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Is the Higgs mechanism a gauge transformation or not?

  1. Nov 21, 2015 #1
    I asked this question to PhysicsStackExchange too but to no avail so far.

    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 [itex]\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}
    [/itex] and want to gauge out the [itex]\xi[/itex] 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 [itex]\rho[/itex] :

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

    As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge [itex]\xi[/itex] 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?

    My problem is that invariance of U(1) transformations means that you end up with the same field in the end, the gradient of the exponent in [itex]U\phi [/itex] is necessary to cancel with the corresponding term of the field 4-vector transformation rule. Any light shed is welcome!
     
  2. jcsd
  3. Nov 21, 2015 #2

    fzero

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    The choice where the phase of the Higgs field is set to zero is a particular gauge fixing, called unitarity gauge. It is possible to choose a different gauge, which can be useful in calculations, as the wiki article suggests. Setting unitarity gauge and then performing a gauge transformation will take you to a configuration where another gauge fixing condition has been imposed. This is the proper interpretation of a gauge choice.

    The advantage of unitarity gauge is that only the physical degrees of freedom remain. The disadvantage is that it is somewhat tricky to organize divergences in loop calculations. In practice, the so-called ##R_\xi##-gauges can be used, where the contributions from unphysical degrees of freedom are organized and accounted for by the renormalization scheme.
     
  4. Nov 21, 2015 #3
    My misconception was cleared when I realized I was trying to express the transformed Lagrangian with respect to [itex]\phi^{'}(=e^{i\theta}\phi)[/itex] and the original [itex]A_{\mu}[/itex] and not, as I should correctly do, with the transformed [itex]A^{'}_\mu[/itex]. It was brought to my attention and this matter cleared out. What I'm doing up there is specifying the transformation one would have to do If he did what I was doing, which is clearly not a gauge transformation!
     
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