# Motivation for classical Yang-Mills theories

I have an assignment to make a monograph on classical Yang-Mills theories for my Electrodynamics II course, but (among other technical doubts) I'm not understanding the purpose of these theories.

At least in my understanding, Yang-Mills theories are used to describe quantum phenomena, so I'd like a motivation or usefulness for their classical counterpart. It seems that they're simply mathematical abstractions or generalizations that don't apply anywhere. Every article I've read so far directly jumps to differential geometry or gauge theory, which I'm not familiar with.

• You need non-Abelian gauge fields if you want to gauge a non-Abelian global symmetry. – SCFT Jun 18 '17 at 4:37
• Can you elaborate? I'm not sure why I'd want that in the first place. – Renan Nobuyuki Hirayama Jun 18 '17 at 4:43
• Possible duplicate: physics.stackexchange.com/q/126978/50583 – ACuriousMind Jun 18 '17 at 12:09
• One motivation to studying classical Yang-Mills theories is that quantization procedure is better done in the path integral formalism, which requires a classical action. – Diracology Jun 18 '17 at 12:52
• @ACuriousMind I had come across that question, but it didn't quite explain to me the specific need for classical Yang-Mills, as did Diracology. – Renan Nobuyuki Hirayama Jun 19 '17 at 4:56

Yang-Mills theories are a type of generalization of electromagnetism, in a sense. An electromagnetic theory (without magnetic monopoles) can be described completely by a vector potential $A_{\mu}$ and obey's the all-important gauge symmetry

$$A_{\mu}\to A_{\mu}+\partial_{\mu}\lambda,$$

where $\lambda$ is an arbitrary differentiable function of space and time. This symmetry is the most important property of electromagnetism.

Now, I'm going to assume you're not familiar with quantum field theory, but perhaps you're familiar with a bit of quantum mechanics. In particular, given a vector field $A_{\mu}$, the Hamiltonian for a charged particle of charge $q$ can be written as

$$H=\frac{1}{2m}(-i\boldsymbol{\nabla}-q\textbf{A})^2+q\phi,$$

where $\textbf{A}$ is the vectorial part of $A_{\mu}$ and $\phi=A_0$ is the electric potential. It can be shown (and you should try this yourself!) that if we transform $A_{\mu}\to A_{\mu}+\partial_{\mu}\lambda$ and $\psi\to e^{i\lambda}\psi$ (note that I am working in units where $\hbar=c=1$), then the Schrodinger equation $H\psi=i\partial_t\psi$ is left unchanged.

The transformation $\psi\to e^{i\lambda}\psi$ should be considered part of this gauge transformation. In fact, in quantum field theories, it is more fundamental than the first definition from above. This transformation is known as a $U(1)$ transformation, since it takes the form $\psi\to g\psi$ where $g$ is a one-dimensional unitary matrix (aka a unimodular complex number).

Yang-Mills theories work in a similar fashion. We consider a wavefunction (or, to be more precise, a field) $\psi$ which may be a vector in both the Hilbert space but also in a separate vector space $V$, and we consider a transformation

$$\psi\to U\psi$$

where $U$ is an element of some continuous group of transformations (more specifically, a compact, semisimple Lie group -- don't worry if you don't know what this means). This transformation will take the form

$$\psi\to e^{i\lambda^at^a}\psi,$$

where $\lambda_a$ are differentiable functions of time and $t^a$ are known as generators of the group in the physics literature, and $a$ ranges from $1$ to $N$, where $N$ is the dimensionality of the group (basically, the number of independent transformations). If we want whatever theory we have to be invariant under this transformation,we introduce a vector potential

$$A_{\mu}\equiv A_{\mu}^{a}t^a,$$

with summation implied over $a$. Note that the components $A_{\mu}^a$ are just numbers, while the $t^a$ are "new" objects that you don't need to take into account in the $U(1)$ case. We will include $A$ in a very similar way that we did in electromagnetism, and we will require it to obey a similar transformation law:

$$A^{a}_{\mu}\to A_{\mu}^{a}+\partial_{\mu}\lambda^a-gf^{a}_{\,\,bc}\lambda^{b}A_{\mu}^c,$$

so as to be compatible with the wavefunction transformation law, where $g$ is a coupling constant, and the constants $f^{a}_{\,\,bc}$ are determined by the structure of the transformation group under consideration, and are thus called structure constants (note also an implied summation over $b$ and $c$), which need to be introduced due to the fact that the $t^a$ generators of the group do not commute with each other (this is why Yang-Mills is often referred to as a non-abelian field theory). This is essentially the super basic outline of Yang-Mills theory.

You are correct: Yang-Mills theory is almost exclusively used in a quantum context. This is because it simply isn't useful as a classical theory to describe the things it is meant to describe. Electromagnetism has the very fortunate property that its low-energy behavior at macro-scales is described simply by its classical equations of motion. Yang-Mills theory is not so lucky. The equations of motion (the YM analogue of Maxwell's equations) simply don't describe nature very well since the Yang-Mills theories included in the standard model (namely the strong interactions -- electromagnetism and the weak interactions are actually part of the same YM theory but are decoupled at low energies and are thus more complicated than a discussion here could do honor) are very strongly coupled at low energies.

There is so much more to go into, and I encourage you to research this wonderful topic for yourself. Without the background in field theories and some mathematical preliminaries, this topic can seem very shrouded in mystery, but just remember: it's all about gauge transformations.

I hope this helped!

[Note to the mathematically inclined reader: throughout this answer I very sloppily used the term "group" to refer to the same object as a Lie group and a Lie algebra. I did this intentionally as to not add more confusion than will definitely be present already.]

• You confuse Lie groups with their algebras. I tried to correct your statements in some places. – Danu Jun 18 '17 at 10:58
• Thanks for the answer, it provided me with some very useful insights. I am a little confused with the Hamiltonian for a charged particle you described: it seems wrong, but I wouldn't know how to correct it. – Renan Nobuyuki Hirayama Jun 18 '17 at 11:18
• @BobKnighton I find it rather odd that you would choose to intentionally write wrong statements (without even a remark indicating it) to avoid confusion. – Danu Jun 18 '17 at 11:59
• There is a remark indicating it at the end of the answer. Remember, OP only cares about the motivation of these theories. If OP wants mathematical precision, literally any other source will provide that. What I am trying to provide is an intuitive physical explanation accessible at the level of a student taking Electrodynamics II and who hasn't seen this mathematical content before. – Bob Knighton Jun 18 '17 at 12:01
• @Danu The disclaimer was always there, I just edited it to be more precise after your comments. You have great points, I just think that the time and space required to give a good mathematical foundation would make this answer much longer at the cost of simplicity. – Bob Knighton Jun 18 '17 at 13:49