QM - Identical Particles, and energy states

1. May 4, 2007

Brewer

1. The problem statement, all variables and given/known data
Five non-interacting particles are placed in a three dimensional harmonic oscillator potential for which the single-particle energy is:
$$E_n = (n_x + n_y + n_z +3/2)\hbar\omega$$

What is the lowest energy of the five-particle system when the particles are:
a) Distinguishable, spinless bosons
b) Identical, spinless bosons
c) Identical fermions each with spin s=1/2
d) Identical fermions each with spin s=3/2

2. Relevant equations
I think effectively just the above equation thats given in the question.

3. The attempt at a solution
From what I can tell so far:
The total energy will be a summation of the (lowest) energies of all the different particles.

For part a) because spin = 0, then the energy of the bosons will be:

$$E = E_n + E_{n'\prime} = (n + n\prime +1)\hbar\omega$$ (this is for 2 particles of different energies, would just add the other 3 terms into the brackets.)

Here the various values of n would be 1->5 as they are the lowest non-identical energy levels (correct?)

b) For this I would use the same equation, just with n=1 for all of them.

c) and d) Again I think I use the same equation, except that this time n will be dependent on the spin of the particle.

Any hints would be appreciated.

Thanks,

Brewer

2. May 5, 2007

nrqed

No.

First, an important detail: the smallest value that "n" may take for the harmonic oscillator is 0, not 1.

Now, if you want the total energy to be the smallest possible, what values of n should you take for all of them? (recall that if they are indistinguishable, there is no condition at all on the allowed values)

Again, the smallest possible value of "n" is zero, not one.
So what do you think the answer is?

Patrick

3. May 5, 2007

Brewer

So for the part with the distinguishable particles - because the energy of a particle is in 3 dimensions, would the lowest energy it could have be 0,0,0 (x,y,z components respectively), then you would have 1,0,0 followed by 0,1,0 and 0,0,1 with a final particle in energy state 1,1,0?

4. May 5, 2007

nrqed

Why not put all of them in the 0,0,0 state?
That would be the state of lowest energy, wouldn't it?

The same can be done for the identical bosons. But that won't be possible for all the fermions.

Patrick

5. May 6, 2007

Brewer

Well surely if the they all have the same energy then they're indistinguishable? I thought would be the point of the question - that the distinguishable ones would each have different energies to the indistinguishable ones?

6. May 6, 2007

nrqed

No, distinguishable vs indistinguishable has nothing to do with the energy of the particles, but only with their intrinsic properties (mass, spin, electric charge, colour quantum number, etc).

To make an analogy, think of the particles as marbles. And the energy levels as being the steps in a staircase. Whether the marbles are distinguishable or not has nothing to do with what step they are on. You can visualize the indistinguishable marbles as being all of the same color, size and texture. The distinguishable ones differ in some way so that you can tell them apart.
The point is that if they are indistinguishable, and I tell you to close your eyes for a few seconds and open them again, you won't be able to tell if I have switched some of them around or not.

Now, this has nothing to do with what step they are on. At least, classically. The surprising aspect of the spin-statistics theorem (which leads to Pauli's exclusion principle) is that in QM, there is a connection.

In the case of indistinguishable particles, you may put them on any step, with no restriction at all. So the lowest energy state will be all of them on the lowest step.

If they are identical bosons, there is still no restriction (the total wavefunction must be symmetric under the exchange of any two of the bosons, but here I am looking at the energy only). Then you may put all of them on the lowest step also.

For spin 1/2 fermion, there is an extra label (i.e. quantum number) on the marbles: spin up or down. If you want, it's as if the marbles come in two different colors, let's say black and white. Pauli's exclusion principle states that there can be no two identical marble on the same step. So to have the lowest energy, you may only put two fermions (one with spin up and one with spin down) on the ground state.

Now, there is the complication of degeneracy. The first excited state is triply degenerate for the 3-d isotropic harmonic oscillator. So it's a bit as if the second step of the staircase contains actually three sections. You can put two of the other fermions (with opposite spins) in one state (let's say n_x =1, n_y=n_z=0) and the third one in a different state or you could put all three in a different state. The end result is still that there will be two fermions with energy $3/2 \hbar \omega p$ and three fermions with energy $5/2 \hbar \omega$ .

For the spin 3/2 fermions, now there are 4 different marbles. It shoudl be clear now how to get the lowest possible energy!

Hope this helps.

Patrick

7. May 6, 2007

Brewer

I haven't looked at this properly as its late, but from a quick glance this seems to make more sense. I have been waiting for someone to give me a simple example for QM for some time. Thank you ever so much!!!