I would like to ask whether fermionic Grassmann fields in a gauge theory path integral (say in QCD) should be chosen to commute or anticommute with ghost and anti-ghost fields. The way most textbooks present it suggests that anti-commutation should be chosen, but I don't know any argument for this. There is also the question of (anti)commutation relations for corresponding operators in the Krein space.

I was asked in the comments where this issue ever came up. It was in derivation of the conserved BRS current in QCD. Let me first explain that in many books it is stated that BRS operator $s$ satisfies graded Leibniz rule with respect to the fermion number, for example $s ( \bar \psi \psi ) = (s \bar \psi ) \psi - \bar \psi s \psi $. On a lecture about anomalies I attended recently it was stated that Leibniz rule graded with respect to the ghost number should be used instead (or at least can be used), so $s ( \bar \psi \psi ) = (s \bar \psi ) \psi + \bar \psi s \psi$ but $s (c^a c^b)=(sc^a) c^b - c^a (s c^b)$. Therefore I am naturally led to consider variations of fields of the form $\phi \to \phi + \epsilon s \phi$, where $\epsilon$ is a Grassmann parameter commuting with $A, \psi, \bar \psi$ but anticommuting with $c^a$ and $\bar c^a$. I found that under these transformations variation of action takes the form $$ \delta S = \int d^4 x (\partial_{\mu} \epsilon) \left[ - F_a^{\mu \nu} D_{\nu} c^a + g \bar \psi \gamma^{\mu} c^a t_a \psi + b^a D^{\mu} c^a - \frac{1}{2} g f_{abc} (\partial^{\mu} \bar c^a) c^b c^c \right]. $$ We see that inside the parenthesis $[]$ we have a conserved current, from now denoted $J_{\mathrm{BRS}}^{\mu}$. After manipulating this current using the equations of motion I found a term $g [\bar \psi, c^a] \gamma^{\mu} t_a \psi$. It turns out that explicit evaluation of the divergence of $J_{\mathrm{BRS}}^{\mu}$ using equations of motion gives zero only if this commutator is taken to be zero. Thus it seems to me that this is the only choice consistent with my choice of the definition of the BRS operator.

Remark I used the Lagrangian $$ \mathcal L = - \frac{1}{4} F^2 + \bar \psi (i \gamma \cdot D - M) \psi + \partial_{\mu} \bar c^a D^{\mu}c^a - A^a_{\mu} \partial^{\mu} b^a + \frac{1}{2} \xi b^2 , $$ with covariant derivative $D=\partial + ig A$.

  • $\begingroup$ Can you write down a term in which this matters? $\endgroup$ – user178876 Dec 23 '17 at 17:09
  • $\begingroup$ Dear @marmot in some calculations (related to BRST operator) I did in QCD I found a term $\bar \psi c^a t_a \psi - c^a \bar \psi t_a \psi$. It vanishes if the ghost field commutes with spinor fields, but not otherwise. $\endgroup$ – Blazej Dec 23 '17 at 17:26
  • $\begingroup$ Thanks! I wonder if you want to add the derivation of the term to the question. $\endgroup$ – user178876 Dec 23 '17 at 17:45
  • $\begingroup$ @Blazej could you elaborate on how this term enters your calculation and why is it necessary? Physically, ghosts and fermions do not couple to each other. In fact, ghosts are merely a perturbative way to accomodate for the correction to the Yang-Mills path integral measure coming from gauge-fixing. Are you using a gauge-fixing condition involving fermions? $\endgroup$ – Prof. Legolasov Dec 23 '17 at 18:10
  • $\begingroup$ Dear @marmot and Solenodon Paradoxus, I edited my question to specify where my problem arises. $\endgroup$ – Blazej Dec 23 '17 at 20:26

It is not completely clear what OP is looking for, but here are some hopefully helpful comments:

  1. Classically (meaning when Planck constant $\hbar\to 0$), two fields $A$ and $B$ are super-commutative $$AB~=~(-1)^{|A||B|}BA,$$ where $|A|$ and $|B|$ denote the corresponding Grassmann-parity. In other words, the classical super-commutator $$[A,B]~\equiv~AB-(-1)^{|A||B|}BA~=~0$$ vanishes.

  2. The super-commutator in quantum theory is typically a quantum deformation of the classical super-commutator.

  3. Note that ghost fields can both be Grassmann-even and Grassmann-odd, depending on the theory.

  4. In principle, one may consider superalgebras with several independent $\mathbb{Z}_2$- or $\mathbb{Z}$-gradings $$|\cdot|_1, \quad \ldots, \quad |\cdot|_n. $$ The super-commutator in such a superalgebra is then defined as $$[A,B]~\equiv~AB-(-1)^{\sum_{i=1}^n|A|_i|B|_i}BA.$$ (For instance, one could consider exterior form degree and usual Grassmann-grading as two independent gradings.)

  5. A theory may allow for different conventions. The main point is that one should be consistent.

  6. Specifically, concerning fermion matter fields $\psi$, Faddeev-Popov ghost field $c$ and antighost fields $\bar{c}$ in Yang-Mills theory, it is possible to consistently set up the BRST formulation using only one type of Grassmann-grading, in which $\psi$, $c$ and $\bar{c}$ are all Grassmann-odd, and pairwise anti-commuting.

  • 1
    $\begingroup$ Dear @Qmechanic mathematically we can have a situation where we have two indepedent Grassmann algebras, generated by $\eta_i$ and $\xi_i$, so that $ \{ \xi_i , \xi_j \} = \{ \eta_i , \eta_j \}=0$ but $[\eta_i, \xi_j]=0$. The question is whether such construction is implemented is quantum field theories. $\endgroup$ – Blazej Dec 23 '17 at 17:31
  • $\begingroup$ Dear @Qmechanic , please note that I edited my question to specify where precisely the problem arises and what led me to believing that $[c^a,\psi]=0$ (on the level of integration variables) is the consistent definition in the approach I am using. $\endgroup$ – Blazej Dec 23 '17 at 20:28
  • $\begingroup$ I accept this answer, even though the question whether convention with BRS operator taken as graded with respect to the ghost number is acceptable is still open for me. However it is not strictly speaking the question I originally asked, and it might be very dificult to give a definite answer. $\endgroup$ – Blazej Dec 24 '17 at 18:49

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.