5
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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$.

$\endgroup$
0
3
$\begingroup$

Depending on the perspective, you can also directly motivate it from the general recipe how you construct or infer any operator in second quantization, given the operator in 'first quantization'. The density operator above is a single-particle operator and the many-body counterpart of the real-space density operator $\rho(x)=|x \rangle \langle x|$ (the dependence on $x$ is understood to be parameteric). The prescription for single-particle operators is simply to 'promote' $|x \rangle \mapsto \Psi^\dagger(x)$ and $\langle x| \mapsto \Psi(x)$. All symmetry requirements for identical bosons / fermions are inherently accounted for in this way.

Just another example for clarity: the position operator $\hat x=\int dx \; x \,|x \rangle \langle x| $ would then become $\hat X= \int dx \; x \,\Psi^\dagger(x) \Psi(x) $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.