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?
 A: 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.
A: 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.
