As I understood, the Higgs field has a non-zero vacuum expectation value (vev) on zero energy.

For example, in the case of the non-quantized electric field, it would be measured in $\dfrac{J}{C}$, i.e. $\dfrac{\textrm{energy}}{\textrm{charge}}$. Thus, it shows the energy in the field per unit charge.

How could it be a simply energy-like value in the case of the Higgs field?


1 Answer 1


When we say field, you shouldn't immediately think electric field. You should instead think of the easier scalar field. Also the electric field is not J/C. That is the voltage potential function. Now look at how the action of the field configuration depends on the field you are looking at. For example, it might be $S=\int d^4 x \partial^\mu \phi \partial_\mu \phi$. We want this to have dimensions that an action needs which is the same as the dimensions of the constant $\hbar$. Now do a dimension count. Usually people ignore $c$ and $\hbar$ so that they can just count powers of $GeV$ and then use $c$ and $\hbar$ to fix it later.

  • $\begingroup$ I know people usually ignore $c$ and $\hbar$, but could you convert the Higgs VEV of 246 GeV to SI units? $\endgroup$
    – Paul
    Mar 31, 2017 at 17:19
  • $\begingroup$ @Paul Here is very well visible the knowledge gap between us, enthusiast laymans , and the physicists :-) This was exactly what I wanted to ask... I think you could make a question from that. I would suggest to ask not only from the Higgs, but from the particle fields in general. But mention your specific interests about to Higgs VEV. $\endgroup$
    – peterh
    Mar 31, 2017 at 17:25
  • $\begingroup$ @peterh It seems the most straightforward way starts with the vev being 246 GeV/$\sqrt{\hbar c}$, which converts to m$^{-1/2}$ kg$^{-1/2}$ s$^{-1}$. That doesn't really make things much clearer. It's nothing at all like newtons per coulomb, but it's at least SI. $\endgroup$
    – Paul
    Mar 31, 2017 at 18:02
  • $\begingroup$ @Paul Right, thanks! Then it may be a good question, how does it come out from an SI viewpoint. $\endgroup$
    – peterh
    Mar 31, 2017 at 18:14

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