这是一份 Imperial帝国理工大学 PHYS97005作业代写的成功案例
expectation value
$$
\varrho(x):=\frac{1}{N}\left\langle\Psi\left|\sum_{i=1}^{N} \delta\left(x_{i}-x\right)\right| \Psi\right\rangle .
$$
Specializing to $\Psi=\Psi^{(0)}$ the density is just the sum of the one-body densities for everyone of the states in the product,
$$
\varrho^{(0)}(x)=\frac{1}{N} \sum_{i=1}^{N}\left|\psi_{i}(x)\right|^{2} .
$$
Its integral over the whole three-dimensional space gives 1 ,
$$
\int \mathrm{d}^{3} x \rho(x)=1
$$
if all single-particle wave functions are normalized to unity.
PHYS97005 COURSE NOTES ：
In either case, with orthogonal or nonorthogonal $\psi_{1}$ and $\psi_{2}$, one verifies that $\int \mathrm{d}^{3} y \varrho(x, y)=\varrho(x)$. Second, if the orbital wave functions are not confined each to a finite domain, the presence of a second particle is felt under all circumstances. For example, assuming them to be plane waves,
$$
\left\langle x \mid \psi_{1}\right\rangle=c \mathrm{e}^{(\mathrm{i} / \hbar) p \cdot x}, \quad\left\langle y \mid \psi_{2}\right\rangle=c \mathrm{e}^{(\mathrm{i} / \hbar q) \cdot y} \text { with } c=(2 \pi \hbar)^{-3 / 2},
$$
the two-body correlation function (5.8) is found to be
$$
C(x, y)=-\frac{1}{2}{1+\cos ((q-p) \cdot(x-y))} .
$$