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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?

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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.

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