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

\[ \begin{align*} {x}_{\mu}{x}^{\mu}=t^{2}-x^{2}-y^{2}-z^{2}=t^{2}-(x^{2}+y^{2}+z^{2}) \end{align*} \]

we can have \(x^{2}\) invariant also by keeping 't' fixed and having \((x^{2}+y^{2}+z^{2})\) fixed by set of rotations. So rotations area included in Lorentz transformation.
for y=z=0, we can write

\[ {x}_{\mu}{x}^{\mu}=t^{2}-x^{2}=(t-x)(t+x)=x_{+}x_{-} \]

Define \(x_{\pm}=t\pm x\) , implies that :

\[ \begin{align*} t &=\frac{x_{+}+x_{-}}{2} \\ x &=\frac{x_{+}-x_{-}}{2} \end{align*} \]

Here \(x_{+}\) and \(x_{-}\) are independent coordinates. Let us ask under what transformation on \(x_{+}\) and \(x_{-}\) such that product \(x_{+}x_{-}={x}_{\mu}{x}^{\mu}\) remains invariant.
We can use the below transformation:

\[ x_{\pm}\longrightarrow e^{ \pm \alpha }x_{\pm} \]

by that we mean that :

\[ \begin{align*} x_{+}\longrightarrow e^{ \alpha } x_{+} \\ x_{-}\longrightarrow e^{- \alpha } x_{-} \\ \end{align*} \]

this implies that

\[ \begin{align*} t\longrightarrow \frac{\left( e^{ \alpha }x_{+} +e^{ -\alpha } x_{-} \right)}{2} \\ t\longrightarrow \frac{\left( e^{ \alpha }(t+x)+e^{ -\alpha }(t-x) \right)}{2}\\ t\longrightarrow \frac{\left( e^{ \alpha }+e^{ -\alpha } \right)}{2}t+\frac{\left( e^{ \alpha }-e^{ -\alpha } \right)}{2}x\\ t\longrightarrow (\cosh \alpha)t+(\sinh \alpha )x \end{align*} \]

So, under Lorentz transformation

\[ \begin{align*} t'=(\cosh \alpha)t+(\sinh \alpha)x \\ x'=(\sinh \alpha)t+(cosh\alpha)x \end{align*} \]

or using matrices we can write that ,

\[ \begin{pmatrix} t' \\ x' \end{pmatrix} =\begin{pmatrix} \cosh \alpha & \sinh \alpha \\ \sinh \alpha & \cosh \alpha \end{pmatrix} \begin{pmatrix} t \\ x \end{pmatrix} \]

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 and 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 we can make any amount of boost. we had

\[ \begin{align*} t' &=\frac{\left( t-\frac{vx}{c^{2}} \right)}{\sqrt{ 1-\frac{v^{2}}{c^{2}} }}=\frac{t-vx}{\sqrt{ 1-v^{2} }} \\ x' &=\frac{x-vt}{\sqrt{ 1-\frac{v^{2}}{c^{2}} }}=\frac{x-vt}{\sqrt{ 1-v^{2} }} \end{align*} \]

above is calculated in natural units where \(c=1 , \hbar=1\)
The transformation matrix then becomes,

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

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 vecctor \(A_{\mu}\) with just \((A_{t},A_{x})\) then the quantity \(A_{-}A_{+}\) is Lorentz invariant.

The field equation for the Klein Gordon field is :

\[ \begin{gather} (\partial^{2}+m^{2})\psi=0 \end{gather} \]

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

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

we do change of variables here :

\[ \begin{align*} \hat{a} &=\hat{q}+i\hat{p}\\ a^{\dagger} &=\hat{q}-i\hat{p} \end{align*} \]

we need to construct quantities of dimension length and momentum from m, \(\omega,\hbar\) . and divide these \(\hat{q}\) and \(\hat{p}\) with corresponding same dimensions to make the operators/quantities (\(a^{\dagger} , a\)) dimensionless.

\[ \begin{align*} q', p' & \propto m^{\alpha}\omega^{\beta}\hbar^{\gamma}\\ & \propto[M]^{\alpha}[T]^{-\beta}[ML^{2}T^{-1}]^{\gamma }\\ & \propto [M]^{\alpha+\gamma}[L]^{2\gamma}[T]^{-\beta-\gamma} Since 'q' has dimensions of length, for that we need \end{align*} \] \[ \begin{align*} \alpha+\gamma &=0 \\ 2 \gamma &=1 \\ -\beta-\gamma &=0 \end{align*} \]

The above equations implies that:

\[ \begin{align*} \gamma &=\frac{1}{2} , & \beta &=-\frac{1}{2}, & \alpha &=-\frac{1}{2}\\ \implies q' &=\sqrt{ \frac{\hbar}{m \omega} } \end{align*} \]

For case of momentum we need ,

\[ \begin{align*} M L T^{-1} \propto [M]^{\alpha+\gamma}[L]^{2\gamma}[T]^{-\beta-\gamma}\\ \\ \implies \alpha+\gamma &=1,& 2\gamma &=1, & -\beta-\gamma &=-1 \\ \implies \gamma &=\frac{1}{2}, & \alpha &=\frac{1}{2}, & \beta &=\frac{1}{2}\\ p' \propto \sqrt{ m \omega \hbar } \end{align*} \]

This provides us the correct forms of \(\hat{a}, \hat{a}^{\dagger}\) .

\[ \begin{align*} \hat{a} &=\frac{\hat{q}}{\hat{q}'}+i \frac{\hat{p}}{\hat{p}'} \\ \hat{a^{\dagger}} &=\frac{\hat{q}}{\hat{q}'}-i \frac{\hat{p}}{\hat{p}'} \\ \implies \hat{a} &=\frac{\hat{q}}{\sqrt{ \frac{2\hbar}{m \omega } }}+i \frac{\hat{p}}{\sqrt{2 m \omega \hbar }} ,& \hat{a}^{\dagger} &= \frac{\hat{q}}{\sqrt{ \frac{2\hbar}{m\omega } }}-i \frac{\hat{p}}{\sqrt{2 m \omega \hbar }} \end{align*} \]

Total Hamiltonian becomes:

\[ H=\frac{p^{2}}{2m} + \frac{1}{2} m w^{2} q^{2} \]

Simplifying we get:

\[ H=\hbar \omega \left( \hat{a}^{\dagger}\hat{a}+\frac{1}{2} \right) \]

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

The reason we had gone throught this exercise is that because the Klein Gordon equation gives us collection of simple harmonic oscillators, all decoupled from each other, so we can quantise teach of those quantum harmonic oscillators and therefore in such way we have quantized the field \(\phi(x)\), remember in QM, we put 't' and 'x ' on different footings but in QFT, we put space and time on same footing.

So object we are insterested in is \(\phi(t, \vec{x})\) , for the moment we will treat time and space on different footing but we will see that the results are relativistically invariant.

let us bound \(\vec{x}\) to size of cube, later we will take size \(L\to \infty\) , then we can accomodate a system as large as we like to use. we need boundary conditions or periodic boundary conditions. i.e.

\[ \begin{align*} \phi(t,x+L,y,z) &=\phi(t,x,y,z)\\ \phi(t,x,y+L,z) &=\phi(t,x,y,z)\\ \phi(t,x,y,z+L) &=\phi(t,x,y,z) \end{align*} \]

Till now we are treating \(\phi\) as some function rather than an operator, we can write \(\phi\) as some vector whose basis set are some exponential function (fourier series).
Let us take \(e^{ i\vec{k}.\vec{x} }\) as some general function, with periodic boundary conditions.

using \(\phi(x,t)=e^{ i\vec{k}.\vec{x} }\) in equation (2) we get ,

\[ \begin{align*} e^{ i\vec{k}.(\vec{x}+L) }=e^{ i\vec{k}.\vec{x} }\\ e^{ i(k_{x}x+k_{y}y+k_{z}z) }=e^{ i(k_{x}(x+L)+k_{y}y+k_{z}z) } \\ e^{ ik_{x}x }=e^{ ik_{x}x }e^{ ik_{x}L }\\ e^{ ik_{x}L }=\cos(k_{x}L)+i\sin(k_{x}L)=1\\ \implies k_{x}L=2\pi m\\ \implies k_{x}=\frac{2\pi m}{L} \\ \text{or, we can say that: }\\ \vec{k}=\frac{2\pi}{L}\left( m_{1},m_{2},m_{3} \right) \end{align*} \]

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

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

where N is some normalisation factor, here the time dependence has to be absorbed in coefficients \(\left( q_{k}(t) \right)\). note that the coefficients are \(\vec{k}\) dependent, so it is separate for each \(\vec{k}\).
we can use this expression and plug it into \(\left( \partial^{2}+m^{2} \right)\phi=0\) , and ask values of \(q_{k}(t)\).

\[ \begin{align*} \implies \left( \partial^{2}+m^{2} \right)\sum_{\vec{k}}q_{\vec{k}}(t)e^{ i\vec{k}.\vec{x} } &=0 \longrightarrow(3)\\ \partial^{2} &={\partial}^{\mu}{\partial}_{\mu}=\left( \frac{\partial}{\partial t},\nabla \right)\\ \begin{pmatrix} \frac{\partial}{\partial t}\\ \\ -\nabla \end{pmatrix} &=\frac{\partial^{2}}{\partial t^{2}}-\nabla^{2}\\ \\ (i) \longrightarrow \partial^{2}q_{k}(t) &=\frac{\partial^{2}}{\partial t^{2}}q_{k}(t)-\nabla^{2}q_{k}(t)\\ (ii) \longrightarrow\partial^{2}e^{ i\vec{k}.\vec{x} } &=(\vec{k}^{2})e^{ i\vec{k}.\vec{x} } \\ \\ \\ \text{Using above in equation (3) we get} \sum_{\vec{k}}\left( \partial^{2}q_{k}(t)e^{ i\vec{k}.\vec{x} }+q_{k}(t)\partial^{2}e^{ i\vec{k}.\vec{x} }+m^{2}q_{k}(t)e^{ i\vec{k}.\vec{x} } \right) &=0\\ \sum_{\vec{k}} \left( \ddot{q}_{k}(t)+(k^{2}+m^{2})q_{k}(t) \right)e^{ i\vec{k}.\vec{x} } =0 \end{align*} \]