# Ladder operators evolution for fermions

For the free Dirac field we have $$\psi(x) = \sum_s\int d\Omega_{m}\frac{1}{\sqrt{2}k_0}\left(b(\mathbf{k},s)u(\mathbf{k},s)e^{-ik\cdot x}+d^\dagger(\mathbf{k},s)v(\mathbf{k},s)e^{+ik\cdot x}\right),$$ and the hamiltonian $$H=\sum_{s}\int d{\Omega}_m\omega(\mathbf{k}) \left(b^\dagger(\mathbf{k},s)b(\mathbf{k},s)+d^\dagger(\mathbf{k},s)d(\mathbf{k},s)\right).$$ I wonder if there is a way of writing down the evolution for the creation and annihilation operators, even though they satisfy the canonical anticommutation relations $$\{a(\mathbf{k},s),a^\dagger(\mathbf{k}',s')\}=k_0\delta(\mathbf{k}-\mathbf{k}')\delta_{ss'}$$ and not the usual commutation relations we are used to while making the calculation of $$a_\alpha(t)=e^{iHt}a_\alpha(0)e^{-iHt}$$ for bosons, which have $H=\sum_{\alpha}\omega_\alpha a^{\dagger}_\alpha a_\alpha$and $[a_\alpha,a^\dagger_{\beta}]=\delta_{\alpha \beta}$.

In the boson case we had $[H,a_\alpha]=-a_{\alpha}$ and $[H,a^\dagger_\alpha]=+a^\dagger_\alpha$ hence $$a_\alpha(t)=e^{-i\omega_\alpha t}a_\alpha(0).$$ Here is it still correct to write $$b_\alpha(t)=e^{iHt}b_\alpha(0)e^{-iHt}?$$ How can I compute the needed commutators?

You can show that, even though the ladder operators satisfy anticommutation relations, the Hamiltonian still satisfies $$[H,b]=-b\quad\text{and}\quad[H,b^\dagger]=b^\dagger.$$ Simply use the fact that $$[xy,z]=x\{y,z\}-\{x,z\}y,$$ just using the definition of (anti-)commutators. So you will get the same time evolution for fermionic ladder operators as for bosonic ones.