Particle density operator in second quantization form The particle-density operator is given by 
$n(\mathbf{x})=\sum_{\alpha}\delta^{(3)}(\mathbf{x}-\mathbf{x}_{\alpha})$, then how to derive its representation in terms of creation and annihilation operators $n(\mathbf{x})=\psi^{\dagger}(\mathbf{x})\psi(\mathbf{x})$?
 A: $\newcommand{\bx}{\mathbf{x}}
\newcommand{\psih}{\hat{\psi}}
$This is worked out on p. 20 of Fetter and Walecka. I'll add just a little extra detail here.
Your particle density operator $n(\bx)=\sum_\alpha\delta(\bx-\bx_\alpha)$ is in first-quantized form and $\hat{n}(\bx)=\psih^\dagger(\bx)\psih(\bx)$ is second-quantized.  
When a general one-body operator is written in first-quantized form $J=\sum_\alpha J(\bx_\alpha)$, the second-quantized form is:
\begin{align}
\hat{J} &\equiv \sum_{rs}\langle r|J|s\rangle \hat{c}_r^\dagger \hat{c}_s\\
 &= \sum_{rs} \int d\bx' \psi^*_r(\bx') J(\bx') \psi_s(\bx') \hat{c}_r^\dagger \hat{c}_s\\
 &= \int d\bx' \psih^\dagger(\bx') J(\bx') \psih(\bx')
\end{align}
The last equality follows from $\psih(\bx)\equiv \sum_s \psi_s(\bx) \hat{c}_s$.
In the case of the particle density operator:
\begin{align}
  \hat{n}(\bx) &= \int d\bx' \psih^\dagger(\bx') \delta(\bx-\bx') \psih(\bx')\\
   &= \psih^\dagger(\bx) \psih(\bx)
\end{align}
Please let me know if you'd like more detail on any of these steps. For somewhat relevant discussion on the relation between first and second-quantized operators, see my answer here: Second quantization. In that answer I only considered the two body interaction operator in detail. I could do the same for (simpler) one body operators if it'd be of help.
