# Time Dependent Wavefunction in Infinite Square Well

1. Mar 25, 2015

### wigglywinks

1. The problem statement, all variables and given/known data
A particle of mass m is confined to a space 0<x<a in one dimension by infinitely high walls at x=0 and x=a. At t=0, the particle is initially in the left half of the well with a wavefunction given by,
$$\Psi(x,0)=\sqrt{\dfrac{2}{a}}$$
for 0<x<a/2

and,
$$\Psi(x,0)=0$$
for a/2 < x < a

Find the particle's time dependent wavefunction $$\Psi(x,t)$$
2. Relevant equations
I think the following equations are relevant (let me know if I don't have them written down correctly),
$$\Psi(x,t)=\sum_n^\infty c_n\psi_n(x) e^{-i E_n t/\hbar}$$
where $$\psi_n(x)=\sqrt{\dfrac{2}{a}}\sin\left(\dfrac{n\pi x}{a}\right)$$
and
$$E_n=\dfrac{n^2\hbar^2 \pi^2}{2ma}$$
and
$$c_n=\int_0^{a/2} \psi_n(x)\Psi(x,0)~dx$$
3. The attempt at a solution
I tried to find the time dependant wavefunction for the left half of the well. I'm not exactly sure what to do about the right half, but since the wavefunction at x=0 is equal to zero on the right half, the probabilities c_n would also be zero, so I'm thinking that the wavefunction is probably zero for the right half (so the full time dependant wavefunction for both halves would be piecewise continuous).

So for the left half of the well, I first tried to find c_n,

$$c_n=\int_0^{a/2}\sqrt{\dfrac{2}{a}}\sin\left(\dfrac{n\pi x}{a}\right)\sqrt{\dfrac{2}{a}}~dx$$
$$c_n=\frac{2}{a}\int_0^{a/2}\sin\left(\dfrac{n\pi x}{a}\right)~dx$$
so I get,
$$c_n=\dfrac{4\sin^2\left(\frac{n\pi}{4}\right)}{n\pi}$$

Then I plug this into

$$\Psi(x,t)=\sum_n^\infty c_n\psi_n(x) e^{-i E_n t/\hbar}$$
and also plugging in the equation for ψ_n(x), to get,
$$\Psi(x,t)=\sum_n^\infty \dfrac{4\sin^2\left(\frac{n\pi}{4}\right)}{n\pi}\sqrt{\dfrac{2}{a}}\sin\left(\dfrac{n\pi x}{a}\right) e^{-i E_n t/\hbar}$$

I tried to compute this on mathematica, but it didn't work (it looks like it probably diverges, but I don't know).

Does anyone know what I did wrong, or if this approach is even correct to begin with?

2. Mar 25, 2015

### soarce

When you wrote the expansion coefficients I think that you obtained $c_n(t=0)$, if you use $c_n$ independent of $t$ then we speak of stationary solutions (or time independent solutions). Unless you are able to solve analitically the time dependent Schroedinger equation you can try to obtain $c_n(t)$ by plugging the series expansion of your wave function into Schroedinger equation with the initial value $c_n(t=0)$ which you calculated.

3. Mar 25, 2015

### PeroK

You haven't done anything wrong.

With the coefficients you have calculated, the initial $|\Psi(x,0)|^2$ will be 0 on the right-hand side, but not for other values of t.

4. Mar 25, 2015

### vela

Staff Emeritus
That's not correct. Did you plug in x=0 accidentally?

5. Mar 25, 2015

### PeroK

The OP has the correct solution. Valid for $0 < x < a$.

The wave function will evolve over time and be non-zero on the right-hand side.

Last edited: Mar 25, 2015
6. Mar 25, 2015

### vela

Staff Emeritus
It's not 0 for all x at t=0 either. That would imply the particle is nowhere.

7. Mar 25, 2015

### PeroK

I
It's only 0 on the Right hand side, as per the definition of the initial function. It's constant probability on the left hand side.

8. Mar 26, 2015

### vela

Staff Emeritus
Oh, I finally figured out what you mean. You mean 0 on the right half of the potential, not the righthand side of the equation.

9. Mar 26, 2015

### vela

Staff Emeritus
Did you try a finite sum of, say, 20 terms? Mathematica might choke on the infinite sum, but it has no problems calculating a partial sum.