# Particle density operator in Fetter and Walecka

In Fetter and Walecka page 20 they say in order to find the second quantized operator $$\hat{J}$$ of the first quantized one body operator $$J$$ you need to calculate $$\hat{J} = \int\mathrm{d}^3x\,\hat{\psi}^{\dagger}(\mathbf{x}) J(\mathbf{x}) \hat{\psi}(\mathbf{x})$$ where $$\hat{\psi}^{(\dagger)}$$ are the field operators.

They then define the number density operator $$n(\mathbf{x}) = \sum_{i=1}^N \delta(\mathbf{x}-\mathbf{x}_i)$$. Now it's not stated what $$\mathbf{x}_i$$ is but presumably it's the location of the $$i$$th particle. Does this statement even make sense in QM/QFT? Since a particle/wavefunction doesn't have a well defined position, how do you define $$\mathbf{x}_i$$? Is it the expectation value $$\langle \psi | \mathbf{\hat{x}}|\psi\rangle$$?

My main question is about the second quantised form however. They clearly state that $$\mathbf{x}$$ is the argument of n, but then compute $$\hat{n}(\mathbf{x}) = \hat{\psi}^{\dagger}(\mathbf{x})\hat{\psi}(\mathbf{x})$$ How does integrating over the delta functions yield $$\mathbf{x}$$? I would think that the 2nd quantised form should be \begin{align*} \hat{n}(\mathbf{x}) &= \int\mathrm{d}^3x\, \hat{\psi}^{\dagger}(\mathbf{x}) n(\mathbf{x})\hat{\psi}(\mathbf{x}) = \sum_{i=1}^N \int\mathrm{d}^3x\,\hat{\psi}^{\dagger}(\mathbf{x})\hat{\psi}(\mathbf{x})\delta(\mathbf{x}-\mathbf{x}_i) \\ &= \sum_{i=1}^N \hat{\psi}^{\dagger}(\mathbf{x}_i) \hat{\psi}(\mathbf{x}_i) \end{align*}

I don't understand why in this answer they simply replace $$\mathbf{x}_\alpha$$ by the integration variable instead of replacing $$\mathbf{x}$$ which is the variable of $$n$$.

You know that in second quantisation you can write a 1-particle operator $$\hat A = \sum_i^N \hat a(i)$$ as $$\hat A = \sum_i^N \hat a(i) = \sum_{rr'} \hat{c}^{\dagger}_r \langle r |\hat a|r'\rangle\hat{c}_{r'}$$
so $$\hat n(\mathbf{x})= \sum_{rr'} \hat{c}^{\dagger}_r \langle r |\delta_3(\mathbf{x}-\hat{\mathbf{x}})|r'\rangle\hat{c}_{r'}$$
If you choose the base $$|\mathbf{x}\rangle$$ you need to switch to the integral and the field operators:
$$\int\mathrm{d}^3x'\mathrm{d}^3x''\,\hat{\psi}^{\dagger}(\mathbf{x}')\langle \mathbf{x}'|\delta_3(\mathbf{x}-\hat{\mathbf{x}})|\mathbf{x}''\rangle \hat{\psi}(\mathbf{x}'')$$ Now let's focus on the matrix element: you need to apply the delta to the ket, getting $$\delta_3(\mathbf{x}-\mathbf{x}'')$$, then you do the internal product, getting a $$\delta_3(\mathbf{x}'-\mathbf{x}'')$$. Now using the properties of the deltas you get to the final result $$\hat{n}(\mathbf{x}) = \hat{\psi}^{\dagger}(\mathbf{x})\hat{\psi}(\mathbf{x})$$