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J. J. Sakurai states in his "Modern Quantum Mechanics", this fact as a theorem ($\pi$ is the parity operator):

Suppose $$[H,\pi]=0$$ and $| n>$ is a nondegenerate eigenket of $H$ with eigenvalue $E_n:$ $$H|n> = E_n|n>;$$ then $|n>$ is also a parity eigenket.

Then as a proof he uses some fact that is specific to the parity operator.

But isn't this trivially true even if we replace $\pi$ with any other operator if $H$ has a nondegenerate spectrum? is there a case where $H$ commutes with an operator and has nondegenerate eigenket, yet the eigenket is not an eigenket of the other operator?

Is this so if $H$ is not diagonalizable?

Why does this deserve the status of a "theorem"?

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  • $\begingroup$ Related: physics.stackexchange.com/q/104674/2451 $\endgroup$
    – Qmechanic ♦
    Mar 20, 2015 at 0:18
  • $\begingroup$ @Qmechanic: Yes, this is why I wondered about why Sakurai made a trivially true statement specifically about "parity operator". It seems obviously true, even if we replace $\pi$ with some other operator. $\endgroup$
    – Quantization
    Mar 20, 2015 at 0:26

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