Consider a Lagrangian $L$ which depends on a number of fields $F_1$, $\cdots$, $F_N$ and their (spacetime) derivatives. Each of those fields $F_n$ is valued in $\mathbb{R}^{k_n}$.

  1. Is the Standard model Lagrangian (or more restrictively, QED, QCD, EW-theory if that's easier to investigate) a special case of this setting ? (What I mean by that is : can we represent all the fields $F_1,\cdots,F_N$ that are independent dynamical variables in the standard model Lagrangian, as valued in an euclidean space (using of course identifications such as $\mathbb{C}=\mathbb{R}^2$, $\mathcal{M}_n(\mathbb{R})=\mathbb{R}^{n^2}$, etc.).

  2. If the answer to the previous question is yes, what choice of $N$ and $(k_n)_{1\leq n\leq N}$ can we take to possibly describe the Lagrangian of the SM ?

  • 2
    $\begingroup$ I'll leave the details of your actual question for someone more qualified to answer, but it's worth noting that your definition of "degrees of freedom" doesn't always agree with the usual definition of "degrees of freedom" used by high-energy physicists. Some of the fields that appear in the SM Lagrangian don't actually appear with time-derivatives ($A_0$ is the classic example), and some other fields are arbitrary due to gauge symmetry. So although you need four real numbers to specify $A^\mu$ at every point in space, the field only has two degrees of freedom in the usual sense. $\endgroup$ – Michael Seifert Jun 2 '16 at 17:55
  • $\begingroup$ Comment to the post (v3): This question seems too broad. Subquestion 3 is related to physics.stackexchange.com/q/256466/2451 and links therein. To reopen this post consider to remove subquestion 3. $\endgroup$ – Qmechanic Jun 2 '16 at 17:58
  • $\begingroup$ Thank you for you comment, I know indeed that my N, and $(k_n)$ are not exactly the "degrees of freedom" used commonly, but I didn't know how to call them really. I think they are still relevant however, and may be related to the classical definition. And I suppose also they will change depending we add a gauge-fixing term or not. $\endgroup$ – Jon-S Jun 2 '16 at 17:59
  • $\begingroup$ I will remove question 3 $\endgroup$ – Jon-S Jun 2 '16 at 18:05
  • $\begingroup$ 1. Yes; 2. Educated guess and verified experimentally $\endgroup$ – Rodriguez Jun 2 '16 at 18:11
  1. Yes but with lots of subtleties to be described in (2).

  2. The main questions to be settled are:

    • whether we count just the fields (configuration space) or fields and their derivatives (phase space)
    • whether we count the fermions along with bosons, or separately (I will count them separately), and whether we double the number for them
    • because the Standard Model has a gauge symmetry, whether we count the unphysical "gauge" degrees of freedom.

OK, let's look at all the fields. First, the fermionic degrees of freedom. They don't really have any configuration or phase space isomorphic to any ${\mathbb R}^N$ because the "classical values" of fermionic values are anticommuting Grassmann numbers that can't be equal to any ordinary real numbers because the latter commute.

If we say that the electron's Dirac field has 4 complex fermionic degrees of freedom (components), the Standard Model's first generation of quarks and leptons has:

  • 1 Dirac field for electron, 1/2 Dirac field for left-handed neutrino, 3 Dirac fields for up quark (3 colors), 3 Dirac fields for down quarks

which is 7.5 Dirac fields or 15 Weyl fields (it becomes 16 if we add right-handed neutrinos, so far not observed), or 30 complex fermionic components per generation. There are 3 generations so there are 90 fermionic (field-like) degrees of freedom.

The counting of the fermions was easy – except that the "number of degrees of freedom" may sometimes be halved or doubled depending on conventions.

The counting of bosons is a bit subtle. The Standard Model has the electromagnetic field which may be said to have four $A_\mu$ off-shell components, a vector. The Standard Model has the gauge group $SU(3) \times SU(2) \times U(1)$ whose dimension is $8+3+1=12$ so it has "12 copies" of the photon's gauge field. You could say it's 48 components. On top of that, there is the Higgs doublet which has 2 complex i.e. 4 real components, so it's 52 off-shell bosonic components in total.

However, this counting overstates the bosonic degrees of freedom because of the gauge symmetry. That makes not just 1 but two components among $A_\mu$ unphysical. A better counting is to count the on-shell degrees of freedom, the types of "particles that can actually be produced". This counting is more independent of the particular way to deal with gauge symmetries etc. – and this on-shell counting is needed to match bosons and fermions in supersymmetry, among other things.

In this counting, there are 2 transverse physical polarizations of a photon, so it is $A_x$ and $A_y$ components only if the photon moves in the $z$-direction, and this 2 is multiplied by 12, so we get 24, and then we add the 4 real Higgs components which don't have any unphysical degrees of freedom. The total is "28 types of particles" – 26 times what a real Klein Gordon field has. One may still count the phase space coordinates to get 56.

This 26 may also be obtained by counting massive bosons. Photons and gluons (1+8 generators) are massless, so they give $9\times 2 = 18$, and $W^+,W^-,Z_0$ are massive, those have three polarizations, so $9$ from the massive bosons. That's 27 in total and add one physical Higgs scalar (the remaining three components of the Higgs were the Goldstone modes eaten by the massive bosons), to get 28 again.

Per every point and counting on-shell physical (after gauge-fixing etc.) polarizations only, the Standard Model has 28 real bosonic fields (plus 28 of their derivatives) plus 90 fermionic complex fields (their derivatives are not counted because the fermions' equations of motion are first-order).


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