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?

  • $\begingroup$ 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. $\endgroup$
    – Paul
    Sep 29, 2018 at 6:47
  • $\begingroup$ @Paul That should be an answer $\endgroup$
    – David Z
    Sep 29, 2018 at 8:30
  • 2
    $\begingroup$ Note that your $\psi_1$ and $\psi_2$ are basically the same thing, as $e^{iz} = \cos(z) + i\sin(z)$ $\endgroup$
    – Noiralef
    Sep 29, 2018 at 8:31
  • $\begingroup$ @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. $\endgroup$
    – Paul
    Sep 29, 2018 at 8:47
  • $\begingroup$ @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. $\endgroup$
    – David Z
    Sep 29, 2018 at 8:52

2 Answers 2


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.


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