So Yang-Mills theory is a non-abelian gauge theory, and we used a lot in QCD calculation.

But what are the distinctions between Yang-Mills theory and QCD? And distinctions between supersymmetric Yang-Mill theories and SUSY QCD?

  • $\begingroup$ More on Yang-Mills theory and terminology: physics.stackexchange.com/q/8686/2451 $\endgroup$
    – Qmechanic
    May 8, 2012 at 11:53
  • 2
    $\begingroup$ I don't understand the question, or what the bounty wants to elicit. QCD is a gauge theory with matter, and SUSY QCD is QCD plus scalar partners for the quarks and a fermionic partner for the gluons. It's like asking "what's the difference between a field theory and electromagnetism?" $\endgroup$
    – Ron Maimon
    Jun 27, 2012 at 4:16
  • $\begingroup$ QCD is a special case of Yang-Mills theory for gauge group $SU(3)$. Yang-Mills generalizes QED, QCD, Electroweak theory, three of the fundamental forces of nature. Yang-Mills truly is one of the great achievements in physics and science. $\endgroup$ May 16, 2020 at 0:21

4 Answers 4


From the beginning of the wikipedia page on Yang-Mills theory (have you read it?):

"Yang–Mills theory is a gauge theory based on the SU(N) group ...

... In early 1954, Chen Ning Yang and Robert Mills extended the concept of gauge theory for abelian groups, e.g. quantum electrodynamics, to nonabelian groups to provide ...

... This prompted a significant restart of Yang–Mills theory studies that proved successful in the formulation of both electroweak unification and quantum chromodynamics (QCD). The electroweak interaction is described by SU(2)xU(1) group while QCD is a SU(3) Yang-Mills theory."

Yang-Mills theoies are a class of (classical) field theories and might be viewed as a generalization of the electromagnetic field theory. What's different between the Yang-Mills theories is the respective gauge group under consideration, but the point is that there are several possible ones.

You can quantize the electromagnetic field theory and you "obtain" quantum electrodynamics. You can also quantize Yang Mills theories and this way you obtain some other specific quantum field theories. One "uses" Yang-Mills theory in the calculations of the different parts of the standard model etc. because the underlying structures are such non-abelian field theories. Notice that when physicists say "that's a Yang-Mills theory" they usually talk about the quantized version already.

For exmaple QCD is a (quantized) SU(3)-Yang-Mills theory with coupling to certain ferimons. The fermions in the Lagrangian are coupled to the bosons via the current term "$j^\mu A_\mu$". The specific (Lie-)group structure (SU(3) in the QCD case) is in particular refleced in the number of gluons (eight) and so on. Like many other physical features, this is determined by group representation theory.

Supersymmetric theories are theories with more features than the usual Yang-Mills theory, which a priori is mostly about the bosonic fields (Photons, W$^{\pm}$/Z-bosons, gluons,...). Supersymmetry relates fermions and bosons.


A Yang-Mills theory has only a gauge field but no associated matter field. Quantum $SU(3)$ Yang-Mills theory describes gluons in the absence of (real or virtual) quarks. Hence, from the phenomenological persective, it is only a toy theory.

QCD is the theory obtained from $SU(3)$ Yang-Mills theory by coupling it to fermionic fields representing quarks.

  • 1
    $\begingroup$ Comment to the answer(v1): It should be noted that several authors allow matter into the definition of Yang-Mills theory. The restriction to only gauge fields with no matter is then called pure Yang-Mills theory. $\endgroup$
    – Qmechanic
    May 9, 2012 at 12:40
  • 1
    $\begingroup$ @Qmechanic: The generally used terminology for a theory with Yang-Mills fields plus matter fields is gauge field theory, and your suggestion would mean that some people use Yang-Mills theory synonymous with nonabelian gauge field theory. Can you give a widely accepted source for this usage? (The OP cites Wikipedia, which defines YM to be what you call pure YM, consistent with my answer.) $\endgroup$ May 9, 2012 at 12:52
  • $\begingroup$ 1. See e.g. Peskin and Schroeder, An introduction to QFT, Section 15.2; or try to Google the term "pure Yang-Mills". 2. The notion of gauge field theory is wider than just pure Yang-Mills plus matter. It also includes e.g. Chern-Simons theory, BF theory, etc. $\endgroup$
    – Qmechanic
    Nov 20, 2012 at 15:09

Yang-Mills theory is the quantization of the following field theory: The Yang-Mills field $A$ is a connection on a $G$-bundle $P \to M$ with semisimple gauge group $G$[1]. The Lagrangian is $$L = -\frac{1}{4}\langle F,F\rangle$$ Here $F$ is the curvature of the connection $A$ and you should think of the bracket as some scalar product. The Lagrangian is a generalization of the Lagrangian for the source-free Maxwell equations, $F$ is a generalization of the covariant field strength tensor. In quantum field theory Yang-Mills fields are force carriers, they are massless vector bosons that mediate interactions between matter fields (various fermions). The term (pure) Yang-Mills theory is used for the above theory, quantization requires the inclusion ghosts, so that additional fermionic matter fields do not hurt either, then one usually speak of Yang-Mill theory with matter.

A supersymmetric generalization is obtained as follows: Consider an additional $\mathfrak{g}$-valued spinor field $\psi$ and the Lagrangian

$$L = -\frac{1}{4}\langle F,F\rangle + \frac{1}{2}\langle \psi,{D}_A\psi\rangle,$$ where $D_A$ is the covariant Dirac operator associated with $A$. This theory is supersymmetric on Minkowskispace if and only if the dimension of spacetime is $d = 3,4,6,10$. This is rather tricky to prove and is related to the fact, that there are 4 normed division algebras: $\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}$. Symbolically the supersymmetry transformation is $\delta A = \epsilon \cdot \psi, \delta \psi = \frac{1}{2}F\epsilon$ for a spinor field $\epsilon$. Physically this theory describes a Yang-Mills gauge field coupled to a massless spinor.

There are several possible modifications of this Lagrangian: Most importantly there is a supersymmetric extension, that allows the inclusion of additional massive matter fields: You can consider a minimally coupled $\sigma$-Model Lagrangian $$L = -\frac{1}{4}\langle F,F\rangle + \frac{1}{2}\|d_A\phi\|^2 -\int_M \phi^*V,$$ where $\phi$ is some section of the associated bundle $X^P = P \times_G X \to M$, $V \colon X \to \mathbb{R}$ a $G$-invariant Potential and $X$ a Riemannian manifold on which $G$ acts by isometries. As above you can now include spinor fields. To extend this to a supersymmetric theory is relatively complicated, but can be done. You can find an exposition for example in "Quantum Fields and Strings" (p. 304ff). (Those models allow you to get spontaneous symmetry breaking, as you need for supersymmetric extensions of the standard model (see below))

QCD is the part of the standard model, that describes the strong interaction. The strong interaction is mediated by a $SU(3)$ Yang-Mills field $A$. This is described by a term

$$-\frac{1}{4}\langle F,F\rangle$$

Then there are 3-generations of fermions: the quarks. Each generation is identical in most respects, they are only related by the Yukawa coupling. Each generation consists of a (complex) spinor $\psi$, with values in some irreducible representation $V$ of the group $G = SU(3) \times SU(2) \times U(1)$. The representation of $SU(2)$ (weak interaction) distinguishes between the left- and right handed parts, the $SU(3)$ represention is simply the fundamental representation, schematically the coupling to the $SU(3)$ Yang-Mills field $G$ (the gluons) is again

$$\frac{1}{2}\langle \psi, D_A \psi\rangle,$$

where again $D_A$ is the covariant dirac operator associated to $G$ and the bracket can be constructed from representation theoretic considerations. The mass terms are the complicated part: You need to combine the (complex) Higgs boson $\phi$, a scalar with values in some other irreducible representation $H$ of $G$ (trivial $\times$ doublet $\times$ hypercharge 1) and its antiparticle, with the left- and right handed parts of $V \oplus \bar{V}$, with "mixing" of all 3 generations. You consider a $3 \times 3$ Matrix $M$ (3 is the number of generations) and add a term that combines all those representations in the only possible way (you need to get the trivial representation, more precisely you want an intertwiner from this large representation in the trivial representation)

This is a huge mess, so usually one studies Yang-Mills Theory with gauge group $SU(N)$ coupled to several generations of fermion fields in the fundamental representation first. Schematically:

$$L = -\frac{1}{4}\langle F,F\rangle + \frac{1}{2}\langle \psi, (D_A + M) \psi \rangle$$

Unfortunately I am not familiar with the term Susy QCD, there are several extensions of the standard model (e.g. MSSM) that all work by adding superpartners to the existing fields (in the way sketched above), such models of course also contain a part that corresponds to QCD and its "super extension".

[1]: For a nice cover $\{U_i\}_{i \in I}$ of $M$ such a connection is described by a family of Liealgebra valued one forms $A_i \in \Omega^1(U_i,\mathfrak{g})$, such that for any collection of functions $g_{ij}\in C^\infty(U_i \cap U_j, G)$, that satisfy the cocycle condition $g_{ij}g_{jk} = g_{ik}$, the following identities hold: $$A_j = g^{-1}A_ig + g^{-1}dg$$
(As stated this does only make sense if $G$ is a matrix group, but transformation rule on the r.h.s is really the sum of the adjoint action on $A_i$ and the pullback of the Maurer-Cartan form.)


This question is kinda like asking "What's the difference between a classical harmonic oscillator and Newton's second Law?" Well, the harmonic oscillator satisfies equations set up by Newton's second Law.

The Yang-Mills theory is more of a (pardon my French) paradigm than a "theory", in the sense Yang-Mills sets up a framework for theories like QCD (as opposed to giving an hypothesis).

As an "input", Yang-Mills requires some specified gauge group. This is usually a "sufficiently nice" Lie group (IIRC, compact, connected, and simply connected).

QCD is a theory that uses the Yang-Mills framework, specifically when we restrict attention to the gauge group SU(3).

What's "nice" about Yang-Mills is it gives us back Maxwell's equations when we plug in U(1) as the Gauge group. So Yang-Mills is a very good machine, indeed!

A good reference for Yang-Mills theory, in my opinion at least, is John Baez's Gauge Fields, Knots, and Gravity.

For super Yang-Mills, as I understand it (and I don't pretend to), you enable the gauge group to be a super Lie group. What makes it "super"? Well, you have a $\mathbb{Z}_{2}$ grading. In other words: you have fermionic guys and bosonic guys (which are odd and even, respectively).

Super Yang-Mills includes Fermions into the model. It appears to me that you simply include a massless fermionic field, and quite interesting stuff happens. My only references on it are:

  1. Pierre Deligne, Daniel Freed, et al, Quantum Fields and Strings: A Course for Mathematicians, vol. I, chapter 6: Supersymmetric Yang–Mills Theories, pp. 299–311.
  2. Green, Schwarz and Witten, Superstring Theory, vol. I, appendix 4.A: Super Yang–Mills Theories, pp. 244–247.
  • $\begingroup$ The gauge group in super Yang-Mills is still an ordinary Lie group. $\endgroup$
    – user1504
    Jun 25, 2012 at 20:16

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