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I was working through some problems today and came across this one:

Consider a particle in an infinitely deep potential 'well'. That is to say: $V(x) = 0$ for $-a/2<x<a/2$ and $V(x)=\infty$ anywhere else.

Show that the functions $\psi_n(x) = A_n \cos(n\pi x/a)$ are eigenfunctions and determine the eigenvalue.

Now from what I previously thought I knew about eigenfunctions is that any eigenfunction of an operator is simply a function $f$ so that the operator ($\hat{V}$) in this case working on that function returns a scalar multiple of the function: $$\hat{V}f = \lambda f.$$ But when trying that here, this doesn't work. Since $A_n$ is just a constant, and $\cos(n\pi x/a)$ will nessecarily be non-zero somewhere with $x>a/2$, there is no way to get a "nice" cosine back; not even with eigenvalue $\lambda=0$.

So then my question is: Somewhere in my reasoning I must be making a mistake, and I suspect it has to do with how I look at eigen functions. Assuming this, how should I look at them instead?

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    $\begingroup$ Your interpretation is fine, just you're using the wrong operator. It's asking you for energy eigenfunctions, so eigenfunctions of the Hamiltonian operator. $\endgroup$ – CDCM Dec 7 '17 at 12:14
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    $\begingroup$ How can I see that an energy eigenfunction is meant? $\endgroup$ – Mitchell Faas Dec 7 '17 at 12:16
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    $\begingroup$ Yes the question was a bit sloppily written. But in QM, one of the most general questions is: I have a potential, what are the eigenfunctions of the Hamiltonian with that potential? Knowing these tells us about e.g. energy levels, and later on about how those states evolve in time. In short the eigenfunctions of the Hamiltonian are of great importance, so I guessed that's what they're asking for. $\endgroup$ – CDCM Dec 7 '17 at 12:22
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The language of the set-piece you quote leaves a lot to be desired. For someone with a reasonable command of QM it's clear what the intention was, but it is still phrased in an ambiguous way and the writer is at fault for any confusion that ensues.

To be clear, the quote is asking you to show that the given wavefunction is an eigenfunction of the hamiltonian, i.e. of the operator $$ H = -\frac{\hbar^2}{2m}\frac{\mathrm d^2}{\mathrm dx^2} + V(x), $$ for the given potential. This is implied by the context, but it isn't spelled out explicitly (where doing so would have taken an extra word or two).


That said, though, the wavefunction you were given is indeed an eigenfunction of the potential operator $\hat V$, since for all $x$ in the configuration space $[-a/2,a/2]$ you have $$ V(x) \psi(x)=0 = 0\psi(x), $$ i.e. the wavefunction returns to itself multiplied by the eigenvalue $\lambda=0$.

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  • $\begingroup$ The task in particular doesn't do a great job at suggesting the configuration space is only $[-a/2, a/2]$. $\endgroup$ – leftaroundabout Dec 7 '17 at 20:05

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