• #PDF
• #Gauge theories ---
• #Generalising the map to other groups ---
• [[#\(SU(2)\) vs \(SO(3)\) ---]]
• #Covariant derivative ---
• #Lie algebra valued gauge potential ---
• #Infinitesimal gauge transformation ---

Prof. Sachindeo Vaidya (CHEP, IISc) | PDF

Previous: Lecture 15

Next: Lecture 17

The principle of minimal coupling can be thought of as the requirement of gauge invariance.

\[ \mathcal{L} \;=\; \bar{\psi}\left(i\slashed{D} - m\right)\psi \] \[ \slashed{D} \;=\; \gamma^\mu\underbrace{\left(\partial_\mu - ieA_\mu\right)}_{D_\mu} \]

\(\mathcal{L}\) is invariant under

\[ \psi \;\longrightarrow\; \psi' \;=\; e^{\,ie\theta(x)}\,\psi(x) \] \[ \bar{\psi} \;\longrightarrow\; \bar{\psi}' \;=\; \bar{\psi}\,e^{-ie\theta(x)} \] \[ A_\mu \;\longrightarrow\; A_\mu' \;=\; A_\mu + \partial_\mu\theta(x) \]

\(\theta(x)\) is an arbitrary function of spacetime. \(\mathcal{L}\) is invariant under this
transformation.

Write this as (\(g\) is a group element)

\[ \psi \;\longrightarrow\; \psi' \;=\; g\,\psi \ ,\qquad \bar{\psi} \;\longrightarrow\; \bar{\psi}' \;=\; \bar{\psi}\,g^{-1} \] \[ e A_\mu' \;\equiv\; e A_\mu^{g} \;=\; g\,(eA_\mu)\,g^{-1} \;-\; i(\partial_\mu g)\,g^{-1} \]

where

\[ g(x) \;=\; e^{\,ie\theta(x)} \]

\(g\) is a function on \(M^{1,3}\) taking values in the group \(U(1)\):

\[ g:\ M^{1,3}\ \longrightarrow\ U(1) \]

i.e. at each point \(x\) we have an element of \(U(1) = e^{ie\theta(x)}\).

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\draw[->,thick] (2.9,0.0) -- (4.3,0.0) node[midway,above] {$g$};
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We want to generalise the map to other groups,

\[ g:\ M^{1,3}\ \longrightarrow\ SU(3),\ SU(2),\ \ldots \text{ and so on.} \]

Consider a set of fields \(\psi_i\) which transform under the \underline{fundamental
representation} of \(SU(N)\), \(i = 1,2,3,\ldots N\).

\(SU(N)\) group is the set of \(N\times N\) unitary matrices with \(\det = +1\):

\[ \left\{U:\ U^{\dagger}U = UU^{\dagger} = \mathbb{I}\ ,\quad |U| = 1\right\} \;=\; SU(N) \]

Example: \(SU(2)\) \(\to\) \(2\times 2\) unitary matrices with determinant \(=1\) (which is the
fundamental representation). We can also represent \(SU(2)\) as a set of \(3\times 3\) matrices,
called the adjoint representation of \(SU(2)\) (cf. \(SO(3)\)).

To describe spin \(2\) we need \((2\times 2+1)\times(2\times 2+1)\), i.e. \(5\times 5\)
representation of \(SU(2)\).

For a representation \(R\), we have a set of fields \(\varphi_\alpha\) with the transformation

\[ \varphi_\alpha' \;=\; D^{(R)}_{\alpha\beta}(g)\,\varphi_\beta \ ,\qquad \alpha = 1,2,3,\ldots N \]

where \(N\) is the dimension of the representation.

\(D^{\frac{1}{2}}(g)\) is a \(2\times 2\) representation, \(D^{j}(g)\) is a \((2j+1)\times(2j+1)\)
representation.

\(SU(2)\) & \(SO(3)\) are completely different groups. The similarity they have is that the Lie
algebras of \(SU(2)\) & \(SO(3)\) are the same:

\[ [J_i,J_j] \;=\; i\epsilon_{ijk}J_k \qquad\longleftrightarrow\qquad [L_i,L_j] \;=\; \epsilon_{ijk}L_k \]

Their difference is visible when we do a finite rotation. Under \(SO(3)\) a vector is the same
after a \(2\pi\) rotation around the \(z\)-axis. Under \(SU(2)\), a vector is the same only after a
\(4\pi\) rotation about the \(z\)-axis; after just a \(2\pi\) rotation the vector gets a minus sign.

\[ (J_1,J_2,J_3) \;\longrightarrow\; \left(\frac{\sigma_1}{2},\frac{\sigma_2}{2},\frac{\sigma_3}{2}\right) \]

these also transform like a vector under rotation.

How does a generator of \(SU(2)\) transform? It should also transform like a vector:

\[ J_i \;\longrightarrow\; g\,J_i\,g^{-1} \;=\; R_{ij}(g)\,J_j \]

The left side is the adjoint action of \(SU(2)\) (which is an \(SU(2)\) operation), and the right
side is \(SO(3)\). For \(g\) & \(-g\) we get the same \(R_{ij}(g)\).

If we want to discuss rotations of a spin \(\frac{3}{2}\) massive particle, we don't want the full
\(SU(4)\), but \(SU(2)\) mapped to some \(4\times 4\) matrices (i.e. the \((3/2)\) representation):

\[ g \;\longrightarrow\; D^{(3/2)}(g) \]

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

If \(\psi' = g\psi\), then (only true for global \(U(1)\), i.e. constant \(g\))

\[ \partial_\mu\psi' \;=\; g\,\partial_\mu\psi \]

(the derivative transforms in the same way). The derivative of the field should transform in
the same way, so

\[ \partial_\mu' \;=\; g\,\partial_\mu\,g^{-1} \]

so that \(\partial_\mu'\psi' = g\partial_\mu g^{-1}g\psi = g(\partial_\mu\psi)\).

Look at the derivative of \(\psi' = g\psi\) for local \(g(x)\):

\[ \partial_\mu\psi' \;=\; (\partial_\mu g)\,\psi \;+\; g\,(\partial_\mu\psi) \qquad\qquad \text{---(1)} \]

\(\partial_\mu\psi\) does \underline{not} transform covariantly.

Let's define the covariant derivative:

\[ D_\mu\psi \;=\; \partial_\mu\psi + A_\mu\psi \;=\; \left(\partial_\mu + A_\mu\right)\psi \;\equiv\; D_\mu(A)\,\psi \]

We have introduced a ``potential'' \(A_\mu\). We choose the transformation rule for \(A_\mu\)
(unless \(g\)) so as to cancel the \((\partial_\mu g)\) term in eqn (1); then \(D_\mu\psi\) will
transform covariantly.

Say \(A_\mu \to A_\mu^{g}\). We require (definition of covariant transformation):

\[ D_\mu(A^{g})\,(g\psi) \;=\; g\,\left(D_\mu(A)\,\psi\right) \] \[ \Rightarrow\qquad A_\mu^{g} \;=\; g\,A_\mu\,g^{-1} \;-\; (\partial_\mu g)\,g^{-1} \]

The term \((\partial_\mu g)g^{-1}\) is Lie algebra valued: if \(t^a\) are generators of \(G\)
(ex. \(SU(N)\)), satisfying

\[ \left[t^a,\ t^b\right] \;=\; i f^{abc}\,t^c \]

with \(f^{abc}\) the structure constants (ex. like \(\epsilon^{abc}\) for \(SU(2)\)), then \(A_\mu\) is
Lie algebra valued. (A point in \((\vec{x},t) = x\) is mapped to some Lie group element.)

Just like \(\varphi(x)\) is a map \(\varphi: M^{3,1}\to\mathbb{R}\), here

\[ g:\ M^{3,1}\ \longrightarrow\ G \]

at each point \(x\) we have a group element \(g\in G\).

Ex: \(SU(2)\),

\[ g(x) \;=\; e^{\,i\frac{\vec{\sigma}}{2}\cdot\hat{n}(x)\,\theta(x)} \] \[ e^{\,i\vec{\sigma}\cdot\hat{n}\theta} \;=\; (\cos\theta)\,I \;+\; i\,\vec{\sigma}\cdot\hat{n}\,\sin\theta \]

Check that \((\partial_\mu g)g^{-1}\) is Lie algebra valued:

\[ (\partial_\mu g)\,g^{-1} \;=\; \sum_i f_{\mu i}(x)\,\frac{\sigma_i}{2} \]

expanding in a linear combination of \(\dfrac{\sigma_i}{2}\) (generators of \(SU(2)\)).

\(g\) and \(\partial_\mu g\) are \underline{not} in the Lie algebra; only \((\partial_\mu g)g^{-1}\in\)
Lie algebra generators.

The group of rotations is \(SO(3)\), which is 3 dimensional: \((\theta,\varphi)\) for \(\hat{n}\) (the
axis) & 1 parameter for the amount of rotation.

\[ g(x) \;=\; e^{-i t^a\theta^a(x)} \;\simeq\; \mathbb{I} - i\,t^a\theta^a(x) \qquad (\theta^a \ll 1) \] \[ g - \mathbb{I} \;=\; -i\,t^a\theta^a(x) \] \[ A_\mu^{g}(x) \;=\; g(x)A_\mu(x)g^{-1}(x) \;-\; (\partial_\mu g(x))\,g^{-1}(x) \] \[ =\; \left(1 - i t^a\theta^a\right)A_\mu\left(1 + i t^a\theta^a\right) \;+\; i\,t^a\partial_\mu\theta^a(x)\times\left(1 + i t^a\theta^a\right) \]

Check it:

\[ =\; A_\mu + i t^a\theta^a - i t^a\theta^a A_\mu + \left(t^a\theta^a\right)^2 + i t^a\partial_\mu\theta^a(x) - (t^a)^2\theta^a\partial_\mu\theta^a(x) \] \[ =\; A_\mu + i\left[A_\mu,\ t^a\right]\theta^a + i\,t^a\partial_\mu\theta^a \] \[ =\; A_\mu \;+\; i\,t^a\underbrace{\left(\partial_\mu\theta^a + f^{abc}A_\mu^b\theta^c\right)}_{ \text{covariant derivative for adj. repn. } (D_\mu\theta)^a} \;+\; O(\theta^2) \]

So, under an infinitesimal gauge transformation, the change in the gauge field is given by

\[ A_\mu^{g} - A_\mu \;=\; i\,t^a\left(D_\mu\theta\right)^a \]