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.(adsbygoogle = window.adsbygoogle || []).push({});

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# Momentum and parity operators

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