# Momentum and parity operators

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1. Mar 31, 2015

### blue_leaf77

Let's denote $\mathbf{p}$ and $\Pi$ as the momentum and parity operators respectively. It's known that $\mathbf{p}$ doesn't commute with $\Pi$, so they do not share the same set of eigenkets (plane wave doesn't have parity). But I just calculated that $[\mathbf{p}^2,\Pi] = 0$, which means kinetic energy and parity operators might share the same eigenkets. While on the other hand, the eigenket of kinetic energy is the same as the eigenket of momentum. Does this mean that commutativity doesn't always guarantee the share of eigenkets? My guess is that this has to do with the degeneracy since the ket $|\mathbf{p}\rangle$ and any other ket $|\mathbf{p}'\rangle$ with $|\mathbf{p}| = |\mathbf{p}'|$ are degenerate eigenkets of kinetic energy.

Last edited: Mar 31, 2015
2. Mar 31, 2015

### Orodruin

Staff Emeritus
Bingo!

That an operator A commutes with both B and C (and thus has common sets of eigenvectors with both) does not mean that B and C commute (nothing states that the sets of eigenvectors are the same), it is only true if A has a non degenerate spectrum.

Further example: Let A be the identity operator and B and C any non-commuting operators.

3. Mar 31, 2015

### blue_leaf77

I just realized that what I wrote in the original post is already discussed in the textbook I'm reading, it's just that I stopped at the point beyond which that exact example is pointed out.

Last edited: Mar 31, 2015