# Nielsen identities and gauge-dependence of effective action

I've been trying to wrap my head around the Nielsen identities and their interpretation, but the available literature has confused me a bit. For instance, this paper states that for the Higgs field, denoted $h(x)$, we have:

$$\xi\frac{\partial\Gamma}{\partial \xi}+\int d^4x\, K[h(x)]\frac{\delta \Gamma}{\delta h(x)}=0$$

where $K$ is some functional computable using perturbation theory, and $\Gamma$ is the effective action. It also claims that:

The Nielsen identities tell us that the gauge dependence of the effective action can be compensated by a local field redefinition.

which one might interpret as saying that the action is gauge independent, but only if we change the fields simultaneously.

On the other hand, other papers (for instance this one) state, somewhat ambiguously, that the effective action is gauge dependent, but not in what sense.

I'm a bit puzzled by all of this, so I'd like a (hopefully) unambiguous answer to the following: is the effective action in general gauge independent (in the sense that $\displaystyle \frac{d\Gamma}{d\xi}=0$, but $\displaystyle \frac{\partial\Gamma}{\partial\xi}\neq 0$), or is it truly gauge dependent (so that $\displaystyle\frac{d\Gamma}{d\xi}\neq 0$), and only its derivatives (1PI correlators), which are observables, are gauge independent? Furthermore, do the Nielsen identities in their most general form hold to all orders in perturbation theory (I'd guess that they do as their derivation doesn't rely on it)?

• hm, 1PI correlators are neither observables nor gauge independent. For example the gluon propagator comes in Landau or whatever gauge – tonydo Dec 13 '16 at 19:18
• What is $\xi$ in this context? Sometimes $\xi$ is used as a parameter in a gauge-fixing term, which obviously induces a gauge-dependence in the effective action. – Neuneck Dec 15 '16 at 13:16
• @Neuneck yes, $\xi$ is the gauge parameter. Can we then immediately conclude that $\frac{\partial\Gamma}{\partial\xi}=\frac{d\Gamma}{d\xi}\neq 0$? The first paper I quoted seems to assume that the field itself can have an implicit gauge dependence, so I'm wondering if we can redefine the field in such a way to exactly cancel the dependence and make is so that $\frac{d\Gamma}{d\xi}= 0$? I see that they only consider the case with one field, and would like to know if considering multiple ones changes anything (my guess would be no, but a proof would be nice). – blueshift Dec 15 '16 at 21:06
• @blueshift Is it the gauge parameter, or a parameter of a gauge-fixing term? There is a huge difference: The gauge parameter determines the size of a gauge transformation, while the parameter in a gauge-fixing term would rather be a Lagrange parameter that removes ambiguities due to gauge-equivalent field configurations. There is a link between the two in that a Lagrangian with a gauge-fixing term obviously has a gauge dependence. – Neuneck Dec 16 '16 at 11:07
• @Neuneck in the original paper by Nielsen, it is stated that "$\xi$ is a parameter fixing the gauge in the class of Fermi gauges", so I guess the second. – blueshift Dec 16 '16 at 15:44