# Eigenvalues of momentum operator

I had a homework problem in my intro QM class, basically asking me to find which of a given set of functions were eigenfunctions of the momentum operator, $$\hat{p_x}$$. For example,

$$\psi_1 = Ae^{ik(x-a)}$$ which is an eigenfunction of $$\hat{p_x}$$, with eigenvalue of $$\hbar k$$. I had another function:

$$\psi_2 = A\cos(kx) + iA\sin(kx)$$ which is also an eigenfunction of $$\hat{p_x}$$, with eigenvalue of $$\hbar k$$.

Now this maybe a basic question, but I am aware that $$p=\hbar k$$, so both eigenvalues are just the momentum, $$p$$. But is it the case that for every one-dimensional function that I can think of: if that function is an eigenfunction of $$\hat{p_x}$$, the corresponding eigenvalue will be $$\hbar k$$? I feel like this sort of makes sense, but I can't quite see why. Can anyone perhaps elaborate on this?

• i believe this is because you can obtain other eigenfunctions of this operator via a change of basis, and the change of basis would leave the eigenvalue invariant. – Paul Sep 29 '18 at 6:47
• @Paul That should be an answer – David Z Sep 29 '18 at 8:30
• Note that your $\psi_1$ and $\psi_2$ are basically the same thing, as $e^{iz} = \cos(z) + i\sin(z)$ – Noiralef Sep 29 '18 at 8:31
• @DavidZ Yes, you're right but I didn't feel inclined at first to give a "full" answer but thought the comment would point the OP in the right direction...Answer added below. – Paul Sep 29 '18 at 8:47
• @Paul Thanks. For the future, as long as what you're posting is along the lines of answering the question (even if it's something like a partial answer or a hint), it should be posted as an answer or not at all. Comments aren't really meant for that sort of thing. – David Z Sep 29 '18 at 8:52

The question as I understand it is asking "Why is the eigenvalue of the same form for all of the 1D eigenfunctions I can think of?".

The simple answer is that you can change the basis used to represent the eigenfunctions but the eigenvalue remains invariant under such transformations.

For example, consider

\begin{align} \hat{p} \psi = p\psi \end{align}

where $$\psi=\frac{1}{\sqrt{2\pi\hbar}}\exp{ipx/\hbar}$$ are the normalised momentum eigenfunctions in position representation, then apply some basis transformation so that $$\psi'= \hat{U}\psi$$ and $$\hat{p}' = \hat{U}\hat{p}\hat{U}^{-1}$$ then the eigenfunction equation becomes

\begin{align} \hat{U}\hat{p}\hat{U}^{-1}\psi' = p\hat{U}\hat{U}^{-1}\psi' = p\psi' \end{align}

Therefore, the eigenvalue remains the same under a change of basis i.e. there are multiple eigenfunctions with the same eigenvalue, as you describe in your question.

Well, if you ignore the conventional meaning of $$k$$ in this context, you can easily enough write a wavefunction like $$\psi_3 = e^{5ikx}$$ which has a momentum eigenvalue of $$5k$$ - of course, in this case $$k$$ stands for a different quantity than it conventionally does. But people generally don't do that.

The thing is, after accounting for multiplication by a constant factor, there's only one eigenfunction of $$\hat{p}_x$$ with a given eigenvalue $$P$$, and that is $$e^{iPx/\hbar}$$. (Note that $$\cos a + i\sin a = e^{ia}$$, so $$\psi_1 = \psi_2$$, more or less.) Then the value of using $$k = P/\hbar$$ is that it lets you write the resulting expression more concisely without obscuring the dependence on $$x$$ or the fact that the exponent is purely imaginary, both of which are important properties of this function to be aware of. Given those constraints, the variable $$k$$ basically has to be proportional to the eigenvalue, and choosing any proportionality constant other than $$\hbar$$ makes the expression less concise.

So, in brief, the reason you will always find that the eigenvalue equals $$\hbar k$$ is that defining $$k$$ in this way makes the formula simpler than any other definition.