In Pathria and Beale Statistical Mechanics section 5.5, the book tries to compute the Partition function of a system of noninteracting, indistinguishable particles confined to a cubical box of volume $V$. The book claims that the density matrix is given by $$\langle \boldsymbol{r}_1,...,\boldsymbol{r}_N|\hat{\rho}|\boldsymbol{r}_1',...,\boldsymbol{r}_N'\rangle$$ and attempts to compute this quantity by inserting a $\sum_{\boldsymbol{k}}|\boldsymbol{k}\rangle\langle\boldsymbol{k}|$ between $\hat{\rho}$ and $|\boldsymbol{r}_1',...,\boldsymbol{r}_N'\rangle$. Here $\boldsymbol{k}$ forms a complete orthonormal basis of the $N$ particle Fock spaces. The book then claims that the partition function is given by $$\int \langle \boldsymbol{r}_1,...,\boldsymbol{r}_N|\hat{\rho}|\boldsymbol{r}_1,...,\boldsymbol{r}_N\rangle\,d^{3N}r.$$

My question is: what is the Hilbert space that $\hat{\rho}$ acts on? My understanding is that the Hilbert space should be the $N$ particle Fock space, but then the question is that when taking the trace to compute the partition function, we can't do $d^{3N}r$ because $d^{3N}r$ includes coordinates that are not symmetrized/antisymmetrized.

However, if the Hilbert space is just the product space, then $|\boldsymbol{k}\rangle$ would not be a complete basis of the Hilbert space.

Add: it appears that Pathria and Beale is just being sloppy. Greiner's thermodynamics and statistical mechanics chapters 10 and 11 discuss this in greater detail, although even Greiner takes some shortcut with the Boson case. After comparing with Greiner, Pathria and Beale definitely should have used the properly (anti)symmetrized position states. It looks like just a coincidence that Pathria and Beale obtains the correct partition function as in Greiner using this sloppy approach.

  • $\begingroup$ I'm sorry but I don't see why $d^{3N}r$ being neither symmetric nor antisymmetric can be a problem. Actually, you can always rewrite volume differentials with wedge products that have antisymmetric properties, but that'd be overcomplicating the matter. Could you clarify a little bit more? $\endgroup$
    – Vicky
    Commented Apr 7, 2022 at 1:42
  • $\begingroup$ My question is if the density operator acts on the Fock space, then when taking the trace, we need to take an orthonormal basis $|\boldsymbol{k}\rangle$ of the Fock space then sum $\sum_{k}\langle k|\hat{\rho}| k\rangle$. $|r_1,...,r_n\rangle$ is not a basis of the Fock space. $\endgroup$ Commented Apr 7, 2022 at 1:49
  • $\begingroup$ The states $\left|r_1,\ldots r_N \right\rangle$ are (anti)symmetrized so the non-(anti)symmetrized contributions from the integrals will be always $0$ $\endgroup$ Commented Apr 7, 2022 at 6:53

1 Answer 1


There's no problem with using a non-symmetric basis for identical particles, only the state must be invariant under particle exchanges. Since the trace of an operator, which is what the partition function is, is basis-independent you can use any basis you like including a non-symmetric one.

The Hilbert space that we're talking about here is the (anti)symmetric product space of $N$ non-relativistic particles

  • $\begingroup$ I would disagree. You have to use a properly (anti)symmetrized basis. In any case Greiner's book answers my confusion nicely. $\endgroup$ Commented Apr 7, 2022 at 19:38

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