# Homework Help: Potential well with inner step, perturbation theory

1. Apr 13, 2012

### physicsjock

hey,

say you have a infinite potential well of length L, in the middle of the well a potential step of potential V and length x. Inside the well is a particle of mass m.

why are the first order energy corrections large for even eigenstates compared to odd ones?

also, say well (without the particle of mass m) you put two spinless non interacting bosons (total symmetric wave function). how could you find the wave functions of the first excited and the ground states, ignoring perturbation?

2. Apr 14, 2012

### fzero

You should sketch the probability distributions for the even and odd (as functions, not necessarily quantum number) unperturbed wavefunctions. What is happening in the vicinity of the small step?

You might review the solution of the step potential for the normal case (without the potential well). When you include the well, you add some boundary conditions.

3. Apr 14, 2012

### physicsjock

So for even functions there is a probability peak in the centre of the step, and for odd functions the probability is zero in the centre of the step, is that right?

So the reason the correction is smaller for odd functions is because there is less probability of the particle being in the step compared to even functions, so less of a correction is required?

Because the first order energy correction term is <n|V|n> and V=0 everywhere inside the well besides at the step?

For the second part, does the additional information "two spinless bosons" just mean two bosons with symmetric spacial wavefunctions? If they're both spinless bosons wouldnt the wavefunctions be the same? Since they don't interact with eachother you treat each seperately and multiply the wavefunctions together?

Thanks for your reply :)

4. Apr 14, 2012

### fzero

Exactly.

Because they are bosons they can be in the same quantum state. They do not necessarily need to be. It is precisely when the two bosons are in different states that we need to take the additional step of symmetrizing the wavefunction.

5. Apr 14, 2012

### physicsjock

oh cool thanks!

so the total wave function is going to be,

$\psi_{T} =\frac{1}{\sqrt{2}} \psi_{\alpha}(x_{1}) \psi_{\beta}(x_{2})+\psi_{\beta}(x_{1}) \psi_{\alpha}(x_{2})$

where α and β are sets of quantum numbers (same of different)

so the only difference between $\psi_{\alpha} and \psi_{\beta}$ in the solutions to the potential step is going the be their energy E, so one will have Eα and the other Eβ.

I went over the solutions to a potential step and they depend on whether E>V or E<V inside the step but all I have been given is that i need to find the solutions for both the ground state and the first excited state.

For the ground state α=β right? and E would be the minimum energy.

so

$\psi_{T} =\frac{2}{\sqrt{2}} \psi_{\alpha}(x_{1}) \psi_{\beta}(x_{2})$

on either side of the step the solutions are ok, but it seems it will get a little messy when you times $\psi_{\alpha}(x_{1}) \psi_{\beta}(x_{2})$. Applying the boundary conditions you do each wave function seperately right, then find the constant, then times both functions together?

And for the first excited state one of the particles stays in the ground state and the other excites to the closest higher state so α≠β so the total wave function will resemble the first equation.

But how can you find the solution inside the step if you arn't sure if E>V or E<V? It looks like it'll get really messy once you find each wave function as well. This is like a sub question of a 4 part past exam question, makes me feel like im over thinking this or something

6. Apr 14, 2012

### fzero

I agree with the above.

You have to be careful with the normalization. $\psi_{\alpha}(x_{1}) \psi_{\alpha}(x_{2})$ is already symmetric and also normalized if the $\psi_\alpha$ are.

You should solve for the $\psi_\alpha$ first, so you don't have to consider products for that part. This is a simplification due to the fact that these particles are noninteracting.

Yes.

You would have to compare the energy of the infinite square well solutions to the height of the step, so there's a few cases to consider. It also seems that there is a degeneracy in the tunneling solutions depending on which side of the box the particle starts out on.

I don't think each case is really messy, but there will be some sorting out of the cases and counting of tunneling solutions. You're probably not over thinking it, this is a problem that rewards good problem solving skills. None of the intermediate steps are too difficult on their own, but putting everything together is challenging.

7. Apr 14, 2012

### Jesssa

hey

this kind of assisted me as well,

thanks

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