Canonical partition of a boson gas [duplicate]

Question

I have a 1D gas made of $N$ particles placed in a harmonic potential well, so the Hamiltonian is:

$$ \mathcal H = \sum_{j=1}^N \left ( \frac{p_j^2}{2m} + \frac{1}{2}m\omega^2 x_j^2 \right )$$

The first part of the exercise asked me to find the canonical partition at temperature $T$ if the particles are distinguishable, then to find the partition if the particles are indistinguishable, but the Maxwell-Boltzmann approximation applies. In both cases this was easy. But now the exercise asks me to find the partition if the particles are identical bosons, and then to show that this is equal to the Maxwell-Boltzmann approximation for large temperatures.

I don't know exactly how to set up the summation to count the states properly when you treat them as bosons. I know we have the states $\epsilon_j = \hbar \omega (j+\tfrac{1}{2})$ and that each of those states will be occupied by $n_j$ bosons and since it's 1D I don't have to worry about degeneracy...but I'm unsure how to continue.

From what I understand, it's nicer to work with fermions and boson in the grand canonical ensemble, but we haven't seen this in class yet.

Thanks!

Answer 1

Bosonic $1D$ $N$ harmonic oscillators allow `exceptionally' a closed form of the canonical partition function:

$Z_{N}=\prod_{n=1}^{N}\frac{q^{1/2}}{1-q^{n}}$ where $q=e^{-\beta \hbar \omega}$

This expression resembles a sort of a grand canonical partition function for a system of bosonic ``phonons" having finite number of possible energy spectra with vanishing chemical potential.

One can derive this expression by noting,

Area of a Young tableux = sum of the length of columns = sum of the length of rows

Answer 2

Let's start by counting the ways in which the states can be counted. For a classical gas of indistinguishable particles, the energy spectrum is continuous and there can be degeneracy. I will assume you know how to derive this from M-B statistics and how to include the effects of overcounting due to indistinguishability.

You essential need to sum over the non-normalized probabilities of all of the different ways in which the bosons can occupy the integer spaced energy levels of the SHO. it is easier to derive in the grand-canonical ensemble because the constraint of a fixed particle number makes the sum harder.

The canonical partition function can be written in the energy basis as,

$$Z_{can} =tr(e^{-\beta H}) = \sum_{\{n(j)\}}exp\left(-\beta \sum_j \hbar\omega(j+\frac{1}{2})n(j)\right) $$

subject to the constraint, $$\sum_j n(j) = N.$$

In practice this sum is difficult, which is why people ordinarily use the grand-partition function to construct partition functions. We can relax this constraint by introducing a chemical potential. Then,

$$Z_{GC} = \prod_{j}\left[ 1-exp\left(\beta \mu - \beta \hbar\omega(j+\frac{1}{2})\right) \right]^{-1}$$