• #Lecture 7 --- Vector fields
  • #Gauge Invariance
• #Lecture 8 --- Scalar QED
  • #Feynman Rules for scalar QED
• #Lecture 9 --- Algebra

Under lorentz transformation ---

\[ \begin{aligned} \text{coordinates} &- \quad x^\mu \longrightarrow x'^\mu = \Lambda^\mu_{\;\nu}\, x^\nu\\ \text{scalar field} &- \quad \varphi(x') = \varphi(x)\\ \text{vector field}&: \quad A^\mu(x) \longrightarrow A'^\mu(x') = \Lambda^\mu_{\;\nu}\, A^\nu(x) \end{aligned} \]

Exercise:

\[ A'_\mu(x') = \left(\Lambda^{-1}\right)^{T\;\nu}_{\mu}\, A_\nu(x) \]

Example: \(\partial_\mu\varphi\) (occur in K.G lagrangian) \(\hookrightarrow\) transform like vector field.

\[ \begin{gathered} \partial_\mu\varphi'(x') = \left(\Lambda^{-1}\right)^{T\;\nu}_{\mu}\,\partial_\nu\varphi(x)\\ \partial^\mu\varphi'(x') = \Lambda^\mu_{\;\nu}\;\partial^\nu\varphi(x) \qquad{\text{with no other fields.}} \end{gathered} \]

We want to study independent vector fields (\(A_\mu\)).
The idea is to find EOM of \(A_\mu\) & write \(\mathcal{L}\) reverse but we can guess it as well.

Guessing a lagrangian density ---

\[ \mathcal{L} = \frac{1}{2}\,\partial_\mu A_\nu\;\partial^\mu A^\nu \qquad(\text{For massless field}) \qquad\left\{\; \begin{aligned} &\text{just like K.G.\ lagrangian}\\ &\varphi\longrightarrow A_\nu \end{aligned}\right. \] \[ \begin{gathered} \text{e.o.m}:\quad \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi)}\right) = \frac{\partial\mathcal{L}}{\partial\varphi}\\ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\right) = \frac{\partial\mathcal{L}}{\partial A_\nu} \;\Rightarrow\; \partial_\mu\left(\partial^\mu A^\nu\right) = 0 \end{gathered} \]

It turns out e.o.m is \(\partial^\mu\partial_\mu A_\nu = 0\) (one independent K.G. eqn for each component).

\[ \boxed{\partial^2 A_\nu = 0} \]

This theory as soon as we quantize, have serious problem.
So to quantize we will first write hamiltonian and impose canonical commutation relations.
To write \(\mathcal{H}\), we require \(\mathcal{L} = \frac{1}{2}\dot{A}_\mu\dot{A}^\mu + \cdots\)

\[ \pi^\mu = \frac{\partial\mathcal{L}}{\partial\dot{A}_\mu} = \dot{A}^\mu \qquad\Bigg|\quad \begin{aligned} \dot{A}^\mu &= \eta^{\mu\nu}\dot{A}_\nu\\ \mathcal{L} &= \frac{1}{2}\dot{A}_\mu\,\eta^{\mu\nu}\,\dot{A}_\nu\\ \pi^\mu = \frac{\partial\mathcal{L}}{\partial\dot{A}_\mu} &= \frac{1}{2}\eta^{\mu\nu}\dot{A}_\nu + \frac{1}{2}\dot{A}_\nu\,\eta^{\nu\mu}\quad\uparrow(\mu\leftrightarrow\nu)\\ &= \dot{A}^\mu \end{aligned} \]

Commutator

\[ \left[A_\mu(t,\vec{x}),\; \dot{A}_\nu(t,\vec{x}')\right] = i\,\eta_{\mu\nu}\,\delta^3(x-x') \]

We will write \(A_\mu(t,\vec{x})\) in expansion of oscillators --- we get ---

\[ \left[a_{\mu,\vec{k}}\,,\; a^\dagger_{\nu,\vec{k}'}\right] = (2\pi)^3\,\eta_{\mu\nu}\,\delta^3(\vec{k}-\vec{k}') \]

Note that \(\eta_{\mu\nu} = \mathrm{diag}(1,-1,-1,-1)\); for \(\mu,\nu\neq 0\): \(\eta_{\mu\nu} = 0\) or \(-1\).
We get

\[ \begin{aligned} \left[a_{i,\vec{k}}\,,\; a^\dagger_{j,\vec{k}'}\right] &= (2\pi)^3\,(-1)\,\delta^3(\vec{k}-\vec{k}')\\ &= -\text{ve}\;! \end{aligned} \]

This is not good as it leads to

\[ a^\dagger_{i,\vec{k}}\,|0\rangle \qquad (i = 1,2,3) \;\;\text{---\,(1)} \]

\(\uparrow\) should create 3 possible types of particle depending upon value of \(i\), with momentum \(\vec{k}\).
Let us find norm of this state --- --- (3)

\[ \begin{aligned} \text{Norm} &\to \langle 0|\,a_{i,k}\; a^\dagger_{i,\vec{k}}\,|0\rangle \qquad\text{using}\; \left\{ \begin{aligned} \left[a_{i,k},\, a^\dagger_{j,\vec{k}'}\right] &= (2\pi)^3\,\delta^3(k-k')\,\eta_{ij}\\ \text{for } i=j:\;\left[a_{i,\vec{k}},\, a^\dagger_{i,k'}\right] &= (2\pi)^3(-1)\,\delta^3(k-k') \end{aligned} \right.\\ &= \langle 0|\,a^\dagger_{i,\vec{k}}\; a_{i,\vec{k}}\,|0\rangle + \langle 0|\left[a_{i,k},\, a^\dagger_{i,\vec{k}}\right]|0\rangle\\ \text{Norm} &= 0 + (-\text{ve})\\ &\quad\text{Norm }<0\text{} \end{aligned} \]

Physically this is not acceptable!

As it means that the theory have negative probabilities and it will be inconsistent.
So we write a slightly better lagrangian

\[ \begin{gathered} \mathcal{L} = -\frac{1}{2}\,\partial_\mu A_\nu\,\partial^\mu A^\nu\\ \text{E.O.M.}\;\longrightarrow\;\partial^2 A_\nu = 0 \qquad(\text{same})\\ \text{quantisation}\;\Rightarrow\;\pi^\mu = \frac{\partial\mathcal{L}}{\partial\dot{A}_\mu} = -\dot{A}^\mu\\ \Rightarrow\quad\left[A_\mu(t,\vec{x}),\, A_\nu(t,\vec{x}')\right] = -i\,\eta_{\mu\nu}\,\delta^3(\vec{x}-\vec{x}')\\ \Rightarrow\quad\left[a_{\mu,\vec{k}}\,,\; a^\dagger_{\nu,\vec{k}'}\right] = -(2\pi)^3\,\eta_{\mu\nu}\,\delta^3(\vec{k}-\vec{k}') \end{gathered} \]

This solves \(\left[a_{i,\vec{k}},\, a^\dagger_{j,k}\right] = +\)ve. But now the problem is transferred to

\[ \begin{gathered} \left[a_{0,\vec{k}}\,,\; a^\dagger_{0,k'}\right] = (2\pi)^3\,(-1)\,\delta(\vec{k}\,\vec{k}') = -\text{ve}\;!\\ \Rightarrow\quad\text{Norm of } a^\dagger_{0,\vec{k}}|0\rangle \;\text{ becomes }-\text{ve!} \end{gathered} \]

We still have inconsistency!
We have a particle (photon) which is described by vector field \(A_\mu(\vec{x},t)\).
Maxwell's E.O.M for \(A_\mu = (\phi, \vec{A})\) are

\[ \partial^\mu F_{\mu\nu} = 0 \;\Rightarrow\; \partial^\mu\left(\partial_\mu A_\nu - \partial_\nu A_\mu\right) = 0 \qquad \begin{aligned} &(\text{set of 4 eqn})\\ &\text{to determine } A_\mu\;(\text{4 components}) \end{aligned} \]

These E.O.M. came from lagrangian

\[ \begin{aligned} \mathcal{L} &= -\frac{1}{4}\,F_{\mu\nu}\,F^{\mu\nu}\\ &= -\frac{1}{4}\left(\partial_\mu A_\nu - \partial_\nu A_\mu\right)\left(\partial^\mu A^\nu - \partial^\nu A^\mu\right)\\ &= -\frac{1}{4}\Big(\underline{\partial_\mu A_\nu\,\partial^\mu A^\nu} - \overline{\partial_\mu A_\nu\,\partial^\nu A^\mu} - \overline{\partial_\nu A_\mu\,\partial^\mu A^\nu} + \underline{\partial_\nu A_\mu\,\partial^\nu A^\mu}\Big)\\ &= -\frac{1}{4}\left(2\,\partial_\mu A_\nu\,\partial^\mu A^\nu - 2\,\partial_\mu A_\nu\,\partial^\nu A^\mu\right)\\ &= -\frac{1}{2}\,\partial_\mu A_\nu\,\partial^\mu A^\nu + \frac{1}{2}\,{\boxed{\partial_\mu A_\nu\,\partial^\nu A^\mu}} \qquad\left\{\;\text{These terms are different}\;\checkmark\right. \end{aligned} \]

This closely resembles to our guessed lagrangian but it has some extra term:

\[ \text{our new guess}\Big]\qquad \mathcal{L} = -\frac{1}{2}\,\partial_\mu A_\nu\,\partial^\mu A^\nu + {\boxed{\dfrac{1}{2}\left(\partial^\mu A_\mu\right)^2}} \qquad{\underline{\text{How?}}} \]

Let us try quantisation with this \(\mathcal{L}\):

\[ \begin{gathered} \pi^\mu = \frac{\partial\mathcal{L}}{\partial\dot{A}_\mu} = -F^{0\mu} = -\left(\dot{A}^\mu - \partial^\mu A^0\right)\\ \Rightarrow\quad \pi^0 = -\left(\dot{A}^0 - \dot{A}^0\right) = 0 \;\;\text{---\,(1)}\\ \Rightarrow\quad A_0(t,x)\;does not have conjugate momentum. \end{gathered} \]

also, \(A_0(t,\vec{x})\) is not a dynamical field.

\[ \begin{aligned} A'_\mu(x) &= A_\mu(x) + \partial_\mu\lambda(x)\\ F'_{\mu\nu} &= \partial_\mu A'_\nu - \partial_\nu A'_\mu\\ &= \partial_\mu\left(A_\nu + \partial_\nu\lambda\right) - \partial_\nu\left(A_\mu + \partial_\mu\lambda\right)\\ &= \partial_\mu A_\nu + \cancel{\partial_\mu\partial_\nu\lambda} - \partial_\nu A_\mu - \cancel{\partial_\nu\partial_\mu\lambda}\\ &= \partial_\mu A_\nu - \partial_\nu A_\mu\\ F'_{\mu\nu} &= F_{\mu\nu} \qquad\Rightarrow\;\mathcal{L}\text{ is invariant under gauge transformation.} \end{aligned} \]

Configuration \(A_\mu\) and \(A_\mu + \partial_\mu\lambda\) has same lagrangian, therefore \(A_\mu\) inherit same dynamics.
where \(\lambda\) = any arbitrary fun. of \(x\).
It is better to call it gauge invariance rather than gauge symmetry.
This is redundancy of description: \(A_\mu(x)\) and \(A_\mu + \partial_\mu\lambda\) are same physical configuration.
We could have choosen \(\mathcal{L}\propto F^2\) where \(F_{\mu\nu}\) is basic field rather than \(A_\mu\), then

\[ \begin{gathered} \text{EOM}\;\Rightarrow\;\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu F)}\right) = \frac{\partial\mathcal{L}}{\partial F} \;\Rightarrow\; 0 = F_{\mu\nu}\\ \Rightarrow\; F_{\mu\nu} = 0 \;\Rightarrow\; \vec{E} = 0 = \vec{B} \end{gathered} \]

Which is not consistent with maxwells eqn. So we have to write/guess right lagrangian in order to get correct E.O.M.

Note: Gauge invariance is only way to have vector fields and get rid of negative norm states.
In eqn (1): \(\pi^0 = 0\) arise from gauge invariance.

• We will see that gauge invariance implies absence of negative norm states.

Note 1:

• Even considering tensor fields rather than vector fields results into occurence of negative norm states. The problem of \(-\)ve norm comes because time & space have different signatures that itself comes from relativity.
• Any theory involving vector or tensor fields must have gauge invariance in order to be consistent. (No consistent quantum theory without gauge invariance, for such fields.)

Note 2: putting mass term \(-\frac{1}{2}m^2 A_\mu A^\mu\) --- but this term is not gauge invariant.
''It suggest that vector fields should be massless.''
(Any particle which is described by a vector field has to be massless)

\{\(massless ness of photon is due to gauge invariance.\)\} except gluons

Unfortunately we know 11 more vector particles, \(W^\pm, Z\) and 8 gluons and none of those is massless. That is due to 2 mechanism which are higgs mechanism and confinement.

\[ \left\{ \begin{aligned} \text{Higgs mechanism} &\;\text{---}\; W^\pm, Z \quad\text{weak interaction}\\ \text{Confinement} &\;\text{---}\; \text{gluons}\quad\text{strong ''} \end{aligned} \right. \]

This will be discussed in end of course.
Note 3: We can perform multiple gauge transformations in different order ---

\[ \begin{gathered} A_\mu \longrightarrow A_\mu + \partial_\mu\lambda_1(x) \longrightarrow \left(A_\mu + \partial_\mu\lambda_2(x)\right) + \partial_\mu\lambda_1(x)\\ \text{(or)}\quad A_\mu \longrightarrow A_\mu + \partial_\mu\lambda_2(x) \longrightarrow \left(A_\mu + \partial_\mu\lambda_1(x)\right) + \partial_\mu\lambda_2(x) \end{gathered} \]

We are getting same answer in RHS.

[This is known as abelian gauge invariance.]

So maxwell lagrangian has abelian gauge invariance.
So we will study/consider

\[ \mathcal{L} = -\frac{1}{4}\,F_{\mu\nu}\,F^{\mu\nu} \qquad\left\{\;\text{where }A_\mu(x)\text{ is field variable rather than }F^{\mu\nu}\text{}\right. \]

We will quantize it, but we know \(\pi^0 = 0\): we can't quantize.
In order to quantize we must fix the gauge.
Here fixing the gauge means, given any configuration of \(A_\mu(x)\) we apply a gauge transformation & bring it to some standard form, after that we are not able to make more gauge transformation.
It will enable us to work with genuine physical degrees of freedom. It will also exhibit for me which d.o.f. can be taken away by gauge transformations, & which remains. ✓
There are many ways to fix the gauge, also we have to make a choice wheather we want to respect manifest lorentz invariance or not, both of these have advantages.
Now we will see two ways of fixing the gauge:

• violates manifest lorentz Invariance (Coulomb gauge)
• Preserves manifest lorentz Invariance (Lorentz gauge)

Finally, all physical results will be lorentz invariant, (Coulomb gauge violates during intermediate steps).

(1) Coulomb gauge:
Lets look at gauge transformation seperately for \(A_0\) and \(A_i\):

\[ \begin{aligned} A_0 &\longrightarrow A_0 + \partial_0\lambda \qquad\text{---\,{(a)}}\\ A_i &\longrightarrow A_i + \partial_i\lambda \qquad\text{---\,{(b)}} \end{aligned} \]

We can make \(A_0 = 0\) (to get rid of \(\pi^0 = 0\)), i.e. choose \(\lambda(x)\) such that \(\partial_0\lambda = -A_0(t,\vec{x})\)

\[ \lambda(t,\vec{x}) = -\int^t dt'\;A_0(t',\vec{x}) \qquad\uparrow\;\text{put back in eq (b) to transform } A_i \]

Using this \(\lambda(t,\vec{x})\), we ''gauge away'' \(A_0\).
But still we can make more gauge transformations which will bring back non zero \(A_0\) (time dependent \(\lambda(x,t)\)). But we only allow for time independent \(\lambda(\vec{x})\) for further work.
If we consider time independent \(\lambda(\vec{x})\) we still have more gauge transformations:

\[ A_i \longrightarrow A_i + \partial_i\lambda(\vec{x}) \] \[ \text{E.O.M}\;\Rightarrow\;\partial^\mu F_{\mu\nu} = 0 \qquad\text{where } \left(A_0 = 0\right) \]

We have two sets of eqn: 1) \(\nu = 0\); 2) \(\nu = j\).

\[ 1)\;\nu = 0 \;\Rightarrow\; \partial^i F_{i0} = 0 \qquad\left(F_{i0} = \partial_i\cancelto{0}{A_0} - \partial_0 A_i\right) \] \[ \begin{gathered} -\partial^i\left(\partial_0 A_i\right) = 0\\ \boxed{\partial_0\left(\partial^i A_i\right) = 0}\;\;\text{---\,(2)}\\ \Rightarrow\quad \partial^i A_i \;\text{ is time independent} \end{gathered} \]

(We first got rid of \(A^0\), by doing gauge transformation; then we further do time independent gauge transformation to such that \(\partial^i A_i\) is time independent.)

\[ \begin{gathered} 2)\;\nu = j:\qquad \text{E.O.M}\;\Rightarrow\;\partial^\mu F_{\mu\nu} = 0\\ \partial^0 F_{0j} + \partial^i F_{ij} = 0\\ \Rightarrow\quad\partial^0\Big(\partial_0 A_j - \partial_j\cancelto{0}{A_0}\Big) + \partial^i\left(\partial_i A_j - \partial_j A_i\right) = 0\\ \Rightarrow\quad\boxed{\;\partial^0\partial_0\,A_j + \partial^i\partial_i\,A_j - \partial_j\left(\partial^i A_i\right) = 0\;}\;\;\text{---\,(3)} \end{gathered} \]

If \(\partial^i A_i\) is set to zero, we get

\[ \begin{gathered} \partial^0\partial_0\,A_j + \partial^i\partial_i\,A_j = 0\\ \left(\partial^0\partial_0 + \partial^i\partial_i\right)A_j = \partial^\mu\partial_\mu\,A_j = 0 \;\;\text{---\,{(2)}} \end{gathered} \]

which is K.G. equation for 3 components of \(\vec{A}\).

\[ \text{i.e.}\quad \left\{ \begin{aligned} \partial^2 A_1 &= 0\\ \partial^2 A_2 &= 0\\ \partial^2 A_3 &= 0 \end{aligned} \right. \qquad\text{---\,{(3)}} \]

This suggest we should set \(\partial^i A_i = 0\) by a gauge transformation.
Under gauge transformation:

\[ \begin{gathered} A_i \longrightarrow A_i + \partial_i\lambda\\ \partial^i A_i \longrightarrow \partial^i A_i + \partial^i\partial_i\lambda\\ \text{For}\quad \partial^i A_i = 0 = \partial^i A_i + \partial^i\partial_i\lambda\\ \Rightarrow\quad \nabla\cdot A + \nabla^2\lambda = 0\\ \nabla^2\lambda = -\nabla\cdot A \qquad\left(\text{Poisson's eqn}\right) \end{gathered} \]

\(\Big(\)Solving for \(\lambda\) from above eqn we set \(\partial^i A_i = 0\Big)\),
and \(\partial^i A_i\) remains zero as \(\partial_0\left(\partial^i A_i\right) = 0\).
Final equations of motion becomes --- (eqn (3) becomes) ---

\[ \Box A_j = 0 \quad\text{(or)}\quad \partial^2 A_j = 0 \quad\text{(or)}\quad \partial^i\partial_i\,A_j = 0, \qquad\text{subject to condn}\quad \partial^j A_j = 0. \]

We have already set \(A_0 = 0\) (to get rid of \(\pi^0=0\)), now one more constraint \(\partial\cdot A = 0\) (\(\partial^j A_j = 0\)) implies there are only two independent oscillators.
That means vector particle has two independent polarisations. (\(A_0\) was non physical we gauged it away very first, then we are left with three with one constraint. So we are left with 2 independent d.o.f of \(A_\mu\).)

• Note that there is an appearent problem b/w lorentz invariance & two polarisations. Four vector has to have 4 components. Gauge invariance removed two degrees of freedom.

Now we will see how these two d.o.f. are removed in a different gauge. Before that we would like to calculate Angular momentum of operators \(J_i\) (\(J_1\,J_2\,J_3\)) by using Noether's theorem and rotational invariance of system.
When we quantize the system we get vacuum state, 1 particle, 2 particle ... \(n\) particle states.
Let's act with \(J^2\) on one particle states. One particle states can not have orbital angular momentum. (There is nothing to orbit around, so it is purely the spin total angular momentum.)
If we find \(J^2\):

\[ J^2\,|1\rangle = j(j+1)\,\hbar^2\,|1\rangle \]

Spin of state \(|1\rangle\) is 1: Vector field describe spin 1 particle.

Lorentz gauge ---
In this gauge we don't distinguish \(A_0\,A_1\,A_2\,A_3\); we make \(A_0\) dynamical and carried along and get rid of at the end.
Starting from

\[ \mathcal{L} = -\frac{1}{2}\,\partial_\mu A_\nu\,\partial^\mu A_\nu + \frac{1}{2}\left(\partial^\mu A_\mu\right)^2 \qquad\hookrightarrow\;\text{this term created gauge invariance!} \]

As a choice of gauge we choose (lorentz gauge)

\[ \partial^\mu A_\mu = 0 \qquad\left(\text{setting 4 divergences to 0 rather than 3 divergences } \partial^i A_i = 0\;(\text{Coulomb gauge})\right) \]

Now;

\[ \begin{gathered} \mathcal{L} = -\frac{1}{2}\,\partial_\mu A_\nu\,\partial^\mu A_\nu\\ \pi^0 = -\dot{A}^0 \qquad\left(\text{But we saw in start of lecture that this leads to negative Norm of } a^\dagger_{0,k}|0\rangle\right) \end{gathered} \]

Given;

\[ \begin{gathered} A_\mu \longrightarrow A_\mu + \partial_\mu\lambda\\ \partial^\mu A_\mu \longrightarrow \partial^\mu A_\mu + \partial^2\lambda\\ \text{if } \partial^2\lambda = 0 \;\text{ then } \partial^\mu A_\mu \text{ is fixed.} \end{gathered} \]

If \(\partial^\mu A_\mu = 0\), then All such gauge transformations \(A_\mu\to A_\mu + \partial_\mu\lambda(x)\) fixes \(\partial^\mu A_\mu = 0\) iff \(\partial^2\lambda = 0\).
If \(\lambda(x)\) satisfies massless K.G eqn then it preserves \(\partial^\mu A_\mu = 0\).

Analysis in momentum space ---
Instead of \(A_\mu(x)\) we have \(\widetilde{A}_\mu(k)\) or \(A_\mu(k)\).
Gauge field: \(\widetilde{A}_\mu(k)\).
Lorentz Gauge condition in \(k\)-space:

\[ k^\mu\,\widetilde{A}_\mu(k) = 0 \;\;\text{---\,(1)} \]

Residual gauge freedom (Remaining gauge transf.):

\[ \widetilde{A}_\mu(k) \longrightarrow \widetilde{A}_\mu(k) + k_\mu\,\lambda(k) \]

\(\partial^2\lambda = 0\) in \(k\) space becomes

\[ k^\mu k_\mu = k^2 = 0 \;\;\text{---\,(2)} \qquad\text{i.e.\ only }\lambda(k)\text{ are allowed where } k^2 = 0. \]

These two conditions remove two out of 4 polarisations. (Removing \(A^0, A^3\) & left with \(A^1\) & \(A^2\) (i.e. 2 transverse polarisations)).
1 particle state \(a^\dagger_{\mu,k}|0\rangle\):

\[ (1)\;\text{set}\;k^\mu\, a^\dagger_{\mu,k}\,|0\rangle \sim 0 \qquad\left(k^\mu a^\dagger_{\mu,k}|0\rangle \text{ should be identified with zero.}\right) \]

\(\downarrow\) To remove component along \(k^\mu\).

\[ \left\{\; \begin{aligned} &\text{Analogy is to have component of } \vec{V} = x\hat{i} + y\hat{j} + z\hat{k}\\ &\text{on } xy \text{ plane; we simply do } \vec{V}\cdot\hat{k} = 0\;\text{ or }\;set z = 0\\ &\text{to have } \vec{v}\,' = x\hat{i} + y\hat{j} \end{aligned} \right. \]

ex: We need \(k^2 = 0\) (to maintain \(k^\mu\cdot\widetilde{A}_\mu(k) = 0\)).
So, choose \(k\) such that \(k^2 = 0\) (light like \(k^\mu\))

\[ \Rightarrow\quad \begin{aligned} k_\mu &= k(1,0,0,1)\\ k^\mu &= k(1,0,0,-1) \end{aligned} \] \[ \begin{gathered} k^\mu\, a_{\mu,k}\,|0\rangle = k\, a^\dagger_{0,k}|0\rangle - k\, a^\dagger_{3,k}|0\rangle = 0\\ \Rightarrow\quad\underline{\;a^\dagger_{0,k}\,|0\rangle \sim a^\dagger_{3,k}\,|0\rangle\;} \end{gathered} \]

Physical states ---
Taking arbitrary linear combination

\[ \xi^\mu\, a^\dagger_{\mu,\vec{k}}\,|0\rangle \]

Scalar QED: Complex scalar field coupled to a vector field.

\[ \begin{gathered} \mathcal{L} = \partial_\mu\varphi^{*}\,\partial^\mu\varphi - m^2\varphi^{*}\varphi\\ \varphi = \varphi_1 + i\varphi_2\,,\qquad \varphi^{*} = \varphi_1 - i\varphi_2 \end{gathered} \] \[ \begin{aligned} \mathcal{L} &= \partial_\mu\left(\varphi_1 - i\varphi_2\right)\partial^\mu\left(\varphi_1 + i\varphi_2\right) - m^2\left(\varphi_1^2+\varphi_2^2\right)\\ &= \partial_\mu\varphi_1\,\partial^\mu\varphi_1 - \cancel{i\,\partial_\mu\varphi_2\,\partial^\mu\varphi_1} + \cancel{i\,\partial_\mu\varphi_1\,\partial^\mu\varphi_2} + \partial_\mu\varphi_2\,\partial^\mu\varphi_2 - m^2\left(\varphi_1^2+\varphi_2^2\right)\\ \mathcal{L} &= \partial_\mu\varphi_1\,\partial^\mu\varphi_1 + \partial_\mu\varphi_2\,\partial^\mu\varphi_2 - m^2\left(\varphi_1^2+\varphi_2^2\right) \end{aligned} \]

which is \(\mathcal{L}\) for two real scalar fields.
Interaction term is \(\lambda\dfrac{\left(\varphi^{*}\varphi\right)^2}{6}\) \(\hookrightarrow\) will be clear later. So that;

\[ \mathcal{L} = \partial_\mu\varphi^{*}\,\partial^\mu\varphi - m^2\,\varphi^{*}\varphi - \lambda\frac{\left(\varphi^{*}\varphi\right)^2}{6} \]

EOM:

\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\varphi^{*})}\right) = \frac{\partial\mathcal{L}}{\partial\varphi^{*}} \;\Rightarrow\; \partial_\mu\partial^\mu\varphi = -m^2 - \frac{\lambda}{3}\left(\varphi^{*}\varphi\right)\varphi \]

i.e.

\[ \text{complex conjugate}\left(\!\!\begin{array}{c}\curvearrowright\end{array}\!\!\right) \left. \begin{aligned} \partial^2\varphi + m^2 + \frac{\lambda\,\varphi^2\varphi^{*}}{3} &= 0\\[1ex] \partial^2\varphi^{*} + m^2 + \frac{\lambda\,\varphi^{*2}\varphi}{3} &= 0 \end{aligned} \;\right\}\; \begin{aligned} &\text{2 set of EOM.}\\ &\text{For } \underline{\varphi\;\&\;\varphi^{*}}. \end{aligned} \]

Our \(\mathcal{L}\) is invariant under Global symmetry.
Initially even this global symmetry (which presently is known as phase symmetry) was called as gauge transformation.
i.e. we rescale field \(\varphi\):

\[ \begin{gathered} \varphi(x) \mapsto e^{\alpha}\,\varphi(x)\\ \varphi^{*}(x) \mapsto e^{\alpha}\,\varphi^{*}(x) \end{gathered} \]

but action/\(\mathcal{L}\) is not invariant so we are forced to have phase transformations

\[ \begin{gathered} \varphi(x) \longrightarrow e^{i\alpha}\,\varphi(x)\\ \varphi^{*}(x) \longrightarrow e^{-i\alpha}\,\varphi(x) \end{gathered} \]

to have invariant \(\mathcal{L}\) under this phase symmetry. (or)
Global gauge transformation (\(\alpha\) = const.).
Under infinitesimal transformation (\(\alpha\ll 1\)):

\[ \begin{aligned} \varphi(x) &\longrightarrow (1+i\alpha)\,\varphi(x) &\Rightarrow\quad \delta\varphi(x) &= i\alpha\,\varphi(x)\\ \varphi^{*}(x) &\longrightarrow (1-i\alpha)\,\varphi^{*}(x) &\Rightarrow\quad \delta\varphi^{*}(x) &= -i\alpha\,\varphi(x) \end{aligned} \]

Conserved current of this symmetry is ---

\[ J_\mu = i\left(\varphi^{*}\,\partial_\mu\varphi - \varphi\,\partial_\mu\varphi^{*}\right) \]

\(\downarrow\) Called as current as it is a 4 vector made from fields.
To see if \(J_\mu\) is conserved, we find:

\[ J_\mu\left(\varphi,\varphi^{*}\right) = J_\mu\left(\varphi',\varphi'^{*}\right) \]

RHS:

\[ \begin{aligned} J_\mu\left(\varphi', \varphi'^{*}\right) &= i\left(\varphi'^{*}\,\partial_\mu\varphi' - \varphi'\,\partial_\mu\varphi'^{*}\right)\\ &= i\left(e^{-i\alpha}\varphi^{*}\,\partial_\mu\left(e^{i\alpha}\varphi\right) - \left(e^{i\alpha}\varphi\right)\partial_\mu\left(e^{-i\alpha}\varphi^{*}\right)\right)\\ &= i\left(\varphi^{*}\,\partial_\mu\varphi - \varphi\,\partial_\mu\varphi^{*}\right)\\ &= J_\mu\left(\varphi,\varphi^{*}\right). \end{aligned} \]

(or)

\[ \partial^\mu J_\mu = 0 \qquad\text{(using EOM)}. \] \[ \begin{aligned} \partial^\mu J_\mu &= \partial^\mu(i)\left(\varphi^{*}\partial_\mu\varphi - \varphi\,\partial_\mu\varphi^{*}\right)\\ &= i\left(\cancel{\partial^\mu\varphi^{*}\,\partial_\mu\varphi} + \varphi^{*}\,\partial^2\varphi - \cancel{\partial^\mu\varphi\,\partial_\mu\varphi^{*}} - \varphi\,\partial^2\varphi^{*}\right)\\ &= i\left(\varphi^{*}\partial^2\varphi - \varphi\,\partial^2\varphi^{*}\right)\\ &= i\left(-\varphi^{*}\left(m^2 + \frac{\lambda}{3}\varphi^{*}\varphi^2\right) + \varphi\left(m^2 + \frac{\lambda}{3}\varphi\left(\varphi^{*}\right)^2\right)\right)\\ &= i\left(-m^2\varphi^{*} - \cancel{\frac{\lambda}{3}\varphi^2(\varphi^{*})^2} + m^2\varphi + \cancel{\frac{\lambda}{3}\varphi\,\varphi^{*2}}\right)\\ &= i\,m^2\left(\varphi - \varphi^{*}\right)\\ &{\;= 0\;! \qquad\left(\text{Complete it}\right).} \end{aligned} \] \[ Q = \int d^3x\;J_\mu = i\int d^3x\left(\varphi^{*}\,\partial_\mu\varphi - \varphi\,\partial_\mu\varphi^{*}\right) \]

Symmetry is generated by \(e^{-i\alpha Q}\):

\[ \begin{gathered} \varphi(x) \longrightarrow e^{-i\alpha Q}\,\varphi(x)\, e^{i\alpha Q}\\ \varphi^{*}(x) \longrightarrow e^{-i\alpha Q}\,\varphi^{*}(x)\, e^{i\alpha Q}\\ \delta\varphi = +i\alpha\left[Q,\varphi\right]\\ \delta\varphi^{*} = -i\alpha\left[Q,\varphi^{*}\right] \end{gathered} \] \[ \begin{aligned} \left[\varphi(t,\vec{x})\,,\; \dot{\varphi}^{*}(t,\vec{y})\right] &= i\,\delta^3(\vec{x}-\vec{y})\\ \left[\varphi^{*}(t,\vec{x})\,,\; \dot{\varphi}(t,\vec{y})\right] &= i\,\delta^3(\vec{x}-\vec{y}) \end{aligned} \]

Symmetry \(\Rightarrow\) current \(\Rightarrow\) Charge \(\curvearrowright\)

If we apply translational invariance; this leads to current \(T_{\mu\nu}\) (energy--momentum tensor) that leads to conserved charge (4-momentum \(P_\mu\)); & four momentum generates back the translational invariance.
Suppose \(\alpha = \alpha(x)\) (local gauge invariance):

\[ \begin{aligned} \partial_\mu\varphi &\longrightarrow \partial_\mu\left(e^{i\alpha(x)}\,\varphi\right)\\ &= e^{i\alpha(x)}\,\partial_\mu\varphi + i\,\partial_\mu\alpha(x)\, e^{i\alpha(x)}\,\varphi\\ &= e^{i\alpha(x)}\left(\partial_\mu\varphi + i\,\partial_\mu\alpha(x)\;\varphi(x)\right) \end{aligned} \] \[ \alpha = \alpha(t,\vec{x}) = \alpha(x)\,,\qquad \varphi = \varphi(t,\vec{x}) = \varphi(x) \]

So under,

\[ \begin{gathered} \varphi \longrightarrow e^{i\alpha(x)}\,\varphi(x)\\ \varphi^{*} \longrightarrow e^{-i\alpha(x)}\,\varphi^{*}(x) \end{gathered} \]

the (kinetic term) \(\mathcal{L}_T = \partial_\mu\varphi\,\partial^\mu\varphi^{*}\) is not invariant!
But; \(\mathcal{L}_m = m\,\varphi^{*}\varphi\) is invariant!
We then have to make \(\mathcal{L}\) which is invariant under local gauge transformation.
We look for term

\[ \mathcal{L}_{\text{kinetic}} = D_\mu\varphi\; D^{*\mu}\varphi^{*} \]

So that

\[ \begin{gathered} D_\mu\varphi \longrightarrow e^{i\alpha(x)}\, D_\mu\varphi\\ D_\mu\varphi^{*} \longrightarrow e^{-i\alpha(x)}\, D_\mu\varphi^{*} \end{gathered} \]

We look for \(D_\mu\) of type ---

\[ D_\mu = \partial_\mu - ie\,A_\mu \] \[ \begin{aligned} D_\mu\varphi \longrightarrow D'_\mu\varphi' &= e^{i\alpha(x)}\,D_\mu\varphi\\ D'_\mu\,e^{i\alpha(x)}\varphi &= e^{i\alpha(x)}\,D_\mu\varphi\\ D'_\mu &= e^{i\alpha(x)}\,D_\mu\,e^{-i\alpha(x)}\\ \partial_\mu - ie\,A'_\mu &= e^{i\alpha(x)}\left(\partial_\mu - ieA_\mu\right)e^{-i\alpha(x)}\\ \cancel{\partial_\mu} - ie\,A'_\mu &= \cancel{\partial_\mu} - i\,\partial_\mu\alpha(x) - ie\,A_\mu \end{aligned} \] \[ \begin{gathered} A'_\mu = \frac{1}{e}\,\partial_\mu\alpha(x) + A_\mu\\[1ex] \delta A_\mu = \frac{1}{e}\,\partial_\mu\alpha(x) \end{gathered} \]

i.e. If we replace \(\partial_\mu \to D_\mu\) & introduce a vector field \(A_\mu\) which transform like

\[ A'_\mu = A_\mu + \frac{1}{e}\,\partial_\mu\alpha(x) \]

then \(\mathcal{L}_{\text{kinetic}} = D_\mu\varphi\; D^{*\mu}\varphi^{*}\) is invariant under local gauge transformations.
So, Invariant \(\mathcal{L}\) becomes; \(\swarrow\) to make real \(\mathcal{L}\).

\[ \mathcal{L} = D_\mu\varphi\; D^{*\mu}\varphi^{*} - m\,\varphi\varphi^{*} - \frac{\lambda}{6}\left(\varphi\varphi^{*}\right)^2 \]

We also know from (Lecture-7) that \(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\) is invariant under local gauge transformations.
This term act as propagation term for \(A_\mu\).
So (scalar QED \(\mathcal{L}\))

\[ \boxed{\;\mathcal{L} = D_\mu\varphi\; D^{*\mu}\varphi^{*} - m^2\,\varphi^{*}\varphi - \frac{\lambda}{6}\left(\varphi^{*}\varphi\right)^2 - \frac{1}{4}\,F_{\mu\nu}\,F^{\mu\nu}\;} \]

The term \(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\) contains derivative of \(A\) which allows \(A\) to propagate to have canonical momenta. Without the term \(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\):

\[ \pi_A = \frac{\partial\mathcal{L}}{\partial\dot{A}(x)} = 0 \]

So we needed lorentz invariant; gauge invariant term which has non zero \(\pi_A\); which was \(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\).

1. First step is to look for propagators; seperate \(\mathcal{L}\) into \[ \mathcal{L} = \mathcal{L}_{\text{free}} + \mathcal{L}_{\text{int}} \]

*To identify \(\mathcal{L}_{\text{free}}\)*; without \(\lambda\) term:

• EOM are linear \(\Rightarrow\) \(\mathcal{L}_{\text{free}}\) shall be quadratic.

So, all 2 field terms are free terms & everything else is interacting term.

\[ \mathcal{L} = +\,\underset{\hookrightarrow\;\text{free term.}}{\varphi_1\varphi_2} + \cdots \]

So,

\[ \mathcal{L}_{\text{free}}\Big)_{\text{scalar QED}} = \underbrace{\partial_\mu\varphi\,\partial^\mu\varphi^{*} - m^2\varphi^{*}\varphi}_{\text{Free K.G.\ theory}} - \underbrace{\frac{1}{4}\,F_{\mu\nu}\,F^{\mu\nu}}_{\text{free gauge theory}} \]

Everything which is left over is \(\mathcal{L}_{\text{int}}\):

\[ \begin{aligned} \mathcal{L}_{\text{int}} &= ie\,A^\mu\left(\varphi^{*}\partial_\mu\varphi - \varphi\,\partial_\mu\varphi^{*}\right) + e^2 A_\mu A^\mu\,\varphi^{*}\varphi - \frac{\lambda}{6}\left(\varphi^{*}\varphi\right)^2 \qquad\left\{\;\text{}(A^\mu)\text{ vector field couples to conserved current }(J_\mu)\text{}\right.\\ &= ie\,A^\mu\,J_\mu + e^2\,A_\mu A^\mu\,\varphi^{*}\varphi - \frac{\lambda}{6}\left(\varphi^{*}\varphi\right)^2 \end{aligned} \]

where;

1. \(A^\mu J_\mu\) is qubic term;
2. \(e^2 A_\mu A^\mu \varphi^{*}\varphi\) involves 4 fields so quartic term;
3. \((\varphi^{*}\varphi)^2\) is also a quartic term.

So,

\[ \begin{gathered} \mathcal{H} = \sum_a \pi_a\,\dot{\phi}_a - \mathcal{L}\\ \mathcal{H} = \pi_A\,\dot{A} + \pi_\varphi\,\dot{\varphi} + \pi_{\varphi^*}\,\dot{\varphi}^{*} - \mathcal{L} \end{gathered} \]

\(\mathcal{H}_{\text{int}}\) involves qubic quartic & higher order terms, & \(\sum p_a\pi_a\) involves only quadratic terms. (Since \(\pi_A = \frac{\partial\mathcal{L}}{\partial\dot{A}}\) & \(\dot{A}\) comes from \(\mathcal{L}_{\text{free}}\).)
So

\[ \begin{aligned} \mathcal{H}_{\text{int}} &= -\mathcal{L}_{\text{int}}\\ &= -ie\,A^\mu J_\mu\; \underset{\substack{\downarrow\\ \text{}-\text{ve sign is also}\\ \text{correct; as } \varphi^{*}\varphi>0\\ A_\mu A^\mu = A_0^2 - \vec{A}^2\\ = -\vec{A}^2\;\text{we can gauge away } A_0.\\ \text{i.e.\ } -e^2A_\mu A^\mu\varphi^{*}\varphi \text{ is +ve quartic term}}}{-\,e^2\,A_\mu A^\mu\,\varphi^{*}\varphi} \;\underset{\substack{\downarrow\\ +\text{ Sign}\;\Rightarrow\;\text{+ve quartic potential}\\ \left(\varphi^{*}\varphi\right)^2 > 0}}{+\,\dfrac{\lambda}{6}\left(\varphi^{*}\varphi\right)^2} \end{aligned} \]

+ve quartic potential is bounded from below.

1. The propagator --- \[ \langle 0|\,T\left(\varphi_0^{\dagger}(x)\;\varphi_0(y)\right)|0\rangle = D_F(x-y) \qquad \text{[diagram below]} \]

   \begin{tikzpicture}[baseline=-2pt]
   \fill (0,0) circle (0.04);
   \fill (2,0) circle (0.04);
   \draw[->] (0,0) -- (1.1,0);
   \draw (1.1,0) -- (2,0);
   \end{tikzpicture}
   

   \[ \langle 0|\,T\;A_\mu(x)\;A_\nu(y)\,|0\rangle = -\eta_{\mu\nu}\;D_F(x-y)\Big|_{m=0} \qquad \text{[diagram below]} \qquad(\text{Lorentz gauge}) \]

   \begin{tikzpicture}[baseline=-2pt]
   \node[left] at (0,0) {$x$};
   \draw[] (0,0) -- (2,0);
   \node[right] at (2,0) {$y$};
   \end{tikzpicture}
   

\(A_0\) & \(A_3\) are removed by fixing the gauge, so for \(A_1\) & \(A_2\) we have

\[ \begin{aligned} \langle 0|\,T\;A_1(x)\;A_1(y)\,|0\rangle &= D_F(x-y).\\ \langle 0|\,T\;A_2(x)\;A_2(y)\,|0\rangle &= D_F(x-y) \end{aligned} \]

(\(A_\mu\) here is free field.)

(2) Second step is to draw interactions ---
For \(\mathcal{L}_{\text{int}}\) term \(\frac{\lambda}{6}\left(\varphi^{*}\varphi\right)^2\): \(e^{+i\int\frac{\lambda}{6}(\varphi^{*}\varphi)^2\,d^4y}\)
(For first order) this term gives

\[ +\frac{i\lambda}{6}\left(\varphi^{*}(y)\,\varphi(y)\right)^2 \]

which is:

\begin{tikzpicture}[>=stealth]
\draw[->] (-1,1) node[above]{$x_1$} node[below right, red]{$\varphi$} -- (-0.1,0.1);
\draw[->] (1,1) node[above]{$x_3$} node[below left, red]{$\varphi^{*}$} -- (0.1,0.1);
\draw[->] (-1,-1) node[below]{$x_2$} node[above right, red]{$\varphi^{*}$} -- (-0.1,-0.1);
\draw[->] (1,-1) node[below]{$x_4$} node[above left, red]{$\varphi$} -- (0.1,-0.1);
\node at (0.25,0) {\footnotesize$y$};
\end{tikzpicture}

Sample calculation:

\[ \frac{\lambda}{6}\int\langle 0|\,T\left(\varphi(x_1)\,\varphi^{*}(x_2)\,\varphi(x_3)\,\varphi^{*}(x_4)\left(\varphi^{*}(y)\,\varphi(y)\right)^2\right)|0\rangle\; d^4y \]

Rule is --- \(\varphi\) can only contract with \(\varphi^{*}\).

\[ = \frac{\lambda}{6}\cdot(2)\cdot(2)\int D_F(x_1-y)\; D_F(x_3-y)\; D_F(x_2-y)\; D_F(x_4-y)\;d^4y\ldots \]

Other quartic term \(-e^2 A_\mu A^\mu\left(\varphi^{*}\varphi\right)\) provides:

\begin{tikzpicture}[>=stealth]
\draw[->] (-1.2,0.9) -- (-0.1,0.1);
\draw[->] (-1.2,-0.9) -- (-0.1,-0.1);
\draw[] (0,0) -- (1.4,0.8);
\draw[] (0,0) -- (1.4,-0.8);
\node[right] at (1.6,0.4) {$\left(-e^2\right)$};
\end{tikzpicture}

Cubic term \(= ie\,A^\mu J_\mu\) (gives) \(= ie\,A^\mu\left(\varphi^{*}\partial_\mu\varphi - \varphi\,\partial_\mu\varphi^{*}\right)\):

\begin{tikzpicture}[>=stealth]
\draw[->] (-1.2,1.1) node[above]{$k_1$} -- (-0.08,0.1);
\draw[->] (-1.2,-1.1) node[below]{$k_2$} -- (-0.08,-0.1);
\draw[] (0,0) -- (1.8,0);
\node[right] at (2.6,0) {$e\left(k_1-k_2\right)_\mu$};
\end{tikzpicture}

Now we can calculate interacting propagators

\[ \langle\Omega|\,T\left(\varphi^{*}(x)\,\varphi(y)\right)|\Omega\rangle \quad\&\quad \langle\Omega|\,T\left(A_\mu(x)\,A_\nu(y)\right)|\Omega\rangle \quad\&\quad n\text{-point functions, in scalar QED.} \]

Sample processes:

\begin{tikzpicture}[>=stealth]
\draw[->] (-1.5,1.2) node[above]{$\varphi$} -- (-0.55,0.45);
\draw[->] (-1.5,-1.2) node[below]{$\varphi^{*}$} -- (-0.55,-0.45);
\draw[fill=blue!10] (0,0) ellipse (0.75 and 0.55);
\draw (-0.4,0.3) -- (0.1,-0.4);
\draw (-0.15,0.45) -- (0.35,-0.25);
\draw (0.1,0.5) -- (0.55,-0.05);
\draw[->] (0.55,0.45) -- (1.5,1.2) node[above]{$\varphi$};
\draw[->] (0.55,-0.45) -- (1.5,-1.2) node[below]{$\varphi^{*}$};
\node[right, text width=6cm] at (2.6,0.4) {$\rightarrow$ \footnotesize represents all possible such processes.};
\node[right, text width=5cm] at (2.6,-0.7) {\footnotesize say, $\varphi\to$ particle, $\varphi^{*}\to$ antiparticle.};
\end{tikzpicture}

[1ex]

\begin{tikzpicture}[>=stealth, scale=0.9]
\draw[->] (-1,1) node[above]{$e^-$} -- (0,0.08);
\draw (0,0.08) -- (1,1) node[above]{$e^-$};
\draw[->] (-1,-1) node[below]{$e^+$} -- (0,-0.08);
\draw (0,-0.08) -- (1,-1) node[below]{$e^+$};
\node at (0,0) {$\times$};
\node[below] at (0.35,-0.15) {\footnotesize$\lambda$};
\node at (1.9,0) {$+$};
\begin{scope}[xshift=4.3cm]
\draw[->] (-1.1,1) node[above]{$e^-$} node[below right]{\footnotesize$k_1$} -- (-0.5,0.1);
\draw[->] (-1.1,-1) node[below]{$e^+$} node[above right]{\footnotesize$k_2$} -- (-0.5,-0.1);
\draw (-0.5,0.1) -- (-0.45,0);
\draw (-0.5,-0.1) -- (-0.45,0);
\draw[] (-0.45,0) -- node[above]{\footnotesize$k_1{+}k_2$} (1.15,0);
\node[below] at (-0.4,-0.15) {\footnotesize$e$};
\node[below] at (1.1,-0.15) {\footnotesize$e$};
\draw[->] (1.15,0) -- (1.8,1) node[above]{$e^-$};
\draw[->] (1.15,0) -- (1.8,-1) node[below]{$e^+$};
\node[below, align=center] at (0.4,-1.2) {\footnotesize$k_1^2 = k_2^2 = 0$ (on shell)\\ \footnotesize$(k_1+k_2)^2 \neq 0$ (off shell)};
\end{scope}
\node at (7.6,0) {$+$};
\begin{scope}[xshift=9.8cm]
\draw[->] (-1,1.2) node[above]{$e^-$} node[below right]{\footnotesize$k_1$} -- (0,0.55);
\draw[->] (0,0.55) -- (1.2,1.2) node[above]{$e^+$};
\node[right] at (0.5,0.95) {\footnotesize$k_3$};
\draw[] (0,0.55) -- node[left]{\footnotesize$\gamma$} (0,-0.55);
\draw[->] (-1,-1.2) node[below]{$e^-$} node[above right]{\footnotesize$k_2$} -- (0,-0.55);
\draw[->] (0,-0.55) -- (1.2,-1.2) node[below]{$e^+$};
\node[right] at (0.5,-0.95) {\footnotesize$k_4$};
\end{scope}
\end{tikzpicture}

(No exchange of \(\gamma\))
\(\downarrow\) \(e^-\) & \(e^+\) annihilated and later photon creates \(e^-\) & \(e^+\).
(exchange of \(\gamma\))
\(\downarrow\) \(e^-\) somewhere emits virtual photon which is absorbed by positron during their propagation.

Fermions (QED) \(\longrightarrow\) Algebra 1 \(\longrightarrow\) Algebra 2

Lorentz Algebra --- (Mnemonic)

\[ \begin{pmatrix} 0 & M_{01} & M_{02} & M_{03}\\ & 0 & M_{12} & M_{13}\\ & & 0 & M_{23}\\ & & & 0 \end{pmatrix} \quad M_{\mu\nu}\;\longrightarrow\; \begin{aligned} &M_{01} \;\text{ boost in } x\\ &M_{02} \;\;\text{''}\;\;\text{''}\; y\\ &M_{03} \;\;\text{''}\;\;\text{''}\; z \end{aligned} \qquad \begin{aligned} &0\to \text{time component}\\ &\hookrightarrow \text{has to be boost} \end{aligned} \] \[ \left. \begin{aligned} &M_{\mu\nu} \text{ is antisymmetric}\\ &\text{so has only 6-independent}\\ &\text{parameters} \end{aligned} \right\} \quad \left. \begin{aligned} &M_{12} \;\text{ Rotation in } x\text{-}y \text{ plane}\\ &M_{23} \;\;\text{''}\quad\text{''}\; y\text{-}z \text{ plane}\\ &M_{13} \;\;\text{''}\quad\text{''}\; x\text{-}z \text{ plane} \end{aligned} \right\} \begin{aligned} &\mu,\nu = 1,2,3\\ &\hookrightarrow\text{space}\\ &(\text{Rotation}) \end{aligned} \]

When we talk about algebra we distinguish two concepts --- one is abstract algebra and another one is actual representation.
We learned that a particle of spin \(J\) has \(2J+1\) states. That means the angular momentum generators \(\vec{J}\cdot\hat{n}\) act on a wave-fun. of particle of spin \(J\), as \((2J+1)\times(2J+1)\) matrix. So for spin \(1/2\) particles we have generators of \(2\times2\) matrix.
So, for particle of spin \(1\) we need \(3\times3\) representation of angular momentum \(\vec{J}\cdot\hat{n}\).
So, for particle of spin \(J\) we need \((2J+1)\times(2J+1)\) representation of angular momentum \((\vec{J}\cdot\hat{n})\).
Note that each such representations of \((J_x, J_y, J_z)\) will obey

\[ [J_x, J_y] = i\,J_z \qquad\text{or}\qquad [J_i, J_j] = i\,J_k\,\epsilon_{ijk} \qquad (\hbar = 1) \]

For higher spin we require higher dimensional matrices. So, there are many kinds/dimensions of matrices satisfying same algebra.
There are many sets of matrices \(S = \{J_x, J_y, J_z\}_{2\times2},\; S_{3\times3}\ldots\) (or \(S=\{M_1, M_2 \ldots M_n\}_{n\times n}\)) satisfying a given algebra, each set is called a representation.
So Algebra is an abstract thing, \(M_{\mu\nu}\) need not be matrices. They only define the algebra. Its like a rule telling to look for matrices which satisfy the rule.
So the analogue of angular momentum algebra \([J_i, J_j] = i\,\epsilon_{ijk} J_k\) is

\[ \boxed{\; [M_{\mu\nu},\, M_{\lambda\rho}] \;=\; \underbrace{M_{\mu\rho}\,\eta_{\nu\lambda} + M_{\nu\lambda}\,\eta_{\mu\rho}}_{\text{1) } \mu\nu \curvearrowright \lambda\rho} \;-\; \underbrace{\left(M_{\mu\lambda}\,\eta_{\nu\rho} + M_{\nu\rho}\,\eta_{\mu\lambda}\right)}_{\text{2) } \mu\nu \curvearrowright \lambda\rho} \;} \quad\longleftarrow\;\text{Lorentz Algebra.} \]

We can also derive angular momentum algebra from lorentz algebra:

\[ \begin{aligned} [M_{23},\, M_{21}] &= \left(M_{21}\,\eta_{33} + M_{33}\,\eta_{21}\right) - \left(M_{23}\,\eta_{21} + M_{31}\,\eta_{23}\right)\\ &= M_{21}(-1) + M_{33}(0) - \left(M_{23}(0) + M_{31}(0)\right)\\ &= -M_{21} \end{aligned} \]

So far we have just written abstract algebra (of Lorentz Algebra). What about its representation --- (16 generators?)
A representation would be set of matrices \(\left(M_{\mu\nu}\right)^{A}_{\;\;B}\) which satisfy the lorentz algebra (above rule).
--- Any such choice of matrices is a representation, and the range of values of \(A\) and \(B\) tells us dimension or size of matrices.
Our favourite example: For \(M_{\mu\nu} = J_x\; J_y\; J_z\),

\[ \begin{gathered} \underline{\text{If}}\quad A,B = 1,2 \quad\text{then matrices are}\quad M_{\mu\nu} = \sigma_1,\ \sigma_2,\ \sigma_3\\ \text{If}\quad A,B = 1,2,3 \qquad M_{\mu\nu} = S_1, S_2, S_3\;\text{---} \end{gathered} \]

*Relation b/w \(\Lambda^{\mu}_{\;\nu}\) & \(M_{\mu\nu}\)* ---
We know, \((x')^{\mu} = \Lambda^{\mu}_{\;\nu}\, x^{\nu}\) (how coordinates & vector transform under lorentz transformation).
--- \(\Lambda^{\mu}_{\;\nu}\) is \(4\times4\) matrix, so it must be a \(4\times4\) representation of lorentz algebra.
--- First of all \(\Lambda^{\mu}_{\;\nu}\) generate finite lorentz transformation while \(M_{\mu\nu}\) generates infinitesimal lorentz transformation. (Because algebra is always among infinitesimal generators).
So we must make a finite lorentz transformation.
Remember, \(M_{\mu\nu}\) is anti-symmetric so has only 6-independent parameters \(\omega_{\mu\nu}\) (3 boost + 3 rotation).
To generate finite lorentz transformation ---

\[ \underbrace{e^{\left(\frac{1}{2}\right)\omega_{\mu\nu} M^{\mu\nu}}}_{\downarrow\;\text{representation}} \qquad \left\{ \begin{aligned} &\text{factor of }\left(\tfrac{1}{2}\right)\text{ comes}\\ &\text{from fact that}\\ &\omega_{01}M^{01} = \omega_{10}M^{10}\\ &\text{since } M^{\mu\nu}\ \&\ \omega^{\mu\nu}\\ &\text{are anti-symmetric.} \end{aligned} \right. \]

No '\(i = \sqrt{-1}\)' in exponential, since these are real things.

For any given representation we can have field which transform under that representation, that is why it is useful for field theory.

Ex:
(1) Representation of lorentz algebra which acts upon scalar fields (or what is infinitesimal transformation on scalar fields). Actually there is no such transformation. \((M_{\mu\nu} = 0)\) (\(M_{\mu\nu}\) is rep\(^{\text{s}}\) of \(M_{\mu\nu}\)),

\[ \begin{gathered} \phi'(x') = \Lambda\,\phi(x)\\ \phi'(x') = 1\cdot\phi(x) \qquad \left(\text{spin } 0 \text{ rep}^{\text{n}}\text{ of }\Lambda\right)\\ \text{or,}\qquad e^{\frac{1}{2}\omega_{\mu\nu} M^{\mu\nu}} = e^{\frac{1}{2}\omega_{\mu\nu}(0)} = e^{0} = 1 \end{gathered} \]

(2) Find rep\(^{\text{s}}\) of lorentz algebra for transformation of vector fields.

\[ A'_{\mu}(x') = \Lambda^{\mu}_{\;\nu}\, A_{\nu}(x) \]

reps: \(\left(M^{\mu\nu}\right)^{\alpha}_{\;\;\beta}\) (or) \(\left(V^{\mu\nu}\right)^{\alpha}_{\;\;\beta}\) (\(V\): vector)

\[ \left(V^{\mu\nu}\right)^{\alpha}_{\;\;\beta} = \eta^{\mu\alpha}\,\delta^{\nu}_{\;\beta} - \eta^{\nu\alpha}\,\delta^{\mu}_{\;\beta} \]

\(\downarrow\) set of 6 matrices

1. \(\left(V^{01}\right)^{\alpha}_{\;\;\beta},\ \left(V^{02}\right)^{\alpha}_{\;\;\beta},\ \left(V^{03}\right)^{\alpha}_{\;\;\beta}\)
2. \(\left(V^{12}\right)^{\alpha}_{\;\;\beta},\ \left(V^{13}\right)^{\alpha}_{\;\;\beta},\ \left(V^{23}\right)^{\alpha}_{\;\;\beta}\)

where each of 6 matrices generates their corresponding transformation. (a) generates Boost (b) generates rotations for vector fields.
Since R.H.S include \(\eta = \begin{pmatrix}1&&&\\&-1&&\\&&-1&\\&&&-1\end{pmatrix}_{4\times4}\) a product with \(4\times4\) matrices, \(V^{\mu\nu}\) has to be \(4\times4\) matrices. For such case where dimension of \((\mu,\nu) = \dim(\alpha,\beta)\) we call it fundamental rep\(^{\text{s}}\).
The finite lorentz transformation thus becomes

\[ \begin{aligned} &= e^{\frac{1}{2}\,\omega_{\mu\nu} V^{\mu\nu}}\\ &= e^{\frac{1}{2}\,\omega_{\mu\nu}\left(\eta^{\mu\alpha}\delta^{\nu}_{\;\beta} - \eta^{\nu\alpha}\delta^{\mu}_{\;\beta}\right)}\\ &= e^{\left(\eta\,\mu\right)^{\alpha}_{\;\;\beta}} \;=\; {\eta^{\alpha\gamma}\,\omega_{\gamma\beta}}\qquad{(\gamma \text{ gets contracted}).} \end{aligned} \]

So \(\Lambda^{\alpha}_{\;\beta} = e^{(\eta\,\omega)^{\alpha}_{\;\;\beta}}\) (spin 1 rep\(^{\text{n}}\))
Check if; \(e^{\eta\omega}\,\eta\left(e^{\eta\omega}\right)^{T} = \eta\) (as \(\Lambda\) satisfies \(\Lambda\eta\Lambda^{T} = \eta\))
We will see that, we will find a representation which will describe spin \(\frac{1}{2}\) particles.
We look for spin \(\frac{1}{2}\) rep\(^{\text{n}}\) of \(M^{\mu\nu}\) which will describe particles
We will look for new rep\(^{\text{n}}\) of lorentz algebra \([M^{\mu\nu}, M^{\lambda\rho}] = \cdots\) we will see that this rep\(^{\text{n}}\) describe or act upon new class of fields which have half integer spin.
One motivation to look for new rep\(^{\text{n}}\) is that so far we have found spin 0 & spin 1

\[ \phi'(x') = 1\,\phi(x) \qquad\qquad V'^{\mu}(x') = \Lambda^{\mu}_{\;\nu}\, V^{\nu}(x) \]

we can add indices to fields \(\longrightarrow\) \(B_{\mu\nu}(x)\) which will transform like

\[ B'^{\mu\nu}(x') = \Lambda^{\mu}_{\;\rho}\,\Lambda^{\nu}_{\;\sigma}\, B^{\rho\sigma}(x) \qquad\text{or}\quad B^{\rho\sigma}(x). \]

But this rep\(^{\text{n}}\) \(\left(\Lambda^{\mu}_{\;\rho}\,\Lambda^{\nu}_{\;\sigma} = \exp\left(\frac{1}{2}\,\omega_{\mu\nu}\,M^{\mu\nu}\right)\right)\) will describe integer spin particles.
Terminology: rep\(^{\text{n}}\) \(= M^{\mu\nu}\) (or) \(\Lambda^{\mu}_{\;\rho}\,\Lambda^{\nu}_{\;\sigma}\) (or) \(B^{\sigma\nu}(x)\)
In Nature we are only interested in spin 0 & spin 1 (as higher spin requires more energy to discover/probe).
But spin \(\frac{1}{2}\) is of crucial interest as \(e^{-}, \mu, \tau, \ldots\) quarks \((u,d,s,c,t,b)\) are spin \(\frac{1}{2}\) particles.

Clifford Algebra ---
The idea here is to change from lorentz algebra to new easier algebra called clifford algebra (it is simpler than lorentz algebra.)
Any rep\(^{\text{s}}\) of clifford algebra automatically satisfies lorentz algebra (i.e. is also a rep\(^{\text{s}}\) of lorentz algebra). So it is an intermediate tool to make reps of lorentz algebra.
In any algebra we have to have abstract generators

\[ \left\{\Gamma^{\mu},\, \Gamma^{\nu}\right\} = 2\,\eta^{\mu\nu} \qquad {\left(\Gamma^{\mu},\ \Gamma^{\nu}\text{ are generators of clifford algebra}\right)} \]

\(\downarrow\)
Notice that it is way simpler than lorentz algebra.
Any representation would be a ''set'' of matrices, \(\left(\gamma^{\mu}_{\;ab}\right)\) satisfying

\[ \begin{gathered} {\gamma^{\mu}_{\;ab}\,\gamma^{\nu}_{\;bc} + \gamma^{\nu}_{\;ab}\,\gamma^{\mu}_{\;bc} = 2\,\eta^{\mu\nu}\,\delta_{ac}}\qquad{\text{(to make } a=c\text{)}}\\ \text{or}\quad \left\{\gamma^{\mu},\, \gamma^{\nu}\right\} = 2\,\eta^{\mu\nu}\cdot I \end{gathered} \]

\(\downarrow\) Since LHS are matrices, RHS should be multiplied with an identity (\(\eta^{\mu\nu}\) = number)
(Q) Why we are interested in clifford algebra?
Ans: There is a theorem why we are interested,
Theorem --- Clifford algebra reps provide lorentz algebra reps via

\[ \dfrac{1}{4}\left[\gamma^{\mu},\, \gamma^{\nu}\right]_{ab} = \left(S^{\mu\nu}\right)_{ab} \qquad \begin{aligned} &{\text{This is called}}\\ &{\hookrightarrow\text{ spinor reps.}} \end{aligned} \]

Find LHS, then, the matrices \(\left(S^{\mu\nu}\right)_{ab}\) satisfy ---

\[ {\underset{\text{This is lorentz algebra}}{\longleftarrow}}\quad \left[S^{\mu\nu},\, S^{\lambda\rho}\right] = \left[S^{\mu\rho}\,\eta^{\nu\lambda} + S^{\nu\lambda}\,\eta^{\mu\rho} - \left(S^{\mu\lambda}\,\eta^{\nu\rho} + S^{\nu\rho}\,\eta^{\mu\lambda}\right)\right] \] \[ {\mu\nu\,,\ \lambda\rho \;-\; \left(\mu\nu\,,\ \lambda\rho\right)} \qquad {\left(\text{hence } S^{\mu\nu}\text{ satisfies lorentz algebra}\right)} \]

Above is easy to remember, but can be derived from using
$\{

\[ \begin{aligned} S^{\mu\nu} &= \tfrac{1}{4}\left[\gamma^{\mu}, \gamma^{\nu}\right]\\ \&\ S^{\lambda\rho} &= \tfrac{1}{4}\left[\gamma^{\lambda}, \gamma^{\rho}\right] \end{aligned} \]

\}$ in above expression.
To put in context of group theory in general, only orthogonal groups have spinor reps. & lorentz group is an orthogonal group.
We were missing lorentz algebra reps of spin \(\frac{1}{2}\) particles & lorentz group is \(SO(3,1)\), \(\Rightarrow\) if we study clifford algebra we will find a new class of reps. That's how we found lorentz reps. of spin \(\frac{1}{2}\) particles.
This allows us to define fields \(\Psi_{a}(x)\), which transform under lorentz transformation as

\[ \Psi_{a} \longrightarrow \Psi'_{a} = \left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,M^{\mu\nu}}\right)_{ab} \Psi_{b} \]

where \(M^{\mu\nu} = S^{\mu\nu}\) here

\[ \Psi_{a} \;\to\; \Psi'_{a} = \left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\right)_{ab} \Psi_{b} \]

& such fields \(\Psi_{a}(x)\) are called spinors.
In next lecture, we will see how to make an eq\(^{\text{n}}\) of motion and an action for such spinors. (which are invariant under lorentz transformation, i.e. the transformation will cancel out in final e.o.m & action.)
Once we have reps of clifford algebra \(\left(\gamma^{\mu}\right)_{ab}\) we can construct \(S^{\mu\nu} = \frac{1}{4}\left[\gamma^{\mu}\,\gamma^{\nu}\right]\) which are also reps of lorentz algebra (or satisfy lorentz algebra)

\[ \text{i.e.}\quad \left[S^{\mu\nu},\, S^{\lambda\rho}\right] = S^{\mu\lambda}\,\eta^{\nu\rho} + S^{\nu\rho}\,\eta^{\mu\lambda} - \left(S^{\mu\rho}\,\eta^{\nu\lambda} + S^{\nu\lambda}\,\eta^{\mu\rho}\right) \]

The reps we get \(\left(S^{\mu\nu}\right)_{ab}\) are called spinor reps.
It happens due to property of \(SO(\;)\) type algebra. Each \(SO(\;)\) type algebra/group has spinor reps. (i.e. we know that lorentz algebra/lorentz group are of orthogonal type \(SO(3,1)\), where \(SO(3)\) is group of rotations, and \(SO(3,1)\) is group of rotation & boost.)
So any rotation group \(\to SO(3)\) or orthogonal group \(O(n)\) has spinor reps.
Spinor reps of rotation group shall satisfy \(\left\{\gamma^{\mu}\,\gamma^{\nu}\right\} = 2\eta^{\mu\nu} I\)

\[ \begin{gathered} \mu = i = 1,2,3\\ \gamma^{\mu} \longrightarrow \sigma^{i}\,,\qquad \eta^{\mu\nu} \longrightarrow \delta^{ij}\,,\qquad \left\{\sigma^{i},\, \sigma^{j}\right\} = 2\,\delta^{ij}\,\underline{I}\\ \hookrightarrow \text{Identity matrix.} \end{gathered} \]

So \(\left\{\sigma_1, \sigma_2, \sigma_3\right\}\) satisfy clifford algebra, it is spinor reps. of rotation group. \((SO(3))\)
Whereas \(\frac{1}{4}\left[\gamma^{\mu},\, \gamma^{\nu}\right] = S^{\mu\nu}\) shall be spinor reps of lorentz group. \((SO(3,1))\)

Remarks --- (i) \(\left(\gamma^{\mu}\right)^{2} = \pm I\)
$\{

\[ \begin{aligned} \left\{\gamma^{\mu}, \gamma^{\nu}\right\} &= 2\eta^{\mu\nu} I\\ \gamma^{\mu}\gamma^{\mu} + \gamma^{\mu}\gamma^{\mu} &= 2\eta^{\mu\mu}\, I\\ 2\left(\gamma^{\mu}\right)^{2} &= 2\,\eta^{\mu\mu}\, I\\ \text{for } \mu = 0\quad \left(\gamma^{\mu}\right)^{2} &= +I\\ \text{for } \mu = 1,2,3\ \left(\gamma^{\mu}\right)^{2} &= -I \end{aligned} \]

.$
(ii) All distinct \(\gamma^{\mu}\) anticommute.

\[ \begin{gathered} \left\{\gamma^{\mu},\, \gamma^{\nu}\right\} = 2\eta^{\mu\nu} I\\ \text{For } \mu \neq \nu \qquad \eta^{\mu\nu} = 0 \quad\Rightarrow\quad \boxed{\left\{\gamma^{\mu},\, \gamma^{\nu}\right\} = 0} \end{gathered} \]

Similarly we can write for \(S^{\mu\nu} = \frac{1}{4}\left[\gamma^{\mu},\, \gamma^{\nu}\right]\)
(1) \(S^{\mu\mu} = 0\)
(2) \(\mu \neq \nu\) \(S^{\mu\nu} = \frac{1}{4}\left(\gamma^{\mu}\gamma^{\nu} - \gamma^{\nu}\gamma^{\mu}\right)\)
But for \(\mu\neq\nu\) \(\left\{\gamma^{\mu}, \gamma^{\nu}\right\} = 0\) \(\Rightarrow\)

\[ \begin{gathered} S^{\mu\nu} = \frac{1}{4}\left(\gamma^{\mu}\gamma^{\nu} + \gamma^{\mu}\gamma^{\nu}\right)\\ \boxed{\;S^{\mu\nu} = \frac{1}{2}\,\gamma^{\mu}\gamma^{\nu}\;} \end{gathered} \]

Now let's introduce this idea in field theory ---
We introduced a field \(\Psi_{a}(x)\), which will have the property that it will transform under lorentz transformations using this matrix \(\left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\right)\)

\[ \Psi_{a}(x) \;\longrightarrow\; \Psi'_{a}(x') = \Big(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\Big)_{\underset{{\text{matrix indices}}}{ab}} \underbrace{\Psi_{b}(x)}_{\hookrightarrow\;\text{column matrix reps of }\Psi(x)} \]

{Just to remind ---
For scalars \(m^{\mu\nu} = 0\);
For vectors \(m^{\mu\nu} = V^{\mu\nu}\);
For spinors \(m^{\mu\nu} = S^{\mu\nu}\)}

*Construction of \(\left(\gamma^{\mu}\right)_{ab}\) matrices using Pauli matrices* ---
We start with pauli matrices satisfying a kind of clifford algebra.

\[ \left\{\sigma^{i},\, \sigma^{j}\right\} = 2\,\delta^{ij} \]

and then we take tensor product : \(\sigma^{i} \otimes \sigma^{j}\)

\[ \begin{aligned} \underline{\text{ex}}\qquad \sigma^{1} \otimes \sigma^{2} &= \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}\\ &= \begin{pmatrix} 0\cdot\sigma^{2} & 1(\sigma^{2})\\ (1)\,\sigma^{2} & 0\,(\sigma^{2}) \end{pmatrix} = \begin{pmatrix} 0 & \sigma^{2}\\ \sigma^{2} & 0 \end{pmatrix}\\ &= \begin{pmatrix} 0 & 0 & 0 & -i\\ 0 & 0 & i & 0\\ 0 & -i & 0 & 0\\ i & 0 & 0 & 0 \end{pmatrix} = \left(\begin{array}{c|c} \text{ 0} & \begin{matrix} 0 & -i\\ i & 0 \end{matrix}\\ \hline \begin{matrix} 0 & -i\\ i & 0 \end{matrix} & \text{ 0} \end{array}\right) \end{aligned} \] \[ \text{---}\quad \underbrace{\left(A \otimes B\right)\left(C \otimes D\right)}_{\text{matrix product}} = \underbrace{\left(AC \otimes BD\right)}_{\text{direct product}} \]

To make \(4\times4\) reps of \(\gamma^{\mu}\) matrices, we need \(\to\) (1) \(\left(\gamma^{\mu}\right)\left(\gamma^{\mu}\right) = \pm I\) (2) & \((\mu\neq\nu)\ \left\{\gamma^{\mu}, \gamma^{\nu}\right\} = 0\)
Start with,

\[ 1)\qquad \sigma_1 \otimes \sigma_1 = \begin{pmatrix} 0 & \sigma_1\\ \sigma_1 & 0 \end{pmatrix} \quad\text{as}\quad \left(\sigma_1 \otimes \sigma_1\right)\left(\sigma_1 \otimes \sigma_1\right) = \sigma_1^{2} \otimes \sigma_1^{2} = I_{2\times2} \otimes I_{2\times2} = I_{4\times4} \]

So, \([\;\sigma_1 \otimes \sigma_1 = \gamma^{0}\;]\)

\[ 2)\qquad \sigma_1 \otimes \sigma_2 \qquad \left(\sigma_1 \otimes \sigma_2\right)\left(\sigma_1 \otimes \sigma_2\right) = \left(\sigma_1^{2} \otimes \sigma_2^{2}\right)_{2\times2} = I \underset{2\times2}{\otimes}\, I_{2\times2} = I_{4\times4} \]

Since we require \(-I_{4\times4}\) for \(\mu = 1,2,3\), we have to multiply with '\(i\)'. So,

\[ \boxed{\;i\,\sigma_1 \otimes \sigma_2 = \gamma^{1}\;} \]

Since \(\left[\sigma_1\, \sigma_1\right] = 0\) but \(\left\{\sigma_1\, \sigma_2\right\} = 0\) \(\Rightarrow\) \(\left\{\sigma_1 \otimes \sigma_1\,,\ \sigma_1 \otimes \sigma_2\right\} = 0\)
(3) We require \(\gamma^{2}\) & \(\gamma^{3}\) such that \(\left(\gamma^{2}\right)^{2} = \left(\gamma^{3}\right)^{2} = -I\) and they should anticommute with each other i.e. \(\left\{\gamma^{0}\, \gamma^{2}\right\} = \left\{\gamma^{0}\, \gamma^{3}\right\} = \left\{\gamma^{1}\, \gamma^{2}\right\} = \left\{\gamma^{1}\, \gamma^{3}\right\} = 0\)
guess --- \(\sigma_1 \otimes \sigma_3\) shall anticommute with all above \(\gamma^{0}\) & \(\gamma^{1}\)
Since \(\left\{\sigma_3,\, \sigma_2\right\} = \left\{\sigma_3,\, \sigma_1\right\} = 0\)
also

\[ \left(\sigma_1 \otimes \sigma_3\right)^{2} = \left(\sigma_1 \otimes \sigma_3\right)\left(\sigma_1 \otimes \sigma_3\right) = \left(\sigma_1^{2} \otimes \sigma_3^{2}\right) = I_{4\times4} \]

again we have to multiply by '\(i\)' to get \(\left(-I_{4\times4}\right)\). So

\[ \boxed{\;i\left(\sigma_1 \otimes \sigma_3\right) = \gamma^{2}\;} \]

(4) We have ran out of \(\sigma\) matrices, we write

\[ \boxed{\;\gamma^{3} = i\left(\sigma_2 \otimes I\right)\;} \quad\text{since its square} = -I_{4\times4} \]

and it anticommute with all \(\gamma^{0}, \gamma^{1}, \gamma^{2}\).
\(\left\{\text{we could have chosen } i\left(\sigma_3 \otimes I\right) \text{ as well!}\right\}\)

\(A \otimes B \;\neq\; B \otimes A\)

We also have \(\gamma_5 = i\,\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}\) with \(\left(\gamma^{\mu}\right)^{2} = +I\) and which anticommute with all \(\gamma^{0}, \gamma^{1}, \gamma^{2}, \gamma^{3}\). Also it is hermitian. \(\left(\left(\gamma_5\right)^{\dagger} = \gamma_5\right)\).
(i) \(\left(\gamma_5\right)^{2} = +I\)

\[ \begin{aligned} \left(\gamma_5\right)^{2} &= i^{2}\,\left(\gamma^{0}\gamma^{1}\,\gamma^{2}\,\gamma^{3}\right)\left(\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}\right) &&\left\{\gamma^{0}\,\gamma^{2}\right\} = 0\;\Rightarrow\; \gamma^{0}\gamma^{3} = -\gamma^{3}\gamma^{0}\\ &&&\text{So when we take } \gamma^{\nu}\text{ to left to }\left(\gamma^{\mu} \neq \gamma^{\nu}\right)\\ &&&\text{take one more minus sign}\\ &= (-1)(-1)^{3}\left(\gamma^{1}\gamma^{2}\gamma^{3}\,\gamma^{1}\gamma^{2}\gamma^{3}\right)\\ &= (-1)^{2}\left(\gamma^{1}\right)^{2}\left(\gamma^{2}\gamma^{3}\,\gamma^{2}\gamma^{3}\right) &&\left(\gamma^{1}\right)^{2} = -I\\ &= (-1)(-1)\cancelto{}{\left(\gamma^{2}\right)^{2}}\left(\gamma^{3}\right)^{2}\\ \left(\gamma_5\right)^{2} &= (-1)(-1)(-1)(-1) \;=\; +I \end{aligned} \]

In 5-d we require 5-gamma matrices in that case \(\gamma_5 = \gamma_0\gamma_1\gamma_2\gamma_3\) which anticommutes with \(\gamma_0\gamma_1\gamma_2\gamma_3\) & \(\gamma_5^{2} = -1\) \(\left(\gamma_5\,\gamma_{5} = \gamma_5^{2}\,\eta^{55},\;\; \eta = (+1,-1,-1,-1,-1)\right)\)

(ii) \(\left(\gamma_5\right)^{\dagger} = \left(i\,\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}\right)^{\dagger} = -i\left(\gamma^{3}\right)^{\dagger}\left(\gamma^{2}\right)^{\dagger}\left(\gamma^{1}\right)^{\dagger}\left(\gamma^{0}\right)^{\dagger} = \gamma\)

Gamma matrices algebra ---

\[ \Big\{\underset{\underline{1}}{\mathbb{I}}\;,\ \underset{+4}{\gamma^{\mu}}\;,\ \underset{+4}{\gamma_5\,\gamma^{\mu}}\;,\ \underset{+6}{\gamma^{\mu}\gamma^{\nu}\ (\text{or } S^{\mu\nu})}\;,\ \underset{+1}{\gamma_5}\Big\} = 16 \text{ such matrices.} \]

Exercise ---

\[ \begin{aligned} \gamma_5\,\underbrace{\gamma^{0}\gamma^{1}}_{S^{01}} &= +\left(i\,\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}\right)\gamma^{0}\gamma^{1}\\ &= -i\,\gamma^{1}\gamma^{2}\gamma^{3}\gamma^{1}\\ &= i\,\gamma^{2}\gamma^{3}\\ &= 2i\,S^{23} \qquad\qquad \left\{\frac{1}{2}\,\gamma^{\mu\nu} = S^{\mu\nu}\right.\\ \gamma_5\, S^{01} &= 2i\, S^{23} \end{aligned} \]

Lets get back ---
We have field \(\Psi_{a}(x)\) which transform like

\[ \Psi'_{a}(x') = \left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\right)_{ab} \Psi_{b} \]

This shall be a 4 component field because gamma matrices are \(4\times4\) matrices.
These are 4 component field but do not transform the same way as 4 component vector field.
These are called spinor field because they transform under spinor representation.

EOM --- We could think that \(\left(\Box^{2} + m^{2}\right)\Psi_{a}(x) = 0\) could be e.o.m but K.G. eq\(^{\text{n}}\) leads to \(-\)ve probability current. So Dirac took an approach to write e.o.m. which are first order in space & time.
The possibility of \(1^{\text{st}}\) order diff. eq\(^{\text{n}}\) arose for the first time, because when we had scalar \(\varphi(x)\) & vector fields \(A_{\mu}(x)\), there was simply no possibility (\(\partial\varphi\) or \(\partial A_{\mu}(x)\),) we don't get anything interesting & ''lorentz invariant''.
Here we have new operator: \(\underline{\left(\gamma^{\mu}\right)_{ab}\,\partial_{\mu}}\) \(\overset{\swarrow}{\;}\) first order in derivatives.
\(\left(\gamma^{\mu}\right)_{ab}\,\partial_{\mu}\,\Psi_{b}(x)\) shall also transform like spinors under lorentz transformation (i.e. to act \(\left(\gamma^{\mu}\right)_{ab}\) on \(\Psi_{b}\), the result shall behave like spinor & thus transform like spinor under L.T.)
Given \(\Psi_{a}\), we can construct a new object ---

\[ \chi_{a} = \left(\gamma^{\mu}\right)_{ab}\,\partial_{\mu}\,\Psi_{b} \qquad\text{---}(1) \]

Now, if \(\Psi \longrightarrow \left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\right)\Psi\)
does \(\chi \longrightarrow \left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\right)\chi\) ?
If it is true that \(\chi\) transform like spinor then we can set \(\chi = 0\) and say that

\[ \boxed{\;\left(\gamma^{\mu}\right)_{ab}\,\partial_{\mu}\,\Psi_{b} = 0\;} \quad\text{is equation of motion.} \]

(Because EOM have to be lorentz Invariant!)
Under lorentz transformation ---

\[ \gamma^{\mu}\,\partial_{\mu}\Psi \;\longrightarrow\; \gamma^{\mu}\underbrace{\left(e^{-\frac{1}{2}\,\omega_{\lambda\rho}\,V^{\lambda\rho}}\right)^{\nu}_{\;\;\mu}}_{}\,\partial_{\nu}\left(e^{\frac{1}{2}\,\omega\, S}\right)\Psi = \left(e^{\frac{1}{2}\,\omega_{\lambda\rho}\,S^{\lambda\rho}}\right)\gamma^{\mu}\,\partial_{\mu}\Psi \]

Since \(A'_{\mu}(x') = \left(\Lambda^{-1}\right)^{T} A_{\mu}(x)\) \(\to\) Transpose

\[ \Lambda^{-1} = \left(e^{-\frac{1}{2}\,\omega_{\lambda\rho}\,V^{\lambda\rho}}\right)^{\nu}_{\;\;\mu} \qquad \left(\text{So,}\quad \partial_{\mu} \to \left(e^{-\frac{1}{2}\,\omega\,V^{T}}\right)\partial_{\mu}\right) \]

if

\[ \left(e^{-\frac{1}{2}\,\omega_{\lambda_1\rho_1}\,V^{\lambda_1\rho_1}}\right)^{\mu}_{\;\;\nu} \left(e^{-\frac{1}{2}\,\omega_{\lambda_2\rho_2}\,S^{\lambda_2\rho_2}}\right)_{ac} \left(e^{\frac{1}{2}\,\omega_{\lambda_3\rho_3}\,S^{\lambda_3\rho_3}}\right)\gamma^{\nu}_{\;cd}\Big._{db} = \left(\gamma^{\mu}\right)_{ab} \qquad {\uparrow\;\text{Matrix indices.}} \] \[ \left\{ \begin{aligned} \lambda_i, \rho_i &= 0,1,2,3\\ i &= 1,2,3 \end{aligned} \right. \]

H.W. (Q) Show that the above identity implies :-

\[ {\boxed{\left[\gamma^{\mu},\, S^{\lambda\rho}\right] = \left(V^{\lambda\rho}\right)^{\mu}_{\;\;\nu}\,\gamma^{\nu}}} \]

So, since \(\chi = \left(\gamma^{\mu}\right)_{ab}\,\partial_{\mu}\,\Psi_{b}\) transform like spinor we can set \(\chi = 0\).
So, we found a lorentz invariant e.o.m ---

\[ \gamma^{\mu}\,\partial_{\mu}\Psi + m\Psi = 0 \qquad {\hookrightarrow\;\text{Because } m\Psi \text{ also transform like } \Psi \text{ under L.T.}} \]

or

\[ {(i)}\,\gamma^{\mu}\,\partial_{\mu}\Psi + m\Psi = 0 \qquad \left(\text{Free Dirac eq}^{\text{n}}\right) \quad\downarrow\quad \left(\text{linear in } \Psi\right)\;{\text{or (quadratic in } \mathcal{L})} \]

\(\downarrow\) will explain later. \(\downarrow\) (lorentz invariant eq\(^{\text{n}}\))
We know \(\gamma^{\mu}\) are in general complex \((\sigma_2 = \text{complex})\). So we must look for complex solution.
i.e. \(\Psi\) should be complex.
Only in reps where \(\gamma^{\mu}\) are real, we can look for \(\Psi\) to be real only.
Since all \(\gamma^{\mu}\) are complex \(\Rightarrow\) \(S^{\mu\nu} = \frac{1}{4}\left[\gamma^{\mu}, \gamma^{\nu}\right]\) shall also be complex.

\[ \Rightarrow\quad \Psi \text{ must be complex} \]

as if we take real \(\Psi\), then \(\left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\right)_{ab}\Psi_{b}\) will be complex after lorentz transformation.

Lagrangian ---
Once we have lagrangian (free (quadratic) lagrangian), then we can think of all higher order interactions to act with. & we will then find an interacting theory.
We need to guess lagrangian ---
Normally, e.o.m obey

\[ \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\varphi\right)}\right) = \frac{\partial\mathcal{L}}{\partial\varphi} \]

If we have e.o.m as \(\left(i\,\gamma^{\mu}\,\partial_{\mu}\Psi + m\Psi\right) = 0\) probably lagrangian density should have \(\Psi^{2}\) term to have above e.o.m.
We want that term to be lorentz invariant. As generally e.o.m remain same if \(\mathcal{L}' = \mathcal{L} + \partial_{\mu}\chi\) \(\hookrightarrow\) total derivative.
but here we just want \(\mathcal{L}\) to be invariant,
Puzzle --- Is \(\Psi^{T}_{a}\Psi_{a}\) lorentz invariant?

\[ \Psi_{a} \longrightarrow \left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}}\right)_{ab}\Psi_{b} \]

So

\[ \begin{aligned} \Psi^{T}_{a}\,\Psi_{a} \;\longrightarrow\;& \left(e^{\frac{1}{2}\,\omega S}\right)^{T}_{ab}\Psi^{T}_{b}\;\left(e^{\frac{1}{2}\,\omega S}\right)_{ac}\Psi_{c}\\ =\;& \left(e^{\frac{1}{2}\,\omega S}\right)^{T}_{ba}\left(e^{\frac{1}{2}\,\omega S}\right)_{ac}\;\Psi^{T}_{b}\,\Psi_{c}\\ =\;& \left(\left(e^{\frac{1}{2}\,\omega S}\right)^{T}\left(e^{\frac{1}{2}\,\omega S}\right)\right)_{bc}\;\Psi^{T}_{b}\,\Psi_{c} \end{aligned} \]

For lorentz invariance we require

\[ \begin{gathered} \left(e^{\frac{1}{2}\,\omega S}\right)^{T} e^{\frac{1}{2}\,\omega S} = I\\ \Rightarrow\quad e^{\frac{1}{2}\,\omega S^{T}}\, e^{\frac{1}{2}\,\omega S} = I \end{gathered} \]

Only possible if \(\left(S^{T} = -S\right)\)
But \(S\) is not antisymmetric, so \(S^{T} \neq -S\) so \(\Psi^{T}_{a}\Psi_{a}\) is not lorentz invariant.
(Q) How do we know '\(S\)' is not antisymmetric?
So the guess should be to check if \(\Psi^{*}_{a}\Psi_{a}\) is lorentz invariant. (the logic is if \(\Psi_{a}\) is complex \(\Psi^{*}_{a}\Psi_{a}\) is real to add in lagrangian)

Digression :- For vector representation of lorentz transformation, we can write it in two ways.

\[ e^{\frac{1}{2}\,\omega V} \qquad\qquad e^{\eta\,\omega} \]

let's write with indices.

\[ \left(e^{\frac{1}{2}\,\omega_{\mu\nu}\,V^{\mu\nu}}\right)^{\alpha}_{\;\;\beta} \qquad\qquad \left(e^{\eta\,\omega}\right)^{\alpha}_{\;\;\beta} \]

expand to \(1^{\text{st}}\) order ---

\[ \left[\,1 + \frac{1}{2}\,\omega_{\mu\nu}\left(V^{\mu\nu}\right)^{\alpha}_{\;\;\beta} + \cdots\right] \qquad\qquad \left[\,1 + \eta^{\alpha\mu}\,\omega_{\mu\beta} + \cdots\right] \] \[ \begin{gathered} \left(1 + \frac{1}{2}\,\omega_{\mu\nu}\left(\eta^{\mu\alpha}\,\delta^{\nu}_{\;\beta} - \eta^{\nu\alpha}\,\delta^{\mu}_{\;\beta}\right) + \cdots\right)\\ \left(1 + \left(\frac{1}{2}\,\omega_{\mu\nu}\,\delta^{\nu}_{\;\beta}\,\eta^{\mu\alpha} - \frac{1}{2}\,\omega_{\mu\nu}\,\eta^{\nu\alpha}\,\delta^{\mu}_{\;\beta}\right) + \cdots\right)\\ \left(1 + \left(\frac{1}{2}\,\omega_{\mu\beta}\,\eta^{\mu\alpha} - \frac{1}{2}\,\omega_{\mu}^{\;\;\alpha}\,\delta^{\mu}_{\;\beta}\right) + \cdots\right) \quad\longleftarrow\;\text{will be equal to} \end{gathered} \]

For transpose: \(1 + \frac{1}{2}\,\omega_{\mu\nu}\left(V^{\mu\nu}\right)^{T\,\alpha}_{\;\;\;\;\beta}\) ; \(1 + \omega^{T}\eta^{T}\)

Is \(\Psi^{*}_{a}\,\Psi_{a}\) lorentz invariant ---

\[ \begin{aligned} \Psi^{*}_{a}\,\Psi_{a} \;\longrightarrow\;& \left(e^{\frac{1}{2}\,\omega S}\right)^{*}_{ab}\Psi^{*}_{b}\,\left(e^{\frac{1}{2}\,\omega S}\right)_{ac}\Psi_{c}\\ =\;& \left(e^{\frac{1}{2}\,\omega S}\right)^{*T}_{ba}\left(e^{\frac{1}{2}\,\omega S}\right)_{ac}\Psi^{*}_{b}\,\Psi_{c}\\ =\;& \left(\left(e^{\frac{1}{2}\,\omega S}\right)^{\dagger}\left(e^{\frac{1}{2}\,\omega S}\right)\right)_{bc}\Psi^{*}_{b}\,\Psi_{c}\\ =\;& \left(\left(e^{\frac{1}{2}\,\omega S^{\dagger}}\right)\left(e^{\frac{1}{2}\,\omega S}\right)\right)_{bc}\Psi^{*}_{b}\,\Psi_{c} \end{aligned} \]

We know, we require \(S^{\dagger} = -S\) for lorentz invariance.

\[ \begin{aligned} S = \frac{1}{4}\left[\gamma^{\mu},\, \gamma^{\nu}\right] \;\Rightarrow\; S^{\dagger} &= \frac{1}{4}\left(\gamma^{\mu}\gamma^{\nu} - \gamma^{\nu}\gamma^{\mu}\right)^{\dagger}\\ &= \frac{1}{4}\left(\left(\gamma^{\nu}\right)^{\dagger}\left(\gamma^{\mu}\right)^{\dagger} - \left(\gamma^{\mu}\right)^{\dagger}\left(\gamma^{\nu}\right)^{\dagger}\right)\\ &= \frac{1}{4}\left[\left(\gamma^{\nu}\right)^{\dagger},\, \left(\gamma^{\mu}\right)^{\dagger}\right] \end{aligned} \]

Is \(\left(\gamma^{\mu}\right)^{\dagger} = \left(\gamma^{\mu}\right)\)?

\[ \left\{ \begin{aligned} \left(\gamma^{0}\right)^{\dagger} &= \left(\sigma_1 \otimes \sigma_1\right)^{\dagger} = \gamma^{0}\\ \left(\gamma^{1}\right)^{\dagger} &= \left(i\,\sigma_1 \otimes \sigma_2\right)^{\dagger} = -\gamma^{1}\\ \text{Similarly}\quad \left(\gamma^{2}\right)^{\dagger} &= -\gamma^{2}\\ \left(\gamma^{3}\right)^{\dagger} &= -\gamma^{3} \end{aligned} \right. \]

\(\gamma^{0}\) is hermitian, \(\gamma^{i}\) is anti-hermitian
So for \(\mu = 0\), \(\nu = 0\): \(S^{\dagger} = S = 0\)
for \(\mu = 0\), \(\nu \neq 0\): \(\left(\gamma^{\nu}\right)^{\dagger} = -\gamma^{\nu}\)

\[ \begin{aligned} S^{\dagger} &= \frac{1}{4}\left[-\gamma^{\nu},\, \gamma^{0}\right]\\ &= \frac{1}{4}\left[\gamma^{0},\, \gamma^{\nu}\right] = S^{0\nu}\\ \left(S^{0\nu}\right)^{\dagger} &= S^{0\nu} \end{aligned} \]

For \(\mu \neq \nu\), \(\mu,\nu = 1,2,3\)

\[ \begin{aligned} \left(S^{ij}\right)^{\dagger} &= \frac{1}{4}\left[-\gamma^{j},\, -\gamma^{i}\right]\\ &= -\frac{1}{4}\left[\gamma^{i},\, \gamma^{j}\right]\\ \boxed{\left(S^{ij}\right)^{\dagger} = -S^{ij}} \end{aligned} \] \[ \left(\gamma^{\mu}\right)^{\dagger} = \left\{ \begin{aligned} \gamma^{\mu}\quad &\mu = 0\\ -\gamma^{\mu}\quad &\mu \neq 0 \end{aligned} \right. \qquad\text{So}\quad S^{\dagger} = -S \quad\forall\ \mu,\nu. \]

We need an operation which treats all \(\gamma^{\mu}\) similar. Some analogue of \(\gamma^{\mu}\) which is always hermitian or anti-hermitian.
So, \(S^{\dagger} = -S\) only for \(\mu, \nu \neq 0\).
So \(\Psi^{*}_{a}\,\Psi_{a}\) is also not lorentz invariant.
Our recent experience with real fields & complex fields does not help us make a lagrangian.
We need something which anticommute with \(\gamma^{i}\) but does not anticommute with \(\gamma^{0}\), which is \(\gamma^{0}\) itself.

\[ \left\{\gamma^{0},\, \gamma^{i}\right\} = 0 \]

We check for \(\overline{\gamma}^{\mu} = \gamma^{0}\left(\gamma^{\mu}\right)^{\dagger}\gamma^{0}\)
$= \{

\[ \begin{aligned} \gamma^{0}\left(-\gamma^{\mu}\right)\gamma^{0} &= \left(\gamma^{0}\right)^{2}\gamma^{\mu} = \gamma^{\mu}\ (\mu\neq0)\\ \gamma^{0}\gamma^{0}\,\gamma^{0} &= \gamma^{0}\qquad \mu = 0 \end{aligned} \]

.$

\[ \overline{\gamma}^{\mu} = \gamma^{\mu} \] \[ \begin{aligned} \Rightarrow\quad \overline{S}^{\mu\nu} = \gamma^{0}\left(S^{\mu\nu}\right)^{\dagger}\gamma^{0} &= \gamma^{0}\,\frac{1}{4}\left[\left(\gamma^{\nu}\right)^{\dagger},\, \left(\gamma^{\mu}\right)^{\dagger}\right]\gamma^{0}\\ &= (-)\,\gamma^{0}\,\frac{1}{4}\left[\left(\gamma^{\mu}\right)^{\dagger},\, \left(\gamma^{\nu}\right)^{\dagger}\right]\gamma^{0}\\ &= (-1)\,\frac{1}{4}\left[\gamma^{0}\left(\gamma^{\mu}\right)^{\dagger}\gamma^{0},\ \gamma^{0}\left(\gamma^{\nu}\right)^{\dagger}\gamma^{0}\right]\\ &= -\frac{1}{4}\left[\gamma^{\mu},\, \gamma^{\nu}\right] \;=\; -S^{\mu\nu} \end{aligned} \] \[ \boxed{\;\overline{S}^{\mu\nu} = -S^{\mu\nu}\;} \]

Finally we found an operation for which \(\overline{S}^{\mu\nu} = -S^{\mu\nu}\)
If we define, \(\overline{\Psi}_{b} = \Psi^{\dagger}_{a}\left(\gamma^{0}\right)_{ab}\) then,

\[ \overline{\Psi}_{a}\,\Psi_{a} \text{ is lorentz invariant.} \] \[ \begin{aligned} \overline{\Psi}\,\Psi \;=\; \Psi^{\dagger}\left(\gamma^{0}\right)\Psi \;\longrightarrow\;& \underbrace{\left(e^{\frac{1}{2}\,\omega S}\right)^{\dagger}\gamma^{0}\left(e^{\frac{1}{2}\,\omega S}\right)}\,\Psi^{\dagger}\Psi\\ =\;& \gamma^{0}\,e^{-\frac{1}{2}\,\omega S}\; e^{\frac{1}{2}\,\omega S}\;\Psi^{\dagger}\Psi\\ =\;& \gamma^{0}\,\Psi^{\dagger}\Psi\\ =\;& \overline{\Psi}\,\Psi \end{aligned} \]

There are probably no other combination which are Lorentz invariant.
So

\[ \begin{aligned} \mathcal{L} &= \overline{\Psi}\left(i\,\gamma^{\mu}\partial_{\mu} - m\right)\Psi\\ &= \Psi^{\dagger}_{a}\,\gamma^{0}_{\;ab}\left(i\,\gamma^{\mu}_{\;bc}\,\partial_{\mu} - m\,\delta_{bc}\right)\Psi_{c}\\ \mathcal{L} &= \overline{\Psi}\left(i\,\not{\partial} - m\right)\Psi \end{aligned} \]

What are equation of motion of this \(\mathcal{L}\)? We can vary \(\Psi\) & \(\overline{\Psi}\) independently.

\[ \frac{\delta\mathcal{L}}{\partial\overline{\Psi}} = \left(i\,\not{\partial} - m\right)\Psi \;=\; \partial_{\mu}\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\overline{\Psi}\right)}\right) = \partial_{\mu}(0) = 0 \]

taking \(\left(\overline{\phantom{x}}\right)\) \(\Rightarrow\)
\(\dfrac{\partial\mathcal{L}}{\partial\Psi} = \overline{\Psi}\left(-i\,\gamma^{\mu}\overleftarrow{\partial}_{\mu} - m\right) = 0\)
Canonical momentum to \(\Psi\)
\(\left(\dot{\Psi}\text{ occurs in }\gamma^{\mu}\partial_{\mu}\text{ at }\mu = 0\right)\)

\[ \Pi_{a} = \frac{\delta\mathcal{L}}{\delta\dot{\Psi}_{a}} = \left(\Psi^{\dagger}_{a}\,\gamma^{0}\right) i\,\gamma^{0} = i\,\Psi^{\dagger}\left(\gamma^{0}\right)^{2} = i\,\Psi^{\dagger}_{a} \]

Weyl representation ---

\[ \gamma^{0} = \sigma^{1} \otimes I = \begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix} \] \[ \gamma^{i} = \begin{bmatrix} 0 & \sigma^{i}\\ -\sigma^{i} & 0 \end{bmatrix} \left\{ \begin{aligned} \gamma^{1} &= i\,\sigma^{2} \otimes \sigma^{1} = i\begin{pmatrix} 0 & -i\sigma^{1}\\ i\sigma^{1} & 0 \end{pmatrix} = \begin{pmatrix} 0 & \sigma^{1}\\ -\sigma^{1} & 0 \end{pmatrix}\\ \gamma^{2} &= i\,\sigma^{2} \otimes \sigma^{2} = i\begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}\otimes\sigma^{2} = \begin{pmatrix} 0 & \sigma^{2}\\ -\sigma^{2} & 0 \end{pmatrix}\\ \gamma^{3} &= i\,\sigma^{2} \otimes \sigma^{3} = i\begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}\otimes\sigma^{3} = \begin{pmatrix} 0 & \sigma^{3}\\ -\sigma^{3} & 0 \end{pmatrix} \end{aligned} \right. \]

Check that \(\left(\gamma^{\mu}\right)^{2} = \left\{\begin{aligned} 1\quad &\mu = 0\\ -1\quad &\mu \neq 0 \end{aligned}\right.\) & \(\left\{\gamma^{\mu}, \gamma^{\nu}\right\} = 0\) \((\mu\neq\nu)\)

\[ \begin{aligned} \gamma_5 &= i\,\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{3}\\ &= i\begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{1}\\ -\sigma^{1} & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{2}\\ -\sigma^{2} & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^{3}\\ -\sigma^{3} & 0 \end{pmatrix}\\ &= i\begin{pmatrix} -\sigma^{1} & 0\\ 0 & \sigma^{1} \end{pmatrix} \begin{pmatrix} -\sigma^{2}\sigma^{3} & 0\\ 0 & -\sigma^{2}\sigma^{3} \end{pmatrix} \qquad \left| \begin{aligned} \sigma_1\sigma_2\sigma_3 &= i\,\sigma_3\sigma_3\\ &= i\,\sigma_3^{2}\\ &= i\,(I) \end{aligned} \right.\\ &= i\begin{pmatrix} \underline{\sigma^{1}\sigma^{2}\sigma^{3}} & 0\\ 0 & -\underline{\sigma^{1}\sigma^{2}\sigma^{3}} \end{pmatrix}\\ &= i\begin{pmatrix} i(I) & 0\\ 0 & -i(I) \end{pmatrix}\\ \gamma_5 &= \begin{pmatrix} -I & 0\\ 0 & I \end{pmatrix} \end{aligned} \]

\(\gamma_5\) is diagonal. Is called as Weyl reps.
\(\left[\text{any reps.\ in which } \gamma_5 \text{ is diagonal}\right.\)
We can find \(S^{\mu\nu} = \frac{1}{4}\left[\gamma^{\mu},\, \gamma^{\nu}\right]\) (for weyl rep.)

\[ \begin{aligned} &= \frac{1}{4}\left(\gamma^{\mu}\gamma^{\nu} - \gamma^{\nu}\gamma^{\mu}\right)\\ &= \frac{1}{4}\left(2\,\gamma^{\mu}\gamma^{\nu}\right) \qquad (\mu\neq\nu) \end{aligned} \]

$\{

\[ \begin{aligned} &\text{for } (\mu\neq\nu)\,,\quad S^{\mu\nu} = \frac{1}{2}\,\gamma^{\mu}\gamma^{\nu}\\ &\text{For } (\mu=\nu)\quad S^{\mu\nu} = 0 \end{aligned} \]

.$

\[ S^{0i} = \frac{1}{2}\,\gamma^{0}\gamma^{i} = \frac{1}{2}\begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix}\begin{pmatrix} 0 & \sigma^{i}\\ -\sigma^{i} & 0 \end{pmatrix} \] \[ S^{0i} = \frac{1}{2}\begin{pmatrix} -\sigma^{i} & 0\\ 0 & \sigma^{i} \end{pmatrix} \qquad {\longrightarrow\;\text{represents boost}} \]

\((i,j = 1,2,3)\)

\[ \begin{aligned} S^{ij} = \frac{1}{2}\,\gamma^{i}\gamma^{j} &= \frac{1}{2}\begin{pmatrix} 0 & \sigma^{i}\\ -\sigma^{i} & 0 \end{pmatrix}\begin{pmatrix} 0 & \sigma^{j}\\ -\sigma^{j} & 0 \end{pmatrix}\\ &= \frac{1}{2}\begin{pmatrix} -\sigma^{i}\sigma^{j} & 0\\ 0 & -\sigma^{i}\sigma^{j} \end{pmatrix}\\ &= \frac{1}{2}\begin{pmatrix} -i\,\epsilon_{ijk}\,\sigma^{k} & 0\\ 0 & -i\,\epsilon_{ijk}\,\sigma^{k} \end{pmatrix} \end{aligned} \] \[ S^{ij} = \left(-\frac{1}{2}\right) i\,\epsilon_{ijk}\,\sigma_{k}\begin{pmatrix} +I & 0\\ 0 & I \end{pmatrix} \qquad \downarrow\;{\text{Represents rotations}} \]

Note that \(S^{0i}\), \(S^{ij}\) (Boost & rotation reps) are block diagonal form \(\left(\begin{array}{c|c}3&0\\\hline0&3\end{array}\right)\) in weyl basis. (i.e. all continuous lorentz transformations are block diagonal). i.e. these are reducible reps of lorentz transformations.
i.e. Top two components only transform among themselves and bottom two components of \(\Psi\) will only transform among themselves.
This shows that 4 component spinors are reducible. (if they are reducible in one basis then they are reducible in all basis, but weyl representation makes it manifest.)

Note --- This is the 4 component representation of lorentz algebra

\[ \Psi = \begin{bmatrix} \Psi_1\\ \Psi_2\\ \Psi_3\\ \Psi_4 \end{bmatrix}\;; \quad\text{it will be reducible if any subset of it transforms within itself under any transformation} \]

and it is reducible under lorentz transformations so under weyl reps.

\[ \begin{pmatrix} \Psi_1\\ \Psi_2\\ 0\\ 0 \end{pmatrix} \longrightarrow \begin{pmatrix} \Psi'_1\\ \Psi'_2\\ 0\\ 0 \end{pmatrix} \] \[ e^{\frac{1}{2}\,\omega_{\mu\nu}\,S^{\mu\nu}} \begin{pmatrix} \Psi_1\\ \Psi_2\\ 0\\ 0 \end{pmatrix} = \begin{pmatrix} \Psi'_1\\ \Psi'_2\\ 0\\ 0 \end{pmatrix} \]

where we will use weyl reps. of \(S^{\mu\nu}\).

\[ \begin{aligned} \left(i\,\not{\partial} - m\right) = \left(i\,\gamma^{\mu}\partial_{\mu} - m\right) &= i\,\gamma^{0}\partial_{0} + i\,\gamma^{i}\partial_{i} - m\,I\\ &= i\begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix}\partial_{0} + i\begin{pmatrix} 0 & \sigma^{i}\\ -\sigma^{i} & 0 \end{pmatrix}\partial_{i} - m\,I\\ &= \begin{pmatrix} -m & i\left(\partial_{0} + \sigma^{i}\partial_{i}\right)\\ i\left(\partial_{0} - \sigma^{i}\partial_{i}\right) & -m \end{pmatrix} \end{aligned} \]

The dirac eq\(^{\text{n}}\) \(\left(i\,\not{\partial} - m\right)\Psi = 0\)
\(\Psi = \begin{pmatrix} \Psi_L\\ \Psi_R \end{pmatrix}\)

\[ \Rightarrow\quad \begin{pmatrix} -m & i\left(\partial_{0} + \sigma^{i}\partial_{i}\right)\\ i\left(\partial_{0} - \sigma^{i}\partial_{i}\right) & -m \end{pmatrix} \begin{pmatrix} \Psi_L\\ \Psi_R \end{pmatrix} \]

where \(L\) & \(R\) label eigen value under \(\gamma_5 = \begin{pmatrix} -I & 0\\ 0 & I \end{pmatrix}\)

\[ \gamma_5\,\Psi = \begin{pmatrix} -\Psi_L\\ \Psi_R \end{pmatrix} \]

\(\Psi_L\) & \(\Psi_R\) transform independently & don't mix under lorentz transformation.
However dirac eq\(^{\text{n}}\) mixes them ---

\[ \begin{gathered} \left(i\,\not{\partial} - m\right)\Psi = 0\\ \begin{pmatrix} -m & i\left(\partial_{0} + \sigma\,\partial_{i}\right)\\ i\left(\partial_{0} - \sigma^{i}\partial_{i}\right) & -m \end{pmatrix} \begin{pmatrix} \Psi_L\\ \Psi_R \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix} \end{gathered} \] \[ \left. \begin{aligned} i\left(\partial_{0} + \sigma^{i}\partial_{i}\right)\Psi_R \;-\; m\,\Psi_L &= 0\\ i\left(\partial_{0} - \sigma^{i}\partial_{i}\right)\Psi_L \;-\; m\,\Psi_R &= 0 \end{aligned} \right\} \quad \begin{aligned} &\text{even dirac eq}^{\text{n}}\\ &\text{can't mix them}\\ &\text{if } m = 0. \end{aligned} \]

For \(m = 0\):
\(\left(\partial_{0} + \sigma^{i}\partial_{i}\right)\Psi_R = 0\;;\qquad \left(\partial_{0} - \sigma^{i}\partial_{i}\right)\Psi_L = 0\)
Suppose we have a particle with right circular polarisation its travelling with some velocity (it is massive), so we can slow it & bring it to rest. If we move it in other way we will see left circular polarisation, so right & left spin can be transformed into each other by just boost.
So it must be that dynamics mixes \(\Psi_L\) & \(\Psi_R\) for massive particle.
For \(m = 0\), we can't bring particle to rest by going in any frame or by any means, its motion can not be reversed. Therefore for \(m = 0\), \(\Psi_L\) & \(\Psi_R\) shall be decoupled in e.o.m.
Sometimes people like to take a P.O.V. that mass is interaction term which flips \(\Psi_L \leftrightarrow \Psi_R\). (in EOM).

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