# Trick with "functional" derivative to evaluate commutators between diagonal hamiltonian and creation fermionic operator

I found a theorem that states that if $$A$$ and $$B$$ are 2 endomorphism that satisfies $$[A,[A,B]]=[B,[A,B]]=0$$ then $$[A,F(B)]=[A,B]F'(B)=[A,B]\frac{\partial F(B)}{\partial B}$$.

Now i'm trying to apply this result using the creation and annihilation fermionics operators $$B=C_k^+$$ and $$A=C_k$$ and the simple diagonal hamiltonian $$F(\not C_k,C_k^+)=H=\sum_k \hbar \omega_k C_k^+C_k$$.

Now i check if my operators satisfies the hypothesis of the theorem and i get $$[A,[A,B]]=[C_k,[C_k,C_k^+]]=-2C_k$$ $$[B,[A,B]]=[C_k^+,[C_k,C_k^+]]=+2C_k^+$$ $$0\neq[A,[A,B]]\neq[B,[A,B]]\neq0$$,

However thinking at $$H$$ as a function of only $$C_k^+$$ and applying the theorem directly

$$[C_k,H]=[C_k,C_k^+]\frac{\partial F(\not C_k,C_k^+)}{\partial C_k^+}=\hbar\omega_kC_k$$,

that is the right result for the commutator, evaluated without using this theorem.

So how i can interpret this fact? Why this works? What i'm missing between the theorem and this application? In the calculation i think at $$H$$ as a function only of $$C_k^+$$, is correct in this case?

I'd like to know what are the most general conditions that allows to use this simple trick to evaluate commutators. Or at least to find a theorem that rules this sort of things. Could anyone help me to understand?

• Hi user251479. Welcome to Phys.SE. That does not seem to be a functional derivative, just a partial derivative. Consider to edit accordingly. Mar 13, 2020 at 18:25
• My professor, during the lecture of many body theory named it "functional derivative" and performed the derivative only in $C_k^+$, i'm confused. Mar 13, 2020 at 18:30
• It could be that your professor implicitly uses DeWitt condensed notation, but that seems irrelevant. Mar 13, 2020 at 18:39
• The hypothesis of the theorem is violated, violently. Instead, it is the anticommutator which is central, here. You must prove new theorems. You must state that your fermion operators are nilpotent. Mar 13, 2020 at 21:35
• I'd like to know what are the most general conditions that allows to use this simple trick to evaluate commutators. Or at least to find a theorem that rules this sort of things. Could anyone help me to understand? Mar 14, 2020 at 13:30

The conditions, $$[A,[A,B]]=[B,[A,B]]=0$$, are needed for commuting operators to satisfy the identity that $$[A,F(B)]=[A,B]\frac{\partial F}{\partial B}$$. It turns out that these conditions are not needed if one works with anticommuting operators A and B (i.e., $$\{A,A\}=\{B,B\}=0$$) instead. To show this, we note that any function $$F(B)$$ constructed from an anticommuting variable has the following expansion, $$F(B)=c_0+c_1B$$ This is in contrast to the case of a function of a commuting variable, where we would instead have $$F(B)=\sum_{n=0}^{\infty}c_nB^n$$. For an anticommuting variable, the higher powers vanish. Using the above expansion, we can see that $$[A,F(B)]=c_1[A,B]$$. Implementing the derivative, $$\frac{\partial F}{\partial B}=c_1$$ gives us the same result, $$[A,F(B)]=c_1[A,B]$$. Thus, in the case of fermionic variables, the identity holds with no extra conditions.
For completeness, let me state the need for the conditions $$[A,[A,B]]=[B,[A,B]]=0$$ in the case of commuting variables. Try calculating the quantity, $$[A,B^n]$$ for general $$n$$. The above condition then allows me to write, $$[A,B^2]=[A,B]B+B[A,B]=2[A,B]B$$ Similarly, $$[A,B^3]=3[A,B]B^2$$, and in the general case, $$[A,B^n]=n[A,B]B^{n-1}$$. Then, we can derive the identity, $$[A,F(B)]=\sum_{n=0}^{\infty}c_n[A,B^n]=[A,B]\sum_{n=1}^{\infty}c_nB^{n-1}=[A,B]\frac{\partial F}{\partial B}$$ which is the statement of the identity you are using.
• the commutator does not vanish: $$[C_k,[C_k,C_k^+]]=[C_k,1-2C_k^+C_k]=2[C_k^+C_k,C_k]=-2C_k$$ using the fermion operator properties $[C_k^+C_k,C_k]=-C_k$ and $\lbrace C_k, C_k^+ \rbrace=1$ Mar 13, 2020 at 18:52
• the anticommutator is the identity $\lbrace C,C^+ \rbrace=1$, so $[C,C^+]=CC^+-C^+C=(1-C^+C)-C^+C$ Mar 13, 2020 at 18:59