I often see that the density operator in $1^{st}$ quantization is defined to be: $$ \hat n(\vec r)=\sum_{i=1}^N \delta(r-\hat r_i)$$

In canonical quantization it is given by $$ \hat n(\vec r)=\hat \psi^\dagger(\vec r)\hat\psi(\vec r)$$ but shouldn't there be a factor of $N$ multiplying since in the definition of the expectation value of $\hat n$, we act with $\delta(r-\hat r_i)$ on the ket $N$ times?


It may be tricky to explain why it is so (because it essentially means explaining the second quantization), but note that this is the case for all the single-particle operators! For example, the kinetic energy of a collection of identical particles in the first quantization is $$\hat{K} = \sum_i\frac{\hat{p}_i^2}{2m},$$ and becomes in the second quantization $$\mathcal{\hat{K}} = \int dx \psi^\dagger(x)\frac{\hat{p}^2}{2m}\psi(x).$$


It depends on the definition of state operators $\hat\psi $ as the number $N$ may be implicit. Since in second quantization the expectation values are taken with respect to the vacuum state $|0 \rangle$ or more formally the Green's function. The state operator $\hat{\psi}$, therefore, plausible to expand in terms of $N$ correlation functions, i.e., $\phi(x_1)\phi(x_2)...\phi(x_n)$, with this at least we acted by $N$ field operators.

  • $\begingroup$ Almost nothing of what you wrote makes sense $\endgroup$
    – lcv
    Mar 27 '20 at 23:31

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