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## Summary:

- Trying to understand the precise mathematical reason as to why a quantum number being conserved means that a system is invariant under the corresponding symmetry operation

## Main Question or Discussion Point

I'm trying to understand the precise reason we claim that a value being conserved means that the system in question is invariant under the corresponding symmetry transformation. Take parity for example. If the parity operator satisfies the commutation relation ##[P, H] = 0## for a given Hamiltonian H, we can show that the expected value of P for a given wave function does not change over time, meaning "parity is conserved". Now from what I understand, if a wave function ##\psi## is an eigenfunction of H that corresponds to a non-degenerate eigenvalue, since ##[P, H] = 0## it must also be an eigenfunction of P. And furthermore, since P is unitary and unitary operators retain the normalization of ##\psi##, the action of P on ##\psi## can at most add an overall phase factor of +1 or -1, meaning that its eigenvalues must satisfy ##|\lambda|## = 1. In this specific case, ##P\psi = \psi' = \pm \psi##, such that ##\psi'## is physically indistinguishable from ##\psi##. In this case, I can see why the system is invariant under a parity transformation, since the corresponding wave function is physically indistinguishable from the original. However, I'm not sure why this would be true in the case of degenerate eigenvalues. If ##\psi## is now an eigenfunction of H that corresponds to a degenerate eigenvalue, since ##[P, H] = 0## it must be composed of a linear combination of eigenfunctions of P, i.e. ##\psi = \alpha |a> + \beta |b>##, where ##|a>## and ##|b>## are eigenfunctions of P. So now if P acts on ##\psi##, ##\psi## will collapse to either ##|a>## or ##|b>##. In this case, wouldn't the resulting wave function be distinguishable from the original? What does the "system being invariant under the action of the parity operator" mean in this case?