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.

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

  • 1
    $\begingroup$ 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? $\endgroup$ – MrLee Apr 17 '14 at 10:51
  • 1
    $\begingroup$ 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. $\endgroup$ – Robin Ekman Apr 18 '14 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.