• #PDF
• [[#Dynamics for \(A_\mu\) ---]]
• #Other possibilities -- Chern--Simons in 2d ---
• #Equations of motion ---
• #Bianchi identity ---
• #Charges & representations ---
• #Matter fields ---
• #Parallel transport ---
• #The path ordered exponential ---

Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 17 | Next: Lecture 19

Recall (electrodynamics, in units \(c=1\), so that \(E = Bc\)):

\[ \mathcal{L}_{\substack{\text{Electro}\\ \text{dynamics}}} \;=\; \frac{1}{2e^2}\left(\vec{E}^2 - \vec{B}^2\right) \;=\; \frac{-1}{4e^2}\,F^{\mu\nu}F_{\mu\nu} \] \[ F^{0i} \;=\; E^i \ ,\qquad F^{ij} \;=\; \frac{1}{2}\epsilon^{ijk}B^k \]

For non-abelian gauge theories:

\[ \mathcal{L} \;=\; -\frac{1}{4g^2}\,F^{a}_{\mu\nu}F^{\mu\nu a} \;=\; -\frac{1}{2g^2}\,\mathrm{Tr}\,F_{\mu\nu}F^{\mu\nu} \]

\(F^{\mu\nu}\) is also valued in the Lie algebra, so `\(a\)' labels the component in Lie algebra
space.

In 2-d \((2+1)\) we can write

\[ \mathcal{L}_{CS} \;=\; -\frac{k}{4\pi}\,\mathrm{Tr} \left(A_\mu\partial_\nu A_\alpha + \frac{2}{3}A_\mu A_\nu A_\alpha\right)\epsilon^{\mu\nu\alpha} \]

Let us write the EOM (4-d) for the above \(\mathcal{L}\):

\[ \frac{\partial L}{\partial q} \;=\; \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right) \qquad\longrightarrow\qquad \frac{\partial\mathcal{L}}{\partial\phi} \;=\; \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) \]

Use

\[ \delta F^{a}_{\mu\nu} \;=\; \left(D_\mu\delta A_\nu - D_\nu\delta A_\mu\right)^{a} \]

to get

\[ \left(D_\mu F^{\mu}_{\ \nu}\right)^{a} \;+\; g^2\,\frac{\delta S_m}{\delta A^{a}_{\nu}} \;=\; 0 \]

where \(S_m\) = matter action.

Recall (Maxwell):

\[ \partial_\alpha F_{\mu\nu} + \partial_\mu F_{\nu\alpha} + \partial_\nu F_{\alpha\mu} \;=\; 0 \]

(the indices are cyclically permuted). We use

\[ D_\alpha F_{\mu\nu} + D_\mu F_{\nu\alpha} + D_\nu F_{\alpha\mu} \;=\; 0 \qquad\longrightarrow\qquad \text{Bianchi Identity} \]

The Bianchi identity is just a statement of the Jacobi identity for covariant derivatives, i.e.

\[ \left[\left[D_\mu,D_\nu\right],D_\alpha\right] + \left[\left[D_\nu,D_\alpha\right],D_\mu\right] + \left[\left[D_\alpha,D_\mu\right],D_\nu\right] \;=\; 0 \]

\(\left\{\text{Jacobi identity}\right\}\)

In QED, for a particle of unit charge `\(e\)' we write the covariant derivative
\(\left(\partial_\mu - ieA_\mu\right)\), and \(\left(\partial_\mu - ineA_\mu\right)\) for charge
\(n e\).

Each different amount of charge corresponds to a different representation of \(U(1)\).

In non-abelian gauge theories, different representations play the role of different charges.
(\(SO(2)\) & \(U(1)\) are isomorphic.)

For a general representation \(R\), we have

\[ D_\mu\psi \;=\; \left(\partial_\mu - i\,t^{a}_{R}A^{a}_{\mu}\right)\psi \]

where \(t^a_R\) is the representation of the generators \(t^a\) in the representation \(R\), and

\[ \psi \;\longrightarrow\; \psi' \;=\; D_R(g)\,\psi \]

The \(A^{a}_{\mu}\) are always the same.

All \(U(1)\), \(SU(2)\), \(SU(3)\) groups are compact groups:

• \(U(1)\) --- electrons
• \(2^2-1 = 3\) generators \(\in SU(2)\) --- weak (non-abelian)
• \(3^2-1 = 8\) generators \(\in SU(3)\) --- strong

with

\[ R(\theta,\varphi,\chi)\,R(\theta',\varphi',\chi') \;\neq\; R(\theta',\varphi',\chi')\,R(\theta,\varphi,\chi) \]

where \(\theta\in[0,\pi]\), \(\varphi\in[0,2\pi]\), \(\chi\in[0,2\pi]\).

Matter fields \(\psi,\varphi\) transform by a representation \(R\):

\[ \psi \;=\; \psi_i\,e_i \]

where \(e_i\) is a basis vector for the \(n\)-dimensional vector space \(V_R\), \(\dim(R) = n_R\).

\[ \psi(x) \;=\; \psi_i(x)\,e_i(x) \]

Under a gauge transformation,

\[ \psi_i' \;=\; g_{ij}\,\psi_j \ ,\qquad g_{ij} = g_{ij}(x) \] \[ \psi_i'(x) \;=\; g_{ij}(x)\,\psi_j(x) \] \[ D_\mu\psi \;=\; 0 \qquad\Rightarrow\qquad \psi \text{ is a covariantly constant field.} \] \[ \left(I\partial_\mu + A_\mu\right)\psi \;=\; 0 \]

with \(A_\mu = A^{a}_{\mu}t^{a}_{R}\), an \((n_R\times n_R)\) matrix (\(n_R\) dimensional).

\[ \left(\partial_\mu + A_\mu\right)U(x) \;=\; 0 \]

In general there may be no solution to this equation, but we can integrate this equation along
a curve \(C\) to get a path dependent solution \(U(x_0,y_0;C)\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[thick] (0,0) .. controls (0.6,1.4) and (1.9,1.6) .. (3.0,2.4);
\filldraw (0,0) circle (1.4pt) node[below left] {$x_0$};
\filldraw (3.0,2.4) circle (1.4pt) node[above left] {$y_0$};
\node[left] at (0.72,1.05) {$C$};
\draw[->] (3.0,2.4) -- (3.9,3.05) node[right] {$C^\mu$};
\node[right,align=left] at (4.3,1.6) {If we take the tangent\\ to this curve\\[2pt]
$C^\mu\left(\partial_\mu + A_\mu\right)U(x)$};
\end{tikzpicture}

Idea for the solution: break up \(C\) into infinitesimal parts of length \(\epsilon\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,decorations.pathmorphing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
\draw[dashed,thick] (0,0) .. controls (0.6,1.4) and (1.9,1.6) .. (3.0,2.4);
\filldraw (0,0) circle (1.2pt) node[below left] {$x_0$};
\filldraw (3.0,2.4) circle (1.2pt) node[above right] {$y_0$};
\foreach \t in {0.15,0.3,0.45,0.6,0.75,0.9}{
  \filldraw ($(0,0)!\t!(0,0)$) circle (0pt);
}
\filldraw (0.42,0.72) circle (1.1pt);
\filldraw (0.95,1.20) circle (1.1pt);
\filldraw (1.55,1.50) circle (1.1pt);
\filldraw (2.20,1.83) circle (1.1pt);
\node[left] at (0.18,0.40) {$\epsilon$};
\node[left] at (0.66,1.00) {$\epsilon$};
\node[above left] at (1.25,1.38) {$\epsilon$};
\end{tikzpicture}

\[ U(x,y;C) \;=\; \left(1 - A_{\mu_1}(x-\epsilon)\,\epsilon^{\mu_1}\right) \left(1 - A_{\mu_2}(x-2\epsilon)\,\epsilon^{\mu_2}\right)\cdots \left(1 - A_{\mu_n}(y)\,\epsilon^{\mu_n}\right) \]

This is an ordered product of factors like
\(\left(1 - A_{\mu_k}(x-k\epsilon)\,\epsilon^{\mu_k}\right)\).

Take the limit \(\epsilon\to 0\), \(N\to\infty\); then

\[ U(x,y;C) \;=\; P\left[\exp\left(-\int_{x,\,C}^{y}A_\mu\,dx^\mu\right)\right] \qquad\longrightarrow\qquad \text{path ordered exponential.} \]

Given \(U\), the equation \(D_\mu\psi = 0\) is solved by \(\psi(y) = U\psi_0(x)\), where \(\psi_0\) is a
constant vector.

----------------- \(\times\) -----------------

(\(x,y\) are four vectors.) If \(y = x+\epsilon\),

\[ U(x,x+\epsilon;C) \;=\; P\,e^{-\int_{x}^{x+\epsilon}A^\mu dx_\mu} \;=\; e^{-A^\mu\epsilon_\mu} \;\simeq\; 1 - A^\mu\epsilon_\mu \]

using \(\displaystyle\int_{x}^{x+\epsilon}f(x)\,dx = f(x)\cdot\epsilon\). So,

\[ U \;=\; \left(1 - A_{\mu_1}(x-\epsilon)\,\epsilon^{\mu_1}\right)\cdots \left(1 - A_{\mu_n}(y)\,\epsilon^{\mu_n}\right) \]