I'm trying to follow Coleman's proof from his lectures "Aspects of Symmetry" on page 200-201. He proofs it is always possible to work in the temporal gauge for a general Yang-Mills-Higgs theory. I shall quickly repeat his argument. Consider some Higgs field, $\phi$, for which the directional covariant derivative vanished on some path $P$:

\begin{equation} \frac{dx^\mu}{ds} D_\mu \phi=0 \Rightarrow \frac{d \phi}{ds}=-\frac{dx^\mu}{ds} A_\mu \phi \end{equation}

where $s$ is the parameter of the path, such that the path starts at the point $x_0$ and ends at $x_1$ for $s \in [0,s_f]$. The solution of this equation is given by:

\begin{equation} g(P) = \mathcal{P} \exp \left( -\int\limits_{P(0)}^{P(s_f)} A_\mu(P(s)) \; \mathrm{d} x^\mu \right) \end{equation}

where $\mathcal{P}$ denotes the path ordering symbol. Furthermore, we can show that transformation properties are given by:

\begin{equation} g(P)' = g(x_1) g(P) g(x_0)^{-1} \end{equation}

Now the proof: ``For any space-time point $x$, define $P_x$ to be the straight-line path from $(\mathbf{x},0)$ to $x$. The desired gauge transformation is defined by:

\begin{equation} g(x)=g(P_x)^{-1} \end{equation}

for, under this transformation:

\begin{equation} g(P_x)' = g(P_x)^{-1} g(P_x) g(P_0)=1 \end{equation}

from which $A_0=0$ follows by differentiation.''

I understand the mathematics before the actual proof, but I find his proof quite confusing (maybe because English is not my first language). From what I understand, he is defining a path $P_x$ at every point $x$ in space-time. Furthermore, $P_x$ is a straight line evolving in time only, i.e. $P_x$ stays at the same point $\mathbf{x}$ in space but evolves with $t$. Is that correct? If so, then $g(P_x)$ is given by:

\begin{equation} g(P_x) = \mathcal{P} \exp \left( -\int\limits_{P(0)}^{P(s_f)} A_0(P_x(s)) \; \mathrm{d} x^0 \right) \end{equation}

and indeed this implies:

\begin{equation} \partial_0 g(P_x)' = A_0' = 0 \end{equation}

If my interpretation is correct until now, then I have the following (perhaps stupid) question:

How do we know that $\phi$ at each space-time point $x$ always obey the first equation I wrote? In other words, the whole proof is based on the idea the $\phi$ satisfied that equation for the path $P_x$, but how do we know that is true?

  • $\begingroup$ Related question: physics.stackexchange.com/q/33133 $\endgroup$
    – twistor59
    Jul 20, 2013 at 7:27
  • $\begingroup$ Hi twistor59, thanks for the link. I had already seen it, but it doesn't really answer my question. In the link, the answer is just given without any reason why. I understand that $A_\mu'=g(P_x)A_\mu (P_x)^{-1}-(\partial_\mu g(P_x))g(P_x)^{-x}=0$ so that the trick works. I would like to understand why it works. For instance, in his lectures "The magnetic monopole 50 years later", he uses this trick three times in total. One time to fix $A_0=0$, then $A_r=0$ and then $A_\theta=0$. So for me it would be interesting to understand where the trick comes from. $\endgroup$
    – Hunter
    Jul 20, 2013 at 14:50
  • $\begingroup$ That should have been: $A_0'=g(P_x) A_0 g(P_x)^{-1} - (\partial_0 g(P_x))g(P_x)^{-1}=0$. Sorry $\endgroup$
    – Hunter
    Jul 20, 2013 at 14:55
  • $\begingroup$ I've stumbled upon this question by accident, and I believe the Higgs field has nothing to do with the gauge choice - the Wilson line operator that makes it possible is not the solution to the equation written down here (how could the algebra-valued $g$ ever be a solution to that, anyway?), but is the solution to the defining equation of the horizontal lift of the path into the gauge bundle. $\phi$ does not influence the allowed gauge choices at all, since it the total Lagrangian is still fully gauge invariant. As twistor's link shows, the only condition is that $A_0$ be time-integrable. $\endgroup$
    – ACuriousMind
    Nov 13, 2014 at 22:21

1 Answer 1


This assumption he mades there in setting up the Wilson (path) dependent line is a trick, to argue that if $\phi$, respective $g\left(\mathcal{P}\right)$ is a true solution of the first order differential equation $\frac{dx^{\mu}}{ds}D_{\mu}\phi=0$, it is unique and indeed has the transformation law $g(P)^{\prime}=g(x_1)g(P)g(x_0)^{−1}$, which he then uses to complete the proof. I can recommend you also to follow the similar discussion about the Wilson Loop in Chapter 15.3, p.491 in the textbook by Peskin & Schroeder, where the argumentation sequence is basically the same.

  • $\begingroup$ Thanks for your reply. You (and Coleman) start of by assuming that $\phi$ satisfies the first order differential equation. But my question is, how do we know it satisfies that equation? $\endgroup$
    – Hunter
    Jul 19, 2013 at 23:47
  • $\begingroup$ We say it satisfies this equation and deduce the consequences out of it. We know it because we say it, we assume that this is the case. Otherwise the result would be something different. $\endgroup$
    – Hansenet
    Jul 19, 2013 at 23:55
  • 1
    $\begingroup$ Similiar computation is here higgs.physics.ucdavis.edu/QFT-III.pdf (p.148) $\endgroup$
    – Hansenet
    Jul 19, 2013 at 23:59
  • $\begingroup$ Thanks for the link, I've looked at it and what he has written seems to make sense, but it is not really related to my original question. I'm sorry to be persistent, but I've the feeling we are doing something wrong by just imposing that $\phi$ must satisfy that equation. Surely there is some underpinning argument for saying that at each point in space-time $\phi$ satisfies that equation for a straight-line path evolving in time? $\endgroup$
    – Hunter
    Jul 20, 2013 at 0:22

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