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PHY 5645: Summary of Lecture 17 (Lenz’s vector and the degeneracy of
the Hydrogen atom bound states)
The extraordinary degeneracy of the hydrogen atom (Kepler problem) is connected to the fact that the classical orbits do not precess. The vector which points from the origin along the semimajor axis of the orbit is a constant of motion. Quantum mechanically, this vector operator, called Lenz’s vector, is
RLenz =
2 m
(p × L − L × p) −
e^2 r
r
which is written in an explicitly Hermitian form. A direct calculation shows that
[RLenz , H] = 0 where H =
p^2 2 m
e^2 r
As we saw before, the fact that the angular momentum L is a constant of motion (i.e. commutes with the Hamiltonian with central forces), lead to 2l + 1-fold degeneracy of the energy levels. Therefore, we expect that the additional conservation law associated with RLenz will lead to additional degeneracy. Now, RLenz · L = L · RLenz = 0.
Moreover,
R^2 Lenz = e^4 +
m
H(L^2 + ¯h^2 ) (1)
It is an unusual feature of the hydrogen atom problem that the Hamiltonian can be written in terms of other constants of motion i.e. in terms if R^2 Lenz and L^2. Since RLenz is a vector under rotations: (^) [ RLenz μ, Lν
] = ih≤¯μνλRLenz λ
also (^) [
RLenz μ, RLenz ν
] = i¯h
− 2 H
m
≤μνλLλ
It is convenient to absorb the factor
√ −^2 mH into the definition of a new vector
K =
√ −m 2 H
RLenz.
This vector operator is Hermitian when acting o the negative energy states, which are the ones we’re interested in. Then
[Kμ, Kν ] = i¯h≤μνλLλ (2) [Kμ, Lν ] = i¯h≤μνλKλ (3) [Lμ, Lν ] = i¯h≤μνλLλ (4)
From Eq.(1) we also have
H = −
e^4 m K^2 + L^2 + ¯h^2
Which means that if we can find the eigenvalues and degeneracies of K^2 and L^2 , we have solved the bound states problem. To decouple the above algebra, lets define
M =
(L + K) ; N =
(L − K)
then
[Mμ, Mν ] = i¯h≤μνλMλ (5) [Nμ, Nν ] = i¯h≤μνλNλ (6) [Mμ, Nν ] = 0. (7)
Notice that both M and N obey commutation relations of the angular momentum. In terms of M and N the Hamiltonian is
H = −
me^4 2
( 2 M2 + 2N^2 + ¯h^2
Since M and N obey angular momentum commutation relations, we know all about their eigenstates and eigenvalues. We can find simultaneous eigenstates of M^2 , Mz , N^2 and Nz which we denote by |M, N , μ, ν〉 where
M^2 |M, N , μ, ν〉 = h¯^2 M(M + 1)|M, N , μ, ν〉 (8) N^2 |M, N , μ, ν〉 = h¯^2 N (N + 1)|M, N , μ, ν〉 (9) Mz |M, N , μ, ν〉 = hμ¯ |M, N , μ, ν〉 (10) Nz |M, N , μ, ν〉 = hν¯ |M, N , μ, ν〉 (11)
where M can take on values 0, 12 , 1 , 32 ,... as can N. In addition, μ = −M, −M + 1,... , M − 1 , M and ν = −N , −N + 1,... , N − 1 , N. Because, RLenz · L = 0, we have K · L = 0 and
M^2 − N^2 = 0.
So the only states which are physical are therefore the ones with M = N.
The eigenstates of H are |M = N , μ, ν〉
and
H|M = N , μ, ν〉 = −
e^4 m 2(4¯h^2 M(M + 1)) + ¯h^2
|M = N , μ, ν〉 = −
e^4 m 2¯h^2 (2M + 1)^2
|M = N , μ, ν〉
where 2M + 1 = 1, 2 , 3 ,... which is the principal quantum number n. For a fixed value of M = N there are 2M + 1 values of μ for each of which there are 2 M + 1 values of ν. Therefore the total degeneracy (not including spin) is (2M + 1)^2 = n^2. Thus, the additional symmetry associated with the conservation of the Lenz vector gives the complete account of the n^2 −fold degeneracy of the hydrogen atom bound state spectrum. Note, that |M = N , μ, ν〉 are eigenstates of Lz , however, they are not eigenstates of L^2.
