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 e_{ijk} by the relations
Because of the definition,
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 a_{i} = S _{j, k} e_{ijk} b_{j} c_{k} and compare a x (b x c) = b(a^{.}c) - c(a^{.}b)
It is also worth noting that in spherical coordinates,
L_{z} = - i hbar d/df Ñ^{2} = (1/r^{2})d/dr (r^{2} d/dr) - L^{2} / ( hbar^{2} r^{2}) |
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 L^{2} and Lz. Clearly L_{z}e^{ imf} = m hbar e^{ imf} (and we know that m is an integer), and our work with the angular part of Ñ^{2} establishes that L^{2}Y_{l}^{m} = hbar^{2} l(l+1). We can also get the eigenvalues algebraically, and in doing so we get a surprise. To work algebraically, we need commutators.
The text shows component by component that
I will get the same result using the
e_{ijk}, as an example of the use of
that symbol.
[L_{i} , L_{p}]
= S_{j,k,q,s}
{e_{ijk}
e_{pqs}
[r_{j}p_{k} , r_{q}p_{s}]}
= S_{j,k,q,s}
{e_{ijk}
e_{pqs}
{r_{j} [p_{k} , r_{q}p_{s}]
+ [r_{j} , r_{q}p_{s}] p_{k} }}
by expanding the commutator
= S_{j,k,q,s}
{e_{ijk}
e_{pqs}
{-i hbar r_{j}
d_{kq} p_{s}
+ i hbar r_{q}
d_{js} p_{k} }}
by evaluating the two commutators
= S_{j,q,s}
[ i hbar [- e_{ijq}
e_{pqs}
r_{j} p_{s}]
+ S_{k,q,s}
[e_{isk}
e_{pqs}
r_{q} p_{k} ]
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:
[L_{i} , L_{p}]
= i hbar S_{j,q,s}
[- e_{ijq}
e_{pqs}
r_{j} p_{s}]
+ i hbar S_{j,q,s}
[e_{iqs}
e_{pjq}
r_{j} p_{s} ]
= i hbar S_{j,q,s}
{r_{j} p_{s}
(-e_{ijq}
e_{qsp}
+ e_{siq}
e_{qpj})}
by factoring r_{j} p_{s}
= i hbar S_{j,q,s}
{r_{j} p_{s}
[-(d_{is}
d_{jp}
-d_{ip}
d_{js})
+(d_{sp}
d_{ij}
-d_{sj}
d_{ip})]}
by using eq. [1] above in both terms
= i hbar S_{j,q,s}
{r_{j} p_{s}
[-(d_{is}
d_{jp})
+(d_{sp}
d_{ij})]}
by combining terms
We can now apply eq. [1] in reverse to get
[L_{i} , L_{p}]
= i hbar S_{j,q,s}
{e_{ipq}
e_{qjs}
r_{j} p_{s}}
and use the definition of L to get
[L_{i} , L_{p}]
= S_{q}
i hbar e_{ipq} L_{q}
[QED, whew!]
With the same technique but much less effort, you can show that
Now commuting, Hermitian operators have the same eigenvectors, each
component of L commutes with L^{2}, and the components
of L do not commute with each other. Hence we can expect to
find simultaneous eigenfunctions of L^{2} and one of
the L_{i} , but not simultaneous eigenfunctions of any
two L_{i} . It is conventional to seek
eigenfunctions of L^{2} and L_{z} = L_{3} .
We seek a state |lm> satisfying
L^{2} |lm>
= l |lm>
L_{z} |lm>
= m hbar |lm>
where the book's m is given by
m= m hbar |
To find l and m, use the same technique we used on the harmonic oscillator. Define
L_{+} = L_{x} + i L_{y}
L_{-} = L_{x} - i L_{y}
It is easy to show that
[L_{z} , L_{±} ]
= ± hbar
L_{±}
[L^{2} , L_{±} ] = 0
Using the second of these,
L^{2} [L_{±} |lm>] = L_{±} L^{2} |lm> = l [L_{±} |lm>]
so that L_{±} |lm> is also an eigenfunction of L^{2} with eigenvalue l . It must also be an eigenfunction of L_{z}, 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,
L_{z}[L_{±} |lm>] = (m ± 1) hbar [L_{±} |lm>]
so that L_{±} |lm> is also an eigenfunction of L_{z} 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.
L_{+} |lm_{t} > = 0
L_{-} |lm_{b} > = 0
Since m_{t} hbar is the largest eigenvalue of L_{z} and m_{b} hbar is the smallest, we would certainly expect that m_{b} = - m_{t} . We would also expect that the eigenvalue of L^{2} would be related to the largest (smallest) eigenvalues of the z-projection of L, L_{z} . So let us see if we can evaluate l in terms of m_{t} and m_{b}. If we can get relations with both, we will be able to verify (or not) that m_{b} = - m_{t} .
It is not hard to show that
L^{2} = L_{-}L_{+} + L_{z}^{2}
+ hbar L_{z}
L^{2} = L_{+}L_{-} + L_{z}^{2}
- hbar L_{z}
Using the first of these on |lm_{t}>, l = hbar^{2} (m_{t}^{2} + m_{t}) = hbar^{2} m_{t} (m_{t} + 1). Using the second, l = hbar^{2} m_{b} (m_{b} - 1). Hence m_{t} (m_{t} + 1) = m_{b} (m_{b} - 1). Although this is a quadratic equation for m_{b} in terms of m_{t} and hence gives two different relations, one of them has m_{b} > m_{t} and may be rejected. The other is the obvious m_{b} = - m_{t} .
Finally, let's simplify our labelling. Let
Then the state |lm> is an eigenstate of L^{} [with eigenvalue hbar^{2} l(l+1) regardless of m] and L_{z} [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> = Y_{l}^{m}(q, f)
(1 0) = (1/2) hbar ( ) (0 -1)
(0 1) = hbar ( ) (0 0)
(0 0) = hbar ( ) (1 0)
(0 1) = (1/2) hbar ( ) (1 0)
(0 -i) = (1/2) hbar ( ) (i 0)
(1) ( 1) [1/sqrt(2)] ( ) , [1/sqrt(2)] ( ) (1) (-1)
(1) ( 1) [1/sqrt(2)] ( ) , [1/sqrt(2)] ( ) (i) (-i)
(1 0) = ( ) (0 0)
(0 0) = (1/2) ( ) (0 1)
(1 1) = (1/2) ( ) (1 1)
(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 |