In my previous question QM Continuity Equation: Many-Body Version for Density Operator? a board member showed me the proof, that the Continuity Equation for single-particle QM can be directly translated to the time-evolution for the number operator $\hat{\psi}^{\dagger}\hat{\psi}$ in the bosonic case.
Now i am curious about what might happen in the fermionic case. To make things easier, i wanted to start with an analogon of the Schrödinger equation, ie a time evolution barely for $\hat{\psi}$ instead of the more difficult one for $\hat{\psi}^{\dagger}\hat{\psi}$.
In the bosonic case, i can easily recover something that looks like the Schrödinger equation.
In the fermionic case, i proceed similiarly, but i have to use the correct commutator for fermions:
The anti-commutator is $ [\hat{\psi}\left(r_{1}\right),\hat{\psi}^{\dagger}\left(r_{2}\right)]_{+}=\hat{\psi}\left(r_{1}\right)\hat{\psi}^{\dagger}\left(r_{2}\right)+\hat{\psi}^{\dagger}\left(r_{2}\right)\hat{\psi}\left(r_{1}\right)=\delta_{r_{1},r_{2}} $
and from this follows $ [\hat{\psi}\left(r_{1}\right),\hat{\psi}^{\dagger}\left(r_{2}\right)]=\delta_{r_{1},r_{2}}-2\hat{\psi}^{\dagger}\left(r_{2}\right)\hat{\psi}\left(r_{1}\right) $
... So the derivation now looks like this: $ \frac{\partial}{\partial t}\hat{\psi}\left(r'\right)=\frac{1}{i\hbar}\left[\hat{\psi}\left(r'\right),\hat{H}\right] $
$ =\frac{i\hbar}{2m}\int\left[\hat{\psi}\left(r'\right),\hat{\psi}^{\dagger}\left(r\right)\triangle\hat{\psi}\left(r\right)\right]dr $
$ =\frac{i\hbar}{2m}\int\left(\left[\hat{\psi}\left(r'\right),\hat{\psi}^{\dagger}\left(r\right)\right]\triangle\hat{\psi}\left(r\right)+\hat{\psi}^{\dagger}\left(r\right)\left[\hat{\psi}\left(r'\right),\triangle\hat{\psi}\left(r\right)\right]\right)dr $
The second term is zero, so that we get
$ =\frac{i\hbar}{2m}\int\left(\delta_{r,r'}-2\hat{\psi}^{\dagger}\left(r'\right)\hat{\psi}\left(r\right)\right)\triangle\hat{\psi}\left(r\right)dr $
and finally
$ \Rightarrow\frac{\partial}{\partial t}\hat{\psi}\left(r'\right)=\frac{i\hbar}{2m}\triangle\hat{\psi}\left(r'\right)-\frac{i\hbar}{m}\int\hat{\psi}^{\dagger}\left(r'\right)\hat{\psi}\left(r\right)\triangle\hat{\psi}\left(r\right)dr $
I guess it is not surprising, that the equation from the bosonic case is not recovered here. After all, fermions must obey the Pauli exlusion principle (PEP) - which is probably enforced by the second term.
However i wonder about the following question: Can the last term be somehow resolved to yield something purely "local", ie without integrals over the whole of space? For example, naively i would have expected, that a simple term like
$ (1-\hat{\psi}^{\dagger}(r')\hat{\psi}(r')) $
mulitplied onto the first term (or onto the bosonic right-hand-side, which is the same) would also have done the job of ensuring the PEP.
