Fourier transform and momentum space

Suppose if we know wavefunction $$\psi(x,0)$$ at an initial time $$t=0$$ and want to find wavefunction at later time t ; for solving such problem first we find $$\phi(p,0)$$ that is initial wave function in momentum space,which we find by using inverse fourier transform and then we use fourier transform to get time dependent wave function $$\psi(x, t)$$ but while doing so we are integrating over all values of $$p$$ (momentum). This method seems to me alright for the case of free particle but for any other case like particle in a box,where momentum is discrete then Is it right to use fourier transform method to find wave function at a later time $$t$$? I have a question based on above problem: If at $$t=0$$, wavefunction is constant for particle in a box in region $$-a then find the complete wavefunction at a later time $$t$$.

I have solved above problem but not by using fourier transform (I will attach a link of the photo of its solution) but if I use fourier transform then I get a differnt result! So my question is which method is more appropriate to use and when to use which method? https://imgur.com/a/iovPvNl

• I think you need to edit your question to include what you mean by the "fourier transform method to find wave function at a later time $t$". There is no such "general" method I know of to do this. The most general method is -- as indicated in the answer below -- to expand the initial state in states of definite energy, and then to multiply each of the individual terms by their individual time-evolution factors $e^{-i E_n t/\hbar}$. Jul 5 '20 at 16:53

The method you describe of writing the wave function in the basis of momentum eigenstates and then applying the time dependence to each term in that basis, will indeed only work for a free particle. The reason for this is that what you really want to do when you study the time dependence of a system is to write it in the basis of energy eigenstates, because it is those that drive the time dependence. The reason it works for a free particle in the way you described is because in this case the momentum operator commutes with the Hamiltonian, and therefore these two operators share a common set of eigenstates.

The momentum operator and the Hamiltonian do not commute for the infinite square well that you are interested in, so the method you described for a free particle will not work in that case. What you need to do is to use the general approach to time dependence in quantum mechanics (for time-independent potentials like the one in the square well). What you do is the following:

1. Solve the eigenvalue equation for the Hamiltonian to find the energy eigenvalues and eigenstates: $$\hat{H}|u_n\rangle=E_n|u_n\rangle.$$
2. Write your initial state $$|\psi(0)\rangle$$ in the basis of eigenstates of the Hamitonian: $$|\psi(0)\rangle=\sum_nc_n|u_n\rangle,$$ where the expansion coefficients are obtained in the usual manner, $$c_n=\langle u_n|\psi(0)\rangle$$.
3. Your state at a later time $$t$$ is now given by: $$|\psi(t)\rangle=\sum_nc_ne^{-iE_nt/\hbar}|u_n\rangle.$$

This is the general approach to solve the time dependence of a system with an arbitrary (time-independent) Hamiltonian. A question you may have is: why is it the Hamiltonian basis that features here (and not the momentum basis, say, that you were initially working with)? The reason is that time dependence in quantum mechanics is governed by the Schrödinger equation: $$i\hbar\frac{d|\psi(t)\rangle}{dt}=\hat{H}|\psi(t)\rangle,$$ and it is the Hamiltonian $$\hat{H}$$ that features in this equation.

If you want specific details of how this works for the infinite square well, I recently discussed it here.

• How do momentum and hamiltonian operator do not commute for particle in a box case? Jul 5 '20 at 10:37
• The Hamiltonian of a particle in a box is not the same as that of a free particle. The Hamiltonian is $\frac{p^2}{2m} + V(x)$ where $V(x)$ is 0 only inside the box, and $\infty$ outside. (The fact that the walls exist is an interaction that has to be taken into account.) This is why you can't find a state of definite energy which is also a state of definite momentum in the box, as you could for a free particle. See my answer here. Jul 5 '20 at 11:18
• @Philip that was helpful.But if we can NOT use fourier transform (in cases other than free particles) then how do we find momentum space wavefunction from real space wave function? Jul 5 '20 at 16:08
• The momentum space wavefunction is always the Fourier Transform of the position space wavefunction. That doesn't change. What this answer explains is that to find the general time dependent solution, you can't simply find the states of definite momentum and "tack-on" the time dependence as you would for the free particle, since the states of definite momentum are not -- in general -- also states of definite energy. It is the states of definite energy that each have simple time evolution behaviour. Jul 5 '20 at 16:51
• @ZeroTheHero, the way I understood the question was that the OP wanted to use the state expressed in the momentum representation to easily calculate the time evolution in a general system, just like they did in the case of a free particle. But this wouldn't work for a general system, so in the answer I explained that what the OP really wants for a general system are the energy eigenstates, not the momentum ones. Do you understand the question in a different way? Jul 5 '20 at 17:31