# Summing over quantum states

For a system of $N$ identical particles we deal in quantum mechanics with wave functions $\langle \{\mathbf{r}_i \} \mid \Psi \rangle=\Psi(\mathbf{r}_1,\dots,\mathbf{r}_N)$ from which determine the probabilities of the particles being at positions $\{\mathbf{r}_1,\dots,\mathbf{r}_N \}$, $$\langle \Psi \mid \Psi \rangle=\langle \Psi \mid \sum_{\mathbf{r}_i}\mid \{\mathbf{r}_i \} \rangle\langle \{\mathbf{r}_i \} \mid \Psi \rangle=\sum_{\mathbf{r}_i}|\Psi(\mathbf{r}_1,\dots,\mathbf{r}_N)|^2=1$$ where $\displaystyle \sum_{\mathbf{r}_i} = \int \prod_{i=1}^Nd^3r_i$.

I cannot see why $\displaystyle \sum_{\mathbf{r}_i} = \int \prod_{i=1}^Nd^3r_i$ holds. Surely we should integrate over the momentum as well? I think this because earlier in my notes it states that:

All microstates should in principle be included in the partition sum over states irrespective of the macroscopic properties of the system

Also why dont we have a factor of $\frac{1}{N!}$ to represent the indistinguishability of particles?

• That has nothing to do with statistical mechanics. $\sum_r \langle \vec r\vert \psi \rangle$ is always $\int \mathrm{d}^3 r \psi(r)$ in the position representation. How would you even integrate $\psi$ over momenta if it doesn't depend on momenta? – ACuriousMind Apr 21 '15 at 9:46
• Apologies, this is from my statistical mechanics course. See I would have thought we would have to integrate over mometum anyway, just the integrand wouldnt effect the integration – Permian Apr 21 '15 at 9:52

I think $$\displaystyle\sum_{\mathbf{r}_i}=\int\prod_{i=1}^Nd^3r_i$$ here is just a notation of your teacher, you don't need to be fussy about it.
And to answer the question about the factor $$\dfrac{1}{N!}$$, you can refer to my answer here for another question https://physics.stackexchange.com/a/462446/223370