# How do fermionic operators transform?

In quantum mechanics, if we have an operator $$\Omega$$, then under the transformation $$T$$, with infinitesimal generator $$G$$ (i.e. $$T(\epsilon)=1-i\epsilon G + \ldots$$), then operator transforms as $$\Omega\rightarrow T^\dagger\Omega T,$$ so, infinitesimally, $$\delta\Omega = T^\dagger\Omega T-\Omega$$ $$=(1+i\epsilon G)\Omega(1-i\epsilon G)-\Omega$$ $$=i\epsilon[G,\Omega],$$ to first order.

Now what happens if, in QFT language, $$\Omega$$ is a fermionic operator? I am looking at BRST symmetry, in particular, on page 19 here [1], it reads

for a general field dependent function $$\Phi$$, $$\delta_B\Phi=i\theta[Q_B, \Phi]_\pm,$$ where $$Q_B$$ is the conserved Noether charge associated with the BRST symmetry, $$[ , ]_− = [ , ]$$ and $$[ , ]_+ = \{ , \}$$, and the sign being minus/plus according as Φ is bosonic/fermionic.

Since $$\Phi$$ is fermionic, the anticommutator is chosen. The anticommutator is used everywhere else I look, e.g. page 74 here [2].

However, I do not understand how the standard quantum mechanical argument above breaks down if $$\Omega$$ is a fermionic operator. Why should $$\delta \Omega$$ be different depending on the bosonic/fermionic nature of $$\Omega$$?

[1] BRST quantization and string theory spectra - Bram M. Wouters

[2] String Theory - R. A. Reid-Edwards

• Are $G$ and $\epsilon$ Grassmann-even or Grassmann-odd? – Qmechanic Aug 31 '20 at 12:15
• @Qmechanic I think I have some confusion here - I don't see why the derivation shown should depend on the Grassman-parity of $G$? – awsomeguy Aug 31 '20 at 12:22

Nothing breaks down per se. It is just that we have to use supernumbers to mathematically describe fermions. The introduction of Grassmann-odd variables has several implication:

• At the classical level, the Poisson-bracket $$\{\cdot,\cdot\}_{PB}$$ is replaced by super-Poisson bracket $$\{\cdot,\cdot\}_{SPB}$$.

• At the quantum level, the commutator $$[\cdot,\cdot]_C$$ is replaced by supercommutator $$[\cdot,\cdot]_{SC}$$.

• The correspondence principle $$[\cdot,\cdot]_{SC}=i\hbar\{\cdot,\cdot\}_{SPB}+{\cal O}(\hbar^2)$$ between classical and quantum mechanics still holds.

• In the Hamiltonian version of Noether's theorem, the relationship between conserved quantity $$Q$$ and infinitesimal symmetry $$\delta~=~-\epsilon\{Q,\cdot\}_{SPB}$$ still holds. Here the infinitesimal variation $$\delta$$ is Grassmann-even while $$\epsilon$$ is an infinitesimal parameter of same Grassmann-parity as $$Q$$.

• For operators, we calculate $$\delta\Phi ~=~(1+i\epsilon Q)\Phi (1-i\epsilon Q)-\Phi~=~i[\epsilon Q,\Phi]_{SC}~=~i\epsilon [Q,\Phi]_{SC}.$$

• We always get the supercommutator to begin with. – Qmechanic Aug 31 '20 at 11:56
• I see - that edit clears up my confusion. One last thing - the other answer to this question talks about the problem that Grassmann numbers don't have a notion of magnitude. Does this not effect the idea of $\epsilon$ being infinitesimal? – awsomeguy Aug 31 '20 at 13:37
• It depends on how one defines infinitesimal. One may argue that Grassmann-odd supernumbers are automatic infinitesimal since they square to zero. – Qmechanic Aug 31 '20 at 13:46
• it has also occurred to me - how do we know that the BRST transformation is going to transform as $T^\dagger \Phi T = (1+i\epsilon Q)\Phi(1-i\epsilon Q)$ for some $T$, or equivalently $Q$? I gather that in most cases (e.g. translations), we can notice that the transformation on the operators is equivalent to the active transformation of the states - but in this case, how do we know that the operator transformation has an equivalent state transformation? – awsomeguy Aug 31 '20 at 13:55
• Should I ask this as a separate question? – awsomeguy Aug 31 '20 at 14:27

Your first formula $$\Omega\to T^{-1} \Omega T$$ describes a finite transformation. The commutator comes from taking the limit $$T(\epsilon)= 1-i\epsilon G+\ldots$$ as $$\epsilon$$ becomes small. There is no such thing as a finite supertransformation as a Grassmann parameter has no notion of being big or small. Consequently there is no super-analogue of $$T^{-1} \Omega T$$.

• In that case - what does it mean to say $Q$ generates the BRST transformation? And how does one show that $\delta \Phi= i\epsilon\{Q, \Phi\}$? – awsomeguy Aug 31 '20 at 13:00
• Or do we just define $Q$ as the conserved charge? In which case I still cannot see why $\delta \Phi = i\epsilon\{Q,\Phi\}$ – awsomeguy Aug 31 '20 at 13:09
• If Grassman parameters have no notion of magnitude, how can $\epsilon$ be infinitesimal? – awsomeguy Aug 31 '20 at 13:20
• It's just a definition. It's not derived from anywhere. It's inspired by Elie Cartan's exterior derivative and its graded-derivation property. – mike stone Aug 31 '20 at 13:21
• So $Q$ is defined so that $\delta \Phi = i\epsilon (Q\Phi + \Phi Q)$? – awsomeguy Aug 31 '20 at 13:23