3
$\begingroup$

I am studying for an exam on quantum mechanics, and have come across something which I don't understand. The problem is:

We have a symmetric potential, i.e. $V(x)=V(-x)$. If the energy eigenvalue is non-degenerate, show that the energy eigenfunction $\psi(x)$ has definite parity.

My interpretation of this:

I understand that a non-degenerate eigenvalue means that no two independent eigenstates have the same eigenvalue. I also know that if two different states correspond to the same physical state, then $\psi(x)=k\psi(-x)$, and so $k=\pm1$, which is what is meant by even or odd parity. I think that what this question means when it says "definite parity" is that we must prove that only one of these values of $k$ is possible.

My attempt:

Suppose $\psi(x)=k\psi(-x)$. Then $$\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)+V(x)\psi(x)=E\psi(x)\\\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}k\psi(-x)+V(-x)k\psi(-x)=Ek\psi(-x)$$

From here, using the transformation $x\to-x$, and cancelling $k$, we get back to the original equation, so I haven't really made any progress, and I don't know what to do next.

My question:

Is my interpretation of definite parity correct? It is probably quite useful to understand the question fully before attempting it. Also, are there any hints anyone can offer for me to solve this? I'd prefer if you didn't give the full solution away straight away so that I can solve it using a hint.

Thanks!

Improve this question
$\endgroup$
4

1 Answer 1

Reset to default
8
$\begingroup$

Yes, that is what 'definite parity' means - it says that $\psi$ is an eigenfunction of the parity operator, without committing to either eigenvalue. Perhaps some examples say it best:

  • $f(x) = x^2$ has definite parity
  • $f(x) = x^3$ has definite parity
  • $f(x) = x^2+x^3$ doesn't.

In terms of the question you've been set, it's important to note that the condition that the energy eigenvalue be non-degenerate is absolutely crucial, and if you take it away the result is in general no longer true. Again, as an example, consider $$\psi(x) = A\cos(kx -\pi/4)$$ as an eigenfunction of a free particle in one dimension: the hamiltonian has a symmetric potential, and yet here sits a non-symmetric wavefunction. Of course, this is because the same eigenvalue, $\hbar^2k^2/2m$, sustains two separate orthogonal eigenfunctions of definite, and opposite, parity, $$\psi_1(x) = A\sin(kx) \ \ \text{and} \ \ \psi_2(x) = A\cos(kx),$$ which takes the eigenspace out of the hypotheses of your theorem.

So, how do you use the non-degeneracy of the eigenvalue? Well, the non-degeneracy tells you that any two eigenfunctions are linearly dependent, and you're only given one wavefunction to start with, so somehow you need to flesh that set out to two, and then move on with the argument.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So could I use the following argument? Let $\psi_1(x)=\psi(x)$. This solves the Schrodinger Equation (SE) with eigenvalue $E$. Let $\psi_2(x)=\psi(-x)$. By using the transformation I described in my question ($x\to-x$), $\psi_2(x)$ also solves SE with the same eigenvalue. Since $E$ is non-degenerate, $\psi_1$ and $\psi_2$ are linearly dependent, i.e. $\psi(x)=k\psi(-x)\implies\psi(x)=k^2\psi(x)\implies k=\pm1\implies\psi(x)$ has definite parity. $\endgroup$
    – John Doe
    Commented May 4, 2017 at 23:14
  • 1
    $\begingroup$ @JohnDoe That's exactly the argument you were asked to provide. $\endgroup$
    – Emilio Pisanty
    Commented May 4, 2017 at 23:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.