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I have read that symmetric potential has even bound ground state, but I don't know how to derive it? The only conclusion I can derive is for even potential I can take real wavefunction.

I also want to ask, if odd bound ground state ever exist? I have never seen any.

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One particle wave function for a bound ground state does not have nodes. This is obviously not the case for the wave functions of several fermions, which necessarily have zeros to satisfy the Pauli principle. Note also that the wave functions of the eigenstates of a scalar potential can also always be chosen real.

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  • $\begingroup$ If we do not have symmetric potential for one particle, then also we will have even wavefunction for ground state? or the wavefunction will be some uneven structure following node theorem? $\endgroup$
    – sawan kt
    Jul 27, 2020 at 2:40
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    $\begingroup$ The wave function does not have to be even, but for a ground state it will not have zeros. Node theorem does not say that the wave function should be symmetric. $\endgroup$
    – Roger V.
    Jul 27, 2020 at 2:43
  • $\begingroup$ Is this exactly correct? It is known that if the solution has no node it must be the ground state, and if solution $A$ has fewer nodes than solution $B$, then $E_A<E_B$, but it is clear that the ground state necessarily has no nodes? $\endgroup$
    – ZeroTheHero
    Jul 27, 2020 at 4:04
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    $\begingroup$ Think for instance $V=(x^2-1)x e^{-x^2}$... it is clear the ground state has no nodes? $\endgroup$
    – ZeroTheHero
    Jul 27, 2020 at 4:14
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    $\begingroup$ The variational approach proves there’s always an even state, but it’s an approximate ground state. $\endgroup$
    – ZeroTheHero
    Jul 27, 2020 at 13:38

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