Check my work please. Particle states.

[frankR]

Problem:

Consider a non-interacting system of 4 particles with each particle having single-particle states with energies equal to 0, e, 2e and 3e. Given that the total energy of the system is 6e, find the number of microstates of the system (and identify the microstates) if the particles are a) distinguishable, b) indistinguishable Bosons and c) indistinguisable Fermions.

For a) I get 3-ways to get 6e and 3*4 ways to get 6e amoung distinguishable particles.

For b) I get 3-ways to get 6e amoung the indistinguishable Bosons.

For c) I get 1-way to get 6e amoung Fermions.

Is this correct?

Also is there such thing as distinguishable Fermions. My guess is by definition of a Fermion, no!

Thanks.

[Janitor]

I agree with your answer for (c): the only possible state in that case is where one fermion has energy 0, another e, another 2e, and the remaining one 3e.

I seem to get more than 3 ways for (b). Here are four (the ordering left-to-right on a row being immaterial since they are indistinguishable):

0 0 3 3
0 1 2 3
1 1 2 2
0 2 2 2

I think I convinced myself that your answer for (a) is an undercount. (In this case the ordering does count.)

[Janitor]

Also is there such thing as distinguishable Fermions. My guess is by definition of a Fermion, no!



This is probably not what you meant, but an example would be a system consisting of a neutrino, an electron, a muon, and a tauon. Four particles, very much distinguishable.

[frankR]

Ahh I missed: 1=e, 2=e, 3=e, 4=3e

So corrected answers:

a) omega = 16 (I permute here by multiplying by 4 right?)
b) omega = 4
c) omega = 1

Thanks.

[frankR]

This is probably not what you meant, but an example would be a system consisting of a neutrino, an electron, a muon, and a tauon. Four particles, very much distinguishable.

Yeah you got me there. I should specified identical particles.

So say all are electrons.

I ask because Prof. Webb specified "indistinguishable Fermions". So that would indicate were not talking about electrons and protons, in which case the Pauli Exclusion Principle doesn't apply, Protons and Electrons can occupy the same state. Unless protons aren't Fermions. :tongue2: :yuck:

[Janitor]

Ahh I missed: 1=e, 2=e, 3=e, 4=3e



Oops. I missed

1 1 1 3

as well, so there are at least 5 states for (b).

[Janitor]

So say all are electrons.

I suppose there are conditions (say strong applied magnetic field, low temperature) where a pair of electrons would be distinguishable, because one would maintain an 'up' state of spin component, and the other a down 'state,' in some sort of metastable situation.

[Janitor]

a) omega = 16 (I permute here by multiplying by 4 right?)


I'm not sure. I just started writing out possible states for particles A B C D, where order counts, and I stopped when I got to 13, since that was more than your first answer of 12. There is probably some elegant way of figuring out the number. The mathematical types here will know.

[frankR]

Okay so I also missed: 1=3e, 2=2e, 3=e, 4=e

Did we get them all this time? :cry: :rofl:

[Janitor]

Did we get them all this time? :cry: :rofl:

It's all my tired brain can find at the moment. I will sleep on it. :zzz:

[Janitor]

I thought about this a bit more today, and here is what I think are the answers to (a) and (b).

0 1 2 3 12
0 2 2 2 4
0 0 3 3 6
1 1 2 2 6
1 1 1 3 4

12 + 4 + 6 + 6 + 4 = 32

The five rows show the five microstates for bosons, so the answer to (b) is 5. The number of permutations of each microstate is shown in bold. The sum is 32, so that is the answer to (a). I am pretty sure that considerations of symmetrization for bosons and antisymmetrization for fermions are not an issue in this particular problem, but you might want to check up on that.

[frankR]

Me and my fellow class mates were able to find 32 as well for a.

Nice work!

You're a janitor that does physics in his spare time? Are you like Will Hunting? :D

[Janitor]

Never saw the movie, I'm afraid.