Part 15, 16, 18
Outline, Schedule

# PHYS 424 Notes

## Part 16

1. Chapter 4
1. Angular Momentum
1. Operators

In classical mechanics the angular momentum is given by L= r x p. We can convert this to quantum mechanics by using QM operators, i.e. p = -i hbar  Ñ. Because of the representation of p, we are not guaranteed that the various components of L commute, and we shall in fact find that they do not. From the commutation relations, we will be able to find the eigenvalues of L; some of these values will correspond to angular momenta which cannot in fact be writted as r x p. We shall call the new kind of angular momentum S. The sum of L and S is conserved in experiments, so we will often use the quantity J = L + S.

Some of the algebra is simplified by defining the quantity eijk by the relations

e123= 1
eijk = - ejik
eijk = ejki
eijk = 0 if any two indices are equal

Because of the definition,

Sk eijk ekmn = dim djn - din djm                   

The proof of this relation is tedious but straight-forward. It is perhaps more plausible if you write the vector cross-product [a = b x c] as       ai = S j, k eijk bj ck and compare a x (b x c) = b(a.c) - c(a.b)

It is also worth noting that in spherical coordinates,

 Lz = - i hbar d/df Ñ2 = (1/r2)d/dr (r2 d/dr) - L2 / ( hbar2 r2)

The first of these is proved in the text; the second can be shown by some straight-forward manipulation of results in the text. Given these expressions, we can see that we already know the eigenvectors and eigenvalues of L2 and Lz. Clearly Lze imf = m hbar e imf (and we know that m is an integer), and our work with the angular part of Ñ2 establishes that L2Ylm = hbar2 l(l+1). We can also get the eigenvalues algebraically, and in doing so we get a surprise. To work algebraically, we need commutators.

2. Commutators

The text shows component by component that

[Li , Lp] = i hbar eipq Lq

I will get the same result using the eijk, as an example of the use of that symbol.

[Li , Lp] = Sj,k,q,s {eijk epqs [rjpk , rqps]}
= Sj,k,q,s {eijk epqs {rj [pk , rqps] + [rj , rqps] pk }} by expanding the commutator
= Sj,k,q,s {eijk epqs {-i hbar rj dkq ps + i hbar rq djs pk }} by evaluating the two commutators
= Sj,q,s [ i hbar [- eijq epqs rj ps] + Sk,q,s [eisk epqs rq pk ] by using the two Kronecker deltas

Now in order that the r and p factors have the same indices in each term, so that they may be factored, we will rename s to q, q to j and k to s in the second term only:

[Li , Lp] = i hbar Sj,q,s [- eijq epqs rj ps] + i hbar Sj,q,s [eiqs epjq rj ps ]
= i hbar Sj,q,s {rj ps (-eijq eqsp + esiq eqpj)} by factoring rj ps
= i hbar Sj,q,s {rj ps [-(dis djp -dip djs) +(dsp dij -dsj dip)]} by using eq.  above in both terms
= i hbar Sj,q,s {rj ps [-(dis djp) +(dsp dij)]} by combining terms

We can now apply eq.  in reverse to get

[Li , Lp] = i hbar Sj,q,s {eipq eqjs rj ps}

and use the definition of L to get

[Li , Lp] = Sq i hbar eipq Lq [QED, whew!]

With the same technique but much less effort, you can show that

[L2 , Li ] = 0

Now commuting, Hermitian operators have the same eigenvectors, each component of L commutes with L2, and the components of L do not commute with each other. Hence we can expect to find simultaneous eigenfunctions of L2 and one of the Li , but not simultaneous eigenfunctions of any two Li . It is conventional to seek eigenfunctions of L2 and Lz = L3 .

3. Algebraic determination of eigenvalues

We seek a state |lm> satisfying

L2 |lm> = l |lm>
Lz |lm> = m hbar |lm>

where the book's m is given by
 m= m hbar
The hbar merely incorporates the units of L into the eigenvalue of Lz .

To find l and m, use the same technique we used on the harmonic oscillator. Define

L+ = Lx + i Ly
L- = Lx - i Ly

It is easy to show that

[Lz , L± ] = ± hbar L±
[L2 , L± ] = 0

Using the second of these,

L2 [L± |lm>] = L± L2 |lm> = l [L± |lm>]

so that L± |lm> is also an eigenfunction of L2 with eigenvalue l . It must also be an eigenfunction of Lz, but the eigenvalue need not be the same. Using the first of the commutators above, which is entirely analogous to the commutator we had in the harmonic oscillator problem,

Lz[L± |lm>] = (m ± 1) hbar [L± |lm>]

so that L± |lm> is also an eigenfunction of Lz with eigenvalue (m ± 1) hbar. Hence, as with the harmonic oscillator, given any eigenstate we can generate more by stepping up (m + hbar) or down (m - hbar) with the ladder operators.

Unlike with the harmonic oscillator, it is obvious that this process cannot go on forever, lest <L2> become smaller than <Lz>2 . There must be states satisfying both

L+ |lmt > = 0
L- |lmb > = 0

Since mt hbar is the largest eigenvalue of Lz and mb hbar is the smallest, we would certainly expect that mb = - mt . We would also expect that the eigenvalue of L2 would be related to the largest (smallest) eigenvalues of the z-projection of L, Lz . So let us see if we can evaluate l in terms of mt and mb. If we can get relations with both, we will be able to verify (or not) that mb = - mt .

It is not hard to show that

L2 = L-L+ + Lz2 + hbar Lz
L2 = L+L- + Lz2 - hbar Lz

Using the first of these on |lmt>, l = hbar2 (mt2 + mt) = hbar2 mt (mt + 1). Using the second, l = hbar2 mb (mb - 1). Hence mt (mt + 1) = mb (mb - 1). Although this is a quadratic equation for mb in terms of mt and hence gives two different relations, one of them has mb > mt and may be rejected. The other is the obvious     mb = - mt .

Finally, let's simplify our labelling. Let

mt = - mb = l
l = hbar2 l(l+1)
Relabel |lm> as |lm>

Then the state |lm> is an eigenstate of L [with eigenvalue hbar2 l(l+1) regardless of m] and Lz [with eigenvalue m hbar]. Draw a plot of the eigenvalues.

The quantity m varies from -l to l in integer steps. The only way for this to happen is for m to be an integer or half an odd integer. Thus l is an integer or half an odd integer. The integer values are expected; the half-integer values are not.

For the integer values of l m, we have verified the angular momentum eigenvalues. The eigenvectors are already known,

<q, f|lm> = Ylm(q, f)

No spatial function can have half-integral values of l m.

2. Spin S
• Options
2. New quantity
3. New type of angular momentum

• Magnetic moments
• m = gS
• H [energy] = - m.B = - gB.S
• Energy can be observed
• Moreover, if J = L + S, (d/dt)<J> = 0 but (d/dt)<L> <> 0 and (d/dt)<S> <> 0
• The half-integer solutions, despite not being r x p, must be included as angular momenta.
• S = 1/2
• States
1. | (1/2) (1/2)> = (1 0)tr
2. | (1/2) (-1/2)> = (0 1)tr
• Operators
1. Sz = (1/2) hbar sz
```              (1   0)
= (1/2) hbar (     )
(0  -1)
```
2. S+
```         (0   1)
=  hbar (     )
(0   0)
```
3. S-
```         (0   0)
=  hbar (     )
(1   0)
```
4. Sx = (1/2) (S+ + S- )
```               (0   1)
=  (1/2) hbar (     )
(1   0)
```
5. Sy = [1/(2i)] (S+ - S- )
```               (0   -i)
=  (1/2) hbar (      )
(i    0)
```

• Eigenstates of spin operators [Sz given above]
1. sx:
```            (1)                 ( 1)
[1/sqrt(2)] ( )  ,  [1/sqrt(2)] (  )
(1)                 (-1)
```
2. sy:
```            (1)                 ( 1)
[1/sqrt(2)] ( )  ,  [1/sqrt(2)] (  )
(i)                 (-i)
```

• Successive measurement of Sz and Sx
• Pz = (1/2)(1 + sz)
```    (1  0)
=  (    )
(0  0)
```
• P-z = (1/2)(1 - sz)
```         (0  0)
= (1/2) (    )
(0  1)
```
• Px = (1/2)(1 + sx)
```         (1  1)
= (1/2) (    )
(1  1)
```
• P-zPxPz
```        (0  0)
= (1/8) (    )
(1  0)
```
<> 0; there is a 1/8 chance of getting (0 1)tr from this operation.
 Last Revised 05/09/05

Part 15, 16, 18
Outline, Schedule