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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:

$\psi_n (x) = \sqrt{(2/a)} sin(n \pi x/a)$

The question now is: How does the ground state 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?

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?

Thanks in advance!

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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

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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 – user17574 Feb 1 '13 at 12:40
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. – Barefeg Feb 1 '13 at 13:14
So the momentum operator in momentum space is simply $p$? – user17574 Feb 1 '13 at 13:27
yes............ – Barefeg Feb 1 '13 at 13:38
ok...........ty – user17574 Feb 1 '13 at 14:08

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