When we have an even potential we say that it has an even and odd parity wavefunctions, cf. e.g. this & this Phys.SE posts.

What about an odd potential?

For example, two delta functions centered around the origin (i.e. one is on the negative side and other on the positive side of the axis) is an even potential and has even and odd solutions.

Now what if they are not centered at the origin (both delta potentials on the same side, either positive or negative side)?

  • $\begingroup$ can you have an odd potential that is bounded from below? What hint does this tell you about the ground state? $\endgroup$ – ZeroTheHero Aug 26 '19 at 13:19
  • $\begingroup$ You mean the energy bounded from below, right? @ZeroTheHero $\endgroup$ – Korra Aug 26 '19 at 13:22
  • $\begingroup$ yes... exactly.. $\endgroup$ – ZeroTheHero Aug 26 '19 at 13:54
  • $\begingroup$ @ZeroTheHero I can say that an even potential's energy is bounded from below as there is an energy called minimum energy (grnd state) $\endgroup$ – Korra Aug 26 '19 at 15:32
  • $\begingroup$ Right... but with an odd potential like $V=x-x^3$, $V$ is not bounded from below. How realistic is this for a 1d potential over $\mathbb{R}$? $\endgroup$ – ZeroTheHero Aug 26 '19 at 15:42

The basic principle is that

  • if the hamiltonian is symmetric in some way,
  • then its eigenfunctions can be chosen in a way that's symmetric with respect to the symmetries of the hamiltonian.

Now, that high-level description is essentially useless without being very precise about what each of those two things really means.

  • By "symmetric in some way", we mean that there exists some symmetry operator $\hat S$ which commutes with the hamiltonian, $$[\hat H, \hat S] = 0.$$ This symmetry operator must be unitary (i.e. $\hat S^\dagger \hat S = \mathbb I$), but other than that there are no restrictions on it, and it can come from a wide class of operations:

    • it can be a rotation or a translation, possibly from a discrete group or possibly depending on a continuous parameter.
    • it can be the parity operator, for which $\hat S|x\rangle = |{-x}\rangle$; if $\hat H$ commutes with this parity operator we say that "it has an even potential".
    • it can be a reflection operator about some other point $R$, so that $\hat S|x\rangle = |{2R-x}\rangle$; this includes the case in which you have two identical delta spikes at arbitrary positions, in which case $R$ is the midpoint between them.

    ... among a wide class of other possible operations.

  • Given the above, you are guaranteed the existence of one basis of eigenfunctions of the hamiltonian which are eigenstates of the symmetry operator $\hat S$. (Note, however, that you are not guaranteed that every eigenbasis of $\hat H$ will consist of eigenvectors of $\hat S$, particularly if the spectrum of the hamiltonian is degenerate.)

    • For some symmetry operations, "being an eigenstate of $\hat S$" can be a fairly complicated property that requires some analysis, such as e.g. when $\hat S$ is a rotation or a translation.

      (As an important note: when we deal with rotationally-invariant hamiltonians, we often have more than one symmetry operator ─ say, we might have three rotation generators $\hat S_1, \hat S_2, \hat S_3$ ─ where those operators don't commute with each other, which complicates things. In these cases, the language shifts over to group representation theory: you have a symmetry group, and you're guaranteed that each eigenspace is/carries a representation of that group. This is the case e.g. with rotations in three dimensions.)

    • If $\hat S$ is a parity operator, though, things are a bit simpler, because this symmetry operator obeys the simple relationship $\hat S^2 = \mathbb I$, and this entails that any eigenvalues $s$ of $\hat S$ must obey the same relationship, $s^2 =1$, which for this case implies that $s = \pm 1$.

      In other words, you're guaranteed eigenfunctions which obey either $\hat S |\psi\rangle = +|\psi\rangle$ (i.e. even functions) or $\hat S |\psi\rangle = - |\psi\rangle$ (i.e. odd functions).

For the case of an off-center parity operator (i.e. $\hat S|x\rangle = |{2R-x}\rangle$) the relationship $\hat S^2 = \mathbb I$ still holds (check!), which means that the consequences of that relationship still hold: you are guaranteed an eigenbasis of even and odd functions with respect to that symmetry operation.

  • $\begingroup$ How do I check what.... You said to check....in your answer. The last part. How do I check that the relationship holds? $\endgroup$ – Korra Aug 26 '19 at 18:25

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