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.
