In case of the gauge-fixed Faddeev-Popov Lagrangian: $$ \mathcal{L}=-\frac{1}{4}F_{\mu\nu}\,^{a}F^{\mu\nu a}+\bar{\psi}\left(i\gamma^{\mu}D_{\mu}-m\right)\psi-\frac{\xi}{2}B^{a}B^{a}+B^{a}\partial^{\mu}A_{\mu}\,^{a}+\bar{c}^{a}\left(-\partial^{\mu}D_{\mu}\,^{ac}\right)c^{c} $$

(for example in Peskin and Schröder equation 16.44)

If you expand the last term (for the ghost fields) you get: $$ \bar{c}^{a}\left(-\partial^{\mu}D_{\mu}\,^{ac}\right)c^{c} = -\bar{c}^{a}\partial^{2}c^{a}-gf^{abc}\bar{c}^{a}\left(\partial^{\mu}A_{\mu}\,^{b}\right)c^{c}-gf^{abc}\bar{c}^{a}A_{\mu}\,^{b}\partial^{\mu}c^{c} $$

And so, the Lagrangian has a term proportional to the second derivative of $c^a$.

In this case, how does one find the classical equations of motion for the various ghost fields and their adjoints?

I found the following equations of motion so far: $$ D_{\beta}\,^{dc}F^{\beta\sigma}\,^{c}=-g\bar{\psi}\gamma^{\sigma}t^{d}\psi+\partial^{\sigma}B^{d}+gf^{dac}\left(\partial^{\sigma}\bar{c}^{a}\right)c^{c} = 0 $$ $$ \sum_{j}\partial_{\sigma}\bar{\psi}_{\alpha,\, j}i\gamma^{\sigma}\,_{ji}-\sum_{\beta}\sum_{j}\bar{\psi}_{\beta,\, j}\left(gA_{\mu}\,^{a}\gamma^{\mu}\,_{ji}t^{a}\,_{\beta\alpha}-m\delta_{ji}\delta_{\beta\alpha}\right)=0 $$ $$ \left(i\gamma^{\mu}D_{\mu}-m\right)\psi=0 $$ $$ B^{b}=\frac{1}{\xi}\partial^{\mu}A_{\mu}\,^{b} $$ $$ \partial^{\mu}\left(D_{\mu}\,^{dc}c^{c}\right)=0 $$ $$ f^{abd}\left(\partial_{\sigma}\bar{c}^{a}\right)A^{\sigma}\,^{b}=0 $$

But it is the last equation that I suspect is false (I saw the equation $ D_\mu\,^{ad} \partial^\mu \bar{c}^d = 0 $ in some exercise sheet (http://www.itp.phys.ethz.ch/education/fs14/qftII/Series7-3.pdf Exercise 3) and I also saw the equation $D^\mu\,^{ad}\partial_\mu B^d = igf^{dbc}(\partial^\mu\bar{c}^b)D_\mu\,^{dc} c^c$ which I don't understand how they were derived.)

EDIT: Thanks to Qmechanic's answer I was able to derive the correct equations of motions (as noted in the comment to that answer) but I still don't know where to "obtain" the last equation I mentioned which connects the auxiliary field with the ghost fields.

  • 1
    $\begingroup$ Comment to the question (v3): Consider including a reference to the last part of the question. Is the exercise sheet electronically available? $\endgroup$
    – Qmechanic
    Commented Jul 4, 2014 at 11:45

1 Answer 1


I) Since total divergence terms do not contribute to Euler-Lagrange (EL) equations, cf. e.g. this Phys.SE post, one could just integrate the Faddeev-Popov $\bar{c}c$ term by part so that there are no more than first derivatives present and the standard form of the EL equations applies.

II) Alternatively, in the presence of higher derivatives, the EL equations get modified with additional terms, see e.g. Wikipedia. Note that special care should be taken in the ordering of Grassmann-odd derivatives.

III) Often in field theory, because of external indices and Grassmann-odd fields, it is quite tedious to use the EL equations directly. It is often simpler to infinitesimally vary the given action $S[\phi]$,

$$\delta S~=~\int d^4x~\text{(EL-eq)}~ \delta\phi(x) +\text{(boundary terms)} . $$ and identify the EL equations as the relevant coefficient functions on the spot, so to speak.

  • $\begingroup$ Dear @Qmechanic, thanks for your response. I followed your first suggestion, changing the Lagrangian into $\mathcal{L}=-\frac{1}{4}F_{\mu\nu}\,^{a}F^{\mu\nu a}+\bar{\psi}\left(i\gamma^{\mu}D_{\mu}-m\right)\psi-\frac{\xi}{2}B^{a}B^{a}+B^{a}\partial^{\mu}A_{\mu}\,^{a}+\left(\partial_{\mu}\bar{c}^{a}\right)\partial^{\mu}c^{a}-gf^{abc} \bar{c}^{a}\left(\partial^{\mu}A_{\mu}\,^{b}\right)c^{c}-gf^{abc}\bar{c}^{a} A_{\mu} \, ^{b}\partial^{\mu}c^{c}$ and got the EoMs for the ghosts: $D_{\sigma}\,^{da}\partial^{\sigma}\bar{c}^{a}$ and $\partial^{\sigma}D_{\sigma}\,^{dc}c^{c}$. $\endgroup$
    – PPR
    Commented Jul 4, 2014 at 10:29
  • $\begingroup$ However, how do you get an equation that has both the ghosts and auxiliary fields?? $\endgroup$
    – PPR
    Commented Jul 4, 2014 at 10:29

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