I'm trying to understand the way that the Higgs Mechanism is applied in the context of a $U(1)$ symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}} $ and want to gauge out the $\xi$ field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\rho$ :

$$ \begin{cases} \phi\rightarrow e^{i\theta}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\ D_{\mu}=\partial_{\mu}+iqA_{\mu} \end{cases} $$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\xi$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$ \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}} $$


$$\begin{cases} \phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because

$$\begin{cases} \phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'} \end{cases} $$ So is this transformation the one that we do or have I wronged somewhere and it can be done correctly via a legit gauge transformation?

  • $\begingroup$ Hi karky, and welcome to Physics Stack Exchange! This is an okay question as is, but I think a couple things are holding it back from being a great question: first, "Am I looking at this the right way?" is kind of vague. What other way do you think you could be looking at it, or what exactly makes you think the way you're looking at it now might not be valid? Also, "Higgs U(1) mechanism question" is not a very good title. If you address the first thing, it will probably suggest a better title. We have some tips on writing good titles. $\endgroup$ – David Z Nov 21 '15 at 11:22
  • $\begingroup$ Thanks for the suggestions, I reworked the title and the question to be more specific. $\endgroup$ – karky Nov 21 '15 at 13:02
  • $\begingroup$ I don't understand your question. When you have $\theta' = 0$, you have exactly gauged away the $\xi$ phase that was there before, i.e. the gauge transformation $\phi\mapsto \mathrm{e}^{-\mathrm{i}\xi}\phi$ leaves you with $\mathrm{e}^{-\mathrm{i}\xi}\phi = \frac{1}{\sqrt{2}}(\rho + v)$. You don't transform by $\theta'$, you transform with $\theta$, and for $\theta = -\xi$, you get the desired form. $\endgroup$ – ACuriousMind Nov 21 '15 at 14:51
  • $\begingroup$ @ACuriousMind that sounds like it could be an answer $\endgroup$ – David Z Nov 21 '15 at 16:59
  • $\begingroup$ @ACuriousMind What confuses me is the fact that if you transform like $$ \begin{cases} \phi\rightarrow e^{-i\xi}\phi\\ A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\left(-\xi\right)=A_{\mu}+\frac{1}{q}\partial_{\mu}\xi \end{cases} $$ then the Lagrangian, as invariant under these transformations, retains its original form for the field $\phi $, but that that is not what we want since then the Lagrangian has an exponential of $\xi$ and the field is not gauged away. $\endgroup$ – karky Nov 21 '15 at 17:01

Let's introduce a bit more notation because I think you're confusing yourself with the $\to$ notation:

Let $\theta : \mathbb{R}^4\to\mathbb{R}$ be any function. Then the gauge transformed fields are \begin{align} \phi^\theta & := \mathrm{e}^{-\mathrm{i}\theta}\phi \\ A^\theta& := A - \frac{1}{q}\mathrm{d}\theta \end{align} and a gauge transformation by $\theta$ is replacing $\phi,A$ by $\phi^\theta,A^\theta$, which is what you denote by $\phi\to\phi^\theta,A\to A^\theta$.

Gauge invariance means that $S[\phi,A] = S[\phi^\theta,A^\theta]$, i.e. the action doesn't change under a gauge transformation, and in this case it also means $L[\phi,A] = L[\phi^\theta,A^\theta]$.1 This holds for this Lagrangian on an abstract level.

Now, you know that you can write $\phi = \mathrm{e}^{\mathrm{i}\xi}\phi_0$, and you want to express the Lagrangian purely in terms of $\phi_0$ without having the $\xi$ in there. This is achieved by fixing the gauge to be $\theta = -\xi$, since $\phi^{-\xi} = \phi_0$. Since the theory is gauge invariant, you have $L[\phi,A] = L[\phi^{-\xi},A^{-\xi}]$.

Your main issue seems to be that the l.h.s. seems to still contain $\xi$ because the $A^{-\xi}$ contains $\xi$. That's indeed a problem that needs to be fixed. There are two ways to do this, I'll only give the short one for now, which unfortunately presupposes $\xi$ to be infinitesimal:

Note that for infinitesimal $\xi$ and $\rho$, $\phi = \phi_0 + \mathrm{i}v\xi$. Now compute the Lagrangian $L[\phi_0+\mathrm{i}v\xi,A]$ and observe that you get a term that looks like2 $\frac{1}{2}e^2v^2(A - \frac{1}{q}\mathrm{d}\xi)^2$. Hence, plugging in $A^{-\xi}$ indeed kills the $\xi$ also in the terms with $A$, since this transforms the term in the bracket into $A^2$ without any $\xi$.

1In general, the Lagrangian can change by a total derivative

2>/sup>"Looks like" because there are several different conventions for how exactly the gauge transformation and the expansion around the VEV look, and I'm not inclined to track which one is in use here.

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  • $\begingroup$ Do you mean $ \mathcal{L}\left[\phi_{0}+iv\xi,A\right]$ in the third line from the end? $\endgroup$ – karky Nov 21 '15 at 18:12
  • $\begingroup$ @karky: Yep, corrected. $\endgroup$ – ACuriousMind Nov 21 '15 at 18:23
  • $\begingroup$ If I asked for the non-infinitesimal way, could you give me a reference? $\endgroup$ – karky Nov 21 '15 at 20:30
  • $\begingroup$ @karky: I looked, but I couldn't find one. I have the rough idea how it is supposed to work, but I'd need to work it out in detail; I'll add it to this answer when I find the time (or a reference). $\endgroup$ – ACuriousMind Nov 21 '15 at 20:58
  • $\begingroup$ I now see what I was thinking wrong. Using your notation, I was trying to express the transformed Lagrangian $[\phi^{\theta},A]$ where instead I should be using $[\phi^{\theta},A^{\theta}]$. Specifically, I was expressing $A^{\theta}$ with respect to $A$ but used the transformed version of $\phi$. I wasted too much time on this misunderstanding, so thanks for your catalytic help! $\endgroup$ – karky Nov 21 '15 at 20:58

No, you got it right. You gauge-transformed to the so-called unitary gauge where the 2-particles of the complex φ doublet, have now reduced just to a real component ρ thereof, while the phase component has mutated into a component of the gauge field $$ \begin{cases} \phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\ A_{\mu}\rightarrow A_{\mu} \end{cases} $$ where "the rest" includes a non-gauge invariant mass term for the gauge field, so it now has 3 components, not 2 because of that: θ' has been reassigned, through a gauge transformation, from the complex scalar to the gauge field.

You are in a given gauge now: if you moved to another gauge, you would not see these degrees of freedom as simply, but, of course, the results for (gauge invariant) amplitudes you'd calculate would be identical.

That's the point: gauge invariance allows transition to a gauge where the physical content of the theory is more transparent.

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