# Derivation for the expectation value of particle density for a pairwise interactions between particles

my question is why is $<\hat{n}(\vec{r})>=n$

I have the Hamiltonian $H_N= \sum_{i}^{N} \frac{P_i^2}{2m}+U(\vec{R_1},\vec{R_2},..,\vec{R_N})$ where $U(\vec{R_1},\vec{R_2},..,\vec{R_N})= \frac{1}{2!}\sum_{i\neq j}^{N} U_2(\vec{R_i},\vec{R_j})$ is a pairwise interaction potential.

By using this you can get the partition function $Z=\sum_{N=0}^{ \infty} \frac{1}{N!}(\frac{e^{\beta \mu}}{(\Lambda_T)^{d}})^{N} (\prod_{i=1}^N[\int d^{d} \vec{R_i}])e^{- \beta U(\vec{R_1},\vec{R_2},..,\vec{R_N})}$

From her if you take the density at position $\vec{r}$ to be $\hat{n}(\vec{r})= \sum_{i=1}^N \delta^{d}(\vec{r}-\vec{R_i})$ you are supposed to get $<\hat{n}(\vec{r})>=n$. When I plug in $\hat{n}(\vec{r})$ I can't seem to derive this result.

When I try I get $<\hat{n}(\vec{r})>=\frac{1}{\sum_{N=0}^{ \infty} \frac{1}{N!}(\frac{e^{\beta \mu}}{(\Lambda_T)^{d}})^{N} (\prod_{i=1}^N[\int d^{d} \vec{R_i}])e^{- \beta U(\vec{R_1},\vec{R_2},..,\vec{R_N})}}*\sum_{N=0}^{ \infty} \frac{1}{N!}(\frac{e^{\beta \mu}}{(\Lambda_T)^{d}})^{N} (\prod_{i=1}^N[\int d^{d} \vec{R_i}])\sum_{i=1}^N \delta^{d}(\vec{r}-\vec{R_i})e^{- \beta U(\vec{R_1},\vec{R_2},..,\vec{R_N})} \rightarrow\frac{1}{\sum_{N=0}^{ \infty} \frac{1}{N!}(\frac{e^{\beta \mu}}{(\Lambda_T)^{d}})^{N} (\prod_{i=1}^N[\int d^{d} \vec{R_i}])e^{- \beta U(\vec{R_1},\vec{R_2},..,\vec{R_N})}}*\sum_{N=0}^{ \infty} \frac{1}{N!}(\frac{e^{\beta \mu}}{(\Lambda_T)^{d}})^{N} e^{- \beta U(\vec{r},\vec{r},..,\vec{r})}$

This is the where I am stuck. I can't seem to figure out where to go after integrating the delta function to get $<\hat{n}(\vec{r})>=n$

Thanks.

my question is why is $<\hat{n}(\vec{r})>=n$
Presumably you are defining $n=N/V$.
In that case, from the definition: $$<\hat n(\vec r)>=\int d^3x_1d^3x_2\ldots d^3x_N\Psi^*(\vec x_1,... \vec x_N)\sum_{i}\delta(\vec x_i-\vec r)\Psi(\vec x_1,... \vec x_N)$$ $$=N\int d^3x_2\ldots d^3x_N |\Psi|^2(\vec r,x_2,\ldots,x_N)$$
On the other hand, since $\Psi$ is normalized: $$\int d^3r\left(d^3x_2\ldots d^3x_N |\Psi|^2(r,x_2,\ldots,x_N)\right)=1$$
But, for a homogeneous system, the part in parenthesis above must be independent of $\vec r$. Just as $<\hat n(r)>$ is actually independent of r. I.e., $$\int d^3r\left(d^3x_2\ldots d^3x_N |\Psi|^2(r,x_2,\ldots,x_N)\right)=1$$ $$V\int\left(d^3x_2\ldots d^3x_N |\Psi|^2(r,x_2,\ldots,x_N)\right)=1$$ $$\int\left(d^3x_2\ldots d^3x_N |\Psi|^2(r,x_2,\ldots,x_N)\right)=1/V$$
Therefore, $$<\hat n(r)>=N\int d^3x_2\ldots d^3x_N |\Psi|^2(\vec r,x_2,\ldots,x_N)$$ $$=N\frac{1}{V}$$