# 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 $$\hat{J} = \int\mathrm{d}^3x\,\hat{\psi}^{\dagger}(\mathbf{x}) J(\mathbf{x}) \hat{\psi}(\mathbf{x}) \tag{2.6}$$ where $$\hat{\psi}^{(\dagger)}$$ are the field operators.

They then define the number density operator $$n(\mathbf{x}) = \sum_{i=1}^N \delta(\mathbf{x}-\mathbf{x}_i)\tag{2.7}.$$ 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 $$n$$.

Let me try to show it step by step.

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 $$\hat{n}(\mathbf{x}) = \hat{\psi}^{\dagger}(\mathbf{x})\hat{\psi}(\mathbf{x})$$

I believe the question remains as to why $$\sum_{i=1}^N \hat{a}(i) = \sum_{rr'}\hat{c}_r^\dagger \langle r| a|r'\rangle \hat{c}_{r'}$$ holds. It is true that in first quantization, there is always the sum that goes as $$\sum_{i=1}^N$$. The crucial observation is mentioned in the discussed Fetter and Walecka book, p9. Instead of counting particles, one counts the number of times a particle has an eigenvalue described by $$i\equiv E_i$$. In this case, the sum is replaced by a sum like $$\sum_{i}\rightarrow\sum_{E_i}$$ and thus $$\sum_{k=1}^N \sum_W \langle E_k|A|W\rangle \propto \sum_{E} \sum_W \langle E|A|W\rangle n_E,$$ where $$n_E$$ countst the number of particles of a certain energy $$E$$. This is almost what one wants in second quantization, as now the ocupattion number $$n_E$$ is explicit in the sum. With a little bit of more work described in the book (which I will not reproduce here), one see that the equality above holds, that is, in second quantization, one is allowed to replace the sum over particles with the matrix elements as writen in the right side above.

You are completely right to observe that the book as an abuse of notation and lacks clarity in the explanation.