I am trying to derive the Feynman rules for $\mathcal{N}=4$ supersymmetric Yang-Mills. The (Euclidean) action that I start with comes from this paper (Wilson Loops in $\mathcal{N}=4$ Supersymmetric Yang-Mills Theory, Erickson et al). For example, let's take the vertices gluon-gluon-gluon, gluon-fermion-fermion and gluon-ghost-ghost, so that I can show where my problem is. The relevant part of the action is:

$$S_\text{relevant}=\frac{1}{2 g^2} \int d^4 x \left[ \frac{1}{2} F^a_{\mu\nu}F^a_{\mu\nu} + \bar{\psi}^a_m i \gamma_\mu f^{abc} A_\mu^b \psi^c_m + \partial_\mu \bar{c}^a f^{abc} A_\mu^b c^c \right]. \tag{1}$$

For the gluon-gluon-gluon vertex, I can rewrite the relevant part of $S$ as:

$$\begin{align}S &= \frac{1}{2 g^2} \int d^4 x \frac{1}{2} 4 f^{abc} \partial_\mu A_\nu^a A_\mu^b A_\nu^c \\ &= \frac{1}{g^2} \int d^4 x f^{abc} \delta_{\nu\rho} \partial_\mu A_\nu^a A_\mu^b A_\rho^c \tag{2} \end{align}$$

where the $4$ comes from the square of the strength tensor giving $4$ such terms. Looking at the possible Wick contractions, I find that the vertex should be (in momentum space):

$$\frac{i}{g^2} f^{abc} \left[ (p-q)_\mu \delta_{\nu\rho} + (q-k)_\nu \delta_{\mu\rho} + (k-p)_\rho \delta_{\mu\nu} \right] \tag{3}$$

with $p$, $q$ and $k$ the momenta of the $3$ gluons. So far, so good, this is also in agreement with the literature.

But things do not turn out so well with fermions. I can rewrite the relevant part of $S$ as:

$$\begin{align} S &=\frac{1}{2 g^2}\int d^4 x\ \bar{\psi}^a_m i \gamma_\mu f^{abc} A_\mu^b \psi^c_m \\ &= \frac{1}{2 g^2} \int d^4 x\ \bar{\psi}^a_m i \gamma_\mu f^{abc} \delta_{m n} A_\mu^b \psi^c_n \tag{4}\end{align}$$

and thus I get that the vertex is:

$$\frac{i}{2 g^2} f^{abc} \gamma_\mu. \tag{5}$$

The factor of $1/2$ looks dubious to me, because everywhere I look in the literature it seems to be absent. On the other hand, the action $(1)$ is usually also written slightly differently, with the factor $1/(2 g^2)$ not always being applied to the fermion sector (while sometimes yes). Is the vertex that I wrote correct?

I thought that maybe the $1/2$ disappears, because the fermions are Majorana here and thus there is no distinction between particles and antiparticles. So I checked the gluon-ghost-ghost vertex, where, as far as I know, a distinction should be made. But I get this vertex:

$$-\frac{i}{2g^2} f^{abc} p_\mu \tag{6}$$

with $p$ the momentum of the antighost, and where the factor $1/2$ persists. Should it be there? Or is it just a matter of convention?


Short answer: The vertices in the OP are wrong. One can read the Feynman vertices from the terms in the Lagrangian only after normalising the quadratic terms to the canonical form $\sim(\partial\varphi)^2$. OP forgot to do this. The correct normalisation is obtained by letting $A\to gA$ and $\psi\to \sqrt 2g\psi$, and the vertices become proportional to $g$ instead of to $1/g^2$.


There are two common conventions for the Yang-Mills action: one where $A$ has dimension of inverse length, and one where $A$ has dimension of inverse length times the dimension of the coupling constant (namely $[g^2]=4-d$). The first one corresponds to $$ F=\mathrm dA+A^2,\qquad\text{and}\qquad L=\frac{1}{4g^2}F^2,\qquad\text{and}\qquad \not D=\not\!\partial+A $$ and the second one to $A\to gA$, with $$ F=\mathrm dA+gA^2,\qquad\text{and}\qquad L=\frac14F^2,\qquad\text{and}\qquad \not D=\not\!\partial+gA $$

The first convention is more natural when doing theoretical work (the main reason is that it highlights that the gauge coupling constant is a property of the gauge fields, and not of the matter fields; the latter couple universally, and are only free to choose their representation). This is precisely what happens in the paper in the OP, because large $N$ arguments become much more transparent in this convention.

The second convention is more natural when doing perturbative calculations (the main reason is that the quadratic terms are $g$-independent, and vertices are proportional to $g$, and so the Feynman rules can be read off directly from the Lagrangian).

One can also include factors of $g$ in the matter fields. This is natural only when doing SUSY, and the reason is that the gauginos are correlated with the gauge fields (a SUSY transformation maps them to each other). In this case, the coupling constant also multiplies the kinetic terms of the gauginos. In other words, mapping from the "natural" convention to the "perturbative" convention requires also redefining $\psi\to \sqrt2g\psi$. This is manifest in equation (42) in the reference, where we see that the kinetic term of the gauge multiplet is $$ \frac{1}{2g^2}\left[\frac12F^2+\bar\psi \not D\psi\right] $$

In order to have canonically normalised ($1/p^2$ and $1/\not p$) propagators, one must perform the rescaling above: $A\to gA$ and $\psi\to \sqrt2g\psi$. After doing this, all factors of $g$ disappear from the quadratic terms, and only appear, multiplicatively, in the cubic and quartic terms. After this redefinition, the Feynman rules become clear: the $A^2dA$ vertex is proportional to $g$, the $A^4$ vertex to $g^2$, the $A\psi^2$ to $g$, etc. The Lorentz and colour indices are straightforward, and so is the combinatorial factor $1/n!$ (recall that a term $\phi_1^{n_1}\phi_2^{n_2}\cdots \phi_j^{n_j}$, with all the $\phi_i$ independent, carries a factor of $1/n_1!n_2!\cdots n_j!$; a Dirac field is independent of its conjugate, while a Majorana field is not).

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  • $\begingroup$ Thank you for your very interesting comments. I am still a little confused, since they say explicitly in the reference that "each vertex carries a factor of $1/g^2$ (p.21, just before eq.$(45)$). Moreover, I had a problem with a $1/2$ in the vertex with the ghosts in the OP. Did you address that issue in your answer? $\endgroup$ – Jxx Sep 13 '19 at 9:07
  • $\begingroup$ My problem is unfortunately still unsolved. I still cannot understand why the $1/2$ should disappear both in the fermion and in the ghost vertices, although the (Majorana) fermions have a combinatorial factor $1/2!$ and the ghosts only a $1$. $\endgroup$ – Jxx Sep 15 '19 at 13:03
  • $\begingroup$ @Jxx I was only paying attention to factors of $g$ to keep the discussion as simple as possible. I now included numerical factors as well. In particular, the correct normalisation for $\psi$ is obtained by $\psi\to \sqrt 2g\psi$, so that the kinetic term becomes $\bar\psi D\psi$. The ghosts require the same rescaling if you want to have the canonical kinetic term $\bar c Dc$. $\endgroup$ – AccidentalFourierTransform Sep 15 '19 at 13:11
  • $\begingroup$ Ah yes, I see now what you are saying. I will try it immediately, but I can see that the rescaling of the $\psi$ field will work. Just one last question though: if I rescale the fields as you prescribe with the $g$, don't I lose the $g^2$ in the propagators? In the reference, their propagators always carry a $g^2$, and I would like to stick with that convention for now. At the end of the day, does the rescaling of $g$ really matter? $\endgroup$ – Jxx Sep 15 '19 at 17:46
  • $\begingroup$ And since the fermions here are Majorana, isn't the rescaling just $\psi\rightarrow g \psi$, since the Majorana Lagrangian has a $1/2$ in its kinetic term? $\endgroup$ – Jxx Sep 15 '19 at 18:02

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