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?
Best Answer
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.