# Stationary state as initial condition for a free particle

I am trying to figure out what will happen to my particle, if it is initially in the ground state of an infinite square well, and suddenly becomes free.

$$V(x) = 0, -\frac{a}{2} \leq x \leq \frac{a}{2}$$ and infinity elsewhere. Suddenly the potential is removed, and I want to see how my probability distributions in x and p will change.

I have worked out that $$\psi(x,0) = \sqrt{\frac{2}{a}} \sin(\frac{\pi x}{a} + \frac{\pi}{2})$$

Which is my initial wavefunction that will change over time once the potential is removed. I thought I could use Plancherel's theorem to find $\phi(p)$ and then use $$\psi(x,t) = \frac{1}{\sqrt{2 \pi \hbar}}\int^{\infty}_{-\infty} dp \phi(p)e^{ipx/\hbar}e^{-ip^2t/2\hbar m}$$

but neither $\psi(x,0)$, nor $\phi(p)$ are normalisable over free space, so how do I proceed?

I have calculated $\phi(p)$ inside the potential well, but I don't think that helps me.

Well I'm still a toddler in QM, but I think $$\psi(x,0)=\sqrt{\dfrac{2}{a}}\sin{(\dfrac{\pi x}{a}+\dfrac{\pi}{2})}$$only in the domain $[-\dfrac{a}{2},\dfrac{a}{2}]$, and zero elsewhere, then it is normalizable and so is $\phi(p)$.
• I have normalised both $\psi(x,0)$ and $\phi(p)$ inside the potential. How do I determine how these evolve once the potential is removed though? I think for my intended method to work, I would need $\psi(x,0)$ to be normalisable over the whole x axis. – jm22b Feb 28 '17 at 11:07
• @Jacobadtr $\psi (x,0)$ does not evolve, it is just the wave function at $t=0$ ( It wouldn't collapse or change instantly once the potential is removed ), you can calculate $\psi(x,t)$ once you get $\phi(p)$ using the formula in your question. – Tofi Feb 28 '17 at 11:28
• $\phi(p)$ is zero outside the well, so does my integral for $\psi(x,t)$ only need to be calculated over that interval? – jm22b Feb 28 '17 at 11:36
• What do you mean by "zero outside the well" ? It is a function of $p$ not $x$. – Tofi Feb 28 '17 at 11:40
You need to take your initial wave function $$\Psi(x,0)=\sqrt{\frac{2}{a}}\sin\left(\frac{\pi}{a}x+\frac{\pi}{2}\right)\, ,$$ then expand this in terms of plane waves, i.e. write it as $$\Psi(x,0)=\sum_p a(p,0) e^{-i p x/\hbar}$$ (it would actually be an integral, not a sum) and then propagate each $p$ component by $a(p,t)=a(p,0)e^{-i E(p)t/\hbar}$.