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I am reading from Quantum Theory, Groups and Representations - Woit. In Chapter 21, on page 237 in discussion of the energy eigenstates of the Coulomb potential, the following figure is presented:

enter image description here

The preceding discussion details how we expect the eigenspaces of the Hamiltonian,

$$H=-\frac{\hbar^{2}}{2 m} \Delta-\frac{e^2}{r}$$

to be representations of $SO(3)$, given that the Hamiltonian possesses $SO(3)$ invariance. But then when the actual energy eigenstates are presented, we observe the well-known fact that there is additional, accidental degeneracy, due to a hidden symmetry. The symmetry group of the Hamiltonian must actually be larger than just $SO(3)$, with $SO(3)$ being a sub-group. My confusion arises from the following sentence,

If the representation of the larger group is reducible when one restricts to the $SO(3)$ subgroup, giving $n$ copies of the $SO(3)$ representation of spin $l$, that would explain the pattern observed here.

I don't understand what is meant by "when one restricts to the $SO(3)$ subgroup", and then how you would use that to explain the observed pattern. Here's my tentative attempt of making sense of it:

Here the eigenspaces are sums of irreducible representations of $SO(3)$, so could it mean that I would look at just the eigenspaces belonging to one value of $l$? And then "the representation of the larger group is reducible" would mean that if I acted with any of the symmetry transformations from the larger group, upon this collection of eigenspaces all having the same $l$, I'd find that they transform in invariant submodules, hence the larger symmetry group is reducible.

I don't feel I have a good grasp of what I'm supposed to think when reading "when one restricts to the [any] subgroup".

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Before Schroedinger published his eponymous equation, Pauli found this QM spectrum and pattern by using this argument. There is a larger symmetry than your angular momentum SO(3), namely SO(4)~SU(2)×SU(2) whose representations are related to n of the spectrum. (For a review see here, or any of the dozens of linked questions above and their links, containing Laplace–Runge–Lenz vector.)

But not all representations of SU(2)×SU(2) enter in the quantization. They have to be of the same dimensionality (and Casimir) for the left and right subgroups. In that case, for the irrep of spin s=0,1/2,1,... we have n= (2s+1), the dimensionality of each spin s. The angular momentum SO(3) is the diagonal subgroup of the above$^\dagger$, so for each energy level characterized by n we have $(2s+1)^2=n^2$ states.

The action of the rotation SO(3) subgroup, then, is the "addition of equal spins" (symmetric Kronecker product of s with s) which is reducible! That means that a representation of SO(4) is a direct sum of representations of its subgroup SO(3); this is the "restriction" alluded to.

For example, writing the dimensionality 2s+1 in boldface, you readily see $$ n=1 \leadsto s=0; \qquad {\mathbf 1}\otimes {\mathbf 1}={\mathbf 1}, ~~~ n^2=1,\\ n=2 \leadsto s=1/2; \qquad {\mathbf 2}\otimes {\mathbf 2}={\mathbf 1}\oplus {\mathbf 3}, ~~~ n^2=4,\\ n=3 \leadsto s=1; \qquad {\mathbf 3}\otimes {\mathbf 3}={\mathbf 1}\oplus {\mathbf 3}\oplus {\mathbf 5}, ~~~ n^2=9,\\ n=4 \leadsto s=3/2; \qquad {\mathbf 4}\otimes {\mathbf 4}={\mathbf 1}\oplus {\mathbf 3}\oplus {\mathbf 5}\oplus {\mathbf 7}, ~~~ n^2=16, $$ etc.

In spectroscopic notation, above, you call $$ {\mathbf 1}\mapsto s; \qquad {\mathbf 3}\mapsto p; \qquad {\mathbf 5}\mapsto d; \qquad {\mathbf 7}\mapsto f, ... $$

Within each n "rung" of the spectrum, SO(3) connects the states of each multiplet in the reduction among themselves, but not to any other multiplet in the reduced sum! To do that, you need one of the other three generators of SO(4) which we did not focus on, but still commute with the hamiltonian, so they bridge to the degenerate relatives multiplets.

NB. In Chapter 41, and everywhere in physics, you'll likely pull the same stunt with the representations of the Lorentz group.


$^\dagger$ That is, for every generator $L_i$ in the spin s representation, a (2s+1)×(2s+1) matrix, the formal analog $R_i$ in the right group factor combines with it to $L_i\otimes 1\!\! 1+ 1\!\!1 \otimes R_i$, in the diagonal subalgebra so(3); the corresponding rotation group element generated by each of these 3 generators is $e^{i\theta L_i}\otimes e^{i\theta R_i}$, hence a (2s+1)²×(2s+1)² matrix acting on (2s+1)²-dimensional state vectors.

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  • $\begingroup$ It isn't clear to me why SO(3) is the diagonal subgroup of SU(2)$\times$SU(2) here. I'm reading the definition of diagonal subgroup for the first time, and it seems that the diagonal subgroup should be two copies of some element of SU(2)? I'm not connecting this to your answer. And then I don't see how the fact of there being $(2s+1)^2 = n^2$ states follows. $\endgroup$ – sylowlover Feb 4 at 23:55
  • $\begingroup$ I added an explanatory footnote. You tensor group elements of the left SU(2) with the same group elements of the right SU(2), to get elements of the full group SO(4) generated by this very specific subalgebra of it generated by mirror twins. You might think of synchronized swimming, if you wish. $\endgroup$ – Cosmas Zachos Feb 5 at 2:30

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