Particle)in)a)central)potential.)The)...
Transcript of Particle)in)a)central)potential.)The)...
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Particle in a central potential. The hydrogen atom Lecture notes 6 (based on CT, Sec4on 6)
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Introduction
Central poten4al: V(r) depends only on the distance r from the origin
The Hamiltonian commutes with the three components of the orbital angular momentum L
The eigenfunc4ons of H can be required to be eigenfunc4ons of L2 and Lz as well
This defines their angular part; the eigenvalue equa4on for H is then a differen4al equa4on in the variable r only
Any isolated system of two interac4ng par4cles can be treated as a central problem
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Stationary states of a particle in a central potential
Consider a spinless par4cle of mass μ , subject to a central force derived from the poten4al V(r) (the center of force is the origin)
The eigenvalue equa4on of the Hamiltonian is:
Spherical coordinates are more suitable to solve the problem:
We look for eigenfunc4ons which are func4ons of r, θ, φ
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Stationary states of a particle in a central potential
We recall that:
From which we can rewrite the Hamiltonian as:
The angular dependence is en4rely contained in the L2 term
We therefore need to solve:
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Stationary states of a particle in a central potential
The three components of L commute with L2 and with all operators which only act on the r dependence
Therefore, the three components of L are constants of the mo4on:
[H,L]=0 and [H,L2]
The components of L do not commute with each other: we can only use L2 and one of the L components: we choose L2 and Lz
Since H, L2 and Lz commute, we can find a basis of the state space E of common eigenfunc4ons to these three observables
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Stationary states of a particle in a central potential
We need to solve the system of differen4al equa4ons:
We already know the eigenfunc4ons of L2 and Lz. The func4ons we are looking for, are the products of a func4on of r and the spherical harmonic
We now need to find R(r), such that the above func4on is eigenfunc4on of H
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Stationary states of a particle in a central potential
When we apply L2 to we get:
is a common factor on both sides, therefore we simplified it
No4ce that the expression for the laplacian is not valid at r=0
We have to make sure that the behavior of the solu4ons R(r) is regular
We look for eigenfunc4ons of an operator Hl which depends on l
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Stationary states of a particle in a central potential
Therefore, we fix l and m. The equa4on to be solved depends only on l: it is the same in the (2l+1) subspaces E(l,m)
We denote the eigenvalue of Hl by Ek,l
The index k represents the various eigenvalues associated with the same value of l
We rewrite the equa4on as:
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Stationary states of a particle in a central potential
We can simplify the equa4on by wri4ng:
Mul4plying both sides by r we get:
This equa4on can be seen as the one for a par4cle which moves in the effec4ve poten4al
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Stationary states of a particle in a central potential
The new term is always posi4ve or zero: the force tends to repel the par4cle from the force center O
It is called centrifugal poten4al
For an a_rac4ve Coulomb poten4al, the poten4al becomes:
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Stationary states of a particle in a central potential
We assume that V(r) approaches infinity less rapidly than 1/r when r approaches 0
We assume that the eigenfunc4on of H behaves like rs at the origin:
Subs4tu4ng it into the equa4on and seang the coefficient of the dominant term to zero we get:
from which:
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Stationary states of a particle in a central potential
Therefore, the two solu4ons of this equa4on behave at the origin as rl or 1/rl+1
The la_er needs to be rejected because it is not a solu4on of
for r=0.
Therefore, the acceptable solu4ons go to zero at the origin
We add the condi4on to the eigenvalue equa4on
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Stationary states of a particle in a central potential
Therefore, the fact that the poten4al is central allows:
-‐ To require the eigenfunc4ons of H to be simultaneous eigenfunc4ons of L2 and Lz, which determines their angular dependence
-‐ To replace the eigenvalue equa4on of H by a differen4al equa4on involving only r and depending on the parameter l
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Stationary states of a particle in a central potential
In principle, the func4ons must be square-‐integrable:
We can separate the variables:
The spherical harmonics are normalized: this reduces to:
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Stationary states of a particle in a central potential
However, if the spectrum of H has a con4nuum part, we only require:
where k is a con4nuous index
It is only because of the behavior of the wavefunc4on for r∞ that the normaliza4on integrals diverge if k=k’
We call k the radial quantum number, l the azimuthal quantum number and m the magne4c quantum number
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Stationary states of a particle in a central potential
The (2l+1) func4ons with k and l fixed and m from –l to l are eigenfunc4ons of H with eigenvalue Ek,l
The level Ek,l is therefore (2l+1)-‐fold degenerate
This degeneracy is due to the fact that H does not contain Lz and it is called essen4al degeneracy
It is also possible that an eigenvalue Ek,l is the same as Ek’,l’
These degeneracies are called accidental
For a fixed value of l, the radial equa4on has at most one solu4on for each Ek,l
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Stationary states of a particle in a central potential
This follows from the condi4on that
Since the radial equa4on is a second-‐order differen4al equa4on, in principle it has two solu4ons but the above requirement eliminates one of them
H, L2 and Lz cons4tute a complete set of commu4ng observables
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Motion of the center of mass and relative motion
Consider a system of two spinless par4cles of mass m1 and m2 and posi4ons r1 and r2
We assume that the forces exerted on these par4cles are derived from a poten4al energy V(r1-‐r2) which depends only on r1-‐r2
This is true for an isolated system
This system can be reduced to a single par4cle placed in a poten4al V(r)
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Motion of the center of mass and relative motion
The operators R1, P1, R2, P2 which describe the posi4ons and momenta of the two par4cles sa4sfy the commuta4on rela4ons:
with analogous expressions along y and z. All observables labeled by the index 1 commute with all those of index 2.
We define the observables (center of mass and rela4ve mo4on):
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Motion of the center of mass and relative motion
The commutators of the new observables are:
With analogous expressions along y and x. All other commutators are 0
Therefore, R, P, RG, PG sa4sfy canonical commuta4on rela4ons
We can interpret {R, P}, {RG, PG} as posi4on and momentum of two fic44ous par4cles
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Motion of the center of mass and relative motion
The Hamiltonian is:
Or equivalently
Therefore, the Hamiltonian is the sum of two terms:
with
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Motion of the center of mass and relative motion
HG and Hr commute with each other and with H
There exists a basis of eigenvectors of H which are eigenvectors of HG and Hr
We look for solu4ons of the system:
which implies with
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Motion of the center of mass and relative motion
We consider the {|rG,r>} representa4on, whose basis vectors are the eigenvectors common to RG and R
In this representa4on, a state is characterized by a wavefunc4on
In this representa4on, RG and R correspond to mul4plying the wavefunc4on by rG and r. PG and P become the differen4al operators
The state space E is the tensor product of E rG and Er
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Motion of the center of mass and relative motion
Therefore we can find a basis in the form
with
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Motion of the center of mass and relative motion
In the {|rG>} and {|r>} representa4ons we have:
The par4cle associated with the center of mass is free:
EG can take any posi4ve value or zero
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Motion of the center of mass and relative motion
The second equa4on concerns the “rela4ve” par4cle
It is subject to a central poten4al V(r)
The total angular momentum is
or equivalently:
where
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The hydrogen atom
The hydrogen atom consists of a proton:
and an electron:
q= -‐ 1.6 x 10-‐19 Coulomb
Their interac4on is electrosta4c
We study the system in the center of mass frame
The center of mass coincides with the proton, the rela4ve par4cle is the electron
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The hydrogen atom
The spectrum of H includes a discrete part (nega4ve eigenvalues) and a con4nuous part (posi4ve eigenvalues)
For E>0, the classical mo4on is not bounded in space: the eigenvalue equa4on has solu4ons for any value of E. The wavefunc4ons are not square integrable
For E<0, the classical mo4on is bounded: the eigenvalue equa4on has solu4ons only for discrete values of E. The wavefunc4ons are square-‐integrable
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The hydrogen atom
We have to find eigenvalues and eigenfunc4ons of the Hamiltonian. We consider E<0. In the {|r>} representa4on it is:
The poten4al is central: the eigenfunc4ons are of the form
We write the radial wavefunc4on as:
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The hydrogen atom
The func4on χ has to sa4sfy the following equa4on:
It is useful to introduce a new (dimensionless) variable η=αr
with:
Besides, we define:
The equa4on becomes:
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The hydrogen atom
The eigenvalues of H will be iden4fied by values of λ corresponding to physically acceptable solu4ons
In terms of λ they read:
We start by inves4ga4ng the asympto4c behavior of χEl
In the limit η∞ the eigenvalue equa4on reduces to:
The only acceptable solu4on has the form:
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The hydrogen atom
In order for the solu4on to be acceptable, QEl(η) has to diverge at most as a power of η
The eigenvalue equa4on for QEl(η) is:
The solu4on to the above equa4on has to behave as a power of η and it has to vanish for η=0
For the above equa4on, the origin is a regular singular point. Its characteris4c exponents are l+1 and –l. Only l+1 is acceptable
Therefore:
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The hydrogen atom
The series has to be truncated, in order for the func4on to have the desired asympto4c behavior
Therefore, all coefficients ck are zero for k>n’, with n’ arbitrary integer number
Therefore, for η∞ the func4on behaves as:
Replacing into the eigenvalue equa4on, and keeping only the dominant terms we have:
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The hydrogen atom
This equa4on has solu4ons only for λ=n’+l+1
Conven4onally we call n=n’+l+1 the “main quantum number”
Once we fix n, we have l≤(n-‐1)
The quan4za4on law for hydrogen atoms is therefore: λ=n and the quan4zed energies are:
We can write the solu4on as:
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The hydrogen atom
Replacing into the eigenvalue equa4on, we find that the coefficients have to sa4sfy:
These are the coefficients of Laguerre polynomials. The radial func4on can therefore be wri_en as:
The Laguerre polynomials are given by:
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The hydrogen atom
The following quan44es are fundamental for the hydrogen atom:
Indeed the first wavefunc4ons are:
The Compton wavelength of the electron is:
a0 is ≃100 4mes the Compton wavelength of the electron
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The hydrogen atom
The degeneracy of the states goes as follows:
For every value of n, there are n-‐1 possible values for l, and for each of them there are 2l+1 values for m. The degeneracy therefore is:
The various energy levels are:
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The hydrogen atom
Behavior of the angular func4ons:
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The hydrogen atom
Behavior of the radial func4ons:
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The hydrogen atom