2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.