# Derivation of partition function for $N$ identical quantum harmonic oscillators

What is the partition function

$$\mathcal Z^{(N)}_\beta(H) : =\mathrm{Tr}\exp(-\beta H) \tag{Z}$$

$$\left(\beta >0\right)$$ for a system of $$N$$ indistinguishable and non-interacting bosons (e.g. harmonic oscillators) with Hamiltonian $$H$$?

Why is $$\mathscr Z^{(N)}_\beta(H) := \frac{\mathcal Z_\beta^{(1)}(h)^N}{N!}\tag{Z^\prime}$$ with the corresponding single particle Hamiltonian $$h$$ and the partition function for a single particle $$\mathcal Z_\beta^{(1)}$$ not the correct partition function? Is it an approximation? If so, under which (physical) circumstances is this a good approximation? How does the factor $$N!$$ arise?

• When I voted to close, your question was only composed of the first part (the first four lines) and, as such, was certainly not conceptual. The question having improved, I have now retracted my close vote. Apr 11 at 11:36
• Factor $N!$ accounts for indistinguishability of particles - discussions can be found in any good statistical physics textbook (see also Gibbs paradox). In quantum case this is already accounted for by the commutation relations... but your oscillators are actually not indistinguishable, as they have different quantum numbers. Apr 11 at 14:35
• @RogerVadim I think this is not correct. What do you mean with different quantum numbers? I have $N$ indistinguishable (and non-interacting) bosons. As you can read in my answer, I've worked with completely symmetrized $N$-particle states - so I am dealing with identical bosons. But this is why I asked the question: I've read more than one time that the second partition function in the question would be the correct one for identical QHO. But this is not the case. Apr 11 at 14:36
• $N$ indistinguishable bosons in one mode is one oscillator. In your partition function there are actually $N$ modes of the same frequency - so they are identical in mathematical sense, but they are physically different modes (e.g., these may be light waves of the same frequency but different direction). Your occupation numbers are labeled by this quantum number: $n_1,n_2,...,n_k$. Apr 11 at 14:40
• I think this is solely a matter of terminology. If you want, then I have $N$ identical bosons in the HO potential... In first quantization my many-body hamiltonian would be of the form $H \sim \sum\limits_{i=1}^N p_i^2 + x^2_i$, i.e. it describes $N$ particles. If you disagree, please feel free to write an answer an elaborate. I'd be interested. Apr 11 at 14:41

The problem with the partition function in $$(\mathrm Z^\prime)$$ is that there the physical states are not counted correctly (cf. the answer by @SolubleFish). However, this partition function can be obtained in a certain limit, as shown in the end. To obtain the correct expression for the partition function, we should start more or less from scratch:

The Hilbert space of $$N$$ identical bosons is given by $$\mathcal H := \vee^N \mathcal H_1$$, where $$\mathcal H_1$$ is the single-particle Hilbert space. If $$h$$ denotes the Hamiltonian for a single particle (e.g. harmonic oscillator), then the Hamiltonian for the system of interest is given by

$$H:= \sum\limits_{i=1}^N h_i \quad , \tag{1}$$

where $$h_i:= \mathbb I \otimes \ldots \otimes h \otimes \ldots \otimes \mathbb I$$ ($$h$$ is at the $$i$$-th position and the total number of factors is $$N$$).

Let $$\{|k\rangle\}_{k \in \mathbb N_0} \subset \mathcal H_1$$ denote the eigenbasis of $$h$$ with $$h |k\rangle = \epsilon_k |k\rangle$$ and

$$|k_1,k_2,\ldots , k_N\rangle := \sqrt{\frac{N!}{n_{k_1}!n_{k_2}!\cdots n_{k_N}!}} S \left(|{k_1}\rangle \otimes |{k_2}\rangle \otimes \ldots \otimes |{k_N}\rangle \right) \quad , \tag{2}$$ with the symmetrization operator $$S:=\frac{1}{N!} \sum\limits_p P$$ and the permutation operator for $$N$$ particles $$P$$. Then a basis in $$\mathcal H$$ is given by (we employ some standard ordering): $$\{|k_1,k_2,\ldots ,k_N\rangle\}_{k_1\leq k_2 \leq \ldots \leq k_N} \quad . \tag{3}$$

It holds that

$$\mathbb I_{\mathcal H} = \sum\limits_{k_1=0}^{\infty}\sum\limits_{k_2=k_1}^{\infty} \cdots \sum\limits_{k_N=k_{N-1}}^\infty |k_1,k_2,\ldots, k_N\rangle \langle k_1,k_2,\ldots ,k_N| \tag{4}$$

and $$\langle k_1,k_2,\ldots ,k_N|k_1^\prime,k_2^\prime,\ldots k^\prime_N\rangle = \delta_{k_1,k^\prime_1}\, \delta_{k_2,k^\prime_2} \cdots\, \delta_{k_N,k^\prime_N} \quad . \tag{5}$$

Moreover, we have $$H|k_1,k_2,\ldots, k_N\rangle = \left(\epsilon_{k_1} +\epsilon_{k_2} + \ldots +\epsilon_{k_N} \right)|k_1,k_2,\ldots, k_N\rangle \quad . \tag{6}$$

The partition function in equation $$(\mathrm{Z})$$ can now be simplified. Indeed, by making use of $$(4)$$-$$(6)$$ we find

\begin{align} \mathcal Z^{(N)}_\beta (H) &= \sum\limits_{k_1=0}^{\infty}\sum\limits_{k_2=k_1}^{\infty} \cdots \sum\limits_{k_N=k_{N-1}}^\infty \exp-\beta\,\left(\epsilon_{k_1} +\epsilon_{k_2} + \ldots +\epsilon_{k_N}\right)\\ &= \sum\limits_{k_1=0}^{\infty} e^{-\beta\epsilon_{k_1}}\sum\limits_{k_2=k_1}^{\infty} e^{-\beta\epsilon_{k_2}}\, \cdots \sum\limits_{k_N=k_{N-1}}^\infty e^{-\beta\epsilon_{k_N}} \quad . \tag{7} \end{align}

For a quantum harmonic oscillator, we have $$\epsilon_k := \hbar \omega \left(k +\frac{1}{2}\right)$$ for some $$\omega >0$$. By defining $$q:= e^{-\beta \hbar \omega}$$ we can hence write

$$\mathcal Z^{(N)}_\beta(H) = q^{N/2} \sum\limits_{k_1=0}^{\infty} q^{k_1}\sum\limits_{k_2=k_1}^{\infty} q^{k_2} \cdots \sum\limits_{k_N=k_{N-1}}^\infty q^{k_N}\quad . \tag{8}$$

To proceed, note that here $$0<|q|<1$$ and thus $$\sum\limits_{k=m}^\infty q^k = \frac{q^m}{1-q} \tag{10} \quad .$$ Consequently, we find \begin{align} \mathcal Z^{(N)}_\beta (H) &= q^{N/2} \sum\limits_{k_1=0}^{\infty} q^{k_1}\sum\limits_{k_2=k_1}^{\infty} q^{k_2} \cdots \sum\limits_{k_{N-1}=k_{N-2}}^\infty q^{k_{N-1}}\sum\limits_{k_N=k_{N-1}}^\infty q^{k_N}\\ &= q^{N/2} \sum\limits_{k_1=0}^{\infty} q^{k_1}\sum\limits_{k_2=k_1}^{\infty} q^{k_2} \cdots \sum\limits_{k_{N-1}=k_{N-2}}^\infty q^{k_{N-1}} \frac{q^{k_{N-1}}}{1-q} \tag{11}\\ &= \frac{q^{N/2}}{1-q} \sum\limits_{k_1=0}^{\infty} q^{k_1}\sum\limits_{k_2=k_1}^{\infty} q^{k_2} \cdots \sum\limits_{k_{N-1}=k_{N-2}}^\infty (q^2)^{k_{N-1}} \\&= \ldots \end{align} which eventually yields

$$\mathcal Z^{(N)}_\beta(H) = q^{N/2} \prod\limits_{i=1}^N \frac{1}{1-q^i} \quad . \tag{12}$$ The reader may formalize this argument as an exercise.

The result coincides with equation $$(14)$$ of Investigations on finite ideal quantum gases and $$(9)$$ of Statistical mechanics and the partitions of numbers.

The second linked paper shows/ states that

$$\mathcal Z^{(N)}_\beta(H) \underbrace{\longrightarrow}_{N^2 \mu\to 0} \mathscr Z_\beta^{(N)}(H) \quad , \tag{13}$$

with $$\mu:=\beta\hbar\omega$$. Alternatively, we can use the following relation, which may be proven by induction:

$$N!= \lim\limits_{\mu \to 0} \frac{\prod\limits_{i=1}^N 1-e^{-\mu i}}{(1-e^{-\mu})^N} = \lim\limits_{\mu \to 0} \frac{\mathcal Z_\beta^{(1)}(h)^N}{\mathcal Z^{(N)}_\beta(H)}\quad. \tag{14}$$

This shows that for a sufficiently small $$\mu^*(N) \gtrsim 0$$ it holds that $$\mathscr Z_{\beta^*}^{(N)}(H) \approx \mathcal Z_{\beta^*}^{(1)}(h)^N \, \frac{\mathcal Z^{(N)}_{\beta^*}(H) }{ \mathcal Z_{\beta^*}^{(1)}(h)^N} = \mathcal Z^{(N)}_{\beta^*}(H) \quad . \tag{15}$$

The reader is encouraged to test this approximation for some finite $$\mu$$ as a function of $$N$$. We find that the larger $$N$$ is, the smaller $$\beta$$, i.e. the higher the temperature, must be, in order to be a good approximation. This is in a qualitatively agreement with the limit taken in $$(13)$$.

All in all, in the high temperature limit, the partition function $$\mathscr Z$$ should be a good approximation for $$\mathcal Z$$, at least in our example.

$$Z'(h,N,\beta) = \mathcal Z^{(1)}_\beta(h)^N$$ is the partition function for $$N$$ distinguishable particles. The Hilbert space for this system is just the $$n$$-fold tensor product of the $$1$$-particle states.
When, the particle are indistinguishable, we have to make sure we are not over-counting states. Dividing the whole partition function by $$N!$$ would be correct if every state was overcounted $$N!$$ times in $$Z'$$ above, or equivalently in the Hilbert space of symmetric states.
But this is not correct : while a generic states is indeed overcounted by a factor of $$N!$$, some states are not. For example, if all the particles are in the same state, there is no overcounting at all.
• What do you think will happen if one uses continuous coordinates in calculating the partition function (i.e. integrate) (of course not the QHO as here)? Then the set of points where some coordinates /labels coincide are of Lebesgue measure zero, no? So the $1/N!$ should be fine?! Apr 10 at 21:53
• To compute the partition function this way, you actually need to perform a path integral over symmetric wave-function (rather than regular integration). In this context, it is not clear that the set of wave-functions who shouldn't get a $1/N!$ factor has measure zero (whatever that means for path integrals). Apr 11 at 7:22