Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Ideal Gas questions

  1. Apr 24, 2012 #1
    Hi everyone, I'm working through an example in my textbook, and it's making very little sense to me. The problem is:

    Let [itex]Z_1(m)[/itex] be the partition function for a single quantum particle of mass m in a volume V. First, calculate the partition function for two of these particles if they are bosons (and then also for fermions).

    So, it tells me earlier in the chapter that, for bosons for example, the wave state has to be symmetric. So, for two particles (a and b) in "plane wave states" [itex]k_1[/itex] and [itex]k_2[/itex] where the energy for a plane wave state k is [itex]E = \frac{\hbar^2 k^2}{2m}[/itex], the total state is [itex]\left|k_1,k_2\right\rangle = \frac{\left|k_1\right\rangle \left|k_2\right\rangle - \left|k_2\right\rangle \left|k_1\right\rangle}{\sqrt{2}}[/itex] (for k1 not equal to k2. For k1=k2, it's just [itex]\left|k_1\right\rangle \left|k_2\right\rangle[/itex])

    I guess I understand that. A plane wave (let's just say 1D, along the x axis) is of the form [itex]e^{ikx}[/itex], so in this case, the total state would be (in the position basis)

    [tex]\left|k_1,k_2\right\rangle = \frac{e^{ik_1x_a}e^{ik_2x_b} - e^{ik_1x_b}e^{ik_2x_a}}{\sqrt{2}}[/tex]

    Is that right?

    Anyway, my real trouble is next. I think I'm trying to do it more rigorously than I have to, but I want to understand this explicitly. So, for bosons, they say

    [tex]Z_2 = tr(e^{-βH})[/tex], where tr() is the trace of a matrix. Now, they never really tell us why this is, earlier in the book. They just seem to introduce it randomly. Could anyone tell me why this is a definition of the partition function?

    Anyway, I'll trust this for now. Going on, they say:

    [tex]tr(e^{-βH}) = \sum_{k_1,k_2} \left\langle k_1,k_2\right|e^{-βH}\left|k_1,k_2\right\rangle = \sum_{k_1>k_2}\frac{\left\langle k_1 \right|\left\langle k_2\right| + \left\langle k_2 \right|\left\langle k_1\right|}{\sqrt{2}}e^{-βH}\frac{\left|k_1 \right\rangle \left| k_2\right\rangle + \left| k_2\right\rangle\left|k_1\right\rangle}{\sqrt{2}} + \sum_{k}\left\langle k \right|\left\langle k\right|e^{-βH}\left|k \right\rangle \left| k\right\rangle[/tex]

    So this is just expanding from before. Now, the Hamiltonian acts on the product kets to the right, and because they both have the same pair of k values, produces the same thing, [itex]e^{-\frac{\beta\hbar^2}{2m}(k_1^2 + k_2^2)}[/itex] for the first sum and [itex]e^{-\frac{2\beta\hbar^2}{2m}k^2}[/itex] for the 2nd one.

    So it seems like we now get

    [tex]\sum_{k_1>k_2}e^{-\frac{\beta\hbar^2}{2m}(k_1^2 + k_2^2)}(\frac{\left\langle k_1 \right|\left\langle k_2\right| + \left\langle k_2 \right|\left\langle k_1\right|}{\sqrt{2}})(\frac{\left|k_1 \right\rangle \left| k_2\right\rangle + \left| k_2\right\rangle\left|k_1\right\rangle}{\sqrt{2}}) + \sum_{k}e^{-\frac{2\beta\hbar^2}{2m}k^2}(\left\langle k \right|\left\langle k\right|)(\left|k \right\rangle \left| k \right\rangle)[/tex]

    So it seems to me that in the first sum, only two of those products (when you foil out the terms inside the parentheses) are nonzero. So that cancels the sqrt(2)'s on the bottom. In the right, it seems like it's just 1. Is this right?

    Then, in the book, they change the indices of the first sum from k1>k2 to k1,k2 and add a factor of 1/2 in front of the sum, as is standard. However, they do the same for the second sum, but I don't see why. The second one is just going over all k, so there is no double counting, right?

    Anyway, both of these sums are now basically some variation of the individual partition function (I see this part easily). So they finally get

    [tex]Z = \frac{1}{2}[Z_1^2(m) + Z_1(m/2)][/tex]

    The fermi one confuses me even more, but I'll be brief with it. Basically, the first step of setting up the PF as the trace of the exponent is the same, and they plug in the antisymmetric state. But here the ones where k1=k2 should be zero, unlike the bosons. Still, they somehow end up with a sum over k, where they mysteriously skip a bunch of steps. Could anyone explain this to me?

    Thanks!
     
  2. jcsd
  3. Apr 26, 2012 #2

    Jano L.

    User Avatar
    Gold Member

    In the beginning of your post, you say that you calculate for bosons and the state has to be symmetric, but you use antisymmetric functions (minus in your formula). You should use sums of products, not their difference.

    It is a definition based on formal analogy from classical statistical physics.

    There the probability that variables of Hamiltonian system are in some small volume element dqdp is

    [tex]
    \rho(p,q) dpdq = \frac{e^{-\beta H(p,q)}}{Z},
    [/tex]

    where Z is some constant. Because total probability is 1, it is immediate that

    [tex]
    Z = \int e^{-\beta H(p,q)} dpdq,
    [/tex]

    This Z, when taken as a function of temperature and volume, is called the partition function.



    In quantum statistical physics, things are very different. The density matrix in the basis of Hamiltonian eigenvectors is supposed to be

    [tex]
    \rho_{kl} = \frac{e^{-\beta E_k}}{A} \delta_{kl},
    [/tex]

    (A is some constant). Considering the meaning of density matrix, its form looks AS IF it corresponded to a system which has probability [itex]e^{-\beta E_k}/A[/itex] that it is in a state described by the k-th eigenvector of the Hamiltonian.

    Since the diagonal sum of the density matrix (total probability) has to be 1, it follows that

    [tex]
    A = \sum_k e^{-\beta E_k} = Tr(e^{-\beta H}).
    [/tex]

    Thus instead of INTEGRATING the PROBABILITY DENSITY, in quantum theory people SUM the PROBABILITIES of eigenstates and the result is called again partition function.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quantum Ideal Gas questions
  1. Quantum question (Replies: 2)

  2. Quantum question (Replies: 2)

Loading...