PHYS 424 Notes

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,

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 L_ze ^imf = m hbar e ^imf (and we know that m is an integer), and our work with the angular part of Ѳ establishes that L²Y_l^m = 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.

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_jp_k , r_qp_s]}
= S_j,k,q,s {e_ijk e_pqs {r_j [p_k , r_qp_s] + [r_j , r_qp_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 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 L_i , but not simultaneous eigenfunctions of any two L_i . It is conventional to seek eigenfunctions of L² and L_z = L₃ .
 

We seek a state |lm> satisfying

L² |lm> = l |lm>
L_z |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 L_z .

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_±
[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 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 L² 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

L² = L_-L₊ + L_z² + hbar L_z
L² = L₊L_- + L_z² - hbar L_z

Using the first of these on |lm_t>, l = hbar² (m_t² + m_t) = hbar² m_t (m_t + 1). Using the second, l = hbar² 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 hbar² 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,