I'm just an aficionado, but my understanding is that in QFT, the photon is an excitation of the electromagnetic field, the electron is an excitation of the electron field, and so on. Is there a quantum field for each elementary particle, and if so, how many?

  • 3
    $\begingroup$ en.wikipedia.org/wiki/Standard_Model $\endgroup$ – Count Iblis Apr 18 '15 at 17:24
  • $\begingroup$ @Count Iblis The article says 61 elementary particles in the Standard Model. So, 61 quantum fields? $\endgroup$ – mick Apr 19 '15 at 14:41
  • 3
    $\begingroup$ I think you are counting the components of the fields separately, it's more natural to count them once. E.g. you don't count an electron twice just because it has two spin degrees of freedom which are both combined as one spinor field, but then in the relativistic treatment you have one Dirac field with four components, two of which correspond to the two spin degree of freedom of the positron. $\endgroup$ – Count Iblis Apr 19 '15 at 15:52
  • 1
    $\begingroup$ You can't that simply count quantum fields. If there is Lorentz invariance, it may seem natural to count Lorentz multiplets (like the different polarizations of the electron) as one. But then if you believe in supersymmetry you'll have to group more. If you believe in string theory there's only one master field. As for the second part of the question: yes, elementary particles are simply excitations of elementary quantum fields, so there is one-one relationship between the two concepts (as long as you make the correct Lorentz groupings, or you will have more fields that particles). $\endgroup$ – David Vercauteren Apr 21 '15 at 9:38

The Standard Model of particle physics is the current well stablished theory for elementary particles. In it, fields are grouped together using symmetries (Lorentz and gauge) in what are called multiplets. Because of it, it is not clear how to count the number of quantum fields there are in it.

For example, we can count for each multiplet how many real degrees of freedom it has. The counting goes as follows (don't worry if you don't understand some of the words, they're just ways of stating how the symmetries act on the multiplet):

  • Complex scalars:
    • Higgs doublet: $SU(2)$ doublet, $4$ d.o.f.
  • Two-component spinors:
    • Right-handed electron ($3$ flavors): singlet, $4$ d.o.f.
    • Right-handed up quarks ($3$ flavors): color-triplet, $12$ d.o.f.
    • Right-handed down quarks ($3$ flavors): color-triplet, $12$ d.o.f.
    • Left-handed lepton doublet ($3$ flavors): $SU(2)$-doublet, $8$ d.o.f.
    • Left-handed quark doublet ($3$ flavors): color-triplet $SU(2)$-doublet, $36$ d.o.f.
  • Massless complex vectors:
    • $U(1)_Y$ gauge boson: singlet, $4$ d.o.f.
    • $SU(2)$ gauge boson: $SU(2)$-triplet, $12$ d.o.f.
    • Gluon: color-octet, $32$ d.o.f.

So there are $19$ multiplets and if we decompose them we get $268$ real fields.

Now, in the Standard Model there is something called spontaneous symmetry breaking. It makes massive the massless fields it contained in principle. This changes the counting and gives us the fields associated with the particles we are more familiar with:

  • The Higgs boson
  • 3 charged leptons: electron, muon and tau.
  • 3 neutral leptons: the neutrinos corresponding to electron, muon and tau.
  • 6 quarks (3 colors each): up, down, charm, strange, bottom and top.
  • The photon
  • 3 electroweak massive bosons: $Z$ and $W^\pm$
  • 8 gluons

Counting this way gives us 37 fields. As you can see, field counting can be very arbitrary as it depends on what you understand by a single quantum field


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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