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  Is the degree of degeneracy a quantum mechanical observable?

+ 1 like - 0 dislike

In quantum mechanics, observable properties correspond to expectation- or eigenvalues of (hermitian) operators.

After measurement (of an eigenvalue) the system is in an eigenstate that corresponds to an eigenfunction of the operator. Sometimes, however an eigenvalue (\(\lambda\)) can not only arise from one certain well-defined state-function but from a whole (Hilbert sub) space that is spanned by n eigenfunctions with the same eigenvalue \(\lambda\) for n>1.

What I am interested in, is if the degeneracy itself is something that can be measured? For example can there exist an operator ,say \(\mathcal{D}_\mathcal{H}\)that measures the degeneracy of a state function\(\Psi_\lambda\) (for a certain operator, for example the Hamiltonian $\mathcal{H})$:

$$ \mathcal{D}_\mathcal{H} \Psi_\lambda = n \Psi_\lambda $$

This question arises from some considerations of special symmetry properties of degenerate states I am investigating. In the course of these works the question arose, if the degeneracy of a state is some physical property or rather only something like an "mathematical artifact" that cannot be directly probed experimentally.

Note: I would expect a possible answer to consider both cases, for the one a "symmetry imposed" degeneracy where the state corresponds to higher-dimensional irreducible representation of the symmetry group of the Hamiltonian and for the other the general thing that is sometimes called accidental degeneracy.

asked Jul 29 in Q&A by Raphael [ revision history ]
edited Jul 29

1 Answer

+ 1 like - 0 dislike

Yes. If there is degeneracy it is lifted under small external perturbations, leading to a multiplet of spectral lines, generally of the multiplicity of the eigenvalue. An exception is if the system has a high amount of symmetry and the perturbation preserves some of it; then the multiplet is smaller and some eigenvalues of the perturbed system are still degenerate. Except in the last case, the origin of the degeneracy does not matter.

answered Jul 29 by Arnold Neumaier (14,557 points) [ no revision ]
Does this work for one single state?Test

If I understand your answer correctly, you are considering something like an ensemble of states? In that case, of course you can simply count the branches into which the state may split. I suppose its equivalent to checking some thermodynamic function of the ensemble, that will tell you about the degrees of freedom. But I was actually interested in the quantum mechanics of a single state of a single system in a single instance. For example you measure some property of the system such that it will be in state that is degenerate, say the energy of the electron in the H-Atom, such that its spin-state will be degenerate. Now can you apply some operator on that state function that gives you eigenvalue 2 and leaves the state function as it is. In this case its obviously the spin-multiplicity, in other cases it will be the angular momentum. But what about the general case? Are there any operators in general that can do the job?

Nothing depends on ensembles. What I stated is just standard degenerate perturbation theory. The invariant subspace in the degenerate case splits under perturbation into 1-dimensional subspaces; an eigenstate turns into a linear combination of nondegenerate eigenstates. The perturbed system therefore oscillates with multiple frequencies close to the single degenerate one, and these oscillations give rise to the measured spectrum.

So if its an observable, then there should be an operator that determines if it acts on the state function. What is this operator then?

The spectrum is not an observable in the operator sense but something computed indirectly from the response of a system of oscillators coupled to the quantum system. The observable frequencies are differences of eigenvalues of the Hamiltonian (Rydberg-Ritz principle). 

So, I summarize: The answer is no.

Theorem: (Neither the spectrum in general nor ) the degree of degeneracy in particular is a quantum mechanical observable of a state (in the operator sense).

Can this be shown somehow? I know calculating the spectrum requires diagonalisation and thats en entirely different buisness than constructing and acting an operator on a wave function. But for degeneracy its a bit more subtle. As in the example of the Hydrogen atom I can simply act the spin operator on the function and use the eigenvalue (say s) to compute the degeneracy = 2s+1. Similarly we can expect for systems with conserved angular momentum to be able to handle it this way. So it doesn't seem so unsensical in the first instance that it wouldn't require some argumentation why its not possible.  

It is not ''an observable'' in the sense of not being described by an operator. But it is ''observable'' in the sense of experimental physics. 

But of course one can handcraft for any degenerate system an operator that counts the degeneracy, for example by mapping the degenerate invariant subspace to a space of spins. This can be done in many ways since Hilbert spaces of the same finite dimension have lots of isomorphisms. But the construction is physically meaningless.

Very nice! Thank you very much that was about what I wanted to know.

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