Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Higgs field is pretty simple for me to understand, you have one field that creates one particle (Higgs boson).

So I continue to assume one field one particle.

Up field creates a up quark. Down field creates a down quark. Strong field creates a gluon. Electron field creates a electron. Electromagnetic field creates a photon. ...

share|improve this question
2  
Electric field creates a electron. Magnetic field creates a photon ---- No, both fields are the same and they are mediated by photons. Electron is just a particle that couples with the field. No matter what, particles and their antiparticles are part of the same field ("part of"="oscisslations in"). I think that neutrinos are part of the same field as their corresponding massive leptions. Dunno about quarks. –  Manishearth Jul 18 '12 at 1:12
    
Can you make a map like this of all the fields because its really confusing me. –  Gert Cuykens Jul 18 '12 at 1:18
    
Yes you can, it's the standard model, and this is the modern definition of an elementary particle, adopted in the mid 1970s. –  Ron Maimon Jul 18 '12 at 1:22
    
@Ron I don't believe that edit was necessary. –  David Z Jul 18 '12 at 1:26
    
@DavidZaslavsky: Is that overstepping the bounds of reasonable edit? It might make some parts of your answer a non-sequitor. –  Ron Maimon Jul 18 '12 at 2:06
show 3 more comments

3 Answers

up vote 8 down vote accepted

You're right that one field basically corresponds to one fundamental particle. To be more precise, there is one type of fundamental particle for each degree of freedom (well... let me not get into that subtlety) in the fields of the standard model. What I mean by that is that you can have linear combinations of fields, but they don't correspond to brand new particles - for example, if $u(x)$ and $d(x)$ are the up and down quark fields, you can have things like $\frac{1}{\sqrt{2}}[u(x) + d(x)]$, but that doesn't make a new particle, it's just a superposition of an up quark and a down quark. This is relevant because e.g. the photon field is actually represented by a linear combination of what you might call two separate fundamental fields.

Off the top of my head, I can think of 58 quantum fields directly included in the standard model, corresponding to the following particles:

  • (6) Left-handed and right-handed electron, muon, and tau lepton
  • (3) Left-handed electron, muon, and tau neutrinos
  • (36) Left-handed and right-handed quarks of six flavors (down, up, strange, charm, bottom, top) and three colors (red, green, blue)
  • (4) Electroweak bosons (W+, W-, Z, photon)
  • (8) Gluons of all non-singlet combinations of two of the three colors
  • (1) Higgs field

I think it's considered likely that there are 3 additional right-handed neutrino fields, although that's not actually confirmed yet. Plus there's a gravity field, corresponding to the (hypothetical) graviton, but that's not part of the standard model. And of course there is the possibility of multiple Higgs fields.

In general, each field takes its name from the corresponding particle. The only exception I can think of is the photon, whose field is sometimes called the electromagnetic field. But to be accurate, the photon field actually corresponds to the electromagnetic vector potential $A^\mu$, and the thing you may be used to hearing called the "electromagnetic field" - the tensor containing $\vec{E}$ and $\vec{B}$ - is actually the derivative of that potential $A^\mu$.

share|improve this answer
1  
Interesting, I thought each lepton flavor had its own field not broken into neutrinos/massive leptons. –  Manishearth Jul 18 '12 at 1:26
    
Nah, they're separate fields. –  David Z Jul 18 '12 at 1:27
1  
They aren't separate--- the left handed electron and the electron neutrino together are two components of a weak doublet. You can't call them separate any more than the red green and blue up quark. They are separated by Higgs mixing only. –  Ron Maimon Jul 18 '12 at 1:33
    
@RonMaimon, I was just asking that at your post. So, does this correspond with the way I have been looking at things which is that, for example, the left handed electron and electron neutrino are, if you will, "different sides of the same coin" where, in this case, "coin" is left handed lepton? –  Alfred Centauri Jul 18 '12 at 1:36
    
Yes. They are two versions of the same field, but the Higgs picks a preferred one to be massive. –  Ron Maimon Jul 18 '12 at 2:06
show 2 more comments

This is the modern definition of an elementary particle--- a particle which is the quantum of a field in the renormalizable Lagrangian. There used to be other definitions, but they are no longer considered fundamental, since the elementary particles sometimes can't be isolated. This means that quarks aren't poles in the S-matrix, as particles were defined by Wigner, and neither are gluons. Also, some particles are massless, and they cannot be nonrelativistic, so they aren't objects with a straightforward forward-in-time position wavefunction description, as particles were defined in the 1930s.

In the modern definition, there is one kind of elementary particle for every field, but the correspondence you write is wrong:

  • Photon: electromagnetic field (which is really a Higgs-induced linear combination, a mixture of a fundamental field and a part of another fundamental field)
  • W/Z bosons: weak gauge field (the Z and the photon are mixed together)
  • Gluon: strong gauge field (not mixed up with anything, but confined)
  • quarks, u/d/c/s/t/b: each of these have a field. Technically the left parts are separate fields from the right part, but they are mixed by the Higgs into massive pairs.
  • leptons e/$\nu_e$/$\mu$/$\nu_\mu$/$\tau$/$\nu_\tau$: Each of these have a field. Again, the left and right parts are fundamentally separate, but mixed by the Higgs. This time, the neutrinos are left-over after the electron-like leptons make three massive pairs.
  • Higgs: in the standard model, it is an elementary scalar, only the radial part is a particle, the rest is absorbed into polarizations of the W's and Z's.

That's all the elementary particles in the standard model.

share|improve this answer
add comment

Note that there are fields which do not have a physical particle associated to them. These are fields that are included for technical reasons. For example, the Faddeev-Popov ghosts fields in the SM.

Furthermore, in QFT in fixed curved backgrounds the particle concept is trickier. In dynamical backgrounds the situation is even worse.

share|improve this answer
1  
The FP ghosts are not necessary, the are perturbation series artifacts in covariant gauges. You can always quantize in axial gauge. All they are are fictitious particles with wrong statistics that cancel out the extra degrees of freedom in covariant gauge boson loops. There are no other such fields in the SM, so it's not "for example", that's it. –  Ron Maimon Jul 18 '12 at 7:04
    
I agree with you, Ron. However, FP ghosts aren't the only non-physical fields. One also has auxiliar non-dynamical fields (like the Nakanishi-Laudrup field) which again appears in some very common formulations of the SM. My answer was a caveat to the identification between physical particles and fields. In the same way, one can say that three components of the SM scalar doublet are not independent particles but longitudinal degrees of freedom of the massive vectorial bosons although in some formulations (i.e., in some gauge fixings) they look as independent particles. –  drake Jul 18 '12 at 16:00
    
FP ghosts are the only non-physical fields. The "Nakashini-Laudrup" field is a stupid optional tack on with no derivative terms, and cannot be interpreted as a particle. You can add a billion o these as you like, without changing the physics--- they correspond to particles with infinite mass. You can add as many infinite mass particles as you like. You cannot say that about the SM doublet really, this is just changing gauge. I don't like this manner of speaking--- you are looking for complications where there are none. –  Ron Maimon Jul 18 '12 at 16:17
    
Maybe I have not expressed myself clearly, Ron. What I was trying to say is that there is not a particle for every field one sees in a path integral o a Hamiltonian. I was not trying to add complications but precision. –  drake Jul 18 '12 at 18:43
    
If you do the path integral over the fewest number of fields, integrating out the auxiliary fields, removing non-interacting sectors, and choosing a physical gauge, then there's one particle per field. Please give the exhaustive list of exceptions: 1. ghosts, 2. no-derivative auxiliaries, 3. completely noninteracting sectors (like right-handed components of neutrinos you sometimes see for a 4-spinor formalism). If you gave the exhaustive list, I would upvote, but if you say "for example this, and sometimes other things" you creates uncertainty and worry for someone who doesn't know it. –  Ron Maimon Jul 18 '12 at 21:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.