# Higgs mechanism and neutral fields

Consider a Lagrangian $L(\phi,A_{\mu})$ with $\phi$ being some scalar field and $A_{\mu}$ some dynamical U(1) gauge field that minimally couples to $\phi$. Under a global U(1) symmetry the field $\phi$ transforms as $$\delta\phi=i\epsilon q \phi.$$ The field $\phi$ is said to be charged (with charge q) under the gauge field $A$.

In a Higgs phase we have that $|\phi(x,t)|\neq 0$. In particular we can fix a gauge so that $|\phi(x,t)|=\Phi(x,t)$ is real. Then we consider small fluctuations $\Phi(x,t)=\Phi_{0}+\delta\Phi$ and integrate them out to obtain an effective theory in which the gauge field A is massive.

My question: It seems to me as though the requirement that $\phi$ is charged enters when integrating out the small flunctuations, because if $\phi$ were neutral (i.e. q=0) there wouldn't be any flucutations that one can integrate out and hence one wouldn't obtain a massive term for the gauge field in the Lagrangian. Is this correct? If not where does the requirement for $\phi$ to be charged enter in the argument? And: Does the requirement for the matter field to be charged with respect to the corresponding gauge field carry over without difficulties to the non-abelian case?

I am looking forward to your responses!

Like you said "$A_\mu$ some dynamical $U(1)$ gauge field that minimally couples to $\phi$". It means that the covariant derivative is : $$D_\mu \phi = \partial_\mu \phi + iqA_\mu\phi$$ with $q$ the $U(1)$ charge of the scalar field. As a consequence, if $\phi$ is not $U(1)$ charged you will not have the second term in the covariant derivative and hence $D_\mu$ is equivalent to the standard derivative $\partial_\mu$. When the scalar field $\phi$ is VEVed with fluctuations around this VEV : $$\phi = \frac{(v+h_1)+ih_2}{\sqrt{2}}$$ and that you compute the kinetic term $(D_\mu \phi)^\dagger D_\mu \phi$, you will get a mass term for $A_\mu$ which is : $$\frac{1}{2}q^2v^2A_\mu A^\mu$$ which is proportional to $q$ the $U(1)$ charge of $\phi$ and this term appears in the contraction of the second terms of $(D_\mu \phi)^\dagger$ and $D_\mu \phi$. If these second terms was not here, i.e. if the scalar field $\phi$ was not $U(1)$ charged, then the gauge field $A_\mu$ can't get a mass.

Finally, the requirement that $\phi$ is $U(1)$ charged enters when you require a non-trivial minimal coupling between $A_\mu$ and $\phi$.

$\textbf{EDIT}$ :

If you've already seen the Glashow-Weinberg-Salam (GWS) Model, the analogy is that the Higgs field is $SU(2)_L\times U(1)_Y$ charged (because it's an $SU(2)$ doublet and it has an hypercharge $Y$). Thus, when the Higgs field is VEVed, the gauge fields acquire a mass.

• Does this construction depend on A being a dynamical gauge field? – MrLee Apr 21 '14 at 8:23
• And why is the mass term not present when $\phi$ is not condensed? – MrLee Apr 21 '14 at 8:54

Let me answer your questions, albeit slightly indirectly. We start with a local $U(1)$ symmetry, i.e. a gauge symmetry, for a Lagrangian describing a scalar, $\phi$, and a gauge boson $A_\mu$. You write global. A global symmetry requries no gauge bosons, because its continuous parameter $\epsilon\neq\epsilon(x)$ commutes with derivatives $\partial_\mu$. The gauge symmetry prohibits a mass for the vector boson. Supposing that the scalar obtains a non-zero vacuum expectation value (VEV)...

• If the scalar is charged under the $U(1)$, the $U(1)$ is said to be spontanouesly broken or hidden, because it is no longer manifest in our Lagrangian. In particular, we now have a massive gauge boson. We do not need to integrate out the fluctuations about the VEV (i.e. the Higgs boson) to see that the gauge boson has obtained a mass.

• If the scalar is not charged under the $U(1)$, the symmetry is unbroken, and the vector boson remains massless. These arguments do transfer to the non-Abelian cases. Indeed, the VEVed part of the Standard Model scalar doublet is electically neutral, so does not break the $U(1)_{em}$ electromagnetism.

• What is your definition of a scalar field $\phi$ being charged under a gauge field $A$? I thought: If say U(1) is the (local) gauge group of your system then this gauge group always containts also a "global" symmetry. This global symmetry is then used to define the charge. Isn't the charge due to a local gauge transformation trivially (i.e. independent of the equations of motion) conserved? – MrLee Apr 17 '14 at 10:51
• The gauge group does not always contain global gauge transformations, or example in the Dirac magnetic monopole there is no global choice of $U(1)$ gauge. This is true generally for monopole and instanton gauge fields. The global $U(1)$ symmetry for e.g. the electron field in the Dirac Lagrangian is strictly speaking unrelated to gauge symmetry. Formally the Dirac field takes values in $S \otimes \mathbb C$ where $S$ is the space of Dirac spinors. The gauge symmetry acts on the latter factor, the global $U(1)$ symmetry on the former. – Robin Ekman Apr 18 '14 at 10:55