7. Angular Momentum

• The order in which you rotate things makes a difference, 12 21

• We can use this to work out commutation relations for the L’s– It can be done more easily directly

• Recall:• Also recall:

• We will calculate the following to second order in :

• If rotations commuted, both sides would be the identity relation

Why it doesn’t commute 7A. Angular Momentum Commutation

ˆ ˆ, expR i n n LR

1 2 1 2R R RR R R R

ˆ ˆ ˆ ˆ, , , ,

ˆ ˆ ˆ ˆ, , , ,

R R R R

R

x y x y

x y x y

R R R R

R R R R

• To second order in :

• Second half is same thing with –

Calculating the Left Side ˆ ˆ ˆ ˆ, , , ,R R R R x y x yR R R R

ˆ ˆ, , exp expx yR R i L i L x yR R

2 2 2 2 2 21 12 21 1x x y yi L L i L L

2 2 2 2121 2x y x x y yi L L L L L L

2 2 2 212

2 2 2 212

ˆ ˆ ˆ ˆ, , , ,

1 2

1 2

x y x x y y

x y x x y y

R R R R

i L L L L L L

i L L L L L L

x y x yR R R R

2 21 ,x yL L 22 2 2 2 2 21 2x y x x y yL L L L L L

2 2 3ˆ ˆ ˆ ˆ, , , , 1 ,x yR R R R L L O x y x yR R R R

• To second order in :

• Second half is same thing with –

Calculating the Right Side

1 0 0

ˆ, 0 cos sin

0 sin cos

xR

ˆ ˆ ˆ ˆ, , , ,R x y x yR R R R

212

212

1 0 0

0 1

0 1

21

2

212

1 0

ˆ , 0 1 0

0 1

yR 21

22 21

22

1 0

ˆ ˆ, , 1

1

x yR R

2 21 12 22 2 2 21 1

2 22 2

1 0 1 0ˆ ˆ, ,

1 1ˆ ˆ, ,

1 1

x y

x y

R R

R R

• To second order in :

• Now match the two sides:

Matching the Two Sides

2

2

1 0

ˆ ˆ ˆ ˆ, , , , 1 0

0 0 1

x y x yR R R R 2ˆ, z=R

2ˆ ˆ ˆ ˆ, , , , exp zR i L x y x yR R R R

2 2 3ˆ ˆ ˆ ˆ, , , , 1 ,x yR R R R L L O x y x yR R R R

2 2 21 , 1x y zL L i L

,x y zL L i L

2ˆ ˆ ˆ ˆ ˆ, , , , ,R R x y x y zR R R R R

2 3ˆ ˆ ˆ ˆ, , , , 1 zR i L O x y x yR R R R

• Generalizing, we have• Define the Levi-Civita symbol

• Then we write:

• We will call any three Hermitian operators Jthat work this way generalized angular momentum

Levi-Civita Symbol

, , , , ,x y z y z x z x yL L i L L L i L L L i L

1 if is an even permutation of

1 if is an odd permutation of

0 otherwiseijk

ijk xyz

ijk xyz

1

1

xyz yzx zxy

xzy zyx yxz

,i j ijk kk

L L i L

,i j ijk kk

J J i J

• What can we conclude just from the commutation relations?• If J commutes with the Hamiltonian, than we can

simultaneously diagonalize H and one component of J– Normally pick Jz

• Define some new operators:• Reverse these if we want:

• These satisfy the following properties– Proof by homework problem

J2 and the raising/lowering operators 7B. Generalized Angular Momentum

,i j ijk kk

J J i J

, 0H J

2 2 2 2x y zJ J J J x yJ J iJ

† 2 2 2 2, , 0 , , , ,z z zJ J J J J J J J J J J J J J

1 12 2,x yJ J J J i J J

Eigenstates

• Since J2 commutes with Jz, we can diagonalize them simultaneously• We will (arbitrarily for now) choose an odd way to write the eigenvalues• Note that j and m are dimensionless• Note that J2 has positive eigenvalues

– We can choose j to be non-negative• We can let J act on any state |j,m to produce a new state J |j,m:

• This new state must be proportional to:

,j m

2 2 2, ,

, ,z

j m j j j m

J j m m j m

J

2 2, 0 , , ,zJ J J J J J J

2 2, ,J j m J j m J J 2 2 ,j j J j m 2 2 ,J j j j m

,zJ J j m

2 ,J j m J

, ,z zJ J J J j m ,J m J j m 1 ,m J j m

, , 1J j m j m

Eigenstates (2)

• To find proportionality, consider

• This expression must not be negative:

• When it is positive, then we have– Choose the phase positive

• Conclusion: given a state |j,m, we can produce a series of other states

• Problem: if you raise or lower enough times, you eventually get |m| > j• Resolution: You must have:

2 2 2, ,

, ,z

j m j j j m

J j m m j m

J

† 2 2, z zJ J J J J J J

2,J j m , ,j m J J j m 2 2, ,z zj m J J j m J 2 2 2j j m m

2 2j j m m 22j j m m j m

, , 1J j m j m

2 2, , 1J j m j j m m j m

, 1 , , 2 , , 3j m j m j m

2 2 0j j m m

max min, 0 ,J j m J j m max

min

m j

m j

Summary• Eigenstates look like• The values of m are

– There are 2j + 1 of them • Since 2j+1 is an integer:

We can use these expressions to write out J’s as matrices of size (2j + 1) (2j + 1):

• First, pick an order for your eigenstates, traditionally• The matrix Jz is trivial to write down, and is diagonal

• The matrix J+ is a little harder, and is just above the diagonal

• You then get J- = J+† and can find Jx and Jy

2 2 2, ,

, ,z

j m j j j m

J j m m j m

J

2 2, , 1J j m j j m m j m

,j m max

min

m j

m j

, 1, 2, ,m j j j j

312 20, ,1, ,j 1 1

2 2,x yJ J J J i J J

, , , 1 , , ,j j j j j j

', ,z zmm

J j m J j m , ,m j m j m mmm

', ,

mmJ j m J j m 2 2 , , 1j j m m j m j m 2 2

, 1m mj j m m

Sample ProblemWrite out the matrix form for J for j = 1

, ,zJ j m m j m

2 2, , 1J j m j j m m j m

• Basis states

• Jz is diagonal:

• J+ just above the diagonal:

1,1 , 1,0 , 1, 1

z mmmmJ m

1

0 0

0 0

0 0zJ

2 2, 1m mmm

J j j m m

0 0

0 0

0 0 0

2

2J

†

0 0 0

2 0 0

0 2 0

J J

1

0

1

2

2

2 2

22

1 1 0 0 2

1 1 1 1 2

Sample Problem (2)

• Now work out Jx and Jy:

• As a check, find J2

1 0 0

0 0 0

0 0 1zJ

12

12

x

y

J J J

J i J J

12

1 12 2

12

0 0

0

0 0

xJ

12

1 12 2

12

0 0

0

0 0

y

i

J i i

i

0 2 0

0 0 2

0 0 0

J

0 0 0

2 0 0

0 2 0

J

Write out the matrix form for J for j = 1

2 2 2 2x y zJ J J J

1 1 1 12 2 2 2

2 2 2

1 1 1 12 2 2 2

0 0 1 0 0

0 1 0 0 1 0 0 0 0

0 0 0 0 1

2

2 0 0

0 2 0

0 0 2

2 2 2j j J 1

Special Cases and Pauli Matrices• The matrices for j = 0 are really simple:

– We sometimes call this the scalar representation

The j = ½ is called the spinor representation, and is important• There are only two states

• Often these states abbreviated

• The corresponding 22 matrices are written in terms of the Pauli matrices

• The Pauli matrices are given by:• Useful formulas:

0x y zJ J J

1 1 1 12 2 2 2, , ,

,

12J σ 0 1 0 1 0

, ,1 0 0 0 1x y z

i

i

1 1 12 2 2, ,z x yJ J J i

• Consider this Hamiltonian:

• All components of L commute with H, because they commute with R2

• It makes sense to choose eigenstates of H, L2 and Lz

• It seems sensible to switch to spherical coordinates:• We write Schrödinger’s equation in spherical coordinates

Spherical Coordinates 7C. Spherically Symmetric Problems

, ,n l m

2 2 21 1

2 2H V V

P R P R

2 2 2, , , , , , , , , , , , , ,n zH n l m E n l m n l m l l n l m L n l m m n l m L

sin cos

sin sin

cos

x r

y r

z r

22

2E V r

2 2 2

2 2 2 2 2

1 1 1sin

2 sin sinE r V r

r r r r

L in Spherical coordinates (1)• We need to write L in spherical coordinates• Start by writing angular derivatives out:

• It’s not hard to get Lz from these equations:

sin cos

sin sin

cos

x r

y r

z r

x y z

x y z

cos cos sin sinr r

x y z

x y z

x y z

sin sin cosr

x y

z y xL XP YP i x yy x

sin cos sini ry x

zL i

L in Spherical coordinates (2)

• Now it’s time to get clever: consider

• And we get clever once more:

sin cos , sin sin , cosx r y r z r

sin cot cos

cos cos sin sinr rx y z

sin sin cosrx y

2 2sin cos cos cos sinr rz x

x z

z x

yL i z xx z

sin cot cosyL i i

cos cot sin

2 2cos sin cos sin sinr r

y z

z y

y z

xL i y zz x

cos cot sinxL i i

Other Operators in Spherical coordinates

• It will help to get the raising and lowering operators:

• And we need L2:• Compare to Schrödinger:

2 2 2 2x y zL L L L

22 2

2 2

1 1sin

sin sin

L

x yL L iL cot cos sin sin cosi i

cotiL e i

2 2 2

2 2 2 2 2

1 1 1sin

2 sin sinE r V r

r r r r

sin cot cosyL i i

cos cot sinxL i i

Solving Spherically Symmetric Problems:• Rewrite Schrödinger’s equation:• Our eigenstates will be

• The angular properties are governed by l and m• This suggests factoring into angular and radial parts:

• Substitute into Schrodinger:• Cancel Y:

2 2

22 2

1

2 2E r V r

r r r

L

, ,n l m

2 2 2, , , , , , , , , , , , , ,n zH n l m E n l m n l m l l n l m L n l m m n l m L

,R r Y r

2 2 2

z

l l

L m

L

2 2 2

z

RY l l RY

L RY m RY

L

2 2 2

z

Y l l Y

L Y mY

L

2 2

22 2

1

2 2ERY rRY RY V r RY

r r r

L

2 2 2

22 22 2

dER rR l l R V r R

r dr r

The Problem Divided:

• It remains to find and normalize R and Y• Note that Y problem is independent of the potential V• Note that the radial problem is a 1D problem

– Easily solved numerically• Normalization:

• Split this up how you want, but normally:

,R r Y r

2 2 2

z

Y l l Y

L Y mY

L

2 2 2

22 22 2

dER rR l l R V r R

r dr r

2 31 d r r 2 22

0 0 0sin ,r dr d d R r Y

2 22

0,R r r dr Y d

2 1 2

0 0 1 0, sin , cos ,f d d f d d f d

2, 1Y d 2 2

01R r r dr

• We will call our angular functions spherical harmonics and label them• We previously found:

• For general angular momentum we know:

• We can easily determine the dependence of the spherical harmonics

• Also, recall that = 0 is the same as = 2

• It follows that m (and therefore l) is an integer

Dependence on , and m restrictions 7D. The Spherical Harmonics

zL i

, , , 1, ,zJ j m m j m m j j j , , 1, ,m mz l lL Y mY m l l l

,mlY

m ml li Y mY

m imlY e ,m im

lY f e

2 imf f e

0,1,2,l

Finding one Spherical Harmonic:

cotiL e i

2 2 1m ml lL Y l l m m Y

• We previously found:• For general angular momentum:

• If we lower m = – l, we must get zero:

• Normalize it:

2 2, , 1J j m j j m m j m

0 llL Y

,m imlY f e

coti ile i f e

coti il d

e e l fd

cotd

f f ld

cotdf

l df

ln ln sinf l k

sin lf N

21 ,l

lY d 22 sin l ilN e d

22 2

0 0sin sinlN d d

222 4 2 !

2 1 !

l lN

l

2 1 !1, sin

2 ! 4l l il

l l

lY e

l

Finding All Spherical Harmonics:

cotiL e i

2 2 1m ml lL Y l l m m Y

• To get the others, just raise this repeatedly:

• Sane people, or those who wish to remain so, do not use this formula• Many sources list them

– P. 124 for l = 0 to 4• Computer programs can calculate them for you

– Hydrogen on my website

2 1 !1, sin

2 ! 4l l il

l l

lY e

l

1

2 2

,,

mlm

l

L YY

l l m m

1

cot ,1

i mle i Y

l m l m

2 1 !1

, cot sin2 ! 4 !

l m

m i l ill l

l l mY e i e

l l m

Properties of Spherical Harmonics: 2 2 2m m

l l

m mz l l

Y l l Y

L Y mY

L

• They are eigenstates of L2 and Lz

• They are orthonormal:

• They are complete; any angular function can be written in terms of them:

• This helps us write the completion relation

*m ml l ll mmY Y d

0

, ,l

mlm l

l m l

f c Y

*, ,m

lm lc Y f d

*

0

, , , ,l

m ml l

l m l

f Y f d Y

0

, , , ,l

m ml l

l m l

f Y Y f d

0

, , cos cosl

m ml l

l m l

Y Y

1

sin

More Properties of Spherical Harmonics: 2 2 2m m

l l

m mz l l

Y l l Y

L Y mY

L

• Recall: parity commutes with L• It follows that

• Hence when you let parity act, youmust be getting essentially the same state

• Recall: L2 is real but Lz is pure imaginary• Take the complex conjugate of our relations above:• This implies • It works out to

2 2 2 2m m ml l l

m m mz l z l l

Y Y l l Y

L Y L Y m Y

L L

1lm m

l lY Y

* *2 2 2

* *

m ml l

m mz l l

Y l l Y

L Y m Y

L

*~m m

l lY Y

*1

mm ml lY Y

~m ml lY Y

The Spherical Harmonics

00

1,

2Y

01

11

3, cos

2

3, sin

2 2

i

Y

eY

0 22

12

22 2

2

5, 3cos 1

4

15, sin cos

2 2

15, sin

4 2

i

i

Y

eY

eY

0 33

1 23

22 2

3

33 3

3

7, 5cos 3cos

4

21, sin 5cos 1

8

105, sin cos

4 2

35, sin

8

i

i

i

Y

eY

eY

eY

• Hamiltonian for hydrogen (SI units):• These operators have commutators:

• Classically, what we do:– Total momentum is conserved– Center of mass moves uniformly– Work in terms of relative position

• Quantum mechanically: Let’s try • Find commutation operators for these

– Proof by homework problem• Find the new Hamiltonian

– Proof by homework problem

Changing Operators 7E. The Hydrogen Atom

22 2

tot 1 21 2 1 2

1 1

2 2ek e

Hm m

P PR R

1 1 2 2, , , all others vanishi j ij i jR P i R P 1 1 2 2

1 21 2

2 1 1 21 2

1 2

,

,

cm cm

m m

m m

m m

m m

R RR P P P

P PR R R P

, ,, , ,

all others vanish

cm i cm j ij i jR P i R P

22 2

tot

1 1

2 2e

cm

k eH

M P P

R

1 2M m m

1 2

1 1 1

m m

• Note that for actual hydrogen, is essentially the electron mass• Split the Hamiltonian into two pieces

• These two pieces have nothing to do with each other– It is essentially two problems

• Hcm is basically a free particle of mass M,– It is trivial to solve

• The remaining problem is effectively a single particle of mass in a spherically symmetric potential

Reducing the problem: 6D to 3D, ,, ,cm i cm j ij i jR P i R P 2

2 2tot

1 1

2 2e

cm

k eH

M P P

R1 2M m m 1 2

1 1 1

m m

tot cmH H H 21

2cm cmHM

P2

21

2ek e

H

PR

21

2cm cmEM

p

tot cmE E E

• Because the problem is spherically symmetric, we will have states• These will have wave functions

• The radial wave function will satisfy:

• Note that m does not appear in this equation, so R won’t depend on it

• We will focus on bound states E < 0

Reducing the problem: 3D to 1D

,mnlm lR r Y r

221

2ek e

H

PR

, ,n l m

22 2 2

22 22 2

ek edER rR l l R R

r dr r r

nlR r

,mnlm nl lR r Y r

Let’s try to approximate behavior at r = 0 and r = :•Large r: keep dominant terms, ignore those with negative powers of r:

•Define a such that:•Then we have•Want convergent •Now, guess that for small r we have•Substitute in, keeping smallest powers of r

•Want it convergent

The Radial Equation: For r Large and Small

22 2 22

2 22 2ek ed

ER rR l l R Rr dr r r

22 2 2 2

22 22 2

ek ed dER R R l l R R

dr r dr r r

~ r aR e ~ , for large r aR e r

2

22E

a

~R r

2 2

2 2 2 2 112 2 eEr r l l r k e r

2 2l l

or = 1l l

~ , for small lR r r

• Factor out the expected asymptotic behavior:– This is just a definition of f(r)

• Substitute in, multiply by 2/2

• Define the Bohr radius:

The Radial Equation: Removing Asymptotic

22 2 2 2

22 2 22 2 2

ek edR rR l l R R

a r dr r r

r aR r f r e

~ , for large r aR e r

22 2

2 2 2 2

21 1r a r a r a r aek ed l le f rfe fe fe

a r dr r r

22 2

2 2 2 2 2

21 1 2 1r a r a r a r a r a r aek ed d l lfe rf e rf e fe fe fe

a r dr ar dr a r r

22 2

2 2 2

21 20ek ed d l l

rf rf f fr dr ar dr r r

2

0 2e

ak e

• Write f as a power series around the origin– Recall that at small r it goes like rl

• Substitute in

• Gather like powers of r:

• On right side, replace i i – 1• On left side, first term vanishes

• These must be identical expressions, so:

The Radial Equation: Taylor Expansion

2 2

2 20

1 2 20

d d l lrf rf f f

r dr ar dr r ra

ii

i l

f f r

~ , for small lR r r

2 2 1 2 2 1

0

2 21 0i i i i

i i i ii l i l i l i l

f i i r f i r l l f r f ra a

2 2 2 1

0

2 21i i

i ii l i l

f i i l l r f i ra a

2 2 2 21

1 0

2 2i ii i

i l i l

f i i l l r f i ra a

012 2

2 2i i

i a af f

i i l l

1

• It looks like we have a solution for any E:– Pick fl to be anything– Deduce the rest by recursion

• Now just normalize everything• Problem: No guarantee it is normalizable• Study the behavior at large i:

• Only way to avoid this catastrophe is to make sure some f vanishes, say fn

Are We Done?i

ii l

f f r

012 2

2 2i i

i a af f

i i l l

r aR r f r e

2

22E

a

12

2i i

i af f

i 1

2ifia 1 2

!

i

if i a

1 2

!

i

i

rf r

i a

2r ae 2r a r aR r e e

012 2

2 20 n n

n a af f

n n l l

0

2 20

n

a a 0a na

2

0 2e

ak e

222

2 22ek e

En

r ae

• Because the exponential beats the polynomial, these functions are now all normalizable– Arbitrarily pick fl > 0

• Online “Hydrogen” or p. 124• Note that n > l, n is positive integer• Include the angular wave functions

• Restrictions on the quantum numbers:• Another way of writing the energy:

• For an electron orbiting a nucleus, is almost exactly the electron mass, c2 = 511 keV

Summarizing Everything

0

1

1

2 20

2

nr na i

nl ii l

ii

R r e f r

i n ff

n i i l l a

2

0 2e

ak e

2 4

2 22ek e

En

2 2 1nlR r r dr

,mnlm nl lR r Y r

1,2,3,

0,1, , 1

, 1, ,

n

l n

m l l l

2 4 2 2

2 2 22 2ek e c

En n

2 1

137.04ek e

c

2

13.6 eVE

n

Radial Wave Functions

0

10 30

2 r aeR r

a

02

20 300

122

r ae rR r

aa

02

21 502 6

r areR r

a

03 2

30 230 00

2 2 21

3 273 3

r ae r rR r

a aa

03

31 500

4 21

627 3

r are rR r

aa

032

32 70

2 2

81 15

r ar eR r

a

04 2 3

40 2 330 0 00

31

4 8 1924

r ae r r rR r

a a aa

04 2

41 250 00

51

4 8016 3

r are r rR r

a aa

042

42 700

11264 5

r ar e rR r

aa

043

43 90768 35

r ar eR r

a

Sample ProblemWhat is the expectation value for R for a hydrogen atom in the state |n,l,m = |4,2,-1?

14,2, 1 42 2 ,R r Y

r

R R 2 3r d r r 23

0r dr d

r

22 3 142 20

,R r r dr d Y

• The spherical harmonics are orthonormal over angles

0

2

2 4 37 00 0

11

20480 12r ar

R e r r dra a

1

042

42 700

11264 5

r ar e rR r

aa

027 7 8 8 9 91 10 0 06 1440

1

20480r aa r a r a r e dr

8 9 101 10 6 144

12 7! 2 8! 2 9!

20480a 021a

> integrate(radial(4,2)^2*r^3,r=0..infinity);

Degeneracy and Other Issues

2 4

2 22ek e

En

1,2,3,

0,1, , 1

, 1, ,

n

l n

m l l l

, ,n l mNote energy depends only on n, not l or m•Not on m because states related by rotation•Not on l is an accident – accidental degeneracyHow many states with the same energy En?•2l + 1 values of m•l runs from 0 to n – 1•Later we will learn about spin, and realize there are actually twice as many statesAre our results truly exact?•We did include nuclear recoil, the fact that the nucleus has finite mass•Relativistic effects

– Small for hydrogen, can show v/c ~ ~ 1/137•Finite nuclear size

– Nucleus is 104 to 105 times smaller– Very small effect

•Nuclear magnetic field interacting with the electron

1

0

2 1n

nl

D E l

2n

Can we apply our formulas to any other systems?•Other atoms if they have only one electron in them•The charge on the nucleus multiplies potential by Z:•Reduced mass essentially still the electron mass•Just replace e2 by e2Z•Atom gets smaller

– But still much larger than the nucleus•Relativistic effects get bigger

– Now v/c ~ Z

Other Nuclei 7E. Hydrogen-Like Atoms

221

2ek e Z

H

PR

2 4 2 2 2 2 2

2 2 2 2

13.6 eV

2 2ek e Z c Z Z

En n n

0 0a a Z

2 2 511 keVec m c

Bizarre “atoms”We can replace the nucleus or the electron with other thingsAnti-muon plus electron •Anti-muon has same charge as proton, and much more mass than electron•Essentially identical with hydrogenPositronium = anti-electron (positron) plus electron•Same charge as proton•Positron’s mass = electron’s mass•Reduced mass and energy states reduced by halfNucleus plus muon•Muon 207 times heavier than electron•Atom is 207 times smaller•Even inside a complex atom, muon sees essentially bare nucleus•Atom small enough that for large Z, muon is partly inside nucleusAnti-hydrogen = anti-proton plus anti-electron•Identical to hydrogen