# Energy States- Bossons & Fermions

1. Jan 19, 2009

### Fourier mn

1. The problem statement, all variables and given/known data
Consider the 3-D infinite potential well (length=L). The energy levels for this system are given by E=(h bar)^2$$\pi^2$$/(2ML^2)*(n(sub x)^2+(n(sub y)^2+(n(sub z)^2)
There are 10 particles in this potential well. What is the lowest energy of this ten-particle state when the particles are:
1. Identical, spinless bosons.
2. Identical, spin 1 bosons.
3. Identical fermions each with s=1/2
4. identical fermions each with spin s=1/2

2. Relevant equations
None

3. The attempt at a solution
I'm a little confused about bosons and fermions properties. If I remember correctly, bosons can be in the same state (all 10 of them) regardless of their spin or spinless. So for a. and b. the energy state would be 1 (the lowest) which all 10 particles are in the same state.
In the case of fermions, only two can be in the same state so the energy states go from 1 to five. I'm not sure how to proceed from here, and how to calculate the energy state. I'll really appreciate any suggestion.

2. Jan 20, 2009

### CompuChip

So I suppose you have seen the derivation of the energy spectrum and you have some idea what the numbers $n_x, n_y, n_z$ mean. In particular you should have seen that they completely define the quantum states, so if you say "this particle has $n_x = 2, n_y = 0, n_z = 4$" I know which state you are talking about.

Now the question asks you first about 10 identical spinless bosons. As you say, they can all be in the same state. So for bosons, you are allowed to all give them the same $n_x, n_y, n_z$. Now you should look at your expression for the energy and find out which values should be taken to get the lowest state.

Fermions on the other hand, cannot occupy the same state. So you can put one in the lowest state (with the same $n_x, n_y, n_z$ as for the bosons) with spin up. You can put one more fermion in that state, with spin down. But then Pauli's exclusion principle does not allow any more fermions there, so they will have to fill up a higher state. Again, look at your expression for the energy, and find the next highest energy level. In fact, you will find three with the same energy, so you can put another 6 fermions in there (three in each state for each spin). Continue this until you have put all the fermions in the lowest possible energy level which is allowed, then sum all the energies.

3. Jan 20, 2009

### Fourier mn

Can I put all of the spinless bosons in the energy state of 0,0,0 (ground state) or it has to be 0,0,1? where do I put the spin of the particle?
For fermions I'll have five energy levels- 0,0,0; 0,0,1; 0,1,1; 1,1,1; 1,2,1; I dont was degeneracy on the sets. Again, where does the spin comes to the picture?

4. Jan 20, 2009

### CompuChip

No, the lowest energy state is (0, 0, 0). And yes, you can put all of them there.

For bosons, does it matter? The energy does not depend on the spin, and whether they all have the same spin or not doesn't matter because you don't have an exclusion principle for bosons. So I don't think it makes any difference.

It's not a matter of what you want. The fact is that (1, 0, 0) is a physically different quantum state than (0, 1, 0) or (0, 0, 1). Of course you can make a system with 1 fermion in the (1, 0, 0) state and one in the (1, 1, 0) state, but it will have a higher energy than one in the (1, 0, 0) and (0, 1, 0) state, as you can easily verify yourself.
So if you want the lowest energy, it's not a matter of what you want :)

If the fermions have spin, then you actually have four quantum numbers: $n_x, n_y, n_z$ and the spin s of the fermion. If we let s take values + and - for spin up and down, respectively, then (0, 0, 0, +) is a different state than (0, 0, 0, -). So now what are the 10 lowest energy states?