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In my university, my professor said that every degeneracy in QM comes from a symmetry of the Hamiltonian. But I'm not sure it is true for all the cases.

My counter-example is as follows:

In the 3D harmonic oscillator, $E_n=\hbar\omega(n+m+l+3/2)$, where $(n,m,l)$ are quantum numbers of $x,y,z$ oscillator.

For $E_n=\hbar\omega(5+3/2)$, there exists possible $(n,m,l)$ like $(1,1,3), (1,3,1), (3,1,1)$. It seems like these 3 degeneracies come from rotational symmetry. However, can we find any symmetry between 2 degeneracies $(1,1,3)$ and $(0,0,5)$?

DanDan面
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ckacl
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1 Answers1

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A simple Google search gave this old article: Haruo Ui, Gyo Takeda, Does Accidental Degeneracy Imply a Symmetry Group? Progress of Theoretical Physics, Volume 72, Issue 2, August 1984, Pages 266–284

The argument in the article proves that there is no maximal symmetry group in their case; In fact, they have found a particular case whereby the degeneracy is not mapped by any symmetry operator. Instead, they found a structure that gets it to work.

Now, do you consider this a symmetry or not? What would be the most general definition of a symmetry in quantum theory?

Maybe you can find an even better counter-example, but this paper is already very informative.

naturallyInconsistent
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