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So I was just reading back through Griffiths' "Introduction to Quantum Mechanics" and solving some of the problems for practice. There is a nice one (problem 2.1c for those playing at home) where you have to show that if the potential energy function $V(x)$ is an even function that $\psi(x)$ can always be taken to be even or odd. This is pretty straightforward, but it got me to thinking about the general restrictions on the wavefunction if $V(-x)=-V(x)$ (ie, it is odd). I decided just to use a one dimensional case. Starting with the time independent Schrodinger equation, and taking $x\rightarrow-x$ gives:


Now use the fact that $V(x)$ is odd, and put the minus sign in an illuminating location:


Now we see that the Schrodinger equation is satisfied if $\psi(-x)=-\psi(-x)=0$.

My suspicion: it is not consistent in the last line to have $V$ defined at $x$ and $\psi$ defined at $-x$. And this method, while it works nicely for the case of even potentials, is just meaningless for the case of odd potentials.

My question: Does the fact that this is meaningless mean that we can't have odd potentials in Quantum mechanics (This seems silly, we can just go into the lab and make some, right?) or does it (more likely) just mean that this method tells us nothing, and that the wavefunctions can be even, odd or neither depending on the problem? Thanks :)

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More on Problem 2.1 in Griffiths, Intro to QM: physics.stackexchange.com/q/44003/2451 –  Qmechanic Mar 3 '13 at 15:57

2 Answers 2

I think you mistake is the following: the Schrodinger equation is satisfied if $\psi(−x)=−\psi(−x)$=0 doesn't mean that it is satisfied only in that case. Non-zero solutions could still occur.

Moreover, physical intuition tells me that there are no particular symmetries in this case. Imagine a potential that is -A for x<0 and A for x>0. Then, looking for a solution with eigenvalue $-A<E<A$, it should have a decaying shape in the right region and an oscillatory behaviour in the left region.

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I suspect you're correct, @Bzazz...To have a self consistent TISE, we need both sides to be a function of $x$ or $-x$ but not a mixture. So my musings above just tell us that $\psi$ can't be even or odd –  Lachy Mar 3 '13 at 14:32

1) On one hand, if $V(x)=V(-x)$ is an even potential, then the reason one may find a generating set $\psi_n$ of even and odd eigenfunction is that the Hamiltonian operator commutes with the parity operator $[H,P]~=~0$, and hence there exists a complete set of common eigenfunctions. (Recall that eigenfunctions of the parity operator $P$ are precisely the even and odd wavefunctions.)

2) On the other hand, if $V(x)=-V(-x)$ is an odd potential (which certainly is a possibility), then the Hamiltonian operator does not commute with the parity operator $[H,P]~\neq~0$, and hence there does not exist a complete set of common eigenfunctions.

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