Slides: Video 8.1.1 Separating for the Text reference ... · 8.1 The hydrogen atom solutions...

8.1 The hydrogen atom solutions

Slides: Video 8.1.1 Separating for the radial equation

Text reference: Quantum Mechanics for Scientists and Engineers

Section 10.4 (up to “Solution of the hydrogen radial wavefunction”).

The hydrogen atom solutions

Separating for the radial equation

Quantum mechanics for scientists and engineers David Miller

Internal states of the hydrogen atom

We start with the equation for the relative motion of electron and proton

We use the spherical symmetry of this equation

and change to spherical polar coordinates From now on, we drop the subscript r in the

operator

2

2

2 HV U E U

r r r r

2


In spherical polar coordinates, we have

where the term in square brackets is the operator we introduced

in discussing angular momentum Knowing the solutions to the angular momentum problem

we propose the separation

22 2

2 2 2 2

1 1 1 1sinsin sin

rr r r r

,U R r Y r

2 2 2,

ˆ /L


The mathematics is simpler using the form

where, obviously

This choice gives a convenient simplification of the radial derivatives

1 ,U r Yr r

r rR r

22

2 2

1 1r rr

r r r r r r


Hence the Schrödinger equation becomes

Dividing byand rearranging, we have

222

2 3

1 1 ˆ, , ,2 2

1 ,H

r r rY L Y Y V r

r r r r

E r Yr

2 3, / 2r Y r

222 2

2 2 2

2 1 1 ˆ ,,H

rr r E V r L Yr r Y


In

in the usual manner for a separation argument the left hand side depends only on rand the right hand side depends only on and

so both sides must be equal to a constantWe already know what that constant is explicitly

i.e., we already know that so that the constant is

222 2

2 2 2

2 1 1 ˆ ,,H

rr r E V r L Yr r Y

2 2ˆ , 1 ,lm lmL Y l l Y 1l l

1l l


Hence, in addition to the eigenequationwhich we had already solved

from our separation above, we also have

or, rearranging

which we can write as an ordinary differential equationAll the functions and derivatives are in one variable, r

2L̂

222

2 2

2 1H

rr r E V r l lr r

22 2

2 2

12 2 H

d r l lV r r E r

dr r


Hence we have mathematical equation

for this radial part of the wavefunctionwhich looks like a Schrödinger wave equation

with an additional effective potential energy term of the form

22 2

2 2

12 2 H

d r l lV r r E r

dr r

2

2

12

l lr

Central potentials

Note incidentally thatthough here we have a specific form for

in our assumed Coulomb potential

the above separation works for any potential that is only a function of r

sometimes known as a central potential

V r

2

4e po e p

eV

r rr r

Central potentials

The precise form of the equation

will be different for different central potentialsbut the separation remains

We can still separate out the angular momentum eigenequation

with the spherical harmonic solutions

22 2

2 2

12 2 H

d r l lV r r E r

dr r

2L̂

Central potentials

Since a reasonable first approximation for more complicated atoms

is to say that the potential is still approximately “central”

approximately independent of anglewe can continue to use the spherical harmonics

as the first approximation to the angular form of the orbitals

and use the “hydrogen atom” labels for theme.g., s, p, d, f, etc.

8.1 The hydrogen atom solutions

Slides: Video 8.1.3 – 8.1.4 Radial equation solutions

Text reference: Quantum Mechanics for Scientists and Engineers

Sections 10.4 starting with “Solution of the hydrogen radial wavefunction”, and 10.5

Note: The section Text reference: Quantum Mechanics for Scientists and Engineers Section 10.4 contains the complete mathematical details for solving the radial equation in the hydrogen atom problem. For this course, not all those details are required and they are consequently not all covered in the online lectures, so the additional detail, in particular on power series solutions in section Text reference: Quantum Mechanics for Scientists and Engineers Section 10.4, is optional for the student.

The hydrogen atom solutions

Radial equation solutions

Quantum mechanics for scientists and engineers David Miller


Using a separation of the hydrogen atom wavefunctionsolutions into radial and angular parts

and rewriting the radial part using

we obtained the radial equation

where we know l is 0 or any positive integer

,U R r Y r

r rR r

22 2 2

2 2

12 4 2 H

o

d r l le r E rdr r r


We now choose to write our energies in the form

where n for now is just an arbitrary real number We define a new distance unit

where the parameter is

2HRyEn

s r

2

2 22 Ho

Ena


We therefore obtain an equation

Then we write

so we get

2

2 2

1 1 04

l ld nds s s

1 exp / 2ls s L s s

2

2 2 1 1 0d L dLs s l n l Lds ds


The technique to solve this equation

is to propose a power series in sThe power series will go on forever

and hence the function will grow arbitrarilyunless it “terminates” at some finite power

which requires thatn is an integer, and

2


1n l


The normalizable solutions of

then become the finite power seriesknown as the associated Laguerre polynomials

or equivalently

2


12 1

10

!1

1 ! 2 1 !

n lql q

n lq

n lL s s

n l q q l

0

!1

! ! !

pqj q

pq

p jL s s

p q j q q


Now we can work back to construct the whole solutionIn our definition

we now insert the associated Laguerre polynomials

where Since our radial solution was

we now have

1 exp / 2ls s L s s

1 2 11 exp / 2l l

n ls s L s s

r rR r (2 / )os na r

1 2 11

1/ 2 exp / 2l lo n lR r na s s L s s

r

2 11 exp / 2l l

n ls L s s

Radial equation solutions - normalization

We formally introduce a normalization coefficient A so

The full normalization integral of the wavefunction

would be

but we have already normalized the spherical harmonicsso we are left with the radial normalization

2 11

1/ 2 exp / 2l lo n lR r na s s L s s

A

,U R r Y r

2

2 2

0 0 0

1 , sinr

R r Y r d d dr

Radial equation solutions - normalization

Radial normalization would be

We could show

so the normalized radial wavefunction becomes

2 2

0

1 R r r dr

22 2 1 21

0

2 !exp

1 !l l

n l

n n ls L s s s ds

n l

1/232 1

1

1 ! 2 2 2 exp2 !

ll

n lo o o o

n l r r rR r Ln n l na na na na

Hydrogen atom radial wavefunctions

We write the wavefunctions using the Bohr radius ao as the unit of radial distance

so we have a dimensionless radial distance

and we introduce the subscriptsn - the principal quantum number, andl - the angular momentum quantum number

to index the various functions Rn,l

/ or a

Radial wavefunctions - n = 1

Principal quantum numbern = 1

Angular momentum quantum number l = 0

0 5 10 15

0.5

1

1.5

2

Radius

R

1,0 2expR

1,0R

0 5 10 150.2

0.2

0.4

0.6

0.8


l = 0

l = 1

Radius

R 2,02 2 exp / 2

4R

2,0R

2,1R

2,16 exp / 2

12R

0 5 10 150.1

0.1

0.2

0.3

0.4


l = 0

l = 1

l = 2Radius

R

3,0

22 3 23 2 exp / 327 9

R

3,0R

3,1R

23,2

2 30 exp / 31215

R

3,16 24 exp / 3

81 3R

3,2R

Hydrogen orbital probability density

n = 1l = 0m = 0

0.3 MMZ

zx

x - zcross-section

at y = 0

1s

1.2 MMZl

zx

1s


n = 1l = 0m = 0

logarithmic intensity scale

x - zcross-section

at y = 0


n = 2l = 0m = 0

2s

0.3 MMZ

zx

x - zcross-section

at y = 0

1.2 MMZl

zx


n = 2l = 0m = 0


2s

x - zcross-section

at y = 0


n = 2l = 1m = 0

2p

2 MMZ

zx

x - zcross-section

at y = 0


n = 3l = 0m = 0

3s

.3 MMZ

zx

x - zcross-section

at y = 0

1.2 MMZl

zx


n = 3l = 0m = 0


3s

x - zcross-section

at y = 0


n = 3l = 1m = 0

3p

2 MMZ

zx

x - zcross-section

at y = 0

1.5 MMZl

zx


n = 3l = 1m = 0


3p

x - zcross-section

at y = 0


n = 3l = 2m = 0

3d

4 MMZ

zx

x - zcross-section

at y = 0


n = 3l = 2m = 1

3d

10 MMZ

zx

x - zcross-section

at y = 0

Behavior of the complete hydrogen solutions

(i) The overall “size” of the wavefunctions becomes larger with larger n

(ii) The number of zeros in the wavefunction is n – 1The radial wavefunctions have n – l – 1 zeros

and the spherical harmonics have l nodal “circles” The radial wavefunctions appear to have an additional

zero at r = 0 for all , but this is already countedbecause the spherical harmonics have at least one nodal “circle” for all

which already gives a zero as in these cases1l

0r

1l

Behavior of the complete hydrogen solutions

In summary of the quantum numbers for the so-called principal quantum number

and We already deduced that l is a positive or zero integerWe also now know the eigenenergies

Given the possible values for n

Note the energy does not depend on l (or m)

1,2,3,n 1l n

2HRyEn

Slides: Video 8.1.1 Separating for the Text reference ... · 8.1 The hydrogen atom solutions...

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