I am trying to understand the partition function of $N$ copies of 1D bosonic harmonic oscillator. $$ Z_N{}^B = q^{\frac{N}{2}} \prod_{n=1}^N \frac{1}{1-q^n}\quad\text{ with }\quad q=e^{-\beta \hbar w}.$$

My trial are follows

For bosonic case, Hilbert space is spanned by states labelled $N$ integers such that, $0\leq k_1 \leq k_2 \leq \cdots k_n \cdots \leq k_N$, The energy states can be \begin{align} H |k_1, \cdots k_N> = \left( \frac{N}{2} + \sum_{n=1}^N k_n \right) | k_1, \cdots k_N> \end{align} Then partition function \begin{align} Z_N{}^B &= tr(q^H) = q^{\frac{N}{2}} \sum_{k_1=0}^{\infty} \sum_{k_2=k_1}^{\infty}\cdots \sum_{k_N=k_{N-1}}^{\infty} q^{\sum_{n=1}^N k_n} \\ & = q^{\frac{N}{2}} \prod_{n=1}^N \frac{1}{1-q^n} \end{align}

what i have trouble with is the step of first line to second line.

First what think about the first term $(1+q+q^2+ \cdots)$, are the case of $k_2=k_3=\cdots k_N=0$, so $\sum_{k_1=0}^{\infty} q^{k_1}=1+q+q^2+\cdots$

then how i can inteprete the second term via $k_2, \cdots?$ $(1+q^2+ \cdots)?$

I found some wrong points in my formula, i correct it. Then it makes sense.


This was meant more as comment, but turned out to be too long.

The key word here is "bosonic":

What you wrote down as $Z_N^B$ in your attempt is the partition function for N identical but distinguishable oscillators, while $Z_N^B$ from the paper is the partition function for N indistinguishable oscillators. Which means the degeneracy factors for the energy levels are different.

The fastest way to sees this is the $N=2$ case. Your attempt gives $Z_2 = q\frac{1}{(1-q)^2}$, whereas the correct result is $Z_2^B = q \frac{1}{1-q^2}\frac{1}{1-q}$, with the different degeneracies compounded in the $\frac{1}{(1-q)^2}$ and $\frac{1}{1-q^2}\frac{1}{1-q}$ factors. But look at the actual degeneracies by re-expanding the series: $$ \frac{1}{(1-q)^2} = (1+q+q^2+q^3+\dots)(1+q+q^2+q^3+\dots) = \\ = 1 + 2q + 3q^2 + 4q^3 + \dots $$ while $$ \frac{1}{1-q^2}\frac{1}{1-q} = (1+q^2+q^4+q^6+\dots)(1+q+q^2+q^3+\dots) = \\ = 1 + q + 2q^2 + 2q^3 + 3q^4 + 3q^5 +\dots $$ The identical unit term corresponds to the unique ground state, but all excited states, even the first one, display different degeneracies.


  • 1st excited state: 2 levels for the distinguishable case, $(n_1=1,n_2=0)$ and $(n_1=0,n_2=1)$,
    1 level only for the indistinguishable case, $(0,1)$.

  • 2nd excited state: 3 levels for the distinguishable case, $(0,2)$, $(1,1)$, and $(2,0)$,
    2 levels only for the indistinguishable case, $(0,2)$ and $(1,1)$, and so on.

As for how to retrieve the bosonic partition function, Sec.III pg.663 in Am.J.Phys.71(7), 661(2003) (U Mass link) will give you the general idea.

  • $\begingroup$ in the process of computing i have something wrong, now i correct it, by the ways thanks for clarifying the concept. Thanks $\endgroup$ – phy_math Nov 27 '16 at 5:12

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