I know that if two hermitian operators commute, then there is complete set of simultaneous eigenstates. But, why does this imply that they have no degeneracy in the spectrum ? For example, if the hamiltonian $\hat H$ and a symmetry operator $\hat Q$ commute,(if $\hat Q$ is only a single symmetry operator) are there no degeneracy in the spectrum?

Or, I guess that it is not the case that the degeneracy is lifted, but we can find another operator that distinguishes the degenerate states. Which is right?

  1. In a new basis of simultaneous eigenbasis of $\hat Q$ and $\hat H$, two degenerate energy eigenstates are not degenerate anymore, or
  2. They are still degenerate, but can be distinguished by $\hat Q$?

1 Answer 1


The second one is the correct viewpoint. Given a Hamiltonian $H$ and another hermitian operator $S$, which commute you can find a simultaneous eigenbasis for both operators. Let's call them $|n,s>$. Acting on these states with $H$ we have $H|n,s>=E_n|n,s>$ for some value(s) of $s$, which means that the eigenspace for the eigenvalue $E_n$ might be more than one-dimensional: $\dim(V_{E_n})=\# \text{Allowed Values of s}$. If this is the case, this means degeneracy regarding the H-Op.

Given any hamiltonian and hilbertspace there is no way by finding other operators to reduce the dimensionality of the eigenspace of $H$. The dimensionality is an algebraic property of $H$ itself.

However, if there is a degeneracy, it is very useful to find an operator, which commutes with $H$, to classify the degeneracy i.e. to learn about it and also to get control about the degenerate eigenspaces i.e. to find a basis in the eigenspaces labelled by a conserved quantum number (eigenvalue of an operator, which commutes with $H$).

  • 1
    $\begingroup$ Okay, so finding maximally commuting set is all about distinguishing states (in the presence of 'unlifted' degeneracy) by commuting operators as much as possible, and giving the states quantum numbers(to label them). Thank you! $\endgroup$ Commented Aug 11, 2021 at 7:38
  • $\begingroup$ Yes , exactly ! $\endgroup$
    – SGGS
    Commented Aug 12, 2021 at 10:25

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