Contents

Lecture 1

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

Q: Why does the subject play central role in particle physics?
A: for that let's 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 occurring also..., 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 distances 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 (\(O(\mathrm{GeV})\)). The breakdown of theory at some decimal place can indicate sign of substructure.

So how do we make the theory?

\[ \begin{aligned} E \to i\hbar\frac{\partial}{\partial t}, \qquad \vec{p} = -i\hbar\nabla, \qquad E^2 &= p^2 + m^2\\ \Rightarrow\quad E^2\psi &= \left(p^2+m^2\right)\psi \end{aligned} \] \[ \begin{aligned} \left(E^2 - p^2 - m^2\right)\psi &= 0 \qquad \psi = \text{wave function}\\ \Rightarrow\quad \hbar^2\left(\frac{\partial^2}{\partial t^2} - \nabla^2\right)\psi + m^2\psi &= 0 \qquad \text{---\,(1) K.G.\ eqn} \end{aligned} \]

General solution is \(\psi(x) = e^{ik_\mu x^\mu}\) (trial soln).
Using \(\psi(x)=e^{ik\cdot x}\) in K.G. eqn we get

\[ \boxed{\hbar^2 k^2 = m^2} \]

Q: What is the energy of a particle having some fixed momentum?
Using above eqn

\[ \begin{aligned} \hbar^2 k^2 &= m^2\\ \hbar^2\left(k_0^2 - \vec{k}^2\right) &= m^2\\ E = \hbar k_0 &= \pm\sqrt{m^2 + \hbar^2 k^2}\\ E &= \pm\sqrt{m^2 + p^2} \;\;\text{---\,(2)}\quad (p=\hbar k) \end{aligned} \]

\(\uparrow\) energy of particle with fixed momentum.
We can find eigen values of \(\hat{E}\) & \(\hat{P}\) operator for wave fun. \(\psi = e^{ik\cdot x}\):

\[ \begin{aligned} \hat{E}\psi &= i\hbar\frac{\partial}{\partial t}\psi = i\hbar (ik_0)\psi = -\hbar k_0 \quad\Rightarrow\quad \boxed{\hat{E}\psi = -\hbar k_0\,\psi}\\ \hat{P}\psi &= -i\hbar\nabla\psi = -i\hbar\left(\frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}\right)e^{i(k_0 t - \vec{k}\cdot\vec{r})}\\ &= -i\hbar\left(-i\vec{k}\right)\psi\\ \boxed{\hat{P}\psi = \hbar\vec{k}\,\psi} \end{aligned} \]

Note --- \(\nabla_i = \dfrac{\partial}{\partial x^i}\), \(x^i\) --- contravariant vector.
Using eqn (2) we have \(\hbar k_0 = \pm\sqrt{\hbar^2k^2+m^2}\).
We have possibility of having negative energy solutions which does not make any sense. In field theory if we consider \(\psi\) as field rather than wave fun. then this problem does not arise.

In quantum mechanics we can't throw away the negative energy solutions. As solution is superposition of all possible solutions.

The idea of fields

We do know the scalar fields (scalar potential \(\phi(x,t)\)) in theory of electrodynamics. But it comes together with vector field called the vector potential.
This field is certainly very well observed experimentally. We have these two classical fields which obey dynamics of:

\[ \left. \begin{aligned} \nabla\cdot\vec{B} &= 0 \qquad & \nabla\times\vec{B} &= +\frac{\partial \vec{E}}{\partial t}\\ \nabla\times\vec{E} &= -\frac{\partial \vec{B}}{\partial t} \qquad & \nabla\cdot\vec{E} &= 0 \end{aligned} \right\}\;\text{In absence of sources and sinks.} \]

where

\[ \vec{E} = -\frac{\partial\vec{A}}{\partial t} - \nabla\varphi \qquad\text{and}\qquad \vec{B} = \nabla\times\vec{A} \]

Relativistic form:

\[ \begin{gathered} A_\mu = (\varphi, -\vec{A}), \qquad A^\mu = (\varphi, \vec{A})\\ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu \quad (\text{anti-symmetric}) \end{gathered} \]

\(F_{\mu\nu}\) has 6 independent parameters.

\[ F_{0i} = -E_i\;; \qquad F_{ij} = \epsilon_{ijk}B_k \]

Relativistic form is ---

\[ \left. \begin{aligned} \partial^\mu F_{\mu\nu} &= 0 \quad\; 4\;\text{eq}^{\text{n}}\\ \epsilon_{\mu\nu\rho\sigma}\,\partial^\nu F^{\rho\sigma} &= 0 \quad\; 4\;\text{eq}^{\text{n}}\text{s} \end{aligned} \right\}\;\text{Total 8 equations.} \]

(1) \(\partial^\mu F_{\mu\nu} = 0 \;\Rightarrow\)

\[ \begin{aligned} \partial^\mu\left(\partial_\mu A_\nu - \partial_\nu A_\mu\right) &= 0\\ \partial^2 A_\nu - \partial^\mu\left(\partial_\nu A_\mu\right) &= 0\\ \partial^2 A_\nu - \partial_\nu\left(\partial^\mu A_\mu\right) &= 0\\ \partial^2 A_\nu - \partial_\nu\left(\partial\cdot A\right) &= 0 \end{aligned} \]

For \(\partial\cdot A = 0\) (Lorentz gauge) we have \(\partial^2 A_\nu = 0\), which is similar to \(\partial^\mu\partial_\mu\varphi = 0\) where \(\varphi\) is scalar field. But here K.G. eqn was wave eqn, but we have similar eqn for fields.

We give up the idea of wave equations for relativistic particles, we are forced to do this by two problems. We replace it with the 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. Now \(\phi(x,t)\) is no longer a wave function but rather is a field. \(\phi\) and \(|\phi\phi^{*}|\) is no longer probability but again some operator 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.

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Also to note. field is an operator where as in QM \(|\psi\rangle\) is a state vector or \(\psi(x,t)\) is component in \(\{|x\rangle\}\) basis.

Lecture 2

Lorentz transformations: all operations which keeps the space time interval \(\delta S^2=t^2-x^2-y^2-z^2\) fixed are called as Lorentz transformations.

\[ \begin{aligned} x_\mu x^\mu &= t^2 - x^2 - y^2 - z^2 \qquad (\text{invariant})\\ &= t^2 - (x^2+y^2+z^2) \end{aligned} \]

We can have \(x^2\) invariant also by keeping \(t\) fixed and have \((x^2+y^2+z^2)\) fixed by set of rotations. So Rotations are included in Lorentz transformation.
For \(y=z=0\):

\[ \begin{gathered} x_\mu x^\mu = t^2 - x^2 = (t-x)(t+x) = x_+ x_-\\ \text{Define } x_\pm = t\pm x \;\;\Rightarrow\;\; t = \frac{x_+ + x_-}{2}, \quad x = \frac{x_+ - x_-}{2} \end{gathered} \]

\(x_\mu x^\mu = x_+ x_-\) = product of two independent coordinates.
Lets ask what are the transformations on \(x_+\) & \(x_-\) such that product \(x_+x_- = x^\mu x_\mu\) remain invariant/fixed.

\[ x_\pm \longrightarrow e^{\pm\alpha} x_\pm \quad\text{will preserve } x_\mu x^\mu. \]

If

\[ \left. \begin{aligned} x_+ &\longrightarrow e^{\alpha} x_+\\ x_- &\longrightarrow e^{-\alpha} x_- \end{aligned} \right\} \;\Rightarrow\; t = \frac{x_+ + x_-}{2} \quad\text{under this transformation} \] \[ \begin{aligned} t &= \frac{e^{\alpha}x_+ + e^{-\alpha}x_-}{2} = \frac{1}{2}\left(e^{\alpha}(t+x) + e^{-\alpha}(t-x)\right)\\ &= \frac{1}{2}\left(\left(e^{\alpha}+e^{-\alpha}\right)t + \left(e^{\alpha}-e^{-\alpha}\right)x\right) = \left(\frac{e^{\alpha}+e^{-\alpha}}{2}\right)t + \left(\frac{e^{\alpha}-e^{-\alpha}}{2}\right)x\\ t &= (\cosh\alpha)\,t + (\sinh\alpha)\,x \end{aligned} \]

so under Lorentz transformation \(t \longrightarrow t(\cosh\alpha) + x(\sinh\alpha)\)

\[ \begin{aligned} \Rightarrow\qquad t' &= (\cosh\alpha)\,t + (\sinh\alpha)\,x\\ x' &= (\sinh\alpha)\,t + (\cosh\alpha)\,x \end{aligned} \]

similarly

\[ \begin{aligned} x = \frac{x_+ - x_-}{2}, \quad x' = \frac{x_+' - x_-'}{2} = \frac{e^{\alpha}x_+ - e^{-\alpha}x_-}{2} = \frac{e^{\alpha}(t+x) - e^{-\alpha}(t-x)}{2} = \left(\frac{e^{\alpha}-e^{-\alpha}}{2}\right)t + \left(\frac{e^{\alpha}+e^{-\alpha}}{2}\right)x \end{aligned} \] \[ \Rightarrow\quad \begin{pmatrix} t' \\ x' \end{pmatrix} = \underbrace{\begin{pmatrix} \cosh\alpha & \sinh\alpha \\ \sinh\alpha & \cosh\alpha \end{pmatrix}}_{\Lambda} \begin{pmatrix} t \\ x \end{pmatrix} \]

Lorentz transformation is mathematically similar to rotation. Since

\[ \begin{vmatrix} \cosh\alpha & \sinh\alpha \\ \sinh\alpha & \cosh\alpha \end{vmatrix} = \cosh^2\alpha - \sinh^2\alpha = 1 \]

Conceptually the boost & rotations are similar but we can't make any amount of rotation, since rotation is bounded by \([0,2\pi]\). But \(\cosh\alpha\) is not periodic which implies we can make any amount of boost.
We had,

\[ t' = \frac{t - vx/c^2}{\sqrt{1-v^2/c^2}} = \frac{t-vx}{\sqrt{1-v^2}}, \qquad x' = \frac{x - vt}{\sqrt{1-(v/c)^2}} = \frac{x-vt}{\sqrt{1-v^2}} \]

The transformation matrix

\[ \Lambda = \begin{pmatrix} \dfrac{1}{\sqrt{1-v^2}} & \dfrac{-v}{\sqrt{1-v^2}}\\[2ex] \dfrac{-v}{\sqrt{1-v^2}} & \dfrac{1}{\sqrt{1-v^2}} \end{pmatrix} \]

One nice thing about this formalism is that the Lorentz invariant quantities are having same number of plus and minus indices, for example we can take any two component vector \(A_\mu\), with just \((A_t, A_x)\) then the quantity \(A_{-}A_{+}\) is Lorentz invariant, and we can see that Lorentz Invariance is just scaling up and scaling down of \(A_+\) and \(A_-\), to make it invariant.

Field Equation

The field equation for the Klein gordon field is:

\[ \hbar^2\partial_\mu\partial^\mu\varphi + m^2\varphi = 0 \quad\text{---\,(1)}\qquad \text{(or)}\quad \boxed{\left(\partial^2+m^2\right)\varphi = 0} \]

treat \(\varphi(x,t)\) as an operator in quantum theory rather than a wave function.
We will start with revision of quantum harmonic oscillator

\[ \begin{aligned} L &= \frac{1}{2}m\dot{q}^2 - \frac{1}{2}m\omega^2 q^2\\ H &= \dot{q}p - L = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2 \end{aligned} \]

We do change of variables,

\[ \begin{aligned} a &= q + ip\\ a^\dagger &= q - ip \end{aligned} \]

We need to construct quantities of dimension length and momentum from \(m, \omega, \hbar\):

\[ \begin{gathered} q' \propto m^\alpha \omega^\beta \hbar^\gamma \qquad\qquad E=\hbar\omega \;\Rightarrow\; [\hbar]=[E][T]\\ \propto [M]^\alpha\,[T^{-1}]^\beta\,[M L^2 T^{-2}]^\gamma\,[T]^\gamma\\ q' \propto [M]^{\alpha+\gamma}\,[L]^{2\gamma}\,[T]^{-\beta-\gamma} \end{gathered} \]

For \(q'\) to be length we need \(2\gamma=1\), \(\alpha+\gamma=0\), \(\beta=-\gamma\)

\[ \begin{gathered} \Rightarrow\; \gamma = 1/2, \;\; \alpha = -1/2, \;\; \beta = -1/2\\ \Rightarrow\; q' = \sqrt{\frac{\hbar}{\omega m}} \end{gathered} \]

For \(q'\) to be momentum we require \(\alpha+\gamma=1\):

\[ \begin{gathered} p \propto (\hbar)^\alpha(\omega)^\beta(m)^\gamma, \qquad MLT^{-1} \propto (ML^2T^{-1})^{\ldots}\\ 2\gamma = 1 \;\Rightarrow\; \gamma = 1/2, \;\alpha = +1/2; \qquad -\beta-\gamma = -1 \;\Rightarrow\; \beta = 1 - 1/2 = +1/2\\ p' = \sqrt{m\omega\hbar} \end{gathered} \]

So,

\[ \begin{aligned} a &= \left(\frac{q}{q'} + \frac{ip}{p'}\right) = q\sqrt{\frac{\omega m}{2\hbar}} + \frac{ip}{\sqrt{2m\omega\hbar}}\\ a^\dagger &= q\sqrt{\frac{m\omega}{2\hbar}} - \frac{ip}{\sqrt{2m\omega\hbar}} \qquad \because\;\text{where }C\text{ is some constant.} \end{aligned} \]

Our Total Hamiltonian becomes---

\[ \begin{aligned} H &= \frac{p^2}{2m} + \frac{1}{2}m\omega^2 q^2\\ H &= \hbar\omega\left(a^\dagger a + \frac{1}{2}\right) \qquad\qquad \nexists\;\text{No state with negative energy.} \end{aligned} \]

We note that \([a,a^\dagger]=1\), assume there is state satisfying \(a|0\rangle=0\).
This assumption ensures we don't end up with minus infinity as energy of lowest state, (also we don't see vacuum decay to photons).

The reason we had gone through this exercise is that because equation (1) gives us collection of simple harmonic oscillators, all decoupled from each other, so we can quantise each of those quantum harmonic oscillators and therefore in such way we have quantised the field \(\varphi\).
Remember in quantum physics we put \(t\) and \(x\) on different footing but in quantum field theory we put time and space on the same footing.

So, object we are interested in is \(\varphi(t,\vec{x})\), for the moment we will treat time and space on different footing but we will see that the result will be relativistically invariant.
Lets bound \(\vec{x}\) to size of cube. Later we will take size of box 'L' to be infinity, then we can accommodate a system as large as we like to use.

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We need boundary conditions/periodic boundary conditions.

\[ \text{i.e.}\qquad \left. \begin{aligned} \varphi(t,\,x{+}L,\,y,\,z) &= \varphi(t,x,y,z)\\ \varphi(t,\,x,\,y{+}L,\,z) &= \varphi(t,x,y,z)\\ \varphi(t,\,x,\,y,\,z{+}L) &= \varphi(t,x,y,z) \end{aligned} \right\}\;\text{---\,(2)} \]

Till now we are treating \(\varphi\) as some function rather than an operator. We can write \(\varphi\) as some vector whose basis set are some exponential function (fourier series).
Let's take \(e^{i\vec{k}\cdot\vec{x}}\) as some general function, with periodic boundary conditions.
Using \(\varphi(x,t) = e^{i\vec{k}\cdot\vec{x}}\) in eqn (2) we get
(RHS is missing time! anyway time is fixed!)

\[ \begin{gathered} e^{i\vec{k}\cdot(\vec{x}+L)} = e^{i\vec{k}\cdot\vec{x}}\\ e^{i(k_x x + k_y y + k_z z)}\cdot e^{ik_xL} \;\text{form:}\quad e^{i(k_x(x+L) + k_y y + k_z z)}\\ e^{ik_x x} = e^{ik_x x}\, e^{ik_x L}\\ \Rightarrow\quad e^{ik_x L} = \cos(k_xL) + i\sin(k_xL) = 1\\ \Rightarrow\quad k_x L = 2\pi m_1, \qquad k_x = \frac{2\pi m_1}{L} \end{gathered} \]

Similarly \(e^{i(k_x x + k_y(y+L) + k_z z)} = e^{i(k_x x + k_y y + k_z z)}\) gives us \(k_y = \dfrac{2\pi m_2}{L}\).
Similarly we get \(k_z = \dfrac{2\pi m_3}{L}\)

\[ \text{(or)}\qquad \vec{k} = \frac{2\pi}{L}\left(m_1, m_2, m_3\right) \]

In fact we have taken all possible functions, as we can expand them in \(e^{i\vec{k}\cdot\vec{x}}\) with some coefficients \((q_k(t))\). Therefore,

\[ \varphi(t,x) = \underbrace{N}_{\substack{\text{Normalisation}\\\text{factor}}} \sum_{\vec{k}} q_k(t)\, e^{i\vec{k}\cdot\vec{x}} \]

\(\hookleftarrow\) expansion of most general configuration \(\varphi(t,\vec{x})\). \(\hookrightarrow\) time dependence has to be taken by \(q_k(t)\), and label \(k\) for '\(q\)' notify the seperate coefficients for each \(\vec{k}\).
We can use this expression and plug it into equation (1) \(\left(\partial^2+m^2\right)\varphi = 0\), and ask values of \(q_k(t)\).

\[ \Rightarrow\quad \left(\partial^2+m^2\right)\sum_{\vec{k}} q_k(t)\,e^{i\vec{k}\cdot\vec{x}} = 0 \qquad\text{---\,(3)} \] \[ \Rightarrow\quad \partial^2 = \partial_\mu\partial^\mu :\; \left(\frac{\partial}{\partial t},\, \nabla\right)\begin{pmatrix}\frac{\partial}{\partial t}\\ -\nabla\end{pmatrix} = \frac{\partial^2}{\partial t^2} - \nabla^2 \] \[ \begin{aligned} \text{(i)}\quad \partial^2 q_k(t) &= \frac{\partial^2}{\partial t^2}q_k(t) - \nabla^2 q_k(t) = \ddot{q}_k(t) - 0\\ \text{(ii)}\quad \partial^2 e^{i\vec{k}\cdot\vec{x}} &= \frac{\partial^2}{\partial t^2}e^{i\vec{k}\cdot\vec{x}} - \left(\frac{\partial^2}{\partial x^2}e^{i\vec{k}\cdot\vec{x}} + \frac{\partial^2}{\partial y^2}e^{i\vec{k}\cdot\vec{x}} + \frac{\partial^2}{\partial z^2}e^{i\vec{k}\cdot\vec{x}}\right)\\ &= k^2 e^{i\vec{k}\cdot\vec{x}} \qquad\qquad \left(k^2 = k_x^2+k_y^2+k_z^2\right) \end{aligned} \]

\(\Rightarrow\) equation (3) reduces to ---

\[ \begin{gathered} \sum_{\vec{k}}\left(\left(\partial^2 q_k(t)\right)e^{i\vec{k}\cdot\vec{x}} + q_k(t)\,\partial^2 e^{i\vec{k}\cdot\vec{x}} + m^2 q_k(t)\,e^{i\vec{k}\cdot\vec{x}}\right) = 0\\ \sum_{\vec{k}}\left(\left(\ddot{q}_k(t) + q_k(t)\,k^2\right)e^{i\vec{k}\cdot\vec{x}} + m^2 q_k(t)\, e^{i\vec{k}\cdot\vec{x}}\right) = 0\\ \sum_{\vec{k}}\left(\ddot{q}_k(t) + q_k(t)\,k^2 + m^2 q_k(t)\right)e^{i\vec{k}\cdot\vec{x}} = 0 \end{gathered} \] \[ \underset{\substack{\text{Infinite set of}\\\text{decoupled harmonic}\\\text{oscillators}}}{\Longrightarrow}\qquad \boxed{\;\ddot{q}_k(t) + \left(k^2+m^2\right)q_k(t) = 0\;}\qquad\text{---\,(4)} \]

Note that this is infinite set of differential equations one for each value of \(\vec{k}\) satisfying \(\vec{k} = \frac{2\pi}{L}(m_1,m_2,m_3)\).
Eq (4) looks like differential eqn of SHO, with frequency of oscillation

\[ \omega_k = \sqrt{k^2+m^2} \qquad (\vec{k}\text{ dependent frequency}) \qquad\Bigg|\;\; \begin{aligned} &\ddot{q} + \omega^2 q = 0\\ &\hookrightarrow \text{SHO eqn of motion.} \end{aligned} \]

We need to know if the \(\varphi(t,x)\) is real or complex. In QM we were forced to live with complex wave functions, but here its upon us to decide if we want to deal with real fields or complex fields. For the moment we choose \(\varphi(t,x)\) to real fields. Later we will extend the result to complex fields.
Which implies---

\[ \begin{gathered} \varphi(t,x) = \varphi^{*}(t,x)\\ \Rightarrow\quad \sum_{\vec{k}} q_k(t)\,e^{i\vec{k}\cdot\vec{x}} = \sum_{\vec{k}} q_k^{*}(t)\, e^{-i\vec{k}\cdot\vec{x}}\\ \boxed{\;q_k^{*}(t) = q_{-k}(t)\;} \end{gathered} \]

So, this is not a real harmonic oscillator, except for case \(\vec{k} = \vec{0} = (0,0,0)\), only for that case \(\left(q_0^{*} = q_0\right)\), else its a complex harmonic oscillator.

\[ \ddot{q}_k(t) + \left(k^2+m^2\right)q_k(t) = 0 \qquad \begin{cases} \text{Real H.O.} & \vec{k}=\vec{0}\\[2ex] \text{Complex H.O.} & \vec{k}\neq\vec{0} \end{cases} \]

\(q_k^{*}\) and \(q_{-k}\) has same frequency since frequency only depend upon \(\vec{k}^2\) \(\left(\omega_k^2 = k^2+m^2\right)\).
We claim that lagrangian for \(\Big(\sum_{\vec{k}} \ddot{q}_k(t) + \underbrace{(k^2+m^2)}_{\omega_k^2} q_k(t) = 0\Big)\) is given by

\[ L = \underbrace{\frac{1}{2}\dot{q}_0^2 - \frac{1}{2}\omega_0^2 q_0^2}_{\vec{k}=\vec{0}} + \sum_{\vec{k}>0}\left(\dot{q}_{-\vec{k}}\,\dot{q}_{\vec{k}} - \omega_{\vec{k}}^2\, q_{-\vec{k}}\, q_{\vec{k}}\right) \qquad{\text{How(?)}} \]

\(\vec{k}>0\) means \(\left(\text{first non zero coefficient in } \vec{k} = \frac{2\pi}{L}(m_1,m_2,m_3)\right)>0\).
In general if \(m_1>0\) we say \(\vec{k}>0\), if \(m_1=0\), we look at sign of \(m_2\), if \(m_2=0\), we look at sign of \(m_3\). If all \(m_1 m_2 m_3\) are zero then \(\vec{k}=\vec{0}\).
The quantisation requires working with hamiltonian rather than lagrangian. So we will find canonical momentum and then do legendre transform to find \(H\).

\[ \begin{aligned} H &= \sum_i \dot{q}_i p_i - L \qquad\qquad \left\{\,p_i = \frac{\partial L}{\partial\dot{q}_i}\right.\\ p_0 &= \frac{\partial L}{\partial \dot{q}_0} = \dot{q}_0 \quad\text{for } \vec{k}=0\\ p_k &= \frac{\partial L}{\partial \dot{q}_k} = \dot{q}_{-k} \qquad \forall\,\vec{k} \end{aligned} \]

So,

\[ \boxed{\;p_k = \frac{\partial L}{\partial\dot{q}_k} = \dot{q}_{-k}\;}\qquad\text{also for } \vec{k}=\vec{0} \] \[ \begin{aligned} H &= \sum p\dot{q} - L\\ &= p_0\dot{q}_0 + \sum_{\vec{k}>0} p_k\dot{q}_k + p_{-k}\dot{q}_{-k} - L\\ &= p_0 p_0 + \sum_{\vec{k}>0}\left(p_k p_k + p_k p_k\right) - \left(\frac{1}{2}\dot{q}_0^{\,2} - \frac{1}{2}\omega^2 q_0^2 + \sum_{\vec{k}>0}\dot{q}_k\dot{q}_{-k} - \omega^2 q_{-k}q_k\right)\\ &= \underline{p_0^2} + \sum_{\vec{k}>0} p_k p_k + p_{-k}p_k - \left(\frac{1}{2}p_0^{\,2} - \frac{1}{2}\omega^2 q_0^2 + \sum_{\vec{k}>0} p_k p_{-k} - \omega^2 q_{-k}q_k\right)\\ &= \frac{p_0^2}{2} + \sum_{\vec{k}>0}\left(\cancel{p_k p_{-k}} - \cancel{p_k p_{-k}} + p_{-k}p_k + \omega^2 q_{-k}q_k\right) + \frac{1}{2}\omega^2 q_0^2 \end{aligned} \] \[ \boxed{\;H = \frac{p_0^2}{2} + \frac{1}{2}\omega_0^2 q_0^2 + \sum_{\vec{k}>0} p_{\vec{k}}\, p_{-\vec{k}} + \omega_{\vec{k}}^2\, q_{-\vec{k}}\, q_{\vec{k}}\;} \]

This Hamiltonian is sum of independent hamiltonians of many harmonic oscillators each one's can be solved by writing them in terms of \(a\) and \(a^\dagger\). That is what we are going to do now.
Let's write canonical commutation relations.
Writing canonical commutation relations make canonical variables \(p_k\) & \(q_k\) Operators which is only possible if field itself is an operator. Therefore imposition of canonical commutation relations makes one field as operators.

\[ \begin{gathered} [q_0, p_0] = i\hbar\\ [q_k, p_k] = i\hbar \qquad\text{or}\qquad [q_{\vec{k}}, p_{\vec{k}}] = i\hbar\\ \boxed{\;[q_{\vec{k}},\, p_{\vec{k}'}] = i\,\delta_{\vec{k},\vec{k}'}\;} \end{gathered} \] \[ \checkmark\quad \begin{aligned} a_{\vec{k}} &= \frac{1}{\sqrt{2\omega_k}}\left(i p_{\vec{k}} + \omega_{\vec{k}}\, q_{\vec{k}}\right)\\ a^\dagger_{\vec{k}} &= \frac{1}{\sqrt{2\omega_k}}\left(-i p^{*}_{\vec{k}} + \omega_{\vec{k}}\, q^{*}_{\vec{k}}\right) = \frac{1}{\sqrt{2\omega_k}}\left(-i p_{-\vec{k}} + \omega_{\vec{k}}\, q_{-\vec{k}}\right) \end{aligned} \qquad\Bigg|\quad \begin{aligned} &\text{using } q^{*}_{k} = q_{-k}\\ &\text{and } p^{\dagger}_{k} = p_{-k}\\ &\vec{k}\to-\vec{k} \end{aligned} \]

Let us check commutation relation ---

\[ \begin{aligned} [a_{\vec{k}},\, a^\dagger_{-\vec{k}}] &= \frac{1}{2\omega_k}\Big(-i^2\underbrace{[p_{\vec{k}},\, q_{\vec{k}}]}^{\hspace{2em}\nearrow 0}\hspace{-1em} + i\omega_k[p_{\vec{k}},\, q_{\vec{k}}] + \omega_k(-i)[q_{\vec{k}},\, p_{\vec{k}}] + \omega_k^2\underbrace{[q_{\vec{k}},\, q_{\vec{k}}]}^{\hspace{2em}\nearrow 0}\Big)\\ &= \frac{1}{2\omega_k}\left(-i^2\omega_k - i^2\omega_k\right) = 1 \end{aligned} \]

in general,

\[ \Rightarrow\quad \boxed{\;[a_{\vec{k}},\, a^\dagger_{-\vec{k}'}] = \delta_{\vec{k},\vec{k}'}\;} \]

Our hamiltonian of infinite decoupled harmonic oscillators

\[ H = \dfrac{p_0^2}{2} + \dfrac{1}{2}\omega_0^2 q_0^2 + \displaystyle\sum_{\vec{k}>0}\left(p_{\vec{k}}\,p_{-\vec{k}} + \omega_k^2\, q_{-\vec{k}}\, q_{\vec{k}}\right) \]

becomes,

\[ \begin{aligned} a_{\vec{k}} &= \frac{1}{\sqrt{2\omega_k}}\left(i p_{\vec{k}} + \omega_{\vec{k}}\,q_{\vec{k}}\right)\\ a^\dagger_{\vec{k}} = \frac{1}{\sqrt{2\omega_k}}\left(-i p^{*}_{\vec{k}} + \omega_{\vec{k}}\, q^{*}_{\vec{k}}\right) \;&\Rightarrow\; a^\dagger_{-\vec{k}} = \frac{1}{\sqrt{2\omega_{-k}}}\left(-i p_{\vec{k}} + \omega_{-\vec{k}}\, q_{\vec{k}}\right) \qquad \left\{\omega_k = \omega_{-k} = \sqrt{k^2+m^2}\right.\\ &= \frac{1}{\sqrt{2\omega_k}}\left(-i p_{-\vec{k}} + \omega_{\vec{k}}\,q_{-\vec{k}}\right) \end{aligned} \] \[ \begin{aligned} a_{\vec{k}} + a^\dagger_{-\vec{k}} &= \frac{1}{\sqrt{2\omega_k}}\left(0 + 2\omega_k q_k\right) \;\Rightarrow\; q_k = \frac{\left(a_{\vec{k}} + a^\dagger_{-\vec{k}}\right)}{\sqrt{2\omega_k}}\\ a_{\vec{k}} - a^\dagger_{-\vec{k}} &= \frac{1}{\sqrt{2\omega_k}}\left(2i p_k + 0\right) \;\Rightarrow\; p_k = \frac{\sqrt{2\omega_k}}{2i}\left(a_k - a^\dagger_{-k}\right) \end{aligned} \]

Using set (4) in original hamiltonian we get

\[ \begin{aligned} H = \frac{1}{2}\frac{\left[\sqrt{2\omega_0}\right]^2}{(2i)^2}\left(a_0 - a_0^\dagger\right)^2 &+ \frac{1}{2}\omega_0^2\frac{1}{2\omega_0}\left(a_0 + a_0^\dagger\right)^2\\ &+ \sum_{\vec{k}>0}\frac{1}{(2i)^2}\left(\sqrt{2\omega_k}\right)^2\left(a_{-k} - a^\dagger_{k}\right)\left(a_k - a^\dagger_{-k}\right)\\ &\qquad + \frac{\omega_k^2}{2\omega_k}\left(a_{-k} + a^\dagger_{k}\right)\left(a_k + a^\dagger_{-k}\right) \end{aligned} \] \[ \begin{aligned} H = -\frac{\omega_0}{4}\Big(\cancel{a_0^2} - a_0 a_0^\dagger - a_0^\dagger a_0 + \cancel{a_0^\dagger a_0^\dagger}\Big) &+ \frac{1}{2}\omega_0^2\frac{1}{2\omega_0}\Big(\cancel{a_0 a_0} + a_0 a_0^\dagger + a_0^\dagger a_0 + \cancel{a_0^\dagger a_0^\dagger}\Big)\\ &+ \sum_{\vec{k}>0}(-)\frac{\omega_k}{2}\Big(\cancel{a_{-k}a_k} - a_{-k}a^\dagger_{-k} - a^\dagger_k a_k + \cancel{a^\dagger_k a^\dagger_{-k}}\Big)\\ &\qquad + \frac{1}{2}\omega_k\Big(\cancel{a_{-k}a_k} + a_{-k}a^\dagger_{-k} + a^\dagger_k a_k + \cancel{a^\dagger_k a^\dagger_{-k}}\Big) \end{aligned} \] \[ \boxed{\;H = \frac{\omega_0}{2}\left(a_0 a_0^\dagger + a_0^\dagger a_0\right) + \sum_{\vec{k}>0}\omega_k\left(a_{-k}\,a^\dagger_{-k} + a^\dagger_k\, a_k\right)\;} \]

Using \([a_k, a^\dagger_k] = 1 \;\Rightarrow\; a_k a^\dagger_{-k} - a^\dagger_{-k} a_k = 1\) we get

\[ a_0 a_0^\dagger - a_0^\dagger a_0 = 1 \;\Rightarrow\; a_0 a_0^\dagger = 1 + a_0^\dagger a_0 \]

but it should be \(\left[a_k\, a^\dagger_{-k}\right] = 1\):

\[ \boxed{{H = \displaystyle\sum_{\text{all }\vec{k}}\omega_{\vec{k}}\, a^\dagger_k a_k + \frac{1}{2}\sum_{\substack{\text{all}\\\vec{k}}}\omega_{\vec{k}}}} \]

For all \(\vec{k}\), \(\sum_{\vec{k}}\omega_k \longrightarrow \infty\) \(\Rightarrow\) ultraviolet divergence!

\[ \text{as } \omega_k = \sqrt{k^2+m^2}, \qquad \sum_k\omega_k = \sum_k\sqrt{k^2+m^2}\longrightarrow\infty. \]

For a field we always subtract the zero-point energy, because all measurements are made with respect to it.

Free Klein-Gordan field---

\[ H = \sum_k \omega_k\, a^\dagger_k\, a_{-k} \]

Lets ask what are states of system, energies and quantum numbers.

(1) States---
Lowest energy state, satisfies

\[ \begin{gathered} a_{\vec{k}}|0\rangle = 0 \qquad \forall\,\vec{k}\\ \text{So,}\quad H|0\rangle = \sum_{\vec{k}}\omega_k\, a^\dagger_k\, a_{-k}|0\rangle = 0 \end{gathered} \]

(2) Excited states---
Most general excited state is

\[ a^\dagger_{k_n}\, a^\dagger_{k_{n-1}}\ldots a^\dagger_{k_2}\, a^\dagger_{k_1}|0\rangle \]

All these creation operators are independent and can be repeated.

(3) What is reasonable interpretation of system?
We first look for momentum operator,

\[ \begin{gathered} \vec{P} = \sum_{\vec{k}}\vec{k}\; a^\dagger_{-k}\, a_k\\ \begin{aligned} \vec{P}\left(a^\dagger_{k_n}\ldots a^\dagger_{k_2}a^\dagger_{k_1}|0\rangle\right) &= \Big(k_n\, a^\dagger_{-k_n}a_{k_n} + \ldots + k_2\, a^\dagger_{-k_2}a_{k_2} + k_1\, a^\dagger_{-k_1}a_{k_1}\Big)\left(a^\dagger_{k_n}\ldots a^\dagger_{k_2}a^\dagger_{k_1}|0\rangle\right)\\ &= \left(k_n + k_{n-1} + \cdots + k_2 + k_1\right)\left(a^\dagger_{k_n}\ldots a^\dagger_{k_2}a^\dagger_{k_1}|0\rangle\right) \end{aligned} \end{gathered} \]

Similarly

\[ \hat{H}\left(a^\dagger_{k_n}\ldots a^\dagger_{k_2}a^\dagger_{k_1}|0\rangle\right) = \left(\omega_{k_1} + \omega_{k_2} + \cdots + \omega_{k_n}\right)\left(a^\dagger_{k_n}\ldots a^\dagger_{k_2}a^\dagger_{k_1}|0\rangle\right) \]

where \(\omega_{k_i} = \sqrt{k_i^2+m^2}\) \(\left(E^2 = p^2+m^2\right)\).
This implies that the state behaves like \(n\)-particle state with momenta \((\vec{k}_1+\vec{k}_2\cdots+\vec{k}_n)\) and energy \((\omega_1+\omega_2+\cdots+\omega_n)\).

Vacuum--- and state \(|0\rangle\) has no momentum and no energy, so we should associate it with vacuum. That is change of terminology, in harmonic oscillator case (in QM), Harmonic oscillator can be in ground state and can be in excited state. In field theory nothing is there, so we call it Vacuum.
Field operators are then simply linear combination of \(e^{i\vec{k}\cdot\vec{x}}\)

\[ \varphi = \sum q_{\vec{k}}\; e^{i\vec{k}\cdot\vec{x}} \]

So as we quantise the field, we get all possible no. of particles as states in the theory. This is not achievable by Schr\"odinger equation.

\[ \left. \begin{aligned} a &\begin{cases}\text{Lowering operator --- QM}\\ \text{Annihilation operator --- QFT}\end{cases}\\ a^\dagger &\begin{cases}\text{Raising operator --- QM}\\ \text{Creation operator --- QFT}\end{cases} \end{aligned} \right\}\;\text{Particle interpretation implies.} \]

Consider two states

\[ a^\dagger_{k_2}\, a^\dagger_{k_1}|0\rangle\,,\qquad a^\dagger_{k_1}\, a^\dagger_{k_2}|0\rangle \]

These states are same. Since \([a^\dagger_{k_1},\, a^\dagger_{k_2}] = 0\)

\[ \begin{gathered} \Rightarrow\quad \left(a^\dagger_{k_1}a^\dagger_{k_2} - a^\dagger_{k_2}a^\dagger_{k_1}\right)|0\rangle = 0\\ \Rightarrow\quad a^\dagger_{k_1}a^\dagger_{k_2}|0\rangle = a^\dagger_{k_2}a^\dagger_{k_1}|0\rangle \end{gathered} \]

Thus, if we interchange two particles, the states are same, such particles are called bosons and follow Bose statistics.
Quantisation of K.G. field gives us particles obeying Bose statistics.

*Spin of such \(n\)-particle state* --- We could construct the angular momentum in field theory (but that is complex way to go), there is simpler way to get spin of state. From statistics those are integer spin particles (since the particles are bosons).
Scalar field does not change under L.T., spin is something that should change under L.T., so spin should be zero.
There is even simpler way to tell spin of those bosons. Since particle with spin forms multiplet, but we don't see any label for those multiplets in general \(n\)-particle states. So these particles are spinless.

\[ \text{2 particle state} = a^\dagger_{k_2}\,\underbrace{a^\dagger_{k_1}|0\rangle}_{}\qquad\text{--- No label for spin} \]

Every \(n\)-particle state is singlet. \(\Rightarrow\) spinless particles. \(\left(\text{Spin}=0\right)\)

If we want to have interactions, we need to put \(\varphi^2 \sim \varphi^3 \sim \varphi^4\ldots\) terms in EOM.

Non linear equations \(\Rightarrow\) Interactions.

Also note that, we don't have negative energy:

\[ E_T = \sum_k \omega_k > 0, \quad\text{since all } \omega_k = \sqrt{k^2+m^2} > 0 \;\;\forall\, k. \]

So the problem of negative energy only arise when we take wavefn. interpretation of \(\varphi(t,x)\). In field repn we don't have negative energy.

''There is room for wavefn. interpretation in field space'' --- QFT by Suvrat.!

Lecture 3 (Sunil Mukhi)

Yesterday what we did, was to try and convert Klein gordon field into its mode \(q_k\).

\[ \varphi(x,t) = N\sum_{\vec{k}} q_{\vec{k}}(t)\, e^{i\vec{k}\cdot\vec{x}} \]

From the fact that \(\left[a^\dagger_k,\, a^\dagger_{k'}\right] = 0\) \(\Rightarrow\) these \(n\) particle states are bosons and since no other label describe states, these are spinless particles.

The values of \(\vec{k}\) take discrete values \(\frac{2\pi}{L}(m_1,m_2,m_3)\), because we took the system in a cubical box of length \(L\). Now let us take limit \(L\to\infty\), then \(\vec{k}\) will take continuous values

\[ H \;\xrightarrow{\;L\to\infty\;}\; \int\frac{d^3k}{(2\pi)^3}\;\omega_{\vec{k}}\; a^\dagger_{-\vec{k}}\, a_{\vec{k}} \]

Commutation relation becomes \(\longrightarrow\)

\[ \left[a_k,\, a^\dagger_{-k'}\right] = (2\pi)^3\,\delta^3(\vec{k}-\vec{k}') \qquad\text{}\hookleftarrow\text{ kronecker delta becomes dirac delta} \]

From here we will only work in this limit (\(L\to\infty\)).

Note that \(a_k = a_k(t)\); \(a^\dagger_{-k} = a^\dagger_{-k}(t)\).
We know,

\[ q_k(t) = \left(a_k(t) + a^\dagger_{-k}(t)\right)\frac{1}{\sqrt{2\omega_k}} \]

Using above,

\[ \varphi(t,x) = \int\frac{d^3k}{(2\pi)^3}\,\left(a_k + a^\dagger_{-k}\right)\frac{1}{\sqrt{2\omega_k}}\; e^{i\vec{k}\cdot\vec{x}} \]

We can also see that this \(\varphi(t,x)\) is real:

\[ \begin{aligned} \varphi^\dagger(t,x) &= \int\frac{d^3k}{(2\pi)^3}\,\frac{\left(a^\dagger_k + a_{-k}\right)}{\sqrt{2\omega_k}}\, e^{-i\vec{k}\cdot\vec{x}}\\ \text{replacing } k\to-\vec{k}:\qquad \varphi^\dagger(t,x) &= \int\frac{d^3k}{(2\pi)^3}\,\frac{\left(a^\dagger_{-k} + a_{k}\right)}{\sqrt{2\omega_k}}\, e^{i\vec{k}\cdot\vec{x}} \qquad\left(\omega_k = \omega_{-k} = \frac{\omega_k}{\sqrt{k^2+m^2}}\right)\\ &= \varphi(t,x) \;\Rightarrow\; \varphi(t,x) \text{ is real} \end{aligned} \]

We can write \(\varphi(t,x)\) as:

\[ \varphi(t,\vec{x}) = \int\frac{d^3k}{(2\pi)^3}\,\frac{1}{\sqrt{2\omega_k}}\left(a_k\, e^{i\vec{k}\cdot\vec{x}} + a^\dagger_k\, e^{-i\vec{k}\cdot\vec{x}}\right)\;\text{---\,(1)} \qquad\text{}\hookleftarrow\text{ (replaced }\vec{k}\to-\vec{k}\text{ since integral remains same.)} \]

We also had \(p_k(t)\) as canonical momentum (from eq (a)\(-\)(b))---

\[ p_k = \frac{-i}{\sqrt{2\omega_k}}\left(a_k(t) - a^\dagger_{-k}(t)\right) \]

So,

\[ \begin{aligned} \pi(t,\vec{x}) &= \int\frac{d^3k}{(2\pi)^3}\; p_k\; e^{i\vec{k}\cdot\vec{x}}\\ &= \int\frac{d^3k}{(2\pi)^3}\,\frac{(-i)}{\sqrt{2\omega_k}}\left(a_k - a^\dagger_{-k}\right)e^{i\vec{k}\cdot\vec{x}}\\ \underset{\substack{\downarrow\\ \text{canonical conjugate}\\ \text{field momentum to }\varphi(t,\vec{x})}}{\pi(t,\vec{x})} &= (-i)\int\frac{d^3k}{(2\pi)^3}\left(a_k\, e^{i\vec{k}\cdot\vec{x}} - a^\dagger_k\, e^{-i\vec{k}\cdot\vec{x}}\right)\;\text{---\,(2)} \qquad\text{}\hookleftarrow\text{ again we flipped }\vec{k}\to-\vec{k}\text{ since integral remains unchanged.} \end{aligned} \]

Lets find

\[ \boxed{\;\left[\varphi(t,x),\,\pi(t,x')\right] = i\hbar\,\delta^3(\vec{x}-\vec{x}')\;}\qquad\text{(Equal time commutation relation)} \]

for \(\hbar = 1\):

\[ \underset{\text{quantised}}{\boxed{\;\left[\varphi(t,x),\,\pi(t,\vec{x}')\right] = i\,\delta^3(x-x')\;}} \qquad\text{ --- This relation is sufficient for telling }\varphi,\pi\text{ are quantised.} \]

But to see energy spectrum, we need equation (1) & (2).

Comment about Normalisation ---
Let us normalise

\[ \begin{gathered} a^\dagger_k|0\rangle = c_k|k\rangle\\ \langle 0|a_k = \langle k|c_k^{*}\\ \langle 0|a_k\, a^\dagger_k|0\rangle = \langle k|c_k^{*} c_k|k\rangle \end{gathered} \]

For

\[ \langle 0|a_{\vec{k}}\, a^\dagger_{\vec{k}'}|0\rangle = (2\pi)^3\,\delta^3(\vec{k}-\vec{k}') \qquad \begin{aligned} &\hookleftarrow\text{ not Lorentz invariant!}\\ &\left\{\begin{aligned} &\text{(as }\vec{k}\text{ is only appearing with sides. }\vec{k}\text{ should mix with }k_0\text{}\\ &\text{to show some Lorentz Invariance.)} \end{aligned}\right. \end{aligned} \]

We change definition, so that we have Lorentz invariant quantity.
Suppose we made boost along \(x^1\longrightarrow\)

\[ \text{Then}\quad \begin{pmatrix} k_0' \\ k_1' \end{pmatrix} = \Lambda\begin{pmatrix} k_0 \\ k_1 \end{pmatrix} \;;\qquad \Lambda = \begin{bmatrix} \cosh\alpha & \sinh\alpha \\ \sinh\alpha & \cosh\alpha \end{bmatrix} \]

Since \(\delta^3(\vec{k}-\vec{l}) = \delta(k_1-l_1)\,\delta(k_2-l_2)\,\delta(k_3-l_3)\) is not invariant. \(\left(\text{i.e. } \delta^3(\vec{k}'-\vec{l}') \neq \delta^3(\vec{k}-\vec{l})\right)\)

\[ \begin{gathered} k_1' - l_1' = \sinh\alpha\,(k_0-l_0) + \cosh\alpha\,(k_1-l_1)\\ \text{we can see,}\quad \delta(k_1'-l_1') \neq \delta(k_1-l_1) \;\Rightarrow\; \delta^3(\vec{k}-\vec{l}) \text{ is not Lorentz Invariant.} \end{gathered} \]

We rather have a beautiful relation ---

\[ \begin{gathered} \delta\left(k_1'-l_1'\right) = \delta\left(\sinh\alpha\,(k_0-l_0) + \cosh\alpha\,(k_1-l_1)\right)\\ \delta\left(k_1'-l_1'\right) = \frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}\;\delta(k_1-l_1) \;\text{---\,(7)}\\ \Rightarrow\quad \delta(k_1'-l_1') \neq \delta(k_1-l_1) \;\text{ because of the factor } \frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}} \text{ being not equal to one.} \end{gathered} \]

Therefore this tells us ---

\[ \boxed{\;\omega_{\vec{k}}\;\delta(\vec{k}-\vec{l})\;\text{ has to be Lorentz invariant}\;} \]

Using eq (7):

\[ \begin{aligned} \omega_{\vec{k}'}\,\delta(\vec{k}'-\vec{l}') &= \cancel{\omega_{\vec{k}'}}\left(\frac{\omega_{\vec{k}}}{\cancel{\omega_{\vec{k}'}}}\,\delta(\vec{k}-\vec{l})\right)\\ &= \omega_{\vec{k}}\,\delta(\vec{k}-\vec{l})\\ \Rightarrow\quad &$\omega_{\vec{k}}\;\delta(\vec{k}-\vec{l})$ is Lorentz invariant! \end{aligned} \]

Therefore we choose the normalisation such that,

\[ \langle\vec{k}|\vec{k}'\rangle = (2\pi)^3\,\underset{\substack{\uparrow\\ \text{depends on us, we could}\\ \text{have just taken }\omega_k}}{2\omega_{\vec{k}}}\;\delta^3(\vec{k}-\vec{k}') \]

This is true if

\[ \boxed{\;|\vec{k}\rangle = \sqrt{2\omega_k}\; a^\dagger_{\vec{k}}\,|0\rangle\;} \]

Note, we have written \(\varphi(t,x)\) as mode expansion which treats \(t,\vec{x}\) on different footings:

\[ \varphi(t,x) = \int\frac{d^3k}{(2\pi)^3}\; q_k(t)\; e^{i\vec{k}\cdot\vec{x}} \]

From QM (heisenberg picture) we can write,

\[ \begin{gathered} \varphi(t,x) = e^{iHt}\,\varphi(0,x)\,e^{-iHt} \;\text{---\,(5)}\\ \text{and}\quad \left. \begin{aligned} a_k(t) &= e^{iHt}\, a_k(0)\, e^{-iHt}\\ a^\dagger_k(t) &= e^{iHt}\, a^\dagger_k(0)\, e^{-iHt} \end{aligned} \right\}\;\text{---\,(6)} \end{gathered} \]

To further solve eq (6), we require \([H, a^\dagger_k]\) & \([H, a_k]\), where

\[ H = \int\frac{d^3k}{(2\pi)^3}\;\omega_k\; a^\dagger_k\, a_k \qquad\left|\quad \begin{aligned} \text{use:}\;\; [A,BC] &= [A,B]C + B[A,C]\\ [AB,C] &= A[B,C] + [A,C]B\\ \text{and}\;\; [a_k\, a^\dagger_{k'}] &= (2\pi)^3\,\delta(k-k') \end{aligned} \right. \]

So,

\[ \begin{aligned} [H,\, a_k] &= -\omega_k\, a_k\\ [H,\, a^\dagger_k] &= +\omega_k\, a^\dagger_k \end{aligned} \]

Eqn (6) becomes,

\[ \begin{aligned} a_k(t) &= e^{iHt}\, a_k(0)\, e^{-iHt} = e^{-i\omega_k t}\, a_k\\ a^\dagger_k(t) &= e^{iHt}\, a^\dagger_k(0)\, e^{-iHt} = e^{i\omega_k t}\, a^\dagger_k \end{aligned} \]

Using

\[ \varphi(0,x) = \int\frac{d^3k}{(2\pi)^3}\,\frac{1}{\sqrt{2\omega_k}}\left(a_k\, e^{i\vec{k}\cdot\vec{x}} + a^\dagger_k\, e^{-i\vec{k}\cdot\vec{x}}\right) \] \[ \begin{aligned} \varphi(t,x) &= e^{iHt}\,\varphi(0,\vec{x})\,e^{-iHt}\\ &= \int\frac{d^3k}{(2\pi)^3}\,\frac{1}{\sqrt{2\omega_k}}\left(e^{-i\omega_k t}\, a_k\, e^{i\vec{k}\cdot\vec{x}} + e^{i\omega_k t}\, a^\dagger_k\, e^{-i\vec{k}\cdot\vec{x}}\right)\\ &\qquad\qquad\text{ }x^\mu = (t,\vec{x})\text{, \quad }k^\mu = (k^0,\vec{k})\text{, \quad }k_\mu = (k_0,-\vec{k})\text{}\\ &= \int\frac{d^3k}{(2\pi)^3}\,\frac{1}{\sqrt{2\omega_k}}\left(a_k\, e^{-i(k_\mu x^\mu)} + a^\dagger_k\, e^{i\,k_\mu x^\mu}\right) \end{aligned} \] \[ \boxed{\;\varphi(t,x) = \int\frac{d^3k}{(2\pi)^3}\,\frac{1}{\sqrt{2\omega_k}}\left(a_k\, e^{-ik\cdot x} + a^\dagger_k\, e^{ik\cdot x}\right)\;} \qquad\text{ Here }a_k\text{, }a^\dagger_k\text{ are time independent!} \]

The important part is \(e^{-ik\cdot x}\) & \(e^{ik\cdot x}\) as these are Lorentz Invariant.
We can regard \(e^{-ik\cdot x}\) as +ve frequency mode as we get \((+\omega_k)\) for \(i\frac{\partial}{\partial t}\left(e^{-ik\cdot x}\right)\) & we call \(e^{+ik\cdot x}\) as \(-\)ve frequency mode

\[ \text{Since}\quad i\frac{\partial}{\partial t}\left(e^{ik\cdot x}\right) = i\frac{\partial}{\partial t}\, e^{i(k^0x^0 - \vec{k}\cdot\vec{x})}, \qquad i(ik^0) = -k^0 = -\omega_k. \]

It does not mean that these corresponds to +ve & \(-\)ve energies, instead of that annihilation operator \(a_k\) multiplies +ve frequency mode and creation operator \(a^\dagger_k\) multiplies \(-\)ve frequency mode.
So, the field operator either creates state with one less particle or creates state with one more particle.

\[ \text{Field}\;\longrightarrow\;\text{change no of particles} \]

In same space it is done:

\[ \begin{aligned} \varphi(t,\vec{x}) &= \int\frac{d^3k}{(2\pi)^3\sqrt{2\omega_k}}\left(b_k\, e^{-i\vec{k}\cdot\vec{x}} + a^\dagger_k\, e^{i\vec{k}\cdot\vec{x}}\right)\\ \varphi^{*}(t,\vec{x}) &= \int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\left(a_k\, e^{-i\vec{k}\cdot\vec{x}} + b^\dagger_k\, e^{i\vec{k}\cdot\vec{x}}\right) \end{aligned} \]

Now we have here this sets of oscillators \((a, a^\dagger)\) and \((b, b^\dagger)\) with equal time commutation relations ---

\[ \begin{gathered} [a_k,\, a^\dagger_{k'}] = (2\pi)^3\,\delta^3(\vec{k}-\vec{k}')\\ [b_k,\, b^\dagger_{k'}] = (2\pi)^3\,\delta^3(k-k')\\ [a_k,\, b_k] = [a_k,\, b^\dagger_k] = [a^\dagger_k,\, b^\dagger_k] = 0 \end{gathered} \]

We will see that charge can be associated with complex fields \(\varphi\) & \(\varphi^{*}\).

\[ \text{Say}\quad \begin{cases} \varphi \longrightarrow +1 \;\;\text{charge}\\ \varphi^{*} \longrightarrow -1 \;\;\text{charge} \end{cases} \]

Then we can see that, if \(a^\dagger_k\) creates particle of charge \(+1\), and \(b_k\) destroys particle of charge \(-1\), then in both cases there is increment of charge by \(+1\) (i.e. both terms create charge that adds charge of \(+1\)).
Similarly, \(\varphi^{*}\) induces charge change of \(-1\).

\[ \left. \begin{aligned} \text{Action of } \varphi \text{ induces}\quad &\Delta q = +1\\ \varphi^{*} \qquad\qquad &\Delta q = -1 \end{aligned} \right\}\; \begin{aligned} &\text{if } a^\dagger_k \text{ produces particle \& } b^\dagger_k\\ &\text{produces anti-particle} \end{aligned} \]

If we say \(a, a^\dagger\) are associated with particles of some charge, then \(b\) & \(b^\dagger\) will be associated with antiparticles of opposite charge. It turns out everything is same except for charge.
(Antiparticles have same spin, mass, other quantum nos as of particles.) They don't travel backward in time! --- have negative mass, spin, energy...
In physics we call particles to those which are more in number, if antiparticles were more than we might call them by term ''particles''.

Propagators --- (Real fields)

Purpose ---
(1) Particles can propagate from one time to future time.
(2) It tests our ability to compute expectation values.

(A) Vacuum expectation values are not that interesting:

\[ \begin{aligned} \langle 0|\varphi|0\rangle &= \langle 0|\varphi(x,t)|0\rangle\\ &= \Big\langle 0\Big|\int\frac{d^3k}{(2\pi)^3\sqrt{2\omega_k}}\left(a_k\, e^{-ik\cdot x} + a^\dagger_k\, e^{ik\cdot x}\right)\Big|0\Big\rangle\\ &= \int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\Big(\underbrace{\langle 0|a_k\, e^{-ik\cdot x}|0\rangle}_{\to\,0} + \underbrace{\langle 0|a^\dagger_k\, e^{ik\cdot x}|0\rangle}_{\to\,0}\Big) \end{aligned} \] \[ \boxed{\;\langle 0|\varphi(t,x)|0\rangle = 0\;}\qquad\text{(In free field theory)} \]

\(\langle 0|\varphi|0\rangle\) could be interesting in interacting theory.

(B) \(\quad\langle 0|\varphi(x)\,\varphi(y)|0\rangle\) --- 2 point function.

\[ \langle 0|\varphi(x^0,\vec{x})\;\varphi(y^0,\vec{y})|0\rangle \]

Note that \(\varphi(y^0,\vec{y})|0\rangle\) creates particle at \(y=(y^0,\vec{y})\);
\(\langle 0|\varphi(x^0,\vec{x})\) annihilates particle at \(x=(x^0,\vec{x})\).
i.e. \(\langle 0|\varphi(x)\varphi(y)|0\rangle\) gives us probability amplitude for particle to be created at \(Y\) and destroyed at \(X\) (here \(x^0 > y^0\)). That is why we call it a propagator. Something (particle) pops out of vacuum at \(y\,(y^0,\vec{y})\), propagates to \(\vec{x}\) and gets destroyed at time \(x^0\).
For \(y^0 > x^0\) it would be more sensible/natural to calculate \(\langle 0|\varphi(y)\,\varphi(x)|0\rangle\), which describe amplitude/probability amplitude for particle to be created at \(\vec{x}\) at time \(x^0\) and gets destroyed at time \(y^0\) at position \(\vec{y}\).

\[ \begin{aligned} \langle 0|&\varphi(x)\varphi(y)|0\rangle\\ &= \int\frac{d^3k}{(2\pi)^3}\frac{1}{\sqrt{2\omega_k}}\Big(\underbrace{\langle 0|a_k\, a_{k'}|0\rangle}_{\to 0}\, e^{-ikx}e^{-ik'y} + \langle 0|a_k\, a^\dagger_{k'}|0\rangle\, e^{-ikx}e^{ik'y}\\ &\qquad\qquad + \underbrace{\langle 0|a^\dagger_k\, a_{k'}|0\rangle}_{\to 0}\, e^{ikx}e^{-ik'y} + \underbrace{\langle 0|a^\dagger_k\, a^\dagger_{k'}|0\rangle}_{\to 0}\, e^{ikx}e^{ik'y}\Big)\\[1ex] &= \int\frac{d^3k}{(2\pi)^3}\int\frac{d^3k'}{(2\pi)^3}\,\frac{1}{2\omega_k}\,\langle 0|a_k\, a^\dagger_{k'}|0\rangle\; e^{-ik\cdot x}\, e^{+ik'\cdot y}\\ &\qquad\qquad\qquad\qquad \begin{aligned} a^\dagger_{k'}|0\rangle &= \frac{1}{\sqrt{2\omega_{k'}}}\,|k'\rangle\\ \langle 0|a_k &= \frac{1}{\sqrt{2\omega_k}}\,\langle k| \end{aligned}\\[1ex] &= \int\frac{d^3k}{(2\pi)^3}\,\frac{d^3k'}{(2\pi)^3}\,\frac{1}{2\omega_k}\,\frac{1}{\left(2\sqrt{\omega_k\,\omega_{k'}}\right)}\,\underline{\langle k|k'\rangle}\; e^{-ik\cdot x}\, e^{ik'\cdot y}\\ &= \int\frac{d^3k}{(2\pi)^3}\,\frac{d^3k'}{(2\pi)^3}\,\frac{1}{2\omega_k}\,\frac{1}{\sqrt{\omega_k\,\omega_{k'}}}\,(2\pi)^3\,\delta^3(k-k')\,\cancel{2\omega_k}\; e^{-ik\cdot x}\, e^{ik'\cdot y}\\ &= \int\frac{d^3k}{(2\pi)^3}\,\frac{1}{2\omega_k}\; e^{ik\cdot(y-x)} \qquad\qquad\left(\text{different }x,y\right) \end{aligned} \]

This motivates us to define

\[ \begin{aligned} D_0(x,y) &= \theta(x^0-y^0)\,\langle 0|\varphi(x)\,\varphi(y)|0\rangle\\ &\qquad + \theta(y^0-x^0)\,\langle 0|\varphi(y)\,\varphi(x)|0\rangle\\ &{\;= \langle 0|\,T\{\varphi(x)\,\varphi(y)\}\,|0\rangle} \end{aligned} \]

This object shall be studied for particle propagation rather than just \(\langle 0|\varphi(x)\varphi(y)|0\rangle\) or \(\langle 0|\varphi(y)\varphi(x)|0\rangle\),
where

\[ \theta(x-y) = \begin{cases} 1 & x>y\\ 0 & x<y \end{cases} \]

Here \(T\{\varphi(x)\,\varphi(y)\} = T\{\varphi(y)\,\varphi(x)\}\) is called time ordered product.
It turns out that \(D_0(x,y)\) (for these variables only) depends upon just the difference '\(x-y\)'.

\[ \text{i.e.}\qquad D_0(x,y) = D_0(x-y) = D_0(x^\mu - y^\mu) \]

Because universe has geometry of translational invariance, we can translate whole system to another place. Therefore \(D_0(x,y)\) shall only depend upon relative position \(x^\mu - y^\mu\) or \(x-y\).

\[ D_0(x-y) = \int\frac{d^4k}{(2\pi)^4}\;\frac{i}{k^2-m^2}\; e^{-i\,k\cdot(x-y)} \qquad \begin{aligned} &\downarrow\\ &\text{pole at } (k^0)^2 - \vec{k}^2 - m^2 = 0\\ &k^0 = \pm\sqrt{\vec{k}^2+m^2}\\ &k_0 = k^0 = \pm\,\omega_{\vec{k}} \end{aligned} \]

We'll see how to go around the pole and calculable DF:

\begin{tikzpicture}[>=stealth, scale=1.1]
\draw[->] (-3.5,0) -- (3.5,0) node[right] {$Re(k_0)$};
\draw[->] (0,-1.6) -- (0,2.2) node[right] {$Im(k_0)$};
\node at (-1.5,0) {$\times$};
\node[above] at (-1.5,0.1) {$-\omega_k$};
\node at (1.5,0) {$\times$};
\node[above] at (1.5,0.1) {$+\omega_k$};
\draw[thick,->] (-3.2,-0.12) -- (-1.9,-0.12);
\draw[thick] (-1.9,-0.12) arc (180:360:0.4 and 0.28);
\draw[thick] (-1.1,-0.12) -- (1.1,-0.12);
\draw[thick] (1.1,-0.12) arc (180:0:0.4 and 0.28);
\draw[thick,->] (1.9,-0.12) -- (3.2,-0.12);
\node[below, align=center] at (-1.7,-0.6) {$\uparrow$\\contour for\\Feynman propagator};
\node[right, align=left, text width=4.2cm, color=blue] at (3.6,1.2) {\footnotesize To get $\pm$ve energy waves we need to go pole below \& above pole respectively};
\end{tikzpicture}

How do we calculate contour integral --- by closing contour and making(?) the arc at infinity does not contribute, then,

\[ \int_{C_F\;\text{open}} = \oint_{\text{closed}} \qquad\text{(so that we can use residue theorem)} \] \[ D_F(x-y) = \lim_{\epsilon\to 0}\int\frac{d^4k}{(2\pi)^4}\;\frac{i}{k^2-m^2+i\epsilon}\; \underset{\substack{\downarrow\\ e^{-ik^0(x^0-y^0)}\, e^{-i\vec{k}\cdot(\vec{x}-\vec{y})}\\ k^0 = -iR\;\text{(LHP)}}}{e^{-i\vec{k}\cdot(x-y)}} \]

We can close the contour in lower half plane (LHP) so that \(k^0\) takes \(-\)ve imaginary values \(\left(k^0 = Re^{i\theta}\right)\):
the factor \(e^{-i(k^0)(x^0-y^0)}\) converges to '0' as \(R\to\infty\). For \((x^0-y^0 > 0)\)

\[ \text{i.e.}\quad$C_F$ can be closed in LHP if $x^0>y^0$; \quad ditto UHP if $x^0<y^0$. \]

For \(x^0 > y^0\) ---

\[ \begin{aligned} \frac{i}{k^2-m^2} = \frac{i}{k_0^2 - \vec{k}^2 - m^2} = \frac{i}{k_0^2 - \omega_k^2} &= \frac{i}{(k_0-\omega_k)(k_0+\omega_k)}\\ &= \left(\frac{i}{k_0-\omega_k} - \frac{i}{k_0+\omega_k}\right)\frac{1}{2\omega_{\vec{k}}} \qquad \begin{aligned} &\hookleftarrow\text{Not useful pole}\\ &\text{since we are closing}\\ &\text{in LHP.} \end{aligned} \end{aligned} \] 
\begin{tikzpicture}[>=stealth, scale=1.0]
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\draw (0,-2.2) -- (0,0.8);
\node at (-1.4,0) {$\times$};
\node[above] at (-1.4,0.15) {$-\omega_k$};
\node at (1.4,0) {$\times$};
\node[above] at (1.4,0.15) {$+\omega_k$};
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\[ \begin{aligned} D_F(x-y) &= \int\frac{d^4k}{(2\pi)^4}\;\frac{i}{k^2-m^2}\; e^{-i\vec{k}\cdot(x-y)}\\ &= \int\frac{dk_0}{2\pi}\int\frac{d^3k}{(2\pi)^3}\left(\frac{i}{k_0-\omega_k} - \frac{i}{k_0+\omega_k}\right)\frac{1}{2\omega_k}\; e^{-ik_0(x^0-y^0)\, +\, i\vec{k}\cdot(\vec{x}-\vec{y})}\\ &\qquad\text{using}\quad\left\{k_\mu(x-y)^\mu = \vec{k}\cdot(x-y)\right\} = k^0(x^0-y^0) - \vec{k}\cdot(\vec{x}-\vec{y})\\ &= \int\underbrace{\frac{\left(-2\pi i\;\mathrm{Res}\,(k_0=\omega_k)\right)}{(2\pi)}\frac{1}{2\omega_k}}_{\text{clockwise (LHP)}}\;\frac{d^3k}{(2\pi)^3}\; e^{-ik(x-y)}\;\Bigg|_{k^0=\omega_k}\\ &= \int\frac{-2\pi i\,(i)}{2\pi\; 2\omega_k}\;\frac{d^3k}{(2\pi)^3}\; e^{-ik(x-y)}\Bigg|_{k^0=\omega_k} \end{aligned} \] \[ D_F(x-y) = \int\frac{d^3k}{(2\pi)^3}\;\frac{1}{2\omega_k}\; e^{-ik(x-y)}\Bigg|_{k^0=\omega_k} \qquad\left(\text{For } x^0>y^0\right) \]

For \(x^0<y^0\) we close the contour in UHP.

\begin{tikzpicture}[>=stealth, scale=1.0]
\draw (-3,0) -- (3,0);
\draw (0,-0.8) -- (0,2.2);
\node at (-1.4,0) {$\times$};
\node[below] at (-1.4,-0.15) {$-\omega_k$};
\node at (1.4,0) {$\times$};
\node[below] at (1.4,-0.15) {$+\omega_k$};
\draw[thick] (-2.8,0.1) -- (-1.75,0.1);
\draw[thick] (-1.75,0.1) arc (180:90:0.35 and 0.22) arc (90:0:0.35 and 0.22);
\draw[thick] (-1.05,0.1) -- (1.05,0.1);
\draw[thick] (1.05,0.1) arc (-180:-90:0.35 and 0.22) arc (-90:0:0.35 and 0.22);
\draw[thick] (1.75,0.1) -- (2.8,0.1);
\draw[thick,<-] (2.8,0.1) arc (0:90:2.8 and 2.0);
\draw[thick] (0,2.1) arc (90:180:2.8 and 2.0);
\end{tikzpicture}
\[ \begin{aligned} D_F(x-y) &= \int\frac{d^4k}{(2\pi)^4}\;\frac{i}{k^2-m^2}\; e^{-i\vec{k}\cdot(x-y)}\\ &= \int\frac{d^3k}{(2\pi)^3}\int\frac{dk^0}{2\pi}\Bigg(\underbrace{\frac{i}{k_0-\omega_k}}_{\text{Not included}} - \frac{i}{k_0+\omega_k}\Bigg)\frac{1}{2\omega_{\vec{k}}}\; e^{-ik^0(x^0-y^0)\,+\,i\vec{k}\cdot(\vec{x}-\vec{y})}\\ &= \int\frac{2\pi i\,(-i)}{(2\pi)\,2\omega_k}\;\frac{d^3k}{(2\pi)^3}\; e^{-i\,k\cdot(x-y)}\;\Bigg|_{k_0=-\omega_k} \end{aligned} \] \[ D_F(x-y) = \int\frac{d^3k}{(2\pi)^3}\;\frac{1}{2\omega_k}\; e^{-i\,k\cdot(x-y)}\Bigg|_{k^0=-\omega_k} \qquad\left(x^0<y^0\right) \]

The reason why feynman propagator is important is that, it is basic ingredient to understand interactions.
Suppose, a theory is weakly interacting, then on average particles with interact with each other once in a while, the more time they interact with each other, lower will be the probability.
Particle interact then propagates freely ... and then they interact and again interact freely.

Process of Interaction \(\approx\) Free propagation + point interactions once in a while.

In between any two interactions particle propagate freely and that is job of feynman propagator (\(D_F(x-y)\) to find amplitude for free theory propagation).

lecture-4-6-sunil-mukhi-qft-1