# Higgs potential shape meaning

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## Main Question or Discussion Point

Hello! I am going to talk here about a real scalar field for simplicity. If we have a Higgs-like potential for a real scalar field, the graph of the potential looks like a section of a "Mexican hat", with a bump at 0 and two absolute minima at, say, ##\pm a##. This is the plot I see in any book talking about spontaneous symmetry breaking. Although I understand how the symmetry is broken I am a bit confused about the graph itself (the actual picture of it). The graph has on the y-axis the potential ##V(\phi)## and on the x-axis the field ##\phi##. I am not sure how to think about the x-axis, as the field is not an independent variable, it depends on the space-time variables i.e. ##\phi = \phi(x,y,z,t)##. So if I read on the graph that ##V(a)## is a minimum, this means means that when the value of the field is ##\phi(x,y,z,t)=a##, the potential reaches a minimum. But what does it mean that ##\phi(x,y,z,t)=a##? The field doesn't take a single value. One can have ##\phi(2,7,\pi,12.4)=a## but ##\phi(e,22,0,12.4)=7*a+25##. Do we assume for this plot that the field is constant everywhere? And if so, why do we do this? The field has a kinetic term, so it can have change in time. Do we do it because we want to see the actual minimum value of the vacuum and hence, as we need minimum energy, we set it constant just to discard the kinetic term? Thank you!

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Orodruin
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The field doesn't take a single value.
If you want it to have the minimal energy configuration, it does.

The potential (density) V is a function of the field value. If you want to plot it, then it makes sense to plot it as a function of the field value. Even if you have a field configuration which is not the global minimum, your potential density at an event only depends on the field value at that event (of course, the potential density may then be different at different spacetime events, but this is not the issue here).

Do we do it because we want to see the actual minimum value of the vacuum and hence, as we need minimum energy, we set it constant just to discard the kinetic term?
Yes. The kinetic terms are non-negative and will be zero when the field is constant. The mimimal energy is therefore obtained for a constant field at the bottom of the potential density.

Hello! I am going to talk here about a real scalar field for simplicity. If we have a Higgs-like potential for a real scalar field, the graph of the potential looks like a section of a "Mexican hat", with a bump at 0 and two absolute minima at, say, ##\pm a##. This is the plot I see in any book talking about spontaneous symmetry breaking. Although I understand how the symmetry is broken I am a bit confused about the graph itself (the actual picture of it). The graph has on the y-axis the potential ##V(\phi)## and on the x-axis the field ##\phi##. I am not sure how to think about the x-axis, as the field is not an independent variable, it depends on the space-time variables i.e. ##\phi = \phi(x,y,z,t)##. So if I read on the graph that ##V(a)## is a minimum, this means means that when the value of the field is ##\phi(x,y,z,t)=a##, the potential reaches a minimum. But what does it mean that ##\phi(x,y,z,t)=a##? The field doesn't take a single value. One can have ##\phi(2,7,\pi,12.4)=a## but ##\phi(e,22,0,12.4)=7*a+25##. Do we assume for this plot that the field is constant everywhere?
Field which isn't constant everywhere means you have "particles" of this field (perturbations of the field). When you look for a vacuum state, of course this shouldn't be the case.

Therefore, yes, you assume that the field does take the same value everywhere in the vacuum. Modulo gauge invariance: field can have seemingly different values at different points, but if they can be made to be the same value everywhere via a gauge transformation, then it's not a state with particles.

The Higgs field cannot be a constant

Orodruin
Staff Emeritus