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I would want to understand why the density operator in second quantization takes the form:

$$\rho_\sigma(\mathbf{r})=\Psi_\sigma^\dagger(\mathbf{r})\Psi_\sigma(\mathbf{r})?$$

Is this a definition or can we derive it from some formula?

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You can find this by noting the the photon potential $A_\mu$ couples to the electromagnetic current $J_\mu$ in the form

$\mathcal{L}_{int} = A^\mu J_\mu. $

Where $J_\mu$ obeys the continuity equation $\partial_\mu J^\mu = 0 $. This prompts us to consider the zero component of $J_\mu$ as the charge density. So we have

$ J^\mu = \bar{\psi} \gamma^\mu \psi $

and so

$ J^0= \bar{\psi} \gamma^0\psi = \psi^\dagger \gamma^0 \gamma^0 \psi = \psi^\dagger \psi $

where we have used $ {\gamma^0}^2 = 1$.

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+1 Solid answer. –  Killercam Apr 4 '13 at 8:35
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