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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

awsomeguy
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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 $$ \delta\Phi ~=~(1+i\epsilon Q)\Phi (1-i\epsilon Q)-\Phi~=~i[\epsilon Q,\Phi]_{SC}~=~i\epsilon [Q,\Phi]_{SC}.$$

For more information, see also e.g. this related Phys.SE post.

Qmechanic
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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 $T^{-1} \Omega T$.

mike stone
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