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Obtaining the propagator for the Faddeev-Popov (FP) ghosts from the path integral language is straightforward. It is simply

$$\langle T(c(x) \bar c(y))\rangle~=~\int\frac{d^4 p}{(2\pi)^4}\frac{i e^{-ip.(x-y)}}{p^2-\xi m^2+i\epsilon}.$$

But I am unable to properly derive it following canonical quantization route.

The problem is that for anti-commuting fields, the time-ordered product is defined as

$$\langle T(c(x) \bar c(y))\rangle=\theta(x^0-y^0)\langle c(x) \bar c(y)\rangle-\theta(y^0-x^0)\langle \bar c(y) c(x)\rangle$$

with a minus sign between the two terms. This minus sign is preventing me from closing the contours in the correct way to obtain the expression in my first equation.

The only way I can save this is to say that FP ghosts are special, and their time-ordered product is defined with a plus sign instead of the minus sign. Is this legitimate? What is the right way to get the Ghost propagator following canonical quantization route?

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  • $\begingroup$ 1. What is $\xi$? If it is $\pm 1$ metric convention, it also affects the $i\epsilon$ term. 2. Have you added an overall phase to $Z[j]$ to absorb the $i$ factor in the kinetic term of the ghost field? 3. Could you write the relevant action/hamiltonian, the Heisenberg picture ghost fields as a function of creation/anhilation operators and the anticommutation relations to see your conventions? $\endgroup$
    – Diego Mazón
    Commented Nov 16, 2012 at 21:27
  • $\begingroup$ By the way, is that a book problem or something you are thinking about? Because I would guess that the questions doesn't make sense. I think that the canonical formalism does not follow directly from the path integral version in FP trick. The closest could be BRST quantization, but there $c$ and $\bar c$ are independent real fields (so the your propagator is 0). I think that in a noncovariant formalism like the canonical makes more sense to fix the temporal gauge instead a covariant one. $\endgroup$
    – Diego Mazón
    Commented Nov 17, 2012 at 1:56
  • $\begingroup$ Last question: how do you deduce the $i\epsilon$ terms from the path integral if you don't know the vacuum wave functional? Or you know it? $\endgroup$
    – Diego Mazón
    Commented Nov 17, 2012 at 2:00
  • $\begingroup$ Another issue: The Hamiltonian for a scalar fermionic field is not bounded from bellow. $\endgroup$
    – Diego Mazón
    Commented Nov 17, 2012 at 2:34
  • $\begingroup$ @drake In the propagator, $\xi$ is the gauge parameter, which comes by adding $\frac{1}{2\xi}(\partial.A)^2$ to the Lagrangian in the BRST approach. I am trying to understand covariant canonical quantization (in the BRST language, of course) so that I may ultimately be able to discuss single-particle states and carry out the LHZ reduction formula for scattering amplitudes. Where can I learn more about this? $\endgroup$
    – QuantumDot
    Commented Nov 17, 2012 at 22:14

1 Answer 1

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The solution to this problem comes from the sneaky fact (Kugo, 1978) that while the FP ghost field is hermitian $c^\dagger (x) = c(x)$, the anti-ghost field is anti-hermitian $\bar c^\dagger (x)=-\bar c (x)$ .

As a result, the plane wave expansion for the ghost/anti-ghost fields (Becchi, 2008), Scholarpedia are:

$$ c^a(x)={1 \over(2 \pi)^{3/2}} \int_{k_0= | \vec k|} {d \vec k \over 2 k_0}( \gamma^a( \vec k)e^{-ik \cdot x}+ (\gamma^a)^\dagger( \vec k)e^{ik \cdot x})$$ $$\bar c^a(x)={1 \over(2 \pi)^{3/2}} \int_{k_0= | \vec k|} {d \vec k \over 2 k_0}( \bar \gamma^a( \vec k)e^{-ik \cdot x}- (\bar\gamma^a)^\dagger( \vec k)e^{ik \cdot x}) \ , $$ with a minus sign between the two terms in mode expansion for the anti-ghost field.

Thus, when evaluating the time ordered correlator (propagator), the minus sign in the plane-wave expansion compensates the minus sign in the definition of the time-ordering shown in my question above. Thus, I am able to derive the standard Feynman propagator for the FP ghost field.

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  • $\begingroup$ Interesting. Weinberg however defines both fields as hermitians in the 2nd volume of his QFT book. I guess he introduces an $i$ factor that makes it real. And this has to make to do with the $i$ in the kinetic term that sometimes is reabsorbed in a redefinition of $Z[j]$. I also think that the minus sign in $\bar c$ makes the Hamiltonian bounded from bellow. What I'm wondering is what happen with the hermiticity of the hamiltonian. Good question and answer! $\endgroup$
    – Diego Mazón
    Commented Nov 18, 2012 at 21:18
  • $\begingroup$ It would be nice if you complete your answer with the full free canonical quantization if you already have everything clear. Essentially: commutation relations between ghost fields, non interacting ghost Hamiltonian in Fock space and vacuum wave functional. $\endgroup$
    – Diego Mazón
    Commented Nov 18, 2012 at 21:22
  • $\begingroup$ By the way, it is obvious from the Heisenberg fields that the Hamiltonian must be $H=\sum_a \int d^3\mathbf{K}~\left(\bar\gamma_\mathbf{K}^{a~\dagger}\gamma_\mathbf{K}^a+\gamma_\mathbf{K}^{a~\dagger}\bar\gamma_\mathbf{K}^a-\text{vacuum energy}\right)$ I asked for an explicit calculation, besides the anticommutation realations and the vacuum wave function. $\endgroup$
    – Diego Mazón
    Commented Nov 23, 2012 at 20:00
  • $\begingroup$ @QuantumDot the link to [Kubo, 1978] seems to be dead. Is it only me? In any case, an alternative link is academic.oup.com/ptp/article/60/6/1869/1846386 $\endgroup$
    – AccidentalFourierTransform
    Commented Jul 4, 2017 at 13:54
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    $\begingroup$ This is a very nice answer that was helpful for me. One possible useful point for others: the ghost term in the Lagrangian is $\partial_\mu \bar{c}\partial^\mu c$. In order for the Lagrangian to be real (i.e. $\mathcal{L}^\dagger=\mathcal{L}$), one sees that either $c$ or $\bar{c}$ must be anti-Hermitian. $\endgroup$
    – astronautgravity
    Commented Jan 20, 2020 at 5:20

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