(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

So, I'm asking for a bit of help before I confuse myself completely.

The question statement is:

Consider a two-dimensional potentialbox

[itex]V(x,y) = 0[/itex] if [itex]0 \leq x \leq a, 0 \leq y \leq 2a[/itex]

and infinity otherwise.

a) Determine the energy eigenstates and energy eigenvalues of a particle in this box. The solutions of the 1D potential well can be considered as known.

b) If we place 3 identical bosons in the box, what will the ground state energy be if we disregard interaction between the bosons.

c) Same as in b), but for 3 identical spin 1/2 fermions.

d) Write down the complete wavefunction (with both spatial and spin parts) for the ground state if two identical fermions with spin 1/2 and without interaction are put in the box.

e) Same as d) but for 3 identical fermions with spin 1/2.

2. Relevant equations

1D potential well equations:

[itex]\psi_n (x) = \sqrt{\frac{2}{a}} {sin(\frac{n \pi x}{a})}[/itex]

[itex]E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2} [/itex]

3. The attempt at a solution

Okay, for a) I did a variable separation and ended up with

[itex]\psi_{n_{x}n_{y}} (x) = \frac{\sqrt{2}}{a} sin(\frac{n_x \pi x}{a})sin(\frac{n_y \pi y}{2a})[/itex]

[itex]E_{n_{x}n_{y}} = \frac{\pi^2 \hbar^2}{2ma^2}(n_{x}^2 + \frac{n_{y}^2}{4}) [/itex]

Then in b)

Since there are three bosons they can all be in the same state, and the lowest state would be for [itex]n_x=n_y=1[/itex], so the total energy would be

[itex]E_{tot} = 3E_{1,1} = \frac{15 \pi^2 \hbar^2}{8ma^2}[/itex]

and for c)

Again, the lowest energy will be for [itex]n_x=n_y=1[/itex], but since only two spin 1/2 fermions can be in that energy at the same time, I'll also have a third particle, which I'm thinking will be in [itex]n_x= 1, n_y=2[/itex], since this will give me a lower energy than [itex]n_x= 2, n_y=1[/itex].

So then, the total energy would be:

[itex]E_{tot} = 2E_{1,1}+E_{1,2} = \frac{10 \pi^2 \hbar^2}{8ma^2} + \frac{8 \pi^2 \hbar^2}{8ma^2} = \frac{9 \pi^2 \hbar^2}{4ma^2}[/itex]

And that's where I'm not completely sure if my reasoning is completely correct, and where I want to confirm. I haven't started d) and e) yet, but I want to confirm this first, and I'm going to continue doing the rest with the assumptions I have above until I get a reply or figure something else out.

Thank you for any help that you may provide!

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# Identical particles in a 2D potential well

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