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(adsbygoogle = window.adsbygoogle || []).push({});

$$ J^\mu = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta\phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^\dagger)} \delta\phi^\dagger = -i \left( \phi^\dagger \partial^\mu \phi - \partial^\mu \phi^\dagger \phi \right) $$

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?

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# Particle # conservation in a spontaneously broken theory

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