Register to reply

Identical particles in a 2D potential well

by Hannisch
Tags: identical, particles, potential
Share this thread:
Oct15-11, 09:02 AM
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

[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!
Phys.Org News Partner Science news on
'Smart material' chin strap harvests energy from chewing
King Richard III died painfully on battlefield
Capturing ancient Maya sites from both a rat's and a 'bat's eye view'
Oct15-11, 01:39 PM
Sci Advisor
HW Helper
PF Gold
P: 11,868
Looks fine!
Oct16-11, 02:33 PM
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 [itex] n_x=n_y=1 [/itex], so that the wavefunctions will be

[itex]\psi_{1,1}^{(1)}(x_1,y_1) = \frac{\sqrt{2}}{a}sin(\frac{\pi x_1}{a})sin(\frac{\pi y_1}{2a})[/itex]

[itex]\psi_{1,1}^{(2)}(x_2,y_2) = \frac{\sqrt{2}}{a}sin(\frac{\pi x_2}{a})sin(\frac{\pi y_2}{2a})[/itex]

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

[itex]\psi (x_1,y_1,x_2,y_2) = \psi^{(1)} (x_1,y_1) \psi^{(2)} (x_2,y_2)[/itex]

So then

[itex]\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})[/itex]

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

[itex]\psi (x_1,y_1,x_2,y_2)\left| singlet \right\rangle[/itex]

And can I put in what the singlet state is? I mean, I know it's

[itex]\left| singlet \right\rangle = \frac{1}{\sqrt{2}} (\left| \uparrow \downarrow \right\rangle - \left| \downarrow \uparrow \right\rangle)[/itex]

so can I put this into the equation?

Oct16-11, 03:00 PM
Sci Advisor
HW Helper
PF Gold
P: 11,868
Identical particles in a 2D potential well

Yes, the complete state is a Cartesian product of the spatial state and the spin state. You typically just write them next to each other as you did or perhaps stick a symbol between them indicating it's a Cartesian product. Do you have any examples in your textbook?
Oct16-11, 03:09 PM
P: 116
I honestly can't find any, the closest I came to it was from my lecture notes, where my teacher wrote


in an example.
Oct16-11, 03:52 PM
Sci Advisor
HW Helper
PF Gold
P: 11,868
It's funny. I just checked two of my books, and I can't find an example either. I'd just do what your professor did and write the two pieces next to each other. It's pretty clear what it means.

Register to reply

Related Discussions
Quantum Mechanics - Identical and Non-Identical Spin 1 particles Advanced Physics Homework 4
Identical particles Quantum Physics 3
Identical particles Advanced Physics Homework 9
Identical particles Quantum Physics 45
Are all particles identical? High Energy, Nuclear, Particle Physics 3