# Identical particles in a 2D potential well

by Hannisch
Tags: identical, particles, potential
 P: 116 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 $V(x,y) = 0$ if $0 \leq x \leq a, 0 \leq y \leq 2a$ 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: $\psi_n (x) = \sqrt{\frac{2}{a}} {sin(\frac{n \pi x}{a})}$ $E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2}$ 3. The attempt at a solution Okay, for a) I did a variable separation and ended up with $\psi_{n_{x}n_{y}} (x) = \frac{\sqrt{2}}{a} sin(\frac{n_x \pi x}{a})sin(\frac{n_y \pi y}{2a})$ $E_{n_{x}n_{y}} = \frac{\pi^2 \hbar^2}{2ma^2}(n_{x}^2 + \frac{n_{y}^2}{4})$ Then in b) Since there are three bosons they can all be in the same state, and the lowest state would be for $n_x=n_y=1$, so the total energy would be $E_{tot} = 3E_{1,1} = \frac{15 \pi^2 \hbar^2}{8ma^2}$ and for c) Again, the lowest energy will be for $n_x=n_y=1$, 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 $n_x= 1, n_y=2$, since this will give me a lower energy than $n_x= 2, n_y=1$. So then, the total energy would be: $E_{tot} = 2E_{1,1}+E_{1,2} = \frac{10 \pi^2 \hbar^2}{8ma^2} + \frac{8 \pi^2 \hbar^2}{8ma^2} = \frac{9 \pi^2 \hbar^2}{4ma^2}$ 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!
 P: 116 Thank you :D Well then, I actually need some help with d) as well, it turns out. Because I know, from c), that the two fermions will be in $n_x=n_y=1$, so that the wavefunctions will be $\psi_{1,1}^{(1)}(x_1,y_1) = \frac{\sqrt{2}}{a}sin(\frac{\pi x_1}{a})sin(\frac{\pi y_1}{2a})$ $\psi_{1,1}^{(2)}(x_2,y_2) = \frac{\sqrt{2}}{a}sin(\frac{\pi x_2}{a})sin(\frac{\pi y_2}{2a})$ Where the exponent on the psi refers to the particle. And this is because if I solve the Schroedinger equation for two particles I can do a variabel (particle) separation as well, and see that $\psi (x_1,y_1,x_2,y_2) = \psi^{(1)} (x_1,y_1) \psi^{(2)} (x_2,y_2)$ So then $\psi (x_1,y_1,x_2,y_2) = \psi_{1,1}^{(1)}(x_1,y_1) \psi_{1,1}^{(2)}(x_2,y_2) = \frac{2}{a^2}sin(\frac{\pi x_1}{a})sin(\frac{\pi y_1}{2a})sin(\frac{\pi x_2}{a})sin(\frac{\pi y_2}{2a})$ I also see that this is a symmetric function when exchanging the particles (if this is indeed correct, which I can't honestly say I'm 100% sure about), so I know that they have to be in the singlet spin state. How on earth do I write this? Can I just write it as $\psi (x_1,y_1,x_2,y_2)\left| singlet \right\rangle$ And can I put in what the singlet state is? I mean, I know it's $\left| singlet \right\rangle = \frac{1}{\sqrt{2}} (\left| \uparrow \downarrow \right\rangle - \left| \downarrow \uparrow \right\rangle)$ so can I put this into the equation?