As we know the eigenfunctions for a particle of mass $m$ in an infinite square well defined by $V(x) = 0$ if $0 \leq x \leq a$ and $V(x) = \infty$ otherwise are:

$$\psi_n (x) = \sqrt{\frac{2}{a}} \sin \left(\frac{n\pi x}{a} \right).$$

How does the ground state wave function look like in momentum space? As far as I recall I have to integrate $\psi_n(x)$ over the whole of space with the extra factor $\frac{e^{(-i p x / \hbar)}} {\sqrt {2 \pi \hbar}}$ (everything for $n = 1$).

In the solutions to this problem they integrated over $-a \leq x \leq a$ while I would've integrated from $0$ to $a$. Am I somehow missing something or is this solution just plain wrong?

A further question: How would I check whether or not my resulting $\psi(p)$ is an eigenstate of the momentum operator? Just slap the momentum operator in front of my function and see if I get something of the form $c \psi(p)$, where $c$ is some constant? Or how does this work?


3 Answers 3


This question is not well-posed from scratch. There is no Momentum Operator for the problem you are considering. Your geometric space is a bounded region of the real axis, so no translation group can be defined and no self-adjoint generator of translation (the momentum observable) exists. The symmetric operator $-i\frac{d}{dx}$ is not essentially self-adjoint on a natural domain like the one of $C^1$ functions vanishing at $0$ and $a$. It admits infinite self-adjoint extensions and there is no preferred physical choice.

  • $\begingroup$ Well, this was a downvote against reality - I'm happy to set things to rights. For more references on this topic, readers can start with Qmechanic's answer here. $\endgroup$ Commented Jun 12, 2017 at 16:51

Let me tweak and discuss a bit the answer of @Prasst, to make sure the student does not simply shrug and walk away because the issue is moot and problematic as per @Valter_Moretti 's answer and @Emilio's link. There is always a whiff of "don't go there!" in such questions, but the better message might be "watch where you are going!". Indeed, you have to fudge the definition of the nonexistent momentum and marvel at where the formal steps take you.

For the first excited state in question, it pays to shift the interval $[0,a]\mapsto [-a/2,a/2]$,
$$\langle x|\psi\rangle=\psi_1(x)=\sqrt{\frac{2}{a}}\cos\left(\frac{\pi}{a}x\right)$$ which is now manifestly symmetric around the origin in x, where it peaks.

The corresponding momentum-space wave function would then be real and symmetric in p, $$\langle p|\psi\rangle=\phi_1(p)= \frac{2 \sqrt{\pi a/\hbar}}{\frac{p^2a^2}{\hbar^2}-\pi^2}~\cos\left(\frac{pa}{2\hbar}\right).$$

So p takes any value on this state, not just the privileged values $p=\pm \hbar \pi/a$ suggested by the energy eigenvalue: It is a wavepacket involving all momentum components! (Contrast this to a plane wave $\psi(x)\sim \exp(ikx)$, so $\phi(p)\sim \delta(p-\hbar k)$, so that $p~\phi(p)=\hbar k~\phi(k)$.)

One sees that $$ \phi_1(0)=- \frac{2 \sqrt{\pi a/\hbar}}{ \pi^2}, \\ \phi_1\left (\pm \frac{\hbar \pi}{a}\right )=-\frac{ \sqrt{\pi a/\hbar}}{ 2\pi}. $$

From symmetry, the expectation of $\hat p$ in this state vanishes, while the expectation of $\hat{ p}^2$ is the energy-privileged value at the peaks, $(\hbar \pi/a)^2 $.

The deep answer above and links clue you into this peculiarity.

  • NB Several of these peculiarities, however, are obviated and bypassed when using the matrix mechanics formulation, which is generically more robust and elegant than the wave picture (but usually technically messier). Admire the spare simplicity of the momentum matrix in Prentis & Ty, Am Jou Phys 82, 583 (2014), which, unlike its square, is not diagonal. Further note the first energy eigenvector is not its eigenvector, as well.

Seems good to me. You are right integrating only from 0 to $a$ because $\psi$ is zero in the region of infinite potential. The solution would be $\psi(p)=\frac{1}{\sqrt{a\pi h}}\int_o^ae^{-ipx/\hbar}\sin(\pi x/a) dx=\frac{\sqrt{a\pi \hbar^3 }}{\pi^2 \hbar^2 - p^2 a^2 } (e^{-ipa/ \hbar }+1)$

As for the other question, that's what one typically does

  • $\begingroup$ Thanks for quickly confirming this! Since the momentum operator is $-i \frac{d}{dx}$ how do I derive my $\psi(p)$ with respect to x?:S $\endgroup$
    – user17574
    Commented Feb 1, 2013 at 12:40
  • $\begingroup$ don't need to, in momentum space the momentum operator is already diagonal. In this space the operator for position $\^{x}$ is the one that is a derivative. $\endgroup$
    – Prastt
    Commented Feb 1, 2013 at 13:14
  • $\begingroup$ So the momentum operator in momentum space is simply $p$? $\endgroup$
    – user17574
    Commented Feb 1, 2013 at 13:27
  • $\begingroup$ ok...........ty $\endgroup$
    – user17574
    Commented Feb 1, 2013 at 14:08

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