# Density, density operator, and number operator in quantum mechanics

Density is the number of particles per unit volume. But in quantum mechanics, the density operator is defined as $$\hat{\rho}(r) =\Psi^\dagger(r) \Psi(r)$$ here $$\Psi(r)$$ is the field operator. The expectation value of this operator is $$\langle \hat{\rho}(r)\rangle=\rho(r)$$ which is number of particles at position $$r$$.

1. Why is it not divided by volume?
2. Is the number operator $$\hat{N}=\Psi^\dagger\Psi$$ and density operator the same things?
• Have you tried to compute the integral $\displaystyle \int \mathrm d x \, \hat\rho (x)$? With that in mind, you should be able to answer your questions. Your question $2$ is not well-defined, tho: What is $\Psi^\dagger \Psi$? Commented Nov 22, 2021 at 8:49
• "number of particles at position $r$" does not make much sense, since $r$ is just a point. The number is an integral (non local) quantity, density is local. Commented Nov 22, 2021 at 9:09

1: $$\rho(r)$$ describes a local density of particle which can in some limit (lattice model) be understood as the number of particle on site "r". In fact the division by the volume is often already contained into the $$\Psi(r)$$ due to the normalization implying that $$\Psi$$ scales with $$\frac{1}{\sqrt{V}}$$
$$\int_V dr\langle\Psi^\dagger\Psi\rangle=N=\langle \hat{N}\rangle$$
$$\hat{N}=\int_V dr \hat{\rho}(r)=\int_V dr \Psi^\dagger(r)\Psi(r)$$