# Infinite potential well

1. May 2, 2014

### unscientific

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

Part (a): Find wavefunction and energy levels.
Part (b): Find a possible wavefunction. Is this wavefunction unique?
Part (c): What is the probability of finding it in the ground state?
Part (d): What's the probability of finding it in the second excited state?

2. Relevant equations

3. The attempt at a solution

Part (a)
I have found the wavefunction and energy levels:

$$\phi = \sqrt{\frac{2}{a}}\sqrt{\frac{2}{a}}sin\left(\frac{n_x \pi x}{a}\right)sin\left(\frac{n_y \pi y}{a}\right)$$

$$E = \frac{\hbar^2}{8ma^2}(n_x^2 + n_y^2)$$

If a = b, there is degeneracy in the first excited state: |nx,ny> |1,0> and |0,1> give the same energy.

Part (b)
$$E = E_0 + \frac{(E_1-E_0)}{4} = \frac{3}{4}E_0 + \frac{1}{4}E_1$$

I'm guessing the wavefunction will be a linear combination of ground and first excited state:

$$|\psi \rangle = \frac{3}{4}|\psi_0\rangle + \frac{1}{4}|\psi_1\rangle$$
$$|\psi \rangle = \sqrt{\frac{2}{a}}\sqrt{\frac{2}{b}}\left[\frac{3}{4} sin(\frac{\pi x}{a})sin(\frac{\pi y}{a}) +\frac{1}{4}sin(\frac{2\pi x}{a})sin(\frac{2\pi y}{a})\right]$$

But the thing is, this wavefunction isn't normalized, as $(\frac{1}{4})^2 + (\frac{3}{4})^2 = \frac{10}{16}$.

2. May 3, 2014

### Fisica

i think that $$|\psi \rangle = \frac{3}{4}|\psi_0\rangle + \frac{1}{4}|\psi_1\rangle$$ its wrong

the correct could be $$|\psi \rangle = (\frac{3}{4})^{1/2}|\psi_0\rangle + (\frac{1}{4})^{1/2}|\psi_1\rangle$$

3. May 3, 2014

### skrat

The expression for $E$ seems to be wrong, or is it just me? Is that $8$ in denominator just a typo or... ? You should check that, because I also think $\pi$ is missing.

Also note that if $n=0$ there is nothing happening. The probability to find a particle is $|\psi |^2=\frac{2}{a}sin^2(\frac{n\pi }{a}x)=0$ if $n=0$ also the energy of state with $n=0$ is zero. Therefore state with $n=0$ does NOT exist. Ground state is $n=1$.

4. May 4, 2014

### unscientific

That's right, because $E = \langle \psi| H | \psi \rangle$.

But it's not unique, right? Because the state can be exapanded as a series of eigenkets: $|\psi\rangle = \sum_n a_n |E_n\rangle$.

Last edited: May 4, 2014
5. May 4, 2014

### unscientific

Sorry it should be $h$ instead of $\hbar$. The correct energy is $E = \frac{h}{8ma^2}$

Yea that's right. I think they usually use $E_0$ to denote ground state but the '0' really means n=1 state.

Is the probability $\frac{3}{4}$ then?

Last edited: May 4, 2014
6. May 4, 2014

### dauto

Also note that the reason the state is not unique is that there is an arbitrary phase ø between the states |ψ> = a|ψ1> + e b|ψ2>.

7. May 4, 2014

### skrat

Close enough. It's actually $E_n=\frac{\hbar ^2\pi ^2n^2}{2ma^2}=\frac{h^2n^2}{8ma^2}$

Yes, to find/measure the particle with $E_0$ (particle in ground state) the probability is $\frac{3}{4}$.

8. May 4, 2014

### unscientific

Yeah you're right. I keep missing stuff out..

Just that for 4 marks? It seems too easy to be true!

And for part (c), the probability for finding the particle in second excited state is....$\frac{1}{4}$? And again that for 3 marks?!

9. May 4, 2014

### skrat

Well the problem says that $E=\frac{3}{4}E_0+\frac{1}{4}E_1$ and coefficient before $E_n$ is the probability to find particle in state $n$.

The problem alsto states that the probability to find a particle in any state with $n\geq 2$ is basically equal to zero. It is trivial to check that, because $3/4 +1/4 = 1$.

10. May 4, 2014

### dauto

There might be a small probability the state might not be E0 or E1 as long as the probability is small enough that a 1000 particle sample might just have missed it.