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Suppose we have a Hamiltonian containing some interaction term $$V = \sum _{\sigma \sigma '\sigma ''\sigma '''}\iint d^3rd^3r'\hat{\psi }_{\sigma}^\dagger (\textbf{r})\hat{\psi }_{\sigma'}^\dagger (\textbf{r}')V_{\sigma \sigma' \sigma'' \sigma'''}(\textbf{r}-\textbf{r}')\hat{\psi }_{\sigma''} (\textbf{r}')\hat{\psi }_{\sigma'''} (\textbf{r})\tag{1}$$ and we wish to find the functional derivative $$\frac{\delta \hat{H}}{\delta \hat{\psi }_s (\textbf{r})}. \tag{2}$$ how do we go about this? The definition of the functional derivative is the coefficient of the $ \hat{\psi }_s (\textbf{r}) $ term in the integrand of the variation of the Hamiltonian. But here we have to integrals, meaning that we have two contributions, given by (dropping the hat but still referring to them as operators) $$ \delta V = \sum _{\sigma _1\sigma _2\sigma _3\sigma _4}\iint d^3rd^3r'\psi ^\dagger _1(r)\psi ^\dagger _2(r')V_{1234}(r-r')\Big[ \left( \psi _3(r') + \delta \psi _3 \right)\left( \psi _4(r) + \delta \psi _4 \right) - \psi _3(r')\psi _4(r) \Big] $$ $$ = \sum _{\sigma _1\sigma _2\sigma _3\sigma _4}\iint d^3rd^3r'\psi ^\dagger _1(r)\psi ^\dagger _2(r')V_{1234}(r-r')\Big[ \delta \psi _3\psi _4(r) + \psi _3(r')\delta \psi _4 + \delta \psi_3\delta \psi _4 \Big]. \tag{3}$$ The next move is to use anti-commutativity to replace $\psi _3(r')\delta \psi _4$ with $-\delta \psi _4\psi _3(r')$ from which we can read off the coefficients. It also gives the expected result when comparing it with the commutator via $$\left[ \hat{\psi }_{s }^\dagger(\textbf{r}) ,\hat{H} \right] = - \frac{\delta \hat{H}}{\delta \hat{\psi }_s (\textbf{r})}.\tag{4}$$

But I'm having a hard time rigorously justifying that last step. What is the meaning of a small variation in the field operator $\delta \hat{\psi }_i (\rho)$? Is it a well defined mathematical/physical entity? Why does it anti-commute with the regular annihilation field operator?

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  1. Note that in supermathematics differentiation comes in two versions: Left (right) differentiation, i.e. acting from left (right), respectively. Functional/variational differentiation similarly comes in two versions: $$ \int \!d^4x ~ \delta \phi(x)~ \frac{\delta_L F}{\delta \phi(x)} ~=~\delta F~=~ \int \!d^4x ~~ \frac{\delta_R F}{\delta \phi(x)}~ \delta \phi(x). \tag{A}$$ Since we only consider infinitesimal variations $\delta F$, it is enough to truncate Taylor series in $\delta \phi$ at linear order.

  2. The left and right derivatives are equal up to a sign factor $$ \frac{\delta_L F}{\delta \phi(x)} ~=~(-1)^{|F|(|\phi|+1)} \frac{\delta_R F}{\delta \phi(x)}.\tag{B} $$ Here $|\cdot|$ denote the Grassmann-parity. We should stress that the infinitesimal variation $\delta \phi$ of a field $\phi$ carries the same Grassmann-parity, i.e. $|\delta \phi|=|\phi|$.

  3. Hence with the canonical anticommutator relation (CAR) $$ \{\hat{\psi}^{\dagger}_{\alpha}({\bf x}), \hat{\psi}_{\beta}({\bf y})\}_{+}~=~\hbar\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1} ,\tag{C}$$ we find OP's eq. (4): $$\frac{1}{\hbar}[\hat{\psi}^{\dagger}_{\alpha}({\bf x}),\hat{H}] ~=~ \frac{\delta_L \hat{H}}{\delta \hat{\psi}_{\alpha}({\bf x})} ~=~ -\frac{\delta_R \hat{H}}{\delta \hat{\psi}_{\alpha}({\bf x})}.\tag{D}$$

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