On the Yang-Mills Lagrangian - 7/26

Introduction

I stopped taking much quantum field theory once we got to quantum chromodynamics. I think at some point things became more conceptual and less algebraic, which is somewhat less to my liking. What I did enjoy was the Yand-Mills Lagrangian. It is a neat use of beginning at very basic first principles and arriving at a physical theory.

I will not discuss much of the group theory; I will actually do my best to avoid it, mainly because I am not that familiar with it. Everything here should be nicely self-contained and one does not need to know group algebra things to proceed.

We set \(c=\hbar=1\) for simplicity, and we use the mostly negative metric convention, \begin{equation} ds^2=\eta^{\mu\nu}x_\mu x_\nu = dt^2-dx^2-dy^2-dz^2. \end{equation}

Gauge Theory Transform

We begin with matter fields \(\psi_i\left(\mathbf{x}\right)\) for \(i=1,2,...N\). Note that each \(\psi_i\) can be a scalar, spinor, tensor, etc. field, for the moment, our indices run over the field index, not the internal indices that may be associated with each \(\psi_i\).

We require that in our theory, our physical observables be invariant under the local transform, \begin{equation}\label{eq: gauge transform} \psi_i\left(\mathbf{x}\right)\rightarrow \psi_i\left(\mathbf{x}\right)’=U_{ij}\left(\mathbf{x}\right)\psi_j\left(\mathbf{x}\right), \end{equation} where the matrix \(\mathbf{U}\) is unitary, \begin{equation} \mathbf{U}^\dagger\mathbf{U}=\mathbf{I}, \end{equation} This is a statement that the inner product \(\left|\psi\right|^2=\left|\psi'\right|^2\) is unchanged under our transform.

Now since the transformation matrix is unitary, we can express \(\mathbf{U}\left(\mathbf{x}\right)\) as a matrix exponential, \begin{equation}\label{eq: U transform exponential form} \mathbf{U}\left(\mathbf{x}\right)=e^{i\alpha(x)^a\mathbf{T}_a}, \end{equation} where each \(\alpha^a(x)\) function is real and each \(\mathbf{T}_a\) matrix is hermitian, \begin{equation} \mathbf{T}_a=\mathbf{T}_a^\dagger. \end{equation} We have done nothing more than express the unitary \(\mathbf{U}\left(\mathbf{x}\right)\) as an exponent raised to a Hermitian matrix and then expressed this Hermitian matrix along the basis with elements \(\mathbf{T}_a\).

Excellent, now there is a good amount of group theory I will sweep under the rug here, and we will content ourselves without proof that our \(\mathbf{T}_a\) basis matrices can be chosen to satisfy the following properties while still spanning the space for all available Hermitian matrices, \begin{equation}\label{eq: T matrix trace} \text{Tr}\left(\mathbf{T} _a\mathbf{T} _b\right)=\frac{1}{2} \delta _{ab}, \end{equation} and, \begin{equation}\label{eq: structure constant commutation relation} \left[\mathbf{T} _a,\mathbf{T} _b\right]=if^{c} _{ab}\mathbf{T} _c. \end{equation} \(f^{ab}_c\) is known as a structure constant.

The Gauge Field and Field Strength Tensor

We have waded through the more nebulous bits that teeter on the edge of abstract mathematics. Onto the more interesting things. Because our observables should be invariant under the transform of Eq. \ref{eq: gauge transform}, so should our Lagrangian. To build such a Lagrangian, we first need a covariant derivative. A covariant derivative \(D_\mu\) transforms in the same way as our matter field, satisfying, \begin{equation}\label{eq: covariant derivative transformation condition} D_\mu\psi\left(\mathbf{x}\right)\rightarrow D’ _\mu\psi’\left(\mathbf{x}\right)=U\left(\mathbf{x}\right)D _\mu\psi\left(\mathbf{x}\right). \end{equation}

Note that from here onward, for clarity, I will be suppressing the indices associated with matter fields and also removing the bold lettering on matrices acting on matter field indices..

We may ask ourselves, what is the problem with an ordinary derivative \(\partial_\mu\)? Consider, \begin{equation}\label{eq: partial_mu failed transform} \partial_\mu\psi\left(\mathbf{x}\right)\rightarrow\partial_\mu\psi’\left(\mathbf{x}\right)=\partial_\mu\left(U\left(\mathbf{x}\right)\psi\left(\mathbf{x}\right)\right)=U\left(\mathbf{x}\right)\partial_\mu\psi\left(\mathbf{x}\right)+\partial_\mu\left(U\left(\mathbf{x}\right)\right)\psi\left(\mathbf{x}\right). \end{equation} Clearly \(\partial_\mu\) does not transform as we hoped, but Eq. \ref{eq: partial_mu failed transform} is suggestive. Let’s try a covariant derivative, \begin{equation}\label{eq: covarient derivative definition} D _\mu =\partial _\mu-i g A _\mu\left(\mathbf{x}\right), \end{equation} where \(g\) is a coupling constant and \(A^a _\mu\left(\mathbf{x}\right)\) is a gauge field that also transforms in a special way under our transformation rule. Employing Eq. \ref{eq: covarient derivative definition},

\begin{equation} D_\mu\psi\left(\mathbf{x}\right)\rightarrow D’_\mu\psi’\left(\mathbf{x}\right)=U\partial _\mu\psi+\partial _\mu U\psi - i g A’ _\mu\left(\mathbf{x}\right) U\psi. \end{equation}

We see our transformation condition, Eq. \label{eq: covarient derivative transformation condition}, is satisfied if, \begin{equation} \partial_\mu\left(U\right) - i g A’ _\mu\left(\mathbf{x}\right) U = - i g A\left(\mathbf{x}\right) U. \end{equation} From this we conclude \(A _\mu\) transforms as, \begin{equation} A _\mu\rightarrow A’ _\mu = U \left(A - i g^{-1} U^{-1} \partial _\mu \left(U\right)\right) U^{-1}. \end{equation} We may simplify by noting, \begin{equation} \partial _\mu\left(U^{-1}U\right)=U^{-1}\partial _\mu U + \partial _\mu\left(U^{-1}\right) U = 0 \rightarrow U^{-1}\partial _\mu\left(U\right) U^{-1} = -\partial _\mu\left(U^{-1}\right), \end{equation} which gives us, \begin{equation} A’ _\mu = U \left(A _\mu + i g^{-1} \partial _\mu\right) U^\dagger, \end{equation} where we have used the unitarity of \(U\).

Because \(A _\mu\) is a matrix in the matter field indices, we should be able to express it in terms of the basis matrices \(T_a\) such that \(A _\mu \left(\mathbf{x}\right) = A^a _\mu\left(\mathbf{x}\right) T_a\). It is easiest to do this by recalling that any transform \(U\) can be compiled as the result of many infinitesimal transforms. So, we will only consider the transform under \(U\left(\mathbf{x}\right)=1+i\alpha^a\left(\mathbf{x}\right)T_a\), for \(\alpha\) small enough to neglect second-order terms. If we can obtain the appropriate transformation for \(A^a _\mu\) in this case, it will be equivalent to the full transform in general. Applying this idea,

\begin{equation} A’ ^a _\mu T_a = \left(1+i\alpha^bT _b\right)\left(A^a _\mu T_a + i g^{-1} \partial _\mu\right) \left(1-i\alpha^cT_c\right) = \left(A^a _\mu +g^{-1} \partial _\mu\alpha^a\right)T_a+iA^a _\mu\alpha^b\left(T _b T _a-T _a T _b\right), \end{equation}

\begin{equation}\label{eq: A matrix component form transform} A’ ^a _\mu = \left(A^a _\mu +g^{-1} \partial _\mu\alpha^a\right) + A^b _\mu \alpha^c f _{bc}^a, \end{equation}

we have used Eq. \ref{eq: structure constant commutation relation} to insert the structure constant.

We almost have all of the ingredients we need for our Lagrangian. Next, we construct the field strength tensor. Since the covariant derivative transforms as \(D_\mu'\psi'=UD_\mu\psi\), it follows that \(D_\mu'=UD_\mu U^\dagger\). From this, we note that the commutator of the covariant derivatives transforms in the same way,

\begin{equation} \left[D _\mu,D _\nu\right]\rightarrow\left[D’ _\mu,D’ _\nu\right]=U\left[D _\mu,D _\nu\right]U^\dagger. \end{equation}

We define the field strength tensor that transforms in the same way, \begin{equation} G_{\mu\nu}=\frac{i}{g}\left[D_\mu,D_\nu\right]=\partial_\mu A_\nu-\partial_\nu A_\mu-ig\left[A_\mu,A_\nu\right]. \end{equation} In terms of the \(T_a\) basis elements, the componenets \(G^a_{\mu\nu}\) can be easily found to be,

\begin{equation}\label{eq: field strength tensor component form} G^a _{\mu\nu}=\partial _\mu A _\nu^a-\partial _\nu A _\mu^a+gA^b _\mu A^c _\nu f^a _{bc}. \end{equation}

We may also ask how the components, \(G^a _{\mu\nu}\), act under a gauge transform. It is possible to use Eq. \ref{eq: field strength tensor component form} to solve for \(G'^a _{\mu\nu}\) directly since we know how \(A _\nu^a\) transforms from Eq. \ref{eq: A matrix component form transform}, but it is easier to recall the matrix transform is straightforward,

\begin{equation} G _{\mu\nu} \rightarrow G’ _{\mu\nu} =U G _{\mu\nu} U^\dagger, \end{equation}

and use the same infinitesimal transformation trick we previously employed,

\begin{equation} {G’}^a _{\mu\nu} T _a=\left(1+i \alpha^b T _b\right) G^a _{\mu\nu} T _a\left(1-i\alpha^c T _c\right) = G^a _{\mu\nu} T _a + G^b _{\mu\nu} \alpha^c f^a _{bc} T_a. \end{equation}

The field strength tensor \(G _{\mu\nu}\) has a host of desirable properties. First, it is a local function of our gauge fields, not a differential operator, which is what we want for our Lagrangian. Second, it is only first-order in the derivatives of the Gauge fields, so we can combine two of these tensors to obtain a quantity that is second-order. To build a Lagrangian, we require Lorentz and gauge invariance. The first of these is easy to do: make sure there are no free indices since scalars are Lorentz invariant. For the second condition, we can take the trace over matter field indices, observing that,

\begin{equation} \text{Tr}\left(G^{\mu\nu}G _{\mu\nu}\right)\rightarrow\text{Tr}\left(G’^{\mu\nu}G’ _{\mu\nu}\right)=\text{Tr}\left( UG^{\mu\nu}U^\dagger UG _{\mu\nu}U^\dagger\right) =\text{Tr}\left(G^{\mu\nu}G _{\mu\nu}\right). \end{equation}

Or, in component form,

\begin{equation} \text{Tr}\left(G^{a\ \mu\nu}G _{b\ \mu\nu}T_aT^b\right) =G^{a\ \mu\nu}G _{b\ \mu\nu}\text{Tr}\left(T _aT^b\right) =\frac{1}{2}G^{a\ \mu\nu}G _{a\ \mu\nu}, \end{equation}

after consulting Eq. \ref{eq: T matrix trace}. We define the gauge group of our Lagrangian,

\begin{equation}\label{eq: Gauge Lagrangian} \mathcal{L} _{\text{gauge}} = -\frac{1}{4} G^{a\ \mu\nu} G _{a\ \mu\nu}, \end{equation}

which describes the interactions between the gauge fields \(A _\mu\left(\mathbf{x}\right)\). This Lagrangian is both gauge invariant and Lorentz invariant; of course, there are other Lagrangians we could write down that have these properties, but we stick with the old adage of keeping things simple, stupid.

For fun, we can write out the equations of motion of Eq. \ref{eq: Gauge Lagrangian} using the Euler-Lagrange equations,

\begin{equation} \partial_\nu\left(\frac{\partial\mathcal{L} _{\text{gauge}}}{\partial\left(\partial _\nu A _{a\ \mu}\right)}\right)-\frac{\partial\mathcal{L} _{\text{gauge}}}{\partial A _{a\ \mu}}=0. \end{equation}

Calculating the variation with respect to \(\partial _\nu A _{a\ \mu}\),

\begin{equation} \frac{\partial\mathcal{L} _{\text{gauge}}}{\partial\left(\partial _\nu A _{a\ \mu}\right)} = -\frac{1}{2} G _{b\ \alpha\beta} \frac{G ^{b\ \alpha\beta}}{\partial\left(\partial _\nu A _{a\ \mu}\right)}, \end{equation} \begin{equation} \frac{G _{b\ \alpha\beta}}{\partial\left(\partial _\nu A _{a\ \mu}\right)} = \delta^{ab}\left(\delta^{\alpha\nu}\delta^{\beta\mu}-\delta^{\alpha\mu}\delta^{\beta\nu}\right), \end{equation} \begin{equation} G _{b\ \alpha\beta} \frac{G ^{b\ \alpha\beta}}{\partial\left(\partial _\nu A _{a\ \mu}\right)} = \delta^{ab}\left(\delta^{\alpha\nu}\delta^{\beta\mu}-\delta^{\alpha\mu}\delta^{\beta\nu}\right)G _{b\ \alpha\beta}=2G^{a \nu \mu}. \end{equation}

\begin{equation} \frac{\partial\mathcal{L} _{\text{gauge}}}{\partial\left(\partial _\nu A _{a\ \mu}\right)} = - G^{a \nu \mu}. \end{equation}

Next, the variation with respect to \(A _{a\ \mu}\),

\begin{equation} \frac{ \partial\mathcal{L} _{\text{gauge}}}{\partial A _{a\ \mu}} = -\frac{1}{2} G _{b\ \alpha\beta} \frac{G ^{b\ \alpha\beta}} {\partial A _{a\ \mu}}, \end{equation}

\begin{equation} \frac{G ^{b\ \alpha\beta}}{\partial A _{a\ \mu}} = g f^{ba} _c \left( \delta^{\alpha\mu} A^{c \beta} - \delta^{\beta\mu} A^{c \alpha} \right), \end{equation} \begin{equation} G _{b\ \alpha\beta} \frac{G ^{b\ \alpha\beta}}{\partial A _{a\ \mu}} = 2 g f^{ba} _c G ^\mu _{b\ \alpha} A^{c \alpha} , \end{equation} \begin{equation} \frac{\partial\mathcal{L} _{\text{gauge}}}{\partial A _{a\ \mu}} = g f^{abc} A _{b\ \nu} G ^{\nu\mu} _c . \end{equation}

The Euler-Lagrange equations yield, \begin{equation} \partial_\nu G^{a \nu \mu} + g f^{abc} A _{b\ \nu} G ^{\nu\mu} _c = 0 \end{equation}

In the case when the gauge fields do not interact, the theory is fully Abelian, and we can set \(g = 0\). We define the abelian field strength tensor, \(F^{\mu\nu}=\partial^\mu A^\nu - \partial^\nu A^\mu\), and our equations of motion greatly simplify to, \begin{equation} \partial_\nu F^{a \nu \mu} = \partial^\nu \partial^\mu A^\nu - \partial^\mu \partial^\nu A^\mu = 0, \end{equation} which we recognize as Maxwell’s equations, as we would expect.

Spinor fields

We have found an adequate term to insert into our Lagrangian that is both Lorentz and gauge invariant, but we have somewhat forgotten about our mass fields. We add these by using the Dirac Lagrangian, replacing the partial derivative with a covariant derivative, \begin{equation} \mathcal{L} _{\text{dirac}} = \sum_i \bar{\psi} _i \left(i \not D - m_i\right)\psi _i, \end{equation} the index \(i\) sums over all the spinor fields in our theory, each one coupling with the gauge field \(A _\mu\). Showing Lorentz invariance is straightforward since, \begin{equation} \bar{\psi}_i\gamma^\mu\psi_i \rightarrow \Lambda^\mu _\nu \bar{\psi}’_i\gamma^\mu\psi’_i, \end{equation} transforms as a Lorentz four vector for the Lorentz transform matrix \(\Lambda^\mu_nu\), then \begin{equation} \bar{\psi}_i\not \partial \psi_i\quad\text{and}\quad\bar{\psi}_i\not A \psi_i, \end{equation} are both Lorentz scalars and invariant under a transform. As for gauge invariance, this is also straightforward to show using the properties of the covariant derivative, \begin{equation} \mathcal{L} _{\text{dirac}} \rightarrow \sum_i \bar{\psi} _i U^\dagger U \left(i \not D - m_i\right)\psi _i = \mathcal{L} _{\text{dirac}}. \end{equation} The equations of motion are nothing more than the Dirac equations for each spinor field \(i\) with an added source term from the gauge field, \begin{equation}\label{eq: dirac lagrangian eom} \left(i\not\partial-m\right)\psi _i = -g\not A \psi _i. \end{equation} Note that Eq. \ref{eq: dirac lagrangian eom} involves couplings between the spinor fields since \(\not A\) is a matrix in matter field indices.

Scalar fields

One final type of field are scalars. These are the easiest. The Lagrangian for scalar fields \(\phi_i\) is, \begin{equation} \mathcal{L} _{\text{scalar}} = \sum_i \left( D _\mu\phi _i\right)^\dagger\left( D^\mu\phi _i\right) - m_i^2\phi_i^\dagger\phi _i. \end{equation} There are no free indices, so the Lagrangian is easily Lorentz invariant. It is also gauge invariant due to our usage of the covariant derivative. The equations of motion are, \begin{equation} \left(D _\mu D^\mu + m^2\right)\phi_i = 0, \end{equation} since \(\phi\) and \(\phi^\dagger\) are indepedent fields.

Yang Mills Lagrangian Summary

Our Lagrangian, adding all terms,

\begin{equation} \mathcal{L} _{\text{Yang Mills}} = -\frac{1}{4} G^{a\ \mu\nu} G _{a\ \mu\nu} + \sum _i \bar{\psi} _i \left(i \not D - m^\psi _i\right)\psi _i + \sum _i \left( D _\mu\phi _i\right)^\dagger\left( D^\mu\phi _i\right) - \left(m_i^\phi\right) ^2\phi_i^\dagger\phi _i. \end{equation}

Where we use, \begin{equation} G^a _{\mu\nu} = \partial _\mu A _\nu^a - \partial _\nu A _\mu^a + gA^b _\mu A^c _\nu f^a _{bc}, \end{equation}

and, \begin{equation} D _\mu = \partial _\mu - i g A^a _\mu T _a. \end{equation} This is the Lagrangian for the Yang-Mills theory. There is a lot more to be said about gauge fixing and propagators and loops and renormalization and whatnot, but we will leave things here for now.

Sources

This was done using my own notes taken during my Quantum Field Theory II class (8.324), taught by Professor Ian Stewart during the 2025 Fall semester, and my Quantum Field Theory III class (8.325), taught by Professor Daniel Harlow during the 2026 Spring semester.