QFT is considered a difficult subject. It is a basic underline formalism which describe laws of nature, we describe laws in terms of fields with each associted particles.

Q:Why does the subject play central role in particle physics?

A: for that let us ask ourselves, how do we describe most fundamental system, i.e. dynamics of particles. So our question is how shall we describe dynamics of the particles? we have so many particles like photons, electrons, quarks.
Before we answer let's ask what do we mean by dynamics of a particle, i.e. What will the particle do? If it's free, it will just propagate, if it is with other particles, it will interact and scattering will occur. So we can ask about the scattering.

Suppose we put so many particles and the attraction makes them live in bound state. Ex. Proton is bound state of "uud". In this case we would like to find energy levels, possible excitation of system we have radiation and chemistry.
The phenomena of scattering is tested in accelerators, (accelerators can be naturally occuring also like cosmic rays. )

We would like to have theory which provides calculation of rate at which scattering happens(we will calculate scattering amplitude which will provide probability of scattering or rate of scattering), which can be tested experimentally.

Strictly speaking quantum field theories are not fundamental theories, these are effective field theories, which only included appropriate degree of freedom to describe physical phenomena occurring at chosen length and energy scale while ignoring the d.o.f. at shorter distance and substructure. This is why the answers to considering proton as fundamental particle agrees with experiment at energy scale where we can't see it's substructure (O(GeV)) The breakdown of theory at some decimal place can indicate sign of substructure.

so how do we make the theory?
We know that QM is key ingredient at low energy scale.

\[ \begin{align*} H \psi = E \psi \\ \frac{\hbar^{2}}{2m} \nabla ^{2} \psi + V(x)\psi=E \psi \\ \end{align*} \]

To summarise : Schrodinger equation describe fixed number of particles, also it is non relativistic(as it does not put space and time on same footing, as it is first order in time and second order in space, in relativity we rotte space and time into each other). For a relativistic equation we need either both second order or both first order, or both having the same order in differential equation.
We can write relativistic version of Schrodinger equation by using relativistic Hamiltonian.

\[ \begin{align*} E \rightarrow i \hbar \frac{\partial}{\partial t} \vec{{p}} {} &=-i \hbar \nabla \\ E^2 &=p^2+m^2 \\ E^2 \psi &= (p^2+m^2)\psi \\ (E^2-p^2-m^2)\psi &=0 \\ \hbar^2 \left(\frac{\partial }{\partial t} ^2-\nabla^2 \right) \psi + m^2\psi &=0\\ \end{align*} \]

General solution is \(\psi(x)=e^{ik^{\mu}.x_{\mu}}\)
using above \(\psi(x)\) in K.G. equation we get: \(\hbar^{2}k^{2}=m^{2}\)

We then ask: what is the energy of the particle having some fixed momentum
using above equation

\[ \begin{align*} h^{2}k^{2} &=m^{2} \\ h^{2}(k_{0}^{2}-\vec{k}^{2}) &=m^{2} \\ E &=\hbar k_{0} =\pm \sqrt{ m^{2}+h^{2}k^{2} } \\ E &=\pm \sqrt{ m^{2}+p^{2} } \\ \end{align*} \]

Now we will find eigen values of the momentum operator for the trial wave function \(\psi(x)=e^{i k_{\mu}x^{\mu}}\), where the momentum operator is simply \(\hat{P_{\mu}}=-i\nabla_{\mu}\)
using eigen value equation we find that :

\[ \begin{align*} \hat{P_{\mu }}\psi(x) &=-i\nabla_{\mu } \psi(x)\\ \hat{P_{\mu }}\psi(x) &=-i . ik_{\mu}\psi(x)\\ \end{align*} \]

So the Eigen value of \(P_{\mu }\) becomes \(k_{\mu}\). Note that \(\partial_{i}=\nabla_{i}=\frac{\partial}{\partial x^{i}}\)
Using equation \(E=\pm \sqrt{ p^{2}+m^{2} }\) ,we get \(k_{0}=\pm \sqrt{ k^{2}+m^{2} }\) ,
we have possibility of having negative energy solutions which does not make any sense. In field theory where we consider \(\psi\) as field rather than probability amplitude (in Q.M.) then this problem does not arise.

But in quantum mechanics we can't throw away negative energy solutions, since the general solution is superposition of all possible solutions.

The idea of fields is not new, we had scalar potential \(\phi(x)\) as well as vector potential \(A_{\mu}(x)\). The field is certainly very well observed experimentally. We have these two classical fields which obey dynamics of :

\[ \begin{align*} \nabla.\vec{B} &=0 && \nabla \times \vec{B} =\mu_{0}J\\ \nabla.\vec{E} &=0 && \nabla \times \vec{E} = -\frac{ \partial \vec{B} }{ \partial t } \end{align*} \]

Where ,

\[ \begin{align*} \vec{E} &=-\frac{ \partial \vec{A} }{ \partial t } -\nabla \phi(x) \\ \vec{B} &=\nabla \times \vec{A} \end{align*} \]

Relativistic Form:

\[ \begin{align*} A_{\mu} &=\left( \phi(x),-\vec{A} \right) && A^{\mu}=\left( \phi(x),\vec{A} \right) \\ F_{\mu \nu} &=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} \end{align*} \]

Here \(F_{\mu \nu}\) has 6 independent parameters.
\(F_{0 i}=-E_{i}\) ; \(F_{ij}=\epsilon_{ijk}B_{k}\) ,
Relativistic form is :

\[ \begin{align*} \partial^{\mu}F_{\mu \nu } &=0 \\ \epsilon_{\mu \nu \lambda \rho}\partial^{\nu}F^{\lambda \rho} &=0 \end{align*} \]

These are in total 8 equations.
Let us expand the first equation:

\[ \begin{align*} \partial^{\mu}F_{\mu \nu} &=0 \\ \partial^{\mu}\left( \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} \right) &=0\\ \partial^{2}A_{\nu}-\partial^{\mu}\partial_{\nu}A_{\mu} &=0\\ \partial^{2}A_{\nu}-\partial_{\nu}(\partial . \vec{A} ) &=0\\ \text{Using lorentz gauge ; which is simply $\partial.\vec{A} =0$} \\ \end{align*} \]

we get an equation of motion for scalar field or the K.G. equation. or \(\nabla^{2}A_{\nu}=0\)
so even fields can obey K.G. equation and thus from here we give up the idea of wave equations for relativistic particles, we are forced to do this by two problems.

We replace it with idea of field equations for the relativistic fields. we will get arbitrary number of particles.
We also have to show that quantising fields gives us particles. from now on \(\phi(x)\) is no longer a wave function but rather is a field.\(\phi(x)\) and \(\left| \phi(x) \right|^{2}\) are no longer waves and probability , but both represents operators that creates particles. In next lecture we will see that this quantum field always creates particles of positive energy and never creates negative energy particles, secondly we can use this field again and again to create as many particles.

Classical particles \(\xmapsto[]{Q.M}\) Quantum particles. (But it is inconsistent with relativity)
Classical fields \(\xmapsto {Q.M}\) Quantum fields \(\xmapsto {Quantisation}\) Quantum particles. (This is consistent with relativity)