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As known, the Higgs field has a 4-th grad "Mexican hat" potential, described by the formula $\nu = -\mu\phi^*\phi+\lambda(\phi^*\phi)^2$:

Higgs potential

Where $\phi$ are the 4-vector of the components of the Higgs field and $\mu$ and $\lambda$ are constant real values.

It means, rotating the Higgs field components doesn't require energy. Thus, its components should be easily rotable on a 4-Sphere with the radius of its current length.

Despite that, the experimental evidence shows that the Higgs field seems very stable in the whole Universe. Why it doesn't rotate even from the slightest quantum fluctuations?

Maybe the space-derivate of the Higgs field has some (probably not really negligible :-) ) energy density?

peterh
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2 Answers2

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The Higgs potential in this two dimensional perspective $$ V(\phi) = -\mu|\phi|^2 + \lambda|\phi|^4 $$ has derivatives $$ \frac{dV(\phi)}{d\phi} = -\mu\phi^* + 2\lambda|\phi|^2\phi^*. $$ That is zero for $\phi = 0$ and $|\phi|^2 = \mu/2\lambda$. This tells us the magnitude of the field, but the field is complex. There is then a degenerate set of values the field may assume $$ \phi = \frac{\mu}{2\lambda}e^{i\phi}. $$ This is the circle at the bottom of the Mexican hat potential. The field is in fact a pair of doublets so this two dimensional picture is expanded to four dimensions

There is this degenerate vacuum of the low energy Higgs field. This is a boson condensate that couples to the electroweak field. The fact there is this degenerate Higgs field configuration is in fact what makes the Higgs mechanism work.

Lawrence B. Crowell
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  • O.k., until the last paragraph I know. The last paragraph: "degenerate vacuum" means that non-zero VeV? Thus the coupling with the electroweak field causes the observed values of $\phi$? Isn't it into the other direction, i.e. $\phi$ causes the electroweak properties (and a significant part of the masses of many particles)? – peterh Jun 15 '16 at 12:18
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My layman answer, after a lot of digging, is this.

The total energy density of the Higgs field is described by this formula:

$H(\phi )={\frac {1}{2}}\left|{\dot {\phi }}\right|^{2}+\left|\nabla \phi \right|^{2}+V(\left|\phi \right|)$

The second term is the sum of the squares of the gradient of the components of the Higgs field. It means, if the Higgs field changes in space, it has an energy density. And nature wants an energy minimum.

peterh
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  • Being a total QFT layman, this answer might be essentially bad, in this case I am ready to delete it asap. – peterh Oct 13 '20 at 18:33