Is well know that if an Hamiltonian $\hat{H}$ commutes with the parity operator then exists a complete system of eigenstates with definite parity. So there will be even and odd states.

I noticed that every definite parity Hamilotnian I have worked with, has the ground state with the same parity of the Hamiltonian. Is this always true? How can I prove it? (maybe with the variational method?)

Morover I noticed the first excited states has always of opposite parity, the second excited state has the opposite and so on.. Is this also always true?

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    $\begingroup$ Well... the ground state has no nodes... how could it have odd parity? Likewise, by the nodal theorem, the first excited state has one node. How could it have even parity? ... $\endgroup$ – Cosmas Zachos Jun 22 '17 at 18:37
  • $\begingroup$ Nodal theorem. $\endgroup$ – Cosmas Zachos Jun 22 '17 at 18:39
  • $\begingroup$ I didn't know that theorem, it cleared a lot of doubts I had. What about 3-D problems? Hamiltonian and ground state have to share the same parity or this is not true in general? $\endgroup$ – skdys Jun 22 '17 at 20:10
  • $\begingroup$ My guess is that it holds for higher D too, as it holds component-by-component, but there might be further plausible conditions to be satisfied for that. It certainly holds for standard systems in 3D. There are spoilsports who manufacture bogus counterexamples which only prove the rule. $\endgroup$ – Cosmas Zachos Jun 22 '17 at 21:21
  • $\begingroup$ You may see this in the ground state of nuclei, which, on its account, lack dipole moments. (Of course, an odd number of nucleons would have the intrinsic parity, only, of the odd-nucleon-out.) $\endgroup$ – Cosmas Zachos Jun 23 '17 at 13:46

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