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Why does the Higgs field have less energy when it's non-zero than when it's zero? There are references to this question on the site, but they are too heavy going for me for a while yet. Anybody want to try a non math (or minimum math ) answer/analogy?

The basis for this question is p35 of Carroll's pop-sci book "The Particle at the End of the Universe" , in which there is a diagram comparing the resting value of fields such as electrons, quarks etc. as zero but the Higgs resting value as some non-zero value. I do understand that the Higgs field is not derived from a gauge theory, and that, afaik, all other fields are.

I simply want an as physical as possible, intuitive picture, rather than maths basis, for the discrepancy between the resting values of all other fields and that of the Higgs.

I self study, so whilst I can follow most of the maths up involved in QM to the Dirac equation level, my knowledge is (very) patchy in parts and sometimes its easier, as a QFT newbie, to get a analogy before diving into the underlying maths. I do fully understand that a physical picture is often misleading, or impossible. If it's not possible to give a physical explanation/picture, then that's perfectly OK and I will keep digging until I understand it on a maths level.

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    $\begingroup$ Because of the Mexican hat shape of the Higgs potential, the minimum energy of the Higgs potential doesn't occur at $\phi = 0$, but at $\phi\neq0$. I don't fully understand the question. $\endgroup$
    – innisfree
    Mar 25 '15 at 11:51
  • $\begingroup$ @innisfree. I have edited my question in, hopefully, a clearer way. Thanks very much $\endgroup$
    – user74893
    Mar 25 '15 at 13:48
  • $\begingroup$ I'm still not sure that the answer you want exists. The Higgs has a non-zero VEV, as described by Higgs mechanism. We don't know the origin of the Higgs potential or parameters in the Higgs potential. $\endgroup$
    – innisfree
    Mar 25 '15 at 14:52
  • $\begingroup$ You might like to look at the logo for this forum... $\endgroup$ Nov 26 '15 at 15:45
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This is more of an extended comment, because I don't know the answer, but hopefully this will be useful.

Any field with a non-zero spin must have to have a vacuum expectation value (VEV) of zero, because any other value would break Lorentz invariance. So spin 0 bosons like the Higgs are the only ones that can have non-zero VEVs.

As for the form of the Higgs potential, the one in the standard model is the only one that is renormalisable. So assuming the Higgs is correctly described by the standard model it has to have a potential of this form.

But none of this explains why the Higgs VEV is non-zero or what the physical interpretation is. I don't know of an explanation for this except that, well, that's the way the universe is.

There is a cliché you'll come across in quantum mechanics:

anything that isn't forbidden is compulsory

and it's not forbidden for the Higgs to have a non-zero VEV. So at the end of the day maybe the Higgs has a non-zero VEV because it can.

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  • $\begingroup$ Why wouldn't a vector boson VEV break Loretnz? It would give a preferred direction, e.g. $<A_\mu>$, wouldn't it? also, the cliche is Gell-Mann's totalitarian principle. $\endgroup$
    – innisfree
    Mar 25 '15 at 11:53
  • $\begingroup$ Yes, agreed.${}{}{}$ $\endgroup$ Mar 25 '15 at 11:59
  • $\begingroup$ I'm not sure about this. $m^2\ge0$ in $V \in m^2\phi^2$ isn't forbidden by symmetries/renormalizability, and would keep $<\phi>=0$.If $<\phi>=0$ isn't forbidden, why isn't it compulsory? This is all a little bit specious, IMHO. $\endgroup$
    – innisfree
    Mar 25 '15 at 14:54
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From the form of the Higgs potential (which is quartic, the famous Mexican hat) you can see that for $\Re \phi =0$ as well as for $\Im \phi=0$ (the real and imaginary parts of the Higgs field), it is sitting on the unstable top of the hat. Thus, a small perturbation would lead it away from $0$. Since the potential has a smaller value away from zero so does the energy. Take a look at this picture for more intuition.

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    $\begingroup$ I don't think this answers the question because it doesn't say why the VEV is not zero. The VEVs of all the other known bosonic fields are zero, so what is it about the Higgs that mkes it different? $\endgroup$ Mar 25 '15 at 11:29
  • $\begingroup$ @john they don't have interactions (i.e. a potential) that favour non-zero VEV $\endgroup$
    – innisfree
    Mar 25 '15 at 11:57
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    $\begingroup$ @innisfree: what I would guess the OP wants to know is why the Higgs potential is the shape it is. Whether that has an answer I'm not sure. $\endgroup$ Mar 25 '15 at 12:01
  • $\begingroup$ @JohnRennie. Hi John, my fault entirely, I should have written my original question more clearly. I hope my edit makes it more understandable regards $\endgroup$
    – user74893
    Mar 25 '15 at 13:56
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    $\begingroup$ Trying to consider other forms of potentials leads to inconsistencies. Indeed, the $\phi^4$ potential is gauge invariant and renormalizable and thus we use this in our effective Lagrangian. Being a $\phi^4$ potential one can try to find the extremea only to see that they are not at 0. This exercise is very simple to do e.g. in two dimensions to convince one self this is true. $\endgroup$
    – Marion
    Mar 25 '15 at 14:35

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