# Schrödinger-like equation for destruction operator: Fermionic Case

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.

• You should be able to derive a Schrödinger-like equation for the field (and a continuity equation for the density) in the fermionic case. The end results will be identical to the bosonic case. You'll need the identity $[A,BC]=\{A,B\}C - B\{A,C\}$. Here I'm using curly braces to denote the anticommutator. Nov 26, 2016 at 21:20
• Thanks.. using the identity suggested by you, i can see that i get the same result as for Bosons... (with almost the same derivation as above). However i am a little bit surprised, that no modification is needed at all to ensure the PEP. So it seems as if the Schrödinger and Continuity Equations will not violate the PEP if the intial conditions did not violate it? Nov 27, 2016 at 21:49
• States representing multiple particles necessarily involve products of multiple field operators. In this case the anticommutation relations automatically enforce the PEP (and more generally the fermionic exchange antisymmetry which has consequences beyond just PEP). Since the Schrödinger equation - and as a consequence, the continuity equation - describe unitary dynamics, they are guaranteed to preserve the anticommutation rules and hence the PEP etc. Nov 28, 2016 at 5:59