Particle # conservation in a spontaneously broken theory

In summary, the conversation discusses the conservation of particle number in a theory with a complex scalar field and global U(1) symmetry, and how this changes in the case of spontaneous symmetry breaking. The abelian Higgs model is used as an example, and it is noted that the particle number can no longer be defined due to the mixing of the field with the gauge field. This leads to the interpretation of the field being "eaten" by the gauge field, resulting in a different number of degrees of freedom. Furthermore, there are no non-trivial conserved charges in this theory due to the absence of global U(1) invariance.
  • #1
EricTheWizard
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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?
 
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  • #2
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.
 
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1. What is Particle # conservation in a spontaneously broken theory?

In a spontaneously broken theory, the symmetry of the system is broken and this can lead to the creation of new particles. Particle # conservation refers to the principle that the total number of particles in a system remains constant, even after the symmetry is broken.

2. How does particle # conservation differ in a spontaneously broken theory compared to a non-broken theory?

In a non-broken theory, the symmetry of the system is preserved and the total number of particles is conserved. However, in a spontaneously broken theory, the symmetry is broken and new particles can be created, thus changing the total number of particles in the system.

3. Why is particle # conservation important in a spontaneously broken theory?

Particle # conservation is important in a spontaneously broken theory because it helps to explain the behavior and interactions of particles in a system. By understanding how the total number of particles is affected by the breaking of symmetry, scientists can better understand the underlying principles of the theory.

4. Can particle # conservation be violated in a spontaneously broken theory?

Yes, particle # conservation can be violated in a spontaneously broken theory. This can occur in certain circumstances, such as during high energy collisions or in the presence of strong external fields. However, this violation is typically only temporary and the total number of particles will eventually be conserved again.

5. How does the Higgs mechanism contribute to particle # conservation in a spontaneously broken theory?

The Higgs mechanism is a key aspect of spontaneously broken theories and plays a crucial role in particle # conservation. It explains how certain particles acquire mass through interactions with the Higgs field, which is responsible for breaking the symmetry of the system. This mechanism helps to ensure that particle # conservation is maintained even after the symmetry is broken.

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