rule for L is D L = 0, ± 1, with L ' = 0 going to L = 0 forbidden.

While the selection rule for J is a strict one, the selection rules for L and S only hold within the LS coupling approximation, in which L and S

are both constants of the motion. This approximation is broken down by the spin-orbit interaction which couples L and S so that they are no

longer exact constants of the motion. In light atoms, this coupling is weak compared to the residual electrostatic interaction and so the LS

coupling approximation is a good one. In heavy atoms, the spin-orbit interaction is strong, LS coupling is no longer a good approximation, and

the selection rules for L and S no longer hold. In particular, we get transitions that change the value of S (they are given the name

‘intercombination transitions’, and the resulting spectral line is called an ‘intercombination line’).

Note that we’ve found these selection rules using simple (but good) qualitative arguments. The selection rules can also be derived from the

theory of tensor operators in quantum mechanics, but that is beyond the scope of this course.

13.5. The natural width of a spectral line

We have supposed that, when an atom decays from a state of energy E2 to one of energy E1 it emits light at, and only at, the angular

frequency Ω0 = HE2 - E1L Ñ. In this picture, the atom emits just a single, perfectly pure, frequency - a perfect monochromatic wave. This

cannot be quite right. We know (think back to Fourier transforms) that a wave is only perfectly monochromatic if it oscillates forever, but our

wave doesn’t do that - its amplitude decays with a characteristic time which is the lifetime of the upper state Τ2, and so the wave must have

an angular frequency spread of approximately 1 Τ2. We can also see this from the uncertainty principle which tells us that if the light is only

emitted for a time Τ2, the uncertainty in the angular frequency will be DΩ = 1 Τ2. As a result, the spectrum of the emitted light will be centred

at Ω0 but will have a width which is the inverse of the lifetime - the shorter the lifetime the broader the spectral line. A full analysis shows

that the frequency distribtution of the emitted light is


52

We have supposed that, when an atom decays from a state of energy E2 to one of energy E1 it emits light at, and only at, the angular

frequency Ω0 = HE2 - E1L Ñ. In this picture, the atom emits just a single, perfectly pure, frequency - a perfect monochromatic wave. This

cannot be quite right. We know (think back to Fourier transforms) that a wave is only perfectly monochromatic if it oscillates forever, but our

wave doesn’t do that - its amplitude decays with a characteristic time which is the lifetime of the upper state Τ2, and so the wave must have

an angular frequency spread of approximately 1 Τ2. We can also see this from the uncertainty principle which tells us that if the light is only

emitted for a time Τ2, the uncertainty in the angular frequency will be DΩ = 1 Τ2. As a result, the spectrum of the emitted light will be centred

at Ω0 but will have a width which is the inverse of the lifetime - the shorter the lifetime the broader the spectral line. A full analysis shows

that the frequency distribtution of the emitted light is

f HΩL µG 2

HΩ - Ω0L2 + HG 2L2,

where G =1

Τ2

+1

Τ1

. Often, the lower level is the ground-state and then G = 1 Τ2. The above lineshape is a Lorentzian distribution. It is easy to

show that its full width at half its maximum height is G (in angular frequency units; G

2 Π in frequency units (Hz)). So the linewidth is broad if

either the upper state or the lower state has a short lifetime.

13.6. Doppler broadening

To study the energy level structure of atoms and molecules it is usual to measure their spectra when they are in the gas phase (so that the

levels are not shifted and broadened by interatomic interactions). The gas phase atoms in the source are excited and the light they emit is

measured with some kind of spectrometer. If an atom is moving towards the spectrometer with speed V the light it emits is Doppler shifted.

The frequency of the light measured by the spectrometer will be Ν = Ν0H1 + V cL, where Ν0 is the frequency in the rest frame of the atom.

The atoms in the source have a spread of velocities given by the Maxwell-Boltzmann distribution,

nHV L µ exp -M V

2

2 kB T

,

where M is the atom’s mass and T the temperature of the source. This distribution translates into a distribution of frequencies:

f HΝL µ exp -M Λ2

2 kB T

HΝ - Ν0L2 ,

where Λ = c Ν0 is the wavelength of the light (in the atom’s rest frame). This lineshape is a Gaussian distribution. It’s easy to show that its

full width at half its maximum height (normally called the Doppler width) is:

∆ΝDopp =8 ln 2 kB T

M Λ2.

In many practical situations, the Doppler width is considerably larger than the natural linewidth of the transition and is the limiting factor in

the ability to resolve one spectral line from another.


è The rate for an electric dipole transition is often zero. This gives us the selection rules.

è In an electric dipole transition, the parity must change. In a magnetic dipole transition, the parity must not change.

è In an electric dipole transition, D J = 0, ± 1 and DMJ = 0, ± 1. J ' = 0 to J = 0 transitions are forbidden, and MJ ' = 0 to MJ = 0 is

forbidden when DJ = 0. These are strict rules.

è In an electric dipole transition, D l = ± 1. This rule holds in the central field approximation.

è In an electric dipole transition, D S = 0 and D L = 0, ± 1 (but 0-0 is not allowed). These rules hold in the LS coupling approximation.

è There can be higher-order transitions - most notably magnetic dipole transitions and electric quadrupole transitions - but their rates are

strongly suppressed relative to an allowed electric dipole transition. They have different selection rules and become important when an

electric dipole transition is forbidden.

è Spectral lines have a natural width which is inversely proportional to the lifetime of the decaying state.

è In spectroscopy experiments, the width of a spectral line is often dominated by Doppler broadening.

13.8. To do

Revise the material in this handout. Useful reading: chapter 3 of Woodgate, sections 2.2, 5.5 and 8.1 of Foot, sections 8.2 and 8.3 of

Thorne.



53

14. Electronic structure of diatomic molecules; bonding

When two atoms approach each other there is a re-distribution of the electron charge distribution which can result in a net attractive force

between the atoms - then they can bind to form a molecule. We will only consider diatomic molecules - molecules made out of just two atoms.

14.1. The Born-Oppenheimer approximation

Molecules have electronic energy levels just like atoms. They have an electronic ground state and a set of electronically excited states, and the

spacing between these is similar to the energy-level spacing in atoms, i.e. of order 10eV. But molecules can do more than atoms - they can

also vibrate and rotate. The vibrational and rotational motion are, of course, quantized. So we have a discrete set of vibrational and rotational

energies. The nuclei are much heavier than the electrons and so they move much more slowly. As a result, it’s possible to separate the

electronic motion from the nuclear motion, effectively treating them as two separate systems. This separation is called the Born-Oppenheimer

approximation. In this approximation, the total wavefunction of the molecule is a product of an electronic function and a nuclear function:

(14.1)Y = Ψe ΧN

Here, Ψe is the “electronic wavefunction” and is a function of the coordinates of the electrons, while ΧN is the “nuclear wavefunction” and is a

function of the coordinates of the nuclei. The nuclear motion can be further separated into an angular part and a radial part (much like in

atoms). The angular part represents the rotation of the molecule and the radial part represents the vibration of the molecule. So we have

(14.2)Y = Ψe Ψv Ψr

where Ψv and Ψr are vibrational and rotational functions respectively. The total energy is just the sum of the individual energies:

(14.3)Etotal = Eelectronic + Evibration + Erotation

Since the nuclei move much more slowly than the electrons, the energy associated with nuclear motion (rotation and vibration) is much

smaller than the electronic energy. It also turns out that the vibrational energy level spacing is much smaller than the rotational energy spacing.

So:

(14.4)Erotation << Evibration << Eelectronic

In the Born-Oppenheimer approximation, we find the electronic eigenfunctions and energies by solving the Schrödinger equation for the

electrons with the nuclei at fixed positions. Of course, the answer we get will depend on how far apart the (fixed) nuclei are. So, we solve the

electronic problem for a whole set of different distances R. This procedure gives us the electronic energies as a function of the internuclear

separation R - these are called the potential curves of the molecule. The figure below is a sketch of one such molecular potential.

Internuclear separation

Ele

ctr

onic

energ

y

14.2. The H2+ molecule

We’ll consider the simplest of all possible molecules - H2+ (singly ionized molecular hydrogen). It is a three-particle system consisting of two

identical nuclei (protons) and a single electron. It nicely illustrates the ideas used to describe the electronic states of a molecule. The coordi-

nates of the particles are shown below. The origin, O, of the coordinate system lies half way along the internuclear axis that joins the two

nuclei A and B, which are separated by a distance R. The coordinates of the two nuclei relative to O are RA and RB respectively. The

coordinate of the electron relative to O is r, and relative to A and B is rA and rB. We have the following relations between these coordinates:

rA = r - RA and rB = r - RB.


54

A BO

R

e-

rrA

rB

RA

RB

Now let’s write down the electronic Hamiltonian for this system - by electronic Hamiltonian we mean the one we get in the Born-Oppen-

heimer approximation where the nuclei are held a fixed distance R apart. The nuclei are not moving, so there is no nuclear kinetic energy

term. We have

(14.5)H`

el =-Ñ2

2 me

Ñr2

-e

2

4 Π Ε0 rA

-e

2

4 Π Ε0 rB

+e

2

4 Π Ε0 R

.

The first term is the electron kinetic energy, the second and third the attraction of the electron to nuclei A and B, and the fourth the repulsion

between the two nuclei.

Suppose the two nuclei are far apart. Then we would expect the electron to be bound to one of the protons - the asymptotic ground-state

solution at large R is simply a ground-state hydrogen atom and a bare proton. In this case, the wavefunction is simply Φ1 sHrAL if the electron is

bound to proton A, or Φ1 sHrBL if the electron is bound to proton B. Here, Φ1 sHrL is the normalized wavefunction of ground state hydrogen:

(14.6)Φ1 sHrL =1

Π a03

12exp -

r

a0

.

There is obviously no difference between the energies of the functions Φ1 sHrAL and Φ1 sHrBL - they are degenerate. We can form linear

combinations of them, and in particular the symmetric and antisymmetric wavefunctions:

(14.7)Ψ+Hr, RL =

1

2

@Φ1 sHrAL + Φ1 sHrBLD,

(14.8)Ψ-Hr, RL =

1

2

@Φ1 sHrAL - Φ1 sHrBLD.

We have selected these particular linear combinations because they satsify the symmetry requirement that the wavefunction be unchanged (up

to a sign) on exchanging the two identical protons (i.e. swapping rA and rB). This is somewhat similar to the way we constructed wavefunc-

tions for helium that were symmetric or antisymmetric under exchange.

Below, on the left, is a plot of the symmetric wavefunction Ψ+ along the molecular axis, with the position of the two nuclei shown by the

dots. It’s easy to make this plot - it’s simply 1

2

@Φ1 sH r - R 2 L + Φ1 sH r + R 2 LD - i.e. a sum of two exponential decay functions centred

at ± R 2, the positions of the two nuclei (each exponential decay is shown by the dashed lines in the plot). On the right is plotted the square

of this same function, Ψ+ 2. This is simply the electron density along the internuclear axis for an electron in the symmetric state Ψ+. What

we find is that, for Ψ+, the electron density at the centre (half way between the two protons) is larger than if the electron is simply bound to

one proton or the other. The two nuclei are attracted toward this enhanced electron density, and this allows them to bind together. So, Ψ+ is

known as a bonding orbital. We see that the electron density is ‘shared’ between the two nuclei, with an excess of electron density near the

mid-point - this is a covalent bond.

Ψ+ ÈΨ+ 2

Similar plots for the antisymmetric wavefunction Ψ- are shown below. Here, we see that the electron density at the mid-point is reduced to

zero as a result of the antisymmetry. There is an absence of negative charge between the two nuclei which therefore repel each other more

strongly. So Ψ- is known as an antibonding orbital.


55

Similar plots for the antisymmetric wavefunction Ψ- are shown below. Here, we see that the electron density at the mid-point is reduced to

zero as a result of the antisymmetry. There is an absence of negative charge between the two nuclei which therefore repel each other more

strongly. So Ψ- is known as an antibonding orbital.

Ψ- ÈΨ- 2

Look at the mathematical expression for the electron density. Using the fact that Φ1 s is real, we get

(14.9)Ψ± 2

=1

2Φ1 sHrAL 2

+1

2Φ1 sHrBL 2

± Φ1 sHrAL Φ1 sHrBL.It’s the last term - the interference term - that makes the electron density different between the symmetric and antisymmetric functions. We

can think of this as constructive or destructive interference of the atomic orbitals in the overlap region between the two nuclei. Constructive

interference in the overlap region leads to enhanced electron density here and thus to bonding. Destructive interference in the overlap region

leads to a reduction in electron density and to anti-bonding.

Now, we haven’t actually found the energy eigenvalues yet! All we really know is that the wavefunctions Ψ± must be the correct eigenfunc-

tions at large R. To find the energies for any value of internuclear distance R, we can use the variational method (see lecture 5). Remember

how it works - we just have to guess a wavefunction, then evaluate the expectation value of the Hamiltonian and know for sure that the

ground-state energy is less than this value. Usually, we leave some free parameters in the guess wavefunction and then vary them to find the

parameters that gives us the minimum expectation value. This is a good way of getting close to the true energy.

For our case of the H2+ molecule, we use the symmetric and antisymmetric wavefunctions Ψ± as our guess wavefunctions. This is a good

idea, because these wavefunctions already have the symmetry that the problem demands. We calculate Ù HΨ+L*H`

el Ψ+ â r with H`

el given by

equation (14.5) and Ψ+ given by equation (14.7) - the answer will be a function of the internuclear distance R - this gives us an approximation

to the true ground state energy as a function of R. Then we repeat the procedure for Ψ- given by equation (14.8). The whole procedure is

perfectly straightforward, though it involves the evaluation of some awkward integrals.

The result of the procedure is plotted below. It shows the approximate electronic energy obtained using this variational method. The solid

curve (labelled E+) is the one obtained using the symmetric wavefunction Ψ+ - we see that as the nuclei approach the energy decreases,

meaning that a bond is formed. At a certain value of R, the energy has a minimum - this is the equilibrium internuclear separation, or bond

length, R0. At values of R < R0 the energy increases again because the two nuclei get too close together and their Coulomb repulsion

dominates. The dashed curve (labelled E-) is the one obtained using the antisymmetric wavefunction Ψ- - in this case, as the nuclei approach

the energy keeps increasing - this is the antibonding orbital.

E+

E-

1 2 3 4 5 6RHa0L

-2

-1

1

2

3

4

Energy HeVL

Notation

We have written the symmetric and antisymmetric functions as Ψ+ and Ψ-, so that it’s easy to remember which is which. However, the

conventional notation is to use the subscripts g and u which stand for “gerade” and “ungerade” (“even” and “odd” in German). So: Ψ+

becomes Ψg while Ψ- becomes Ψu.

14.3. Symmetry in diatomic molecules

A diatomic molecule has cylindrical symmetry about the internuclear axis. The internuclear axis defines a special axis in a molecule, and we

make this our z-axis. The projection of the angular momentum onto the internuclear axis - characterized by the quantum number ml - is a

constant of the motion in a diatomic molecule. So the ml quantum number has a special status. It cannot make any difference to the energy

whether the electrons are going around the internuclear axis in the clockwise or anticlockwise direction, so there’s a degeneracy with respect

to ± ml. We introduce a new quantum number Λ = ml , and states are labelled according to their value of Λ. The notation is the same as that

used for atoms, except greek letters are used instead - so for Λ = 0, 1, 2, 3 ... we use the letters Σ, Π, ∆, Φ...


56

A diatomic molecule has cylindrical symmetry about the internuclear axis. The internuclear axis defines a special axis in a molecule, and we

make this our z-axis. The projection of the angular momentum onto the internuclear axis - characterized by the quantum number ml - is a

constant of the motion in a diatomic molecule. So the ml quantum number has a special status. It cannot make any difference to the energy

whether the electrons are going around the internuclear axis in the clockwise or anticlockwise direction, so there’s a degeneracy with respect

to ± ml. We introduce a new quantum number Λ = ml , and states are labelled according to their value of Λ. The notation is the same as that

used for atoms, except greek letters are used instead - so for Λ = 0, 1, 2, 3 ... we use the letters Σ, Π, ∆, Φ...

Molecular orbitals are now labelled according to the value of Λ, the atomic orbital used to construct the molecular one, and the letter g or u to

denote the symmetry. Thus, the ground-state bonding orbital for H2+ is labelled Σg 1 s - Σ means that Λ = 0, the subscript g tells us it’s the

“gerade” symmetry, and the 1s tells us that it’s the 1s atomic orbital we’ve used to construct the molecular orbital. The antibonding orbital is

labelled Σu* 1 s. Here, we add the asterisk to denote the fact that this is an antibonding orbital.

14.4. Building up other diatomic molecules

Molecular orbitals are filled up according to the Pauli exclusion principle, with each orbital accommodating two electrons with opposite spin

directions. So, now we can build some molecules.

è H + H. When two hydrogen atoms come together both electrons go into the Σg 1 s orbital with opposite spins. The H2 molecule is stable

because both electrons are in bonding orbitals - this is why hydrogen gas is a diatomic gas.

è He + He. When two ground-state He atoms come together we have 4 electrons altogether, all of them in 1s states. Two of the electrons go

into the Σg 1 s orbital, but the other two have to go into the Σu* 1 s antibonding orbital. The net effect is to give a total binding energy very

close to zero. This is why helium gas is a monatomic gas.

è O + O. The ground-state configuration of oxygen is 1 s2 2 s

2 2 p4. Now a 2p atomic orbital can give six molecular orbitals: they are Σg 2 p,

Σu* 2 p, Πg 2 p and Πu

* 2 p with the Π orbitals being doubly degenerate because we can have ml = ± 1. Three of the six are bonding orbitals,

and the other three antibonding. We have to accommodate 8 electrons from atomic p orbitals in total. The first 6 can go into the 3 bonding

orbitals. The other two go into one of the antibonding orbitals. So - 3 filled bonding orbitals, 1 filled antibonding orbital, giving a net of two

bonds. So the O2 molecule is stable, having a “double bond”. This is why oxygen gas is a diatomic gas.


è The Born-Oppenheimer approximation is based on the fact that the nuclei, being much heavier than the electrons, move much more

slowly than the electrons.

è In the Born-Oppenheimer approximation, the total wavefunction of the molecule is written as a product of electronic, vibrational and

rotational wavefunctions: Y = Ψe Ψv Ψr. The total energy is the sum of the individual energies: Etotal = Eelectronic + Evibration + Erotation .

è To find the electronic wavefunction and energy, the nuclei are fixed at a separation R. Then the nuclear kinetic energy term is absent in

the electronic Hamiltonian. The electronic energy is a function of R - these are the potential energy curves of the molecule.

è Molecular orbitals are built up from linear combinations of atomic orbitals.

è For a homonuclear molecule, there has to be a symmetry through the origin, so we have symmetric (“gerade”) and antisymmetric

(“ungerade”) states.

è In the H2+ molecule, the symmetric orbital is a bonding orbital. The bond is formed because of the excess electron density in the region

between the nuclei. The antisymmetric orbital is an antibonding orbital because there is an absence of electron density in the region

between the nuclei.

14.6. To do

Revise the material in this handout. Useful reading is sections 5.1 and 5.2 of Thorne.



57

15. Vibrational and rotational structure of diatomic molecules

In the last lecture, we considered the Born-Oppenheimer approximation which tells us that the nuclear motion and the electronic motion can

be treated separately. Then we looked at the electronic part of the problem. In this lecture, we look at the nuclear motion. As we will see, this

gives us vibrational and rotational energy levels.

Let’s begin with a sketch of the procedure involved in the Born-Oppenheimer separation of the electronic and nuclear motion. We begin with

the Schrödinger equation for the diatomic molecule,

(15.1)H`

Y = E Y.

The full Hamiltonian H`

contains the the nucletar kinetic energy T`

N , the electronic kinetic energy, T`

e, and the Coulomb potential for every

charge interacting with every other charge, V :

(15.2)H`

= T`

N + T`

e + V .

For a diatomic molecule, with nuclei of charge ZA e and ZB e, mass MA and MB, and having N electrons each labelled by the index i, these

terms in the Hamiltonian are

(15.3)T`

N = -Ñ2

2 ΜÑR

2, T`

e = -Ñ2

2 me

âi=1

N

Ñi2, V =

ZA ZB e2

4 Π Ε0 R

-e

2

4 Π Ε0

âi=1

N ZA

riA

+ZB

riB

+e

2

4 Π Ε0

âall pairs

1

rij

.

Here, Μ is the reduced mass of the two nuclei, given by

(15.4)Μ =MA MB

MA + MB

.

Now we write the wavefunction as a product of an electronic part and a nuclear part:

(15.5)YHri, RL = ΨeHri, RL ΧN HRL.To find the electronic wavefunction, the electrons are fixed in place with a separation R. The nuclear kinetic energy then disappears from the

Hamiltonian, and the equation we have to solve is

(15.6)IT` e + V M ΨeHri, RL = EeHRL ΨeHri, RL.This equation is considerably easier to solve than the one where the nuclei are also allowed to move. The methods that are used are similar to

the ones used to find the energy levels of an atom, e.g. the self-consistent field method. Solving this equation we get EeHRL - the electronic

energy as a function of internuclear distance. This is called the potential energy curve.

It can then be shown that the nuclear wavefunctions satisfy the following equation:

(15.7)-Ñ2

2 ΜÑR

2+EeHRL ΧN HRL = E ΧN HRL.

This is just a Schrödinger equation for the nuclear motion, where the potential that appears is the electronic energy as a function of R, EeHRL.So the solution to the electronic part of the problem appears as the potential for the nuclear motion. This makes good sense. In order for the

nuclei to move away from the equilibrium separation R0 to some other R, they have to provide the energy necessary to change the electronic

energy from EeHR0L to EeHRL - which is just a way of saying that the nuclei are moving in the potential EeHRL.15.1. Rotational structure

The potential EeHRL depends only on the internuclear distance R. It does not depend on the angular coordinates of the nuclei. So, once again,

we have a central potential. We know that whenever we have a potential that is not a function of the angular coordinates, we can separate the

wavefunction into an angular part and a radial part. So, in the usual way, we write equation (15.7) in spherical coordinates HR, Q, FL, and

write the wavefunction as a product of a radial function f HRL and an angular part gHQ, FL:(15.8)ΧN HR, Q, FL =

1

R

f HRL gHQ, FL.As always for a central potential, the angular solutions are the spherical harmonics, gHQ, FL = YJ ,M HQ, FL. They are angular momentum

eigenfunctions satisfying

(15.9)J` 2

YJ ,M = JHJ + 1L YJ ,M , Jz

`YJ ,M = M YJ ,M .

Then, we end up with the following radial equation:

(15.10)-Ñ2

2 Μ

â2

â R2

+JHJ + 1L Ñ2

2 Μ R2

+ EeHRL - E f HRL = 0

Next, we make the approximation that the molecular bond is quite rigid, and so the nuclear separation stays close to its equilibrium value R0.

This is called the rigid rotor approximation. It allows us to replace R by R0 in the second term of the above equation, so that this term is now

just a constant. So the equation becomes


58

(15.11)-Ñ2

2 Μ

â2

â R2

+Er + EeHRL - E f HRL = 0

where

(15.12)Er =Ñ2

2 Μ R02

JHJ + 1L.

This is an expression for the rotational energy of the molecule, in this rigid rotor approximation. Note that (apart from the quantum number J

and the constant Ñ), the rotational energy depends only on the reduced mass of the nuclei and on the bond length.

We can see why this is the rotational energy from a very simple model - a stick and ball model. We model the molecule as two balls (mass

MA and MB) on the ends of a rod of length R0. Recall from classical mechanics that the energy of a rigidly rotating body is

(15.13)Er =L

2

2 I

,

where L is the angular momentum, and I is the moment of inertia. It’s easy to show that the moment of inertia for our two balls on a stick is

I = Μ R02, where Μ is the reduced mass. Using this, and the quantization of angular momentum which replaces L2 by JHJ + 1L Ñ2 (integer J),

and we immediately have the rotational energy given by equation (15.12).

Let’s look at an example - the NaCl molecule. It has a reduced mass of Μ = 13.9 atomic mass units, and a bond length of R0 = 0.24 nm. This

gives us, Er h = 6.3 JHJ + 1L GHz, where we’ve expressed the result as a frequency. The plot below shows the ladder of rotational levels in

the molecule, expressed in frequency units, with each level labelled by its value of J . The spacing of the two lowest levels is 12.6 GHz. This

is in the microwave region of the electromagnetic spectrum. This is typical for the rotational states of all molecules. So transitions between

rotational states of a diatomic molecule are in the microwave part of the spectrum.

0

1

2

3

4

J

Frequency HGHzL

15.2. Vibrational structure

We are left needing to solve the radial equation - equation (15.11). The equation contains the potential curve EeHRL, and so the solutions will

of course depend on the form of this potential. Whatever its form, the equation can always be solved numerically (i.e. on a computer) and the

energy levels in the potential found - these are the vibrational energy levels of the molecule.

Let’s look at a typical potential energy curve once again:


59

R0 R

EeHRL

Whatever the exact form of the potential energy curve, we know that it has a turning point at R0. Let’s make a Taylor expanasion of EeHRLaround R0:

(15.14)EeHRL » EeHR0L +â Ee

â RR0

HR - R0L +1

2

â2Ee

â R2

R0

HR - R0L2+ ...

The first derivative is zero at the turning point, and so the second term in the above expression is zero. So, around the turning point we have

(15.15)EeHRL » EeHR0L +1

2k HR - R0L2.

where

(15.16)k =â2

Ee

â R2

R0

.

This is simply telling us that, close enough to the turning point, the potential depends quadratically on the distance from the equilibrium point.

This is the potential of a harmonic oscillator, where k is the “spring constant”. In other words, provided the molecule doesn’t stretch too

much, it behaves like two masses connected by a spring (quantized, of course).

Substituting equation (15.15) into equation (15.11), and defining the vibrational energy, Ev by E = EeHR0L + Ev + Er we have

(15.17)-Ñ2

2 Μ

â2

â R2

+1

2k HR - R0L2

- Ev f HRL = 0,

which is the equation for a quantum harmonic oscillator. You’ve studied the solutions to this equation before, so we can skip directly to the

solution. In particular, the energy eigenvalues are

(15.18)Ev = Hv + 1 2L h f0,

where v = 0, 1, 2 ... is the vibrational quantum number, and

(15.19)f0 =1

2 Π

k

Μ.

is the frequency of the oscillations. So we get a ladder of equally spaced vibrational energies, separated by the energy h f0, and with the

lowest level being 1 2 h f0 above the zero-point of the potential.

Let’s look again at the example of the NaCl molecule. For this molecule, f0 = 1.1 ´ 1013 Hz. This is the fundamental vibrational frequency of

the molecule. The vibrational frequencies of other molecules are similar. Notice how the vibrational frequency is much larger than the

rotational frequency (which for most molecules is in the 1010 - 1011 Hz range), but much smaller than a typical electronic energy spacing

(which is usually in the 1014 - 1015 Hz range). Also, the vibrational frequency is small compared to the typical depth of the electronic

potential - for NaCl, h f0 = 0.045 eV whereas the potential depth is 4.3eV. This means that there are a very large number of vibrational

energy levels inside the potential, and our harmonic-oscillator approximation is a very good approximation for all the low-lying levels.

The figure below shows schematically the bottom of the potential energy curve, and the low-lying vibrational energy levels within it.


60

1

2h f0

h f0

RR0

EeHRL

v=0

v=1

v=2

v=3

v=4

v=5

v=6

v=7

v=8

If we go far enough up the vibrational ladder, we notice that the potential is not harmonic any longer. For this reason, the high-lying vibra-

tional levels are no longer equally spaced. The spacing between the levels decreases as the vibrational quantum number increases. This is

illustrated in the figure below which shows the vibrational energy levels for the ground-state molecular potential of the LiH molecule.

2 4 6 8 10 12

-20 000

-15 000

-10 000

-5000

0

5000

10 000

RHa0L

Energ

yHcm

-1

L


è The rotational energies of a diatomic molecule are given by Er =Ñ2

2 Μ R02

JHJ + 1L, where Μ is the reduced mass, R0 is the bond length,

and J is the rotational quantum number (J = 0, 1, 2 ...L. The rotational spacing increases as J increases.

è Rotational eigenfunctions are spherical harmonics. Rotational energy spacings are in the microwave region, typically 10-100 GHz.

è The low-lying vibrational energies of a diatomic molecule are given by Ev = Hv + 1 2L h f0 where f0 is the vibrational frequency and v is

the vibrational quantum number (v = 0, 1, 2 ...). The vibrational frequency is related to the reduced mass and the spring constant via

f0 =1

2 Π

k

Μ. The vibrational spacing is a constant for the low-lying levels.

è Vibrational eigenfunctions are those of a harmonic oscillator. Vibrational energy spacings are in the infra-red, typically 10 THz.

è Electronic energy spacings are in the visible and ultra-violet region, typically 100-1000 THz.

15.4. To do

Revise the material in this handout. Useful reading is sections 5.3 & 5.4 of Thorne.

Do question 3 from problem set 5.


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