# Higgs Field

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1. May 29, 2014

### RobTheHoff

Is the Higgs field 4 dimensional?

2. May 29, 2014

### vasilis

Yes, it's an SU(2) doublet with two complex components, or equivalently 4 real components.

3. May 29, 2014

### Bill_K

There is only one Higgs field, and it is a scalar (a Lorentz invariant). It has the same value everywhere, and to all observers.

The other three fields that vasilis refers to are not observed. They are absorbed and become part of the W and Z bosons when those particles acquire a mass.

4. May 29, 2014

### Staff: Mentor

Unless we excite it in the LHC?

5. May 29, 2014

### Bill_K

Even in the LHC, the Higgs field has the same value, namely v = 246 GeV. Producing an excitation (Higgs boson) does not change the value of the field.

If v ever changed, the masses of all the fermions and gauge bosons would change, since they are proportional to it. The weak coupling constants depend on it also. And obviously this does not happen!

6. May 29, 2014

### Staff: Mentor

Sorry, I don't understand that. How can a field get excited and not change at all - at the same time?
246 GeV is the vacuum expectation value. We don't have a vacuum if there is a Higgs boson.

7. May 30, 2014

### nikkkom

Hijacking the thread a bit, I have a question about Higgs vev. It's stated that it is 246 GeV.

I'm puzzled by the unit used. GeV is energy unit. So, there is 246 GeV of Higgs field... in what? Cubic meter? Cubic millimeter? Cubic planck lenght?
Why unit is not, say, GeV/(mm^3)?

8. May 30, 2014

### Bill_K

It's not the energy OF anything, it merely serves to set the scale. All the particle masses are proportional to v.

For example, MW = ½ g v, where g is the (dimensionless) weak coupling constant. MW = 80 GeV and g = 0.65, so v comes out to 246 GeV.

9. May 30, 2014

### Orodruin

Staff Emeritus
By this logic, electromagnetic waves should not have any field strength since the vev of the field is zero and producing excitations (photons) would not change the field value.

10. May 30, 2014

### Bill_K

See, mfb? This is why your comment was counterhelpful. Maybe you'd like to try explaining to Orodruin the difference between a macroscopic field and its quantum excitation.

11. May 30, 2014

### ChrisVer

Also the masses change with energy (yukawa coupling constant is a running constant)?

12. May 30, 2014

### ofirg

Yes, but only one of the components has a non zero vacuum expectation value (vev) and physical particle like excitations.

The other three have zero vev and their particle excitations are not physical scalar like excitations.

It is important to note that the physical mass (pole mass) is a single number. The running mass which varies with energy (or more correctly, scale) is a parameter used in perturbation theory with the scale chosen to improve convergence.

I believe Bill_K means that if you look on the expectation value of a zero vev field in a state with an excitation from that field ( such as looking at the expectation value of the higgs field without the vev in a state with a higgs boson) the expectation value is still zero. You need to build states which are a superposition of states with different particle numbers for the field to have non zero expectation value.

13. May 30, 2014

### vasilis

In regard to the confusion about the exciting the Higgs,

By definition, a particle is a quantum flluctuation about a vacuum, i.e pick a classical field configuration of minimal energy, that is the vev of your field, consider small fluctuations about it and quantize them (have them obey Bose or fermi-Dirac statistics). This is precisely what is meant by exciting the vacuum with a single creation operator. Note, in QFT the fields appearing in the lagrangian should always be understood as a two-term structure: the vev plus the fluctuation.

Since the vacuum must be Lorentz invariant, the only field which can have a non-zero vev is the (scalar) Higgs field. (Of course we can have non-zero vevs for non-abelian gauge fields, e.g. Instanton backgrounds and etc, but for our discussion we can ignore these). So, for most fields the vev piece is set to zero and what really appears in the lagrangian is just the fluctuation which you would like to quantize.

Now, the essential point is that the background is fixed at the vevs of your fields and it never changes. Thus, we have zero vevs for all the fields but the Higgs. Particles should be thought as small (quantum) fluctuations propagating in this background with their mass parameters depending on the vev of the Higgs field. For all these to work out nicely though, we must couple the "massive" fields to the Higgs field, which appears in L with a zero vev at the moment, then add a spontaneously-symmetry-breaking potential term in the Lagrangian which picks up the vev of the Higgs and thereafter we need to consider a shift of the Higgs in order to make sure that the theory will be quantized on a true (stable) vacuum. Note that initially there are no mass terms for these fields. After the shift, the massive fields acquire a mass term from the vev piece as well as some other coupling terms with the Higgs. So, the mass terms do not involve any interactions with the "quanta" of the Higgs field but rather they contain an explicit factor of its vev. In other words, all the particles feel heavy when they propagate in the presence of this non-zero vev.

Higgs excitations do not alter the vev and hence, neither the mass parameters. They only turn on interactions among the massive fields and the Higgs bosons with a yukawa coupling parameter. You can probe these interactions at different energy scales and determine the value of the yukawa coupling at these energies. You can also determine the mass parameters and then given e.g. Mass of W = g.v/2 you can determine v=vev, and confirm that it's independent of the probe scale.