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I am doing an introductory quantum mechanics course, I have been told that the momentum space wave function is essentially the Fourier transform of the position-space wave function. I.e. $$ \Phi(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{-ipx/{\hbar}}\Psi(x,t) dx. $$ I was initially wondering if you could express the wave function for a specific state in the same way, i.e. can the ground state for a system be expressed in momentum space by: $$ \Phi_1(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}e^{-ipx/{\hbar}}\Psi_1(x,t) dx. $$ I am aware that this is a general expression true for any general function, however in response to one of the comments it seems in my situation it does not hold. (The final question is in bold at the bottom)

The reason I am asking, Is that I was deriving the ground state momentum space wave function for an infinite square well (in 1D) (centred at the origin), i.e. $$V(x)= \begin{cases} 0 & -\frac{a}{2}\leq x\leq\frac{a}{2} \\ \infty & \text{elsewhere} \end{cases} $$ where the well has length $a$.

I know the position wave functions can be expressed for both even and odd n: $$\Psi_{\text{even},n}(x)=\sqrt{\frac{2}{a}}\cos\left(\frac{n\pi x}{a}\right)e^{-iE_nt/\hbar},$$ $$\Psi_{\text{odd},n}(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)e^{-iE_nt/\hbar} \, .$$ From here, I attempted to find the ground state momentum space wave function. I substituted n=1 for the ground state (so used the sin relation) and then plugged this into $$ \Phi_1(p,t)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\frac{a}{2}}^{\frac{a}{2}}e^{-ipx/{\hbar}}\Psi_1(x,t) dx \, . $$ After some integration and simplification I got to the answer: $$\Phi_1(p,t)=\frac{1}{\sqrt{\pi \hbar a}} e^{-iE_1t/\hbar} \left(\frac{2i \hbar a^2 p}{a^2p^2-\pi^2 \hbar^2}\right)\cos\left(\frac{pa}{2\hbar}\right) \, .$$ I have found online the general form to be $$\Phi_n(p,t)=\sqrt{\frac{a\pi}{\hbar}}\frac{ne^{-iE_n t/\hbar}}{(n\pi)^2-\left(\frac{ap}{\hbar}\right)^2}\cdot 2e^{-ipa/2\hbar} \left[\cos\left(\frac{pa}{2\hbar}\right)+i\sin\left(\frac{pa}{2\hbar}\right)\right] \, ,$$ so from their derivation $$\Phi_1=\sqrt{\frac{a\pi}{\hbar}}\frac{e^{-iE_n t/\hbar}}{\pi^2-\left(\frac{ap}{\hbar}\right)^2}\cdot 2e^{-ipa/2\hbar} \left[\cos\left(\frac{pa}{2\hbar}\right)\right].$$ These are not the same, and after seeing the comment below stating that a Fourier series would be the correct way to compute this problem I attempted to plot the probability density of my solution and it gave a very strange (so probably incorrect) form.

I was hoping someone could explain to me why I cannot use the Fourier transform in this situation, I imagine I am misinterpreting what the Fourier transform is even doing.

George
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    Your first equation is a general rule. Why should it not work for the specific case of your second equation? – flaudemus Mar 02 '19 at 22:38
  • I am doubting it, as I was deriving the momentum space wavefunction for an infinite square well's ground state and got an answer. I found a solution to the problem online, where they derived the generalised momentum wave function and then substituted n=1 and it didn't give my answer. I'm fairly confident my solution is mathematically correct (after I have assumed the above Fourier transform relation), I have checked the maths using Wolfram. I thought there may be some condition that you can only use the Fourier transform in certain places. It could be that the online solutions are wrong. – George Mar 02 '19 at 22:53
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    This question has nothing to do with quantum mechanics. If you have any function $f(x)$ you can represent it as a Fourier integral $$ f(x) = \int_{\infty}^\infty \tilde{f}(k) \exp(i k x) \frac{dk}{2\pi} $$ where $$\tilde{f}(k) = \int_{-\infty}^\infty f(x) \exp(-i k x) , dx , .$$ This mathematical fact works just as well for quantum wave functions as anything else. Since $f$ is general, it works for any function including $\Phi_0$, $\Phi_1$, etc. – DanielSank Mar 02 '19 at 23:13
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    (continued) However, Fourier integrals only work for functions defined over the entire real axis. Functions defined over an interval, like for the square well, require Fourier series. In order to actually answer your question, please edit it to explain what you're really asking, which is why the Fourier integral didn't work for the square well. The way the post is written now, you're actual question is hidden in the comments. – DanielSank Mar 02 '19 at 23:15
  • @DanielSank Thanks for your response, I have updated the question now :) – George Mar 03 '19 at 10:29
  • @DanielSank Your comments should definitely be an answer. – BioPhysicist Mar 03 '19 at 10:48
  • Related: What's the deal with momentum in the infinite square well? – Emilio Pisanty Mar 03 '19 at 12:27
  • @AaronStevens I'm not so sure. When I wrote those comments, it seemed that the real question was vague and in the comments. I prefer to have a well-formed question before attempting an answer. – DanielSank Mar 03 '19 at 15:01
  • @George Why do your $\Psi_{\text{even},n}$ and $\Psi_{\text{odd},n}$ have energy $E_1$? – DanielSank Mar 03 '19 at 15:03
  • @DanielSank That was a typo – George Mar 04 '19 at 12:50
  • In $$\Phi_n(p,t)=\sqrt{\frac{a\pi}{\hbar}}\frac{ne^{-iE_n t/\hbar}}{(n\pi)^2-\left(\frac{ap}{\hbar}\right)^2}\cdot 2e^{-ipa/2\hbar} \left[\cos\left(\frac{pa}{2\hbar}\right)+i\sin\left(\frac{pa}{2\hbar}\right)\right] , ,$$ you can recompose $\left[\cos\left(\frac{pa}{2\hbar}\right)+i\sin\left(\frac{pa}{2\hbar}\right)\right]=e^{ipa/2\hbar}$ and this would cancel your other exponential: I doubt this expression is correct. – ZeroTheHero Mar 04 '19 at 14:13
  • @ZeroTheHero I think that expression, much like the position wavefunctions has different forms for even and odd n. I.e. I think a better way to write it would be as two different wave functions: $$\Phi_{\text{odd},n}(p)=\sqrt{\frac{a\pi}{\hbar}}\frac{ne^{-iE_n t/\hbar}}{(n\pi)^2-\left(\frac{ap}{\hbar}\right)^2}\cdot 2e^{-ipa/2\hbar} \left[\cos\left(\frac{pa}{2\hbar}\right)\right], $$ $$\Phi_{\text{even},n}(p)=\sqrt{\frac{a\pi}{\hbar}}\frac{ne^{-iE_n t/\hbar}}{(n\pi)^2-\left(\frac{ap}{\hbar}\right)^2}\cdot 2e^{-ipa/2\hbar} \left[i\sin\left(\frac{pa}{2\hbar}\right)\right], .$$ (I may be wrong) – George Mar 04 '19 at 14:24

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