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

  • $\begingroup$ Are $G$ and $\epsilon$ Grassmann-even or Grassmann-odd? $\endgroup$ – Qmechanic Aug 31 '20 at 12:15
  • $\begingroup$ @Qmechanic I think I have some confusion here - I don't see why the derivation shown should depend on the Grassman-parity of $G$? $\endgroup$ – 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 $$ \delta\Phi ~=~(1+i\epsilon Q)\Phi (1-i\epsilon Q)-\Phi~=~i[\epsilon Q,\Phi]_{SC}~=~i\epsilon [Q,\Phi]_{SC}.$$

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

  • $\begingroup$ We always get the supercommutator to begin with. $\endgroup$ – Qmechanic Aug 31 '20 at 11:56
  • $\begingroup$ 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? $\endgroup$ – awsomeguy Aug 31 '20 at 13:37
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    $\begingroup$ It depends on how one defines infinitesimal. One may argue that Grassmann-odd supernumbers are automatic infinitesimal since they square to zero. $\endgroup$ – Qmechanic Aug 31 '20 at 13:46
  • $\begingroup$ 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? $\endgroup$ – awsomeguy Aug 31 '20 at 13:55
  • $\begingroup$ Should I ask this as a separate question? $\endgroup$ – 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 $T^{-1} \Omega T$.

  • $\begingroup$ In that case - what does it mean to say $Q$ generates the BRST transformation? And how does one show that $\delta \Phi= i\epsilon\{Q, \Phi\}$? $\endgroup$ – awsomeguy Aug 31 '20 at 13:00
  • $\begingroup$ Or do we just define $Q$ as the conserved charge? In which case I still cannot see why $\delta \Phi = i\epsilon\{Q,\Phi\}$ $\endgroup$ – awsomeguy Aug 31 '20 at 13:09
  • $\begingroup$ If Grassman parameters have no notion of magnitude, how can $\epsilon$ be infinitesimal? $\endgroup$ – awsomeguy Aug 31 '20 at 13:20
  • $\begingroup$ 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. $\endgroup$ – mike stone Aug 31 '20 at 13:21
  • $\begingroup$ So $Q$ is defined so that $\delta \Phi = i\epsilon (Q\Phi + \Phi Q)$? $\endgroup$ – awsomeguy Aug 31 '20 at 13:23

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