# Boson or fermion density matrix in path-integral Monte Carlo

In path-integral Monte Carlo literature the thermal density matrix of $$N$$-particle boson or fermion system is written as symmetrized or antisymmetrized sum $$\rho_\mathrm{B,F}(R,R',\beta)=\frac1{N!}\sum_P(\pm1)^P\rho(R,PR',\beta),$$ where $$P$$ are all permutations of the $$N$$-particle coordinate vector, and $$\rho$$ is the density matrix of distinguishable particles. The example of such formula for bosons is Eq. (2.27) in D.M. Ceperley, Rev. Mod. Phys. 67, 279 (1995).

I think the $$1/N!$$ factor should not be here, as evident from the simplest two-particle case. Consider, for example, the ''fermionic'' trace $$\mathrm{Tr}_\mathrm{F}\rho$$ of the two-particle density matrix over antisymmetrized states $$\Psi_{ij}(\mathbf{r}_1,\mathbf{r}_2)=\frac{\varphi_i(\mathbf{r}_1)\varphi_j(\mathbf{r}_2)-\varphi_j(\mathbf{r}_1)\varphi_i(\mathbf{r}_2)}{\sqrt2},$$ where $$\{\varphi_i(\mathbf{r})\}$$ is the complete set of single-particle states. Writing it as $$\mathrm{Tr}_\mathrm{F}\rho=\sum_{ij}\langle\Psi_{ij}|\rho|\Psi_{ij}\rangle\\=\sum_{ij} \int d\mathbf{r}_1d\mathbf{r}_2d\mathbf{r}_1'd\mathbf{r}_2'\:\frac{\varphi_i^*(\mathbf{r}_1')\varphi_j^*(\mathbf{r}_2')-\varphi_j^*(\mathbf{r}_1')\varphi_i^*(\mathbf{r}_2')}{\sqrt2}\\ \times\langle\mathbf{r}_1'\mathbf{r}_2'|\rho|\mathbf{r}_1\mathbf{r}_2\rangle \frac{\varphi_i(\mathbf{r}_1)\varphi_j(\mathbf{r}_2)-\varphi_j(\mathbf{r}_1)\varphi_i(\mathbf{r}_2)}{\sqrt2}$$ and taking into account the completeness condition $$\sum_i\varphi_i^*(\mathbf{r}')\varphi_i^*(\mathbf{r})=\delta(\mathbf{r}-\mathbf{r}'),$$ we arrive at $$\mathrm{Tr}_\mathrm{F}\rho=\int d\mathbf{r}_1d\mathbf{r}_2\:\{\langle\mathbf{r}_1\mathbf{r}_2|\rho|\mathbf{r}_1\mathbf{r}_2\rangle-\langle\mathbf{r}_1\mathbf{r}_2|\rho|\mathbf{r}_2\mathbf{r}_1\rangle\}.$$ Or we can write it as the trace of fermion density matrix $$\langle\mathbf{r}_1\mathbf{r}_2|\rho_\mathrm{F}|\mathbf{r}_1\mathbf{r}_2\rangle=\langle\mathbf{r}_1\mathbf{r}_2|\rho|\mathbf{r}_1\mathbf{r}_2\rangle-\langle\mathbf{r}_1\mathbf{r}_2|\rho|\mathbf{r}_2\mathbf{r}_1\rangle.$$

We see that the factor $$1/2$$ is absent here, because the completeness conditions in the calculations above resulted in 4 terms (2 direct and 2 exchange ones), so the normalization factor $$1/2$$ has been cancelled. So, at least for two particles, the correct formula should be $$\rho_\mathrm{B,F}(R,R',\beta)=\sum_P(\pm1)^P\rho(R,PR',\beta),$$ without the $$1/N!$$ factor. Perhaps there is some similarity with the Hartree-Fock approximation, where the Coulomb interaction energy of two particles is a sum of direct and exchange energies, and not half-sum of them.

Am i right or there is some error in my calculations?