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How big is the standard model?

  1. Mar 16, 2013 #1
    How many internal degrees of freedom does it have? In other words, how many 4d real functions do we have to specify in configuration space (i mean the space we use in Path integration) in order to specify a state?

    It is difficult to state the question, so, in order to be more precise, the answer is an integer number, i tried to sum but then I found difficulties. I mean, there are 6 kind of leptons fields but they count as 12 because of spin (And what about the antiparticles, they make it 24?)Then we have 3 quarks but we have to multiply them by three? And then the gauge bosons, the higgs, has anyone ever made the math?

  2. jcsd
  3. Mar 16, 2013 #2


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    6 leptons, 6 x 3 quarks, 8 gluons, W, Z, photon and Higgs. That's 36?
  4. Mar 16, 2013 #3
    But why do you count each lepton as one 4d funtion. Don't you have to count it as 2 because of spin? (Or 4 because of antiparticles?) the same for the gauge bosons? And the higgs, do i have to count it as 1 4d function or as 4 (i think its one because the other 3 real components are forced to be zero, right?)

  5. Mar 16, 2013 #4
    Let's count the degrees of freedom before electroweak symmetry breaking, since there are the same number of d.o.f. before and after.

    The left-handed leptons are three SU(2) doublets, each with two spin components: 3x2x2 = 12 d.o.f.

    The right-handed leptons are three SU(2) singlets, each with two spin components: 3x1x2 = 6 d.of.

    The left-handed quarks are three SU(2)-doublet SU(3)-triplets, each with two spin components: 3x2x3x2 = 36 d.o.f.

    The right-handed quarks are six SU(2)-singlet SU(3)-triplets, each with two spin components: 6x1x3x2 = 36 d.o.f.

    The U(1) gauge boson has two polarizations = 2 d.o.f.

    The SU(2) gauge bosons have two polarizations and come in three different kinds: 2x3 = 6 d.o.f.

    The SU(3) gauge bosons have two polarizations and come in eight different kinds: 2x8 = 16 d.o.f.

    The Higgs is a complex SU(2) doublet with 2x2 = 4 d.o.f.

    Adding everything up, 12 + 6 + 36 + 36 + 2 + 6 + 16 + 4 = 118 d.o.f.

    I suppose, though, that we really ought to add in right-handed neutrinos, which I think would be three SU(2) singlets, each with two spin components: 3x2 = 6 new d.o.f.

    So with massive neutrinos we have 118 + 6 = 124 d.o.f. in the standard model.

    (Edit: originally I forgot to add in the 4 in the totals!)
    Last edited: Mar 16, 2013
  6. Mar 16, 2013 #5
    Awesome answer, thank you very much!
  7. Mar 16, 2013 #6
    The leptons are complex as the higgs right? Shouldnt you multiply by 2 like you did with the higgs?

    Why the right handed quarks count as Six And the left ones as 3? Why does that not happen with the leptons?

  8. Mar 16, 2013 #7


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    124. You forgot the Higgs, did you?
  9. Mar 16, 2013 #8
    The Dirac fields are all complex, which you might expect to double the number of degrees of freedom, but spinor fields obey the Dirac equation, which cuts this in half, so it's a wash. In detail:

    Consider a complex scalar field ##\phi##. It represents two particles: a scalar and its antiparticle. We can see this when we write out the Fourier expansion of a complex scalar field in terms of creation and annihilation operators:

    ##\phi(x) = \int d^4 p \left ( a(p) e^{i p x} + b(p)^\dagger e^{-i p x} \right )##

    Here ##a^\dagger## creates a particle and ##b^\dagger## creates an antiparticle. So indeed ##\phi## represents two particles.

    Now consider a 4-component Dirac spinor ##\psi_\alpha##. It represents four particles: a spin-up fermion, a spin-down fermion, a spin-up antifermion, and a spin-down antifermion. But since ##\psi_\alpha## is complex in the sense that ##\psi_\alpha^\dagger \neq \psi_\alpha##, why does it not represent eight particles, two per component? If we wrote down the most general four-component complex field, it would indeed represent eight particles. The most general four-component complex field can be written

    ##\psi_\alpha(x) = \int d^4 p \sum_{i = 1}^4 \left ( u^{(i)}_\alpha(p) a^{(i)}(p) e^{i p x} + \bar{v}^{(i)}_\alpha(p)b^{(i)}(p)^\dagger e^{-i p x} \right )##

    where the ##u^{(i)}_\alpha##, i = 1, 2, 3, 4 are four linearly independent vectors in Dirac space. There are eight creation operators here, ##{a^{(i)}}^\dagger## and ##{b^{(i)}}^\dagger## for i = 1, 2, 3, 4.

    This looks like the usual mode expansion of a Dirac field except we seem to be summing over four spins. This is because two of the spins go away when we impose the Dirac equation

    ##(i \gamma^\mu_{\alpha\beta} \partial_\mu - m \delta_{\alpha\beta}) \psi_\beta(x) = 0##

    I've written out the Dirac indices explicitly in the Dirac equation to emphasize that in addition to being a differential equation, it is also a matrix equation in Dirac space. When you apply this to the Fourier-expanded version of the field above, you get some conditions on ##u## and ##v##:

    ##(\gamma^\mu_{\alpha \beta} p_\mu + m \delta_{\alpha \beta}) u^{(i)}_{\beta}(p) = 0##

    ##(\gamma^\mu_{\alpha \beta} p_\mu - m \delta_{\alpha \beta}) v^{(i)}_{\beta}(p) = 0##

    It turns out that each of these matrix equations only has two linearly independent solutions. So the index ##i## can only run from 1 to 2. We've lost half of our original eight degrees of freedom to the Dirac equation, which is why the Dirac field ##\psi_\alpha## only represents four degrees of freedom.

    Of course, for the counting I did earlier we have to remember that before electroweak symmetry breaking the left- and right-handed components of each Dirac spinor are separate fields, each carrying two of the degrees of freedom.

    There are equal numbers of left- and right-handed quarks, it's just that the left-handed fields are organized into SU(2) doublets but the right-handed ones aren't. The left-handed quarks form three SU(2) doublets [(u, d), (c, s), (t, b)] while the right-handed quarks are six separate SU(2) singlets [u, d, c, s, t, b]. You'll note that we end up with equal numbers of degrees of freedom for the left-handed and for the right-handed quarks (36 in each case), as we must.

    The same thing happens with leptons, except that before neutrinos were known to have mass you could get away with claiming that there were no such things as right-handed neutrinos. That eliminates half the right-handed leptons. However, if neutrinos have Dirac masses then there must be right-handed neutrinos. I think if neutrinos have Majorana masses then there still might be no right-handed neutrinos?
  10. Mar 16, 2013 #9
    Oops, yes I did!
  11. Mar 17, 2013 #10
    Excelent Duck, thank you very much!
  12. Mar 17, 2013 #11


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    Note that the 2DHM has then a nice power, 128 dof. And thus the MSSM has 128+128; while a really minimal SSM, considering susy gauge supermultiplets instead of the higgs, had 126+126... and you could add a graviton/gravition to the bag.
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