# Particle # conservation in a spontaneously broken theory

1. Aug 31, 2015

### EricTheWizard

So consider a theory of a complex scalar field $\phi$ with a global U(1) symmetry $\phi \rightarrow e^{i\theta} \phi$. The theory admits a conserved current
$$J^\mu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^\dagger)} \delta\phi^\dagger = -i \left( \phi^\dagger \partial^\mu \phi - \partial^\mu \phi^\dagger \phi \right)$$
which gives us the conserved charge (or particle number in this case)
$$N = \int d^3x\, J^0 = -i \int d^3x\, \left( \phi^\dagger \dot{\phi} - \dot{\phi}^\dagger \phi \right)$$
Now this is fine when the fields involved are complex (due to the U(1) symmetry) and die off exponentially as $r \rightarrow \infty$, but what happens when there is spontaneous symmetry breaking? Is particle number still conserved? Does the concept of particle number still make sense?

Consider now an abelian Higgs model:
$$\mathcal{L} = \partial_\mu H^\dagger \partial^\mu H - \lambda \left( |H|^2 - v_0^2/2 \right)^2$$
This clearly has a U(1) symmetry, so at the outset we might assume the particle number is conserved analogous to the $\phi$ model above. However, due to the potential, the Higgs takes on a vacuum expectation value (vev), spontaneously breaking the U(1) symmetry. If we parametrize the Higgs in the unitary gauge so as to expand around the new vacuum
$$H = \frac{1}{\sqrt{2}} e^{i\theta/v_0} (h + v_0)$$
where $h$ and $\theta$ are real scalar fields, the lagrangian becomes (ignoring the massless Goldstone bosons)
$$\mathcal{L} = \frac{1}{2} \partial_\mu h \partial^\mu h - \frac{\lambda}{4} \left( h^2 + 2v_0 h \right)^2 = \frac{1}{2} \partial_\mu h \partial^\mu h - \frac{1}{2} m_h^2 h^2 - \lambda v_0 h^3 - \frac{\lambda}{4} h^4$$
where $m_h^2 = 2 \lambda v_0^2$. The U(1) symmetry in this new parametrization is now equivalent to $\theta \rightarrow \theta + \alpha$, and you can see the broken lagrangian is invariant under this change (the terms I ignored were derivatives of $\theta$ and are clearly invariant under a constant shift). If we just substitute this into the particle number integral we derived before, we get
$$N = -i \int d^3x\, \left( H^\dagger \dot{H} - \dot{H}^\dagger H\right) = \int d^3x\, \dot{\theta} \left( h + v_0 \right)^2$$
However, this integral is now divergent because of the $v_0^2 \dot{\theta}$ term which (may or may not) die out at infinity. If $\theta=\text{const}$, everything is fine, but if we say the $\theta$ field only has a simple linear time dependence $\theta = \omega t$ this becomes a constant. Does this mean the Higgs particle number cannot be defined? Or can this be saved by simply dropping the constant in the same way we drop infinite energy shifts from a Hamiltonian? The issue I have with the second one is that when we shift the Hamiltonian, it's by some universal constant that is the same throughout the theory. Here, the constant $\omega$ may depend on the system under consideration and is not universal to the theory. If there is any more complicated time dependence of $\theta$ than linear, this just compounds the issue. Is this suggesting that the argument of the $H$ field is fixed at infinity (which kind of makes sense) and I should just forget about this term altogether? Any insights?

2. Sep 1, 2015

### fzero

If you include the gauge field in your Lagrangian, you'll find that $\theta$ mixes with the gauge field via quadratic terms like $A^\mu \partial_\mu \theta$. The usual interpretation of this is to note that the field redefinition (gauge transformation) $A_\mu' = A_\mu + \partial_\mu \theta/q$ completely removes $\theta$ from the Lagrangian. Here $q$ is the charge of the Higgs field, perhaps other numerical factors will appear depending on conventions. This is interpreted as $\theta$ being absorbed ("eaten") by the gauge field to give the 3rd polarization of the massive gauge field. So the 2 d.o.f. of the massless gauge field and 2 d.o.f. of the complex scalar become 3 d.o.f. of a massive gauge field and 1 d.o.f. of a real massive scalar. There are no non-trivial conserved charges in this theory because there is no longer any global $U(1)$ invariance. Explicitly, one finds $N=0$ because $H$ is real in this gauge.