# Quantum well, time evolution of probabilities

1. Aug 6, 2016

### frerk

hey,
this thread might be a bit longer, but I have already calculated everything and I am quiet sure that it is right :)
Just need someone who confirms me or not :)

1. The problem statement, all variables and given/known data

Time evolution of probabilities. An electron inside a quantum well of length L is at time t = 0 in the state
$$\psi(x,0) =\frac{1}{ \sqrt{L} } ( \sin( \frac{\pi x}{ L} ) + \sin( \frac{2 \pi x}{L} ))$$

Outside the well $$\psi(x,0) = 0.$$ The state $$\psi(x,0)$$ is thus a superposition of two stationary states$$\psi_{1}(x) \ and \ \psi_{2}(x)$$ with the energies $$E_{1} = \frac { \pi^2\hbar^2 }{2mL^2} \ and \ E_{2} = 4 E_{1}$$

Calculate the porbability that the electron is in the interval (0, L/2) at the time $$t = \pi \hbar / E_{1}$$

3. The attempt at a solution

So, that is the homework, now I show you how i get a result:

$$\psi(x,t) = \psi(x,0) e^{-iHt/\hbar} [ \psi_1(x,0) + \psi_2(x,0) ] e^{-iHt/\hbar} =\psi_1(x,0)e^{-iE_1 t/\hbar } + \psi_2(x,0)e^{-iE_2 t/\hbar }$$

$$\Rightarrow | \psi(x,t)|^2 = |\psi_1(x,0)e^{-iE_1 t/\hbar } + \psi_2(x,0)e^{-iE_2 t/\hbar }|^2$$
$$= | \psi_1(x,0)|^2 + |\psi_2(x,0)|^2 + \psi_1(x,0)^* \psi_2(x,0) e^{i(E_1 - E_2) t / \hbar } + \psi_2(x,0)^* \psi_1(x,0) e^{i ( E_2 -E_1) t / \hbar}$$
Now we put this into the integral:
$$\int_0^b \! | \psi(x,t)|^2 \, d x$$ where b = L/2 and we geht the result
$$0,5 + \frac{2}{3\pi} e^{i(E_1 E_2)t/\hbar} + \frac{2}{3\pi} e^{-i(E_1 E_2)t/\hbar}$$
Now I insert: $$t = \pi\hbar/E_1 = \pi \hbar 2mL^2 / \pi^2 \hbar^2 = 2mL^2/(\pi\hbar)$$
$$E_1 = \pi^2\hbar^2 / (2mL^2)$$
$$E_2 = 4E_1 = 2\pi^2 \hbar^2 / (mL^2)$$

And after doing that, we find out that the probability that the electron in in the interval (0,L/2) for the specific time is 0,076.

Thank you :)

Last edited: Aug 6, 2016
2. Aug 6, 2016

### blue_leaf77

Is the wavefunction at $t=0$ what you are already given? It has an incorrect unit, for 1D system like this the unit of the wavefunction should be $m^{-0.5}$.

3. Aug 6, 2016

### frerk

Yes it is already given. I forgot the root over the "L". Sorry for that, i corrected it.