Prof. Sachindeo Vaidya (CHEP, IISc) | PDF
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We need to define the covariant derivative because derivatives transform covariantly under a
\[
\psi' \;=\; g\,\psi
\]
then
\[
\partial_\mu\psi' \;=\; (\partial_\mu g)\,\psi \;+\; g\,\partial_\mu\psi
\]
We require \(\partial_\mu\psi' = g\,(\partial_\mu\psi)\). So define
\[
D_\mu\psi \;=\; \left(\partial_\mu + A_\mu\right)\psi \;\equiv\; D_\mu(A)\,\psi
\]
with \(A_\mu\) the \underline{gauge potential}.
Let, under a local gauge transformation,
\[
D_\mu \to D_\mu' \ ,\qquad \psi \to \psi' = g\psi \ ,\qquad A \to A'
\]
such that \(\partial_\mu\psi' = g(\partial_\mu\psi)\). We require that
\[
D_\mu(A^{g})\,(g\psi) \;=\; g\,D_\mu(A)\,\psi
\]
\[
\underset{\text{(III)}}{A_\mu^{g}} \;=\; \underset{\text{(I)}}{g\,A_\mu\,g^{-1}}
\;-\; \underset{\text{(II)}}{(\partial_\mu g)\,g^{-1}}
\]
\((\partial_\mu g)g^{-1}\) is a Lie algebra valued field (at each \(x=(\vec{x},t)\)).
\(t^a\) are generators of \(G\), satisfying
\[
\left[t^a,\ t^b\right] \;=\; i f^{abc}\,t^c
\]
Since term (II) is Lie algebra valued, therefore terms (I) & (III) are Lie algebra valued.
Ex: \(SU(2)\), with \(\left[\dfrac{\sigma_i}{2},\dfrac{\sigma_j}{2}\right] = i\epsilon_{ijk}\dfrac{\sigma_k}{2}\),
\[
g(x) \;=\; e^{\,i\vec{\sigma}\cdot\hat{n}\,\theta(x)}
\;=\; \cos\theta(x)\,(I) \;+\; i\,\vec{\sigma}\cdot\hat{n}\,\sin\theta(x)
\]
Since \(\vec{\sigma}\) is a \(2\times2\) matrix \(\Rightarrow\) \(g(x)\) is a \(2\times2\) matrix.
Let's find:
\[
\partial_\mu g \;=\; -\sin\theta(x)\,\partial_\mu\theta(x)\,I
\;+\; i\,\vec{\sigma}\cdot\hat{n}\,\cos\theta(x)\,\partial_\mu\theta(x)
\;+\; i\,\vec{\sigma}\cdot\partial_\mu\hat{n}\,\sin\theta(x)
\]
\[
(\partial_\mu g)g^{-1} \;=\; \Big(-\sin\theta\,\partial_\mu\theta\ I
+ i\vec{\sigma}\cdot\hat{n}\cos\theta\,\partial_\mu\theta
+ i\vec{\sigma}\cdot\partial_\mu\hat{n}\,\sin\theta\Big)
\cdot\Big(\cos\theta - i\vec{\sigma}\cdot\hat{n}\sin\theta\Big)
\]
\[
=\; -\sin\theta\,\partial_\mu\theta\,\cos\theta
\;+\; i\,\vec{\sigma}\cdot\hat{n}\,\cos^2\theta\,\partial_\mu\theta
\;+\; i\,\vec{\sigma}\cdot\partial_\mu\hat{n}\,\sin\theta\cos\theta
\]
\[
\qquad +\; i\,\sin^2\theta\,\partial_\mu\theta\ \vec{\sigma}\cdot\hat{n}
\;+\; \left(\vec{\sigma}\cdot\hat{n}\right)^2\sin\theta\cos\theta\,\partial_\mu\theta
\;+\; \vec{\sigma}\cdot\hat{n}\left(\vec{\sigma}\cdot\partial_\mu\hat{n}\right)\sin^2\theta
\]
using $\left(\vec{\sigma}\cdot\hat{n}\right)^2 = \sigma_i\hat{n}_i\sigma_j\hat{n}_j
\[
=\; -\cancel{\sin\theta\cos\theta\,\partial_\mu\theta}
\;+\; i\,\vec{\sigma}\cdot\hat{n}\,\partial_\mu\theta
\;+\; i\,\vec{\sigma}\cdot\partial_\mu\hat{n}\,\sin\theta\cos\theta
\;+\; \cancel{\sin\theta\cos\theta\,\partial_\mu\theta}
\;+\; i\,\epsilon_{ijk}\,\sigma_k\left(\partial_\mu\hat{n}\right)_i\hat{n}_j
\]
\[
=\; \frac{\sigma_i}{2}\left(\text{stuff}\right)_i \;\in\; \mathfrak{g}
\]
So with \(a = 1,2,\ldots \dim(G)\) (\(a\) runs over the number of generators of the group), for
\[
\tilde{A}_\mu(x) \;=\; A_\mu^{a}(x)\,t^a
\]
\[
g(x) \;=\; e^{-it^a\theta^a(x)} \;=\; 1 - i\,t^a\theta^a(x)
\qquad \text{for } \theta^a \ll 1
\]
Compare with:
\[
e^{-i\vec{\sigma}\cdot\hat{n}\theta(x)} \;=\; \cos\theta(x)\,I
\;-\; i\left(\vec{\sigma}\cdot\hat{n}\right)\sin\theta(x)
\]
For \(\theta^a \ll 1\),
\[
A_\mu^{a\,g} \;=\; g\,A_\mu\,g^{-1} \;-\; (\partial_\mu g)\,g^{-1}
\]
\[
=\; \left(1 - it^a\theta^a\right)A_\mu^{a}\left(1 + it^b\theta^b\right)
\;-\; \partial_\mu\left(1 - it^a\theta^a(x)\right)\left(1 + it^b\theta^b(x)\right)
\]
\[
=\; \left(A_\mu^{a} - it^a\theta^a A_\mu^{a}\right)\left(1 + it^b\theta^b\right)
\;+\; i\,t^a\partial_\mu\theta^a(x) \;-\; t^a t^b(\partial_\mu\theta^a)\theta^b
\]
\[
=\; A_\mu^{a} + i\,A_\mu^{a}t^b\theta^b - it^a\theta^a A_\mu + t^a\theta^a A_\mu^{a}t^b\theta^b
+ i\,t^a\partial_\mu\theta^a(x) - t^a t^b(\partial_\mu\theta^a)\theta^b
\]
\[
=\; A_\mu^{a} \;+\; i\left[A_\mu^{a}t^a,\ t^b\right]\theta^b \;+\; i\,t^b\partial_\mu\theta^b
\]
\[
=\; A_\mu \;+\; i\,A_\mu^{a}\left[t^a,t^b\right]\theta^a \;+\; i\,t^b\partial_\mu\theta^b
\]
\[
=\; A_\mu \;+\; i\,t^a\underbrace{\left(\partial_\mu\theta^a + i f^{abc}A_\mu^{b}\theta^c\right)}_{
\text{covariant derivative for adj. rep.}}
\]
The adjoint representation is
\[
\left\{\left(T^a\right)_{bc} \;=\; -i f_{abc}\right\}
\]
For \(SU(2)\),
\[
\left(T^a\right)_{bc} \;=\; -i\,\epsilon_{abc}
\]
i.e. \(T^1_{\ 23} = -i = T^2_{\ 31} = T^3_{\ 21}\),
\[
T^1 \;=\; \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & i\\ 0 & -i & 0\end{pmatrix}
\]
Similarly we can find \(T^2\) & \(T^3\).
The defining representation & the adjoint representation have fundamental importance in the
So,
\[
A_\mu^{g} - A_\mu \;\simeq\; i\,t^a\left(D_\mu\theta^a\right)
\]
\(\hookrightarrow\) the covariant derivative of the adjoint field.
\(\mathcal{L}\) for \(A_\mu\) comes from the field strength \(F_{\mu\nu}\), which is defined as
\[
F_{\mu\nu} \;\equiv\; \left[D_\mu,\ D_\nu\right]
\;=\; \left[\partial_\mu + A_\mu,\ \partial_\nu + A_\nu\right]
\]
\[
=\; -i\,t^a\left(\partial_\mu A_\nu^{a} - \partial_\nu A_\mu^{a} + f^{abc}A_\mu^{b}A_\nu^{c}\right)
\;=\; -i\,t^a\,F^{a}_{\mu\nu}
\]
We can check that:
\[
F_{\mu\nu}(A^{g})\,(g\psi) \;=\; \left[D_\mu(A^{g}),\ D_\nu(A^{g})\right](g\psi)
\;=\; g\left[D_\mu(A),\ D_\nu(A)\right]\psi \;=\; g\,F_{\mu\nu}(A)\,\psi
\]
So,
\[
F_{\mu\nu}(A^{g}) \;=\; g\,F_{\mu\nu}(A)\,g^{-1}
\]
\(F\) transforms as an adjoint, or \(F\) transforms covariantly under a gauge transformation.
\[
\mathrm{Tr}\left(t^a t^b\right) \;=\; \frac{1}{2}\delta^{ab}
\]
\[
t^a \;\xrightarrow{\ g\ }\; g\,t^a\,g^{-1}
\qquad \text{adjoint action on the generators } t^a
\]
\[
F_{\mu\nu}^{a} \;\xrightarrow{\ g\ }\; g\,F_{\mu\nu}^{a}\,g^{-1}
\qquad \text{adjoint action}
\]
\[
t^a \;\longrightarrow\; g\,t^a\,g^{-1} \;=\; T^{ba}(g)\,t^b
\]
where \(T^{ba}(g)\) is the matrix representation of the adjoint action.