# Is there a rest frame for the Higgs boson?

If there is a non-zero expectation value for the Higgs boson even in "vacuum", since the Higgs boson has a mass unlike photons, then I would expect it to have a rest frame.

So why doesn't a non-zero expectation value for the Higgs boson not only break electroweak symmetry, but also break Lorentz symmetry?

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Damn good question, I say. +1 –  user346 Mar 19 '11 at 17:15

Dear John, the vacuum doesn't contain any physical Higgs bosons; it only contains a "Higgs condensate". So the vacuum doesn't break the Lorentz symmetry.

The potential energy energy for the Higgs field looks like $$V(h) = \frac{m_H^2}{2} (h-v)^2 + O((h-v)^3)$$ for $h$ near $v$, so it is minimized by $h(x,y,z,t)=v$. Note that $V(h)$ is a Lorentz scalar so when added to the Lagrangian, it preserves the Lorentz symmetry.

The nonzero value of $h$ at each point is what breaks the electroweak symmetry because $h$ is a component of a doublet whose only symmetry-invariant point is $h=0$. But because this configuration is constant, it doesn't pick any preferred reference frame. There are no physical particles in the vacuum state.

Physical Higgs bosons may be added and they're energy quanta that allow $h$ to deviate away from $h=v$ - more than the uncertainty principle requires. Those quanta carry energy and momentum $(E,p)$ that satisfies $E^2 - p^2 = m_H^2$. That's the energy one has to add to the energy of the vacuum state; the latter state is free of particles and its energy may be chosen to be zero.

(Let me not discuss the cosmological constant issues here because they have nothing to do with this question.)

When it comes to the difference between the vacuum, which has no particles, and states with particles (Higgs bosons), you may define a new field, $h'=h-v$, and this field will be expanded around zero just like the electromagnetic fields or other fields. The vacuum will be at $h'=0$ and it may perhaps make it easier to understand why there's no Lorentz-symmetry breaking in the vacuum. The $SU(2)\times U(1)$ symmetry will act "non-linearly" on $h'$ - in a way that can be easily deduced from $h'=h-v$.

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