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Homework Help: Fermi/Boson statistics partition function/probabilities

  1. May 17, 2013 #1
    1. The problem statement, all variables and given/known data

    If you have a three energy level system, with energies 0, A, B where B>A, which consists of only two particles what is the probability that 1 of the particles is in the ground state? What about if two of them are in the ground state? Do this using both fermi and boson statistics.

    2. Relevant equations

    Using Fermi-statistics,

    Using Boson statistics,
    Z=1+exp(-β*A)+exp(-2 β*A)+exp(-β*B)+exp(-2 β*B)+exp(-β*(A+B))
    3. The attempt at a solution

    For 1 particle in the ground state,

    In fermi statistics there are two arrangments which satisfy this condition,
    when 1 particle is in the ground state and the other in A, and when the other is in B.

    P = Pg(Pa+Pb) = (1/Z)(exp(-β*A)/Z + exp(-β*B)/Z)

    Does that look correct?

    Would be a similar answer for boson statistics, just with the different partition function.

    For both particles in the ground state


    P = Pg1 Pg2 = (1/Z)(1/Z) = Z-2

    This doesn't look correct, the answer should be zero because fermions cant be in the same energy level..

    Can anyone see what I have done wrong?

    Boson is the same answer with the different partition function, though its correct that the answer is non-zero here.

    Have I done the probabilities incorrectly?
  2. jcsd
  3. May 18, 2013 #2


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    The 1-particle case looks good. You can check it for the limits β->0 and β->infinity to see if it gives the expected results.

    Your partition function includes both particles, you should not "use it twice".
    Just look for terms in it which correspond to both particles in the ground state.
  4. May 18, 2013 #3
    Thanks for your reply,


    So it should be

    For both particles in the ground state


    P =(1/Z)

    When T->0 this probability should be zero right?

    But when you actually take the limit as T->0 , i.e. β->infinity, you'll see that P tends to infinity, not zero.

    That's why I'm a bit confused, intuitively there should be 0 probability of two fermions residing in the same state right?
  5. May 18, 2013 #4


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    That is wrong.
    Which states in Z correspond to "both particles in the ground state"?
    Use those states to do the same as you did for the 1-particle case.
    For fermions, the probability has to be 0 independent of T.
  6. May 18, 2013 #5
    In the fermion case, the state doesn't exist..

    Is that what you mean? The probability is zero for that reason.

    So it's simply P = 0, as the state does not exist using fermi statistics.

    For bosons though it does exist, and it is when both atoms are at E=0, which means that the the probability is in fact what I wrote before (incorrectly as the fermi case)

    P=(e^0/Z) (e^0/Z)=(1/Z)(1/Z)

    This is the right way to do it, right?

    I just feel like 'saying' P=0 for the fermi case isn't concrete enough, I thought there would be a more formal, mathematical way it would come out.
  7. May 18, 2013 #6


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    No. You do not have "two different Z". You have to consider both particles at the same time, and your partition function does this already.
    You already considered this as one of multiple cases for the previous problem.

  8. May 18, 2013 #7
    The way I've been doing this is consistent with how I did the first question, multiplying the probability that one particle is in the ground state by the probability the other is in the ground state. I'm not sure what you mean by "two different Z's", the two Z's are the same, (1/Z)(1/Z) = 1/Z^2.

    Do you mean that the probability that both particles are in the ground straight is just 1/Z ?

    To me that looks like the probability that one particle is in the ground state and the other is in any other state (0, A or B).

    I think I have done the first parts wrong, one particle in the ground state.

    For example in the Fermi case there are two states which have one particle in the ground state, so the probability is:

    P = [exp(-β*A) + exp(-β*B)]/Z

    Without the extra factor of (1/Z)

    I think this is correct, after rereading the definition of P.

    Also, after rereading the definition, it seems correct that the probability both particles are in the ground state is 1/Z.

    Was just trying Boltzmann statistics as well just to compare the probabilities at the end,

    I'm not very confident with probabilities because they can get kind of confusing, i think this is right though, using Boltzmann statistics:

    Z=1+2exp(-bA) + exp(-2bA) + 2exp(-bB) + exp(-2bB) + 2exp(-b(A+B))

    Similar to the Boson statistics just with some extra factors of 2, since some of the states are now distinguishable.

    So the probability for one particle to be in the ground state is:

    P = 2exp(-bA)/Z + 2exp(-bB)/Z

    Does that seem better now?
    Last edited: May 19, 2013
  9. May 19, 2013 #8


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    Oh sorry, I missed the additional /Z in the first post. They are wrong.

    For bosons, right.

    How did Boltzmann statistics come in now? We do not have the classical limit.
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