Is the Higgs mechanism a gauge transformation or not? ( $U(1)$ context )

Question

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

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

so

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

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?

Answer 1

Just to add to Cosmas' answer. I think people are confusing gauge transformation and gauge fixing. What we are doing here is gauge fixing, which means that we constrain the fields from an arbitrary choice $(\phi\in\mathbb{C}, A_\mu)$ to a gauge fixed choice $(\phi_0\in\mathbb{R}, A_\mu)$. The point of doing this is that we remove redundancy due to gauge symmetry, in the sense that every point $(\phi\in\mathbb{C}, A_\mu)$ can be reached exactly once by doing gauge transformation on the constrained (gauge fixed) field space. More specifically, suppose we want to reach the point $(\phi_0 e^{i\xi}, A_\mu)$ (where $\phi_0\in\mathbb{R}$), we should start from the gauge fixed point $(\phi_0, A_\mu+\frac{1}{q}\partial_\mu\xi)$ then do the gauge transformation with the transformation parameter $\theta=\xi$ (following convention of @karky). In summary, when eliminating the goldstone boson $\xi$ in abelian Higgs model, what we are doing is gauge fixing, NOT gauge transformation. Any gauge transformation leaves the lagrangian invariant so there is no hope that the field $\xi$ may be eliminated.

Answer 2

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.

Answer 3

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 $\xi$.

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^{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.}