I'm trying to come up with an expression for the partition function of a system of spin-1/2 ideal gas particles on a line of length $L$. The total number of particles $N$ is fixed, with $N = N_\uparrow + N_\downarrow$. Here, $N_\uparrow$ is the number of spin-up particles and $N_\downarrow$ is the number of spin-down particles in a particular microstate.

I have the following Hamiltonian for the particles of mass $m$.

$$H = \sum_{i=1}^{N}{\frac{(p_i + \beta s_i)^2}{2m} -b s_i} $$

Here, $s_i = 1$ for the spin-up $N_\uparrow$ particles and $s_i = -1$ for the spin-down $N_\uparrow$ particles. $\beta$ and $b$ are constants.

I'm trying to use the Hamiltonian to write down the energy for the spin-up and spin-down particles so I can write down the partition function. If I expand the Hamiltonian, I get:

$$H = \sum_{i=1}^{N}{\frac{p_i^2}{2m} + \frac{p_i \beta s_i}{m} - b s_i + \beta^2} $$

How do I find the energy of the two sets of spin particles from this and use it to come up with the partition function? Is the energy of the particles just?

$$E_\uparrow = \frac{p_i^2}{2m} + \frac{p_i \beta}{m} - b + \beta^2$$ $$E_\downarrow = \frac{p_i^2}{2m} + \frac{-p_i \beta}{m} + b + \beta^2$$

How would one evaluate this in the canonical partition function:

$$Z = \sum_{\mu_s}e^{-\beta H}$$

where $\mu_s$ is the summing over all microstates. I'm not sure how to evaluate this.

  • $\begingroup$ Are you neglecting the magnetic interaction between particles, $H=-\sum_{\langle i,j \rangle} J_{ij} s_{i} s_{j}$ ? $\endgroup$ Commented Nov 6, 2013 at 23:16
  • $\begingroup$ No, I don't have this term in the Hamiltonian. I'm just trying to treat the orbital motion of the particles classically. $\endgroup$ Commented Nov 7, 2013 at 0:04

1 Answer 1


First of all write down an explicit expression for the summation over all microstates.

Edit Since you're treating the system classically this includes an integral over phase-space & a summation over all possible spin-configurations.

$$ \sum_{\mu_s} = \sum_{\{s_i\}}\int\frac{d^Np d^Nq}{h^{3N}N!}$$

The second thing is to realize that your Hamiltonian is non-interacting and the canonical density $e^{-\beta H} $ is just a product of one-particle Hamiltonians $$ e^{-\beta H} =\prod_{i=1}^Ne^{-\beta h_i} $$ where of course $$ h_i = {\frac{(p_i + \gamma s_i)^2}{2m} -b s_i} $$ (I renamed $\beta$ to $\gamma$, because you don't mean the inverse temp. here)

So you have to evaluate
$$ \frac{1}{N!}\sum_{\{s_i\}}\prod_{i=1}^N\int\frac{dp_i dq_i}{h^{3}} e^{-\beta h_i} = $$

Because $H$ ist non-interacting, the N-particle phase-space-integral factorizes into N integrations over a 1-particle phase-space. Similarily one may interchange the spin-summation and the product (convince yourself that this is true! e.g that one ends up with the same terms, whether you sum over spin first or not.)

That means, instead of summing over all the many-body microstates, one first sums over the possible configurations of a single particle and accounts for the fact, that there are many afterwards. Additionally all $h_i$ are equivalent. They each just carry a different but redundant index:

$$ Z = \frac{1}{N!}\prod_{i=1}^N\sum_{s_i=\pm1}\int\frac{dp_i dq_i}{h^{3}} e^{-\beta h_i} = \frac{1}{N!}\left(\sum_{s=\pm1}\int\frac{dp dq}{h^{3}} e^{-\beta h}\right)^N$$ where $h$ is $h_i$ but without an index, because the reference is not to a specific particle anymore.


You'll have to think about, what to do with the momentum integration. I haven't calculated the result, but it might be you won't end up with a solution in closed form. There might be an approximation needed to do the momentum summation. Edit I think a gaussian integration will do the trick. /Edit Let us know what you end up with!

  • $\begingroup$ @vectorize7891 Did you make progress? Do you need further clarification? $\endgroup$
    – Nephente
    Commented Nov 10, 2013 at 11:15
  • $\begingroup$ Thank you greatly for following up on this! I have made some progress. Don't you also need to sum over positions as well? The only way I could evaluate this is for large $N$. Another concern I had is, for large $N$, you don't necessarily know that $N_\uparrow$ AND $N_\downarrow$ will both be large, do you? $\endgroup$ Commented Nov 10, 2013 at 23:20
  • $\begingroup$ @vectorize7891 You are right regarding the volume-integration. i've updated my answer. I think it will suffice to consider large $N$ at some point, without making reference to $N_\uparrow$ or $N_\downarrow$. Were do they come in during your calculation? Although in the limit of zero external field, their expectation values will be $N/2$, an external field will drive them appart. $\endgroup$
    – Nephente
    Commented Nov 11, 2013 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.