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Why is it that the field configuration that minimizes the Higgs potential is said to be a constant one (meaning $\phi(x) = \phi_0$)? Is it a physical reasoning, which kind of makes sense, since a non flat ground state of the field would be rather strange? Or is it a mathematical reasoning that the field configuration which minimizes the potential must be a constant one? If so how is it done? (It is not obvious to me right now, since the fields are functions of space-time (no quantization yet)).

I have read this and links therein but it did not really answered my question.

Qmechanic
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AlmostClueless
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1 Answers1

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The ground state (or vacuum state) minimizes the energy, not just the potential. A solution of the field equations with non-constant field $\phi$ (satisfying $|\phi(x)|=\phi_{0}$, so that the potential is minimized) will still have kinetic energy associated with it—making it automatically not the vacuum.

That’s all there is to say about finding the vacuum state of a scalar field. However, things get more complicated for fields with higher spin, like gauge fields. For a gauge field $A^{\mu}$, there are certain kinds of spacetime dependence that do not contribute to the energy, because they the energy only depends on the field strength $F^{\mu\nu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$. A gauge transformation can make $A^{\mu}(x)$ spacetime dependent, without affecting $F^{\mu\nu}=0$. In certain cases, such as with nonabelian gauge fields, this can mean that there are different vacuum states, with different spacetime-dependent $A^{\mu}$, which nonetheless all minimize the energy.

Buzz
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