Can the operator field Dirac equation be expressed as Heisenberg's equation? The Dirac equation of the operator spinor field is:
$$(i\gamma ^{\mu}\partial _{\mu} -m)\psi =0$$
where $\psi$ is interpreted to be a quantum field.
I'm wondering, can this be derived from the Heisenberg equation?
$$\frac{d\psi}{dt}=\frac{-i}{h}[\psi, H]$$.
I'm in doubt because the above equation has a commutator while the Dirac quantisation involves anti-commutators.
 A: Suppose you want to change $\psi$ by a "tiny" amount $\delta \psi$. This $\delta \psi$ has to fulfill the same anti-commutation relations as the $\psi$, in particular
$$
\{ \delta \psi, \psi \} = 0
$$
You can generate such a $\delta \psi$ by use of the commutator (not the anticommutator): Suppose for example that
$$
H = \left(\delta \psi\right) \pi
$$
with
$$\{\psi, \pi \} = i
$$
$\psi$ and $\pi$ satisfy the usual anticommutation relations!
Then we can compute:
$$
-i[\psi, H] = -i[\psi, \left(\delta \psi\right) \pi] = -i \{\psi, \delta \psi \} \pi - -i\delta \psi \{\psi, \pi \} = \delta \psi
$$
Here we use the properties of commutators and anticommutators (which hold in general). Your Hamiltonian (or any other observable that you will use to generate transformations) will in general look different (or involve sums or integrals over more degrees of freedom), but this calculation should illustrate that one can generate anticommuting operators, using the usual commutator from the Heisenberg equation.
