Monopoles - 10 - Dirac Monopoles in Quantum Mechanics - Part - 6

Leaving the superscript indices, we substitute for $$L = 2^{-\frac{S+M}{2}} z^{S/2} (2-z)^{M/2} V$$ Doing the necessary differentiation and putting it in our primary differential equation we get, $$ 2^{-(S+M)/2}z^{S/2} (2-z)^{M/2}\left[(2z-z^2)\\\left(V''+ \frac{S}{2}\frac{S-2}{2z^2}V+\frac{M}{2}\frac{M-2}{2(2-z)^2}V-\frac{SM}{2}\frac{1}{z(2-z)}V+\frac{S}{z}V'-\frac{M}{(2-z)}V'\right)\\+ 2(1-z)\left(\frac{S}{2z}V-\frac{M}{2(2-z)}V+V'\right)+\left(\lambda{V} - \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{2z-z^2}V\right)\right]= 0 $$

Rewriting as (by cancelling the common term assuming it is not equal to zero), $$ (2z-z^2)V'' + V' \left[2(1+S)-z(2+S+M)\right]+ [arithmetic \,\,simplification]V = 0 $$

Arithmetic Simplification worked out separately as,

$$ V\left[\frac{S(S-2)}{4}\frac{2-z}{z} +\frac{M(M-2)}{4}\frac{z}{2-z}-\frac{SM}{2}\\+\frac{S(1-z)}{z} - \frac{M(1-z)}{2-z}+\lambda-\frac{\left(m+\frac{n}{2}(2-z)\right)^2}{2z-z^2}\right]V$$

which becomes (bracket and V is not important), $$ \lambda - \frac{SM}{2} + \\ \left[\frac{S(S-2)(2-z)^2+M(M-2)z^2+4S(1-z)(2-z)-4Mz(1-z)\\- 4m^2 - n^2(2-z)^2-4mn(2-z)}{4(2z-z^2)}\right]$$

with common terms of numerator and denominator,

$$\lambda-\frac{SM}{2}+\\\left[\frac{(2-z)\left(2S^2-4S-S^2z+2Sz+4S-4Sz)\right)+\\z\left(M^2z-2Mz-4M+4Mz\right)-4m^2-n^2(2-z)^2-4mn(2-z)}{4(2z-z^2)}\right] $$

it gives, $$\lambda-\frac{SM}{2}+\\\left[\frac{(2-z)2z(-S)+(2-z)(2S^2-S^2z)+2z(2-z)(-M)+\\z^2M^2-\left(4m^2+n^2(2-z)^2+4mn(2-z)\right)}{4(2z-z^2)}\right]$$

which gets more simplified by the substitution, $$ M^2 = |m|^2\\ S^2 = |m+n|^2$$ as,

$$ \lambda -\frac{SM}{2}-\frac{S}{2}-\frac{M}{2}+\\\left[\frac{\left(-2S^2z(2-z)+S^2z(2-z)+2S^2(2-z)+\\z^2M^2-4m^2-4n^2-n^2z^2+4zn^2-8mn+4mnz\right)}{4(2z-z^2)}\right]$$

and $$\lambda - \frac{\left(S^2+S+M+SM\right)}{2}+\\\left[\frac{2S^2z-S^2z^2-2S^2z+4S^2+z^2M^2-4m^2-4n^2-n^2z^2+4zn^2-8mn+4mnz}{4(2z-z^2)}\right]$$

$$ \lambda - \frac{\left(S^2+S+M+SM\right)}{2}+\\\left[\frac{-S^2z^2+4n^2+4m^2\pm8mn+z^2m^2-4m^2-4n^2-8mn-n^2z^2+4n^2z+4mnz}{4(2z-z^2)}\right]$$

gives, $$\lambda-\frac{\left(S^2+S+M+SM\right)}{2}+\\\left[\frac{-m^2z^2-n^2z^2+z^2m^2-n^2z^2+4n^2z+4mnz\mp2mnz^2\pm8mn-8mn}{4(2z-z^2)}\right]$$

It gives, $$ \lambda-\frac{\left(S^2+S+M+SM\right)}{2} +\left[\frac{4mnz\mp2mnz^2+2n^2(2z-z^2)\pm8mn-8mn}{4(2z-z^2)}\right]$$$\rightarrow$$$\lambda - \frac{\left(S+M+SM\right)}{2}+\\ \left[\frac{-S^24z+2S^2z^2+4mnz\mp2mnz^2+4n^2z-2n^2z^2\pm8mn-8mn}{4(2z-z^2)}\\ \leftrightarrow\frac{2m^2z^2+2n^2z^2\pm4mnz^2-4m^2z-4n^2z\mp8mnz+\\4mnz\mp2mnz^2+4n^2z-2n^2z^2\pm8mn-8mn}{4(2z-z^2)}\right]$$$$\lambda-\frac{\left(S+M+SM\right)}{2}+\left[\frac{2m^2z^2-4m^2z+4mnz-8mn\pm4mnz^2\mp8mnz\mp2mnz^2\pm8mn}{4(2z-z^2)}\\\leftrightarrow\frac{-2m^2(2z-z^2)+4mnz\mp8mnz-8mn\pm8mn\pm4mnz^2\mp2mnz^2}{4(2z-z^2)}\right]$$Using the fact $m^2=M^2$ and assuming m and n are positive i.e.$|m+n|^2 = (m+n)^2 = m^2 +n^2 +2mn$ and not "-2mn", then we have,

$$\lambda-\frac{\left(S+M+SM\right)}{2}+\\\left[\frac{-2M^2(2z-z^2)+2mn(z^2-2z)}{4(2z-z^2)}\right]$$

which gives our final equation as, $$ (2z-z^2)V''+\left[2(1+s)-z(S+M+2)\right]V'+ \left[\lambda- \frac{\left((S+M)(1+M)+nm\right)}{2}\right]V=0$$

[You can reduce some two to three steps without bringing out $S^2$ term]

Now, we need to proceed with this differential equation again using power series method for the final solution.

Path Integral formulation - Part - 3 - Free Particle and Schrodinger's time evolution equation

Of  course, for any problem all the paths are not going to be counted one by one, but it will be used a simple plausible way.
To understand how this works, it is conventional to start with the most general free particle problem whose Lagrangian is \[L = \frac{m\dot{x}^2}{2}\,\,\tag{2.1}\]
The technique is to split the action into two parts of which one of them is the classical action and the other is treated as the variational part. For this, it is defined the arbitrary path as \[x(t) = x_c(t)+y(t)\,\,\,\tag{2.2}\] where $x_c(t)$ represents the actual classical path. Substituting this in our Lagrangian and expanding it in terms of Taylor expansion, \[L(\dot{x}) = L(\dot{x}_c(t)+\dot{y}(t)) = L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\,\tag{2.3}\] The expansion is exact since L is quadratic in $\dot{x}$. Thus, it is possible to write the action integral as, \[S = \int_{t_1}^{t_2} \,dt \left(L(\dot{x}_c(t))+\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} + \left.\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}} \right\vert_{\dot{x}_c} \dot{y}^2\right)\,\,\tag{2.4}\]
The first term can be denoted as,
\[ S_c = \int_{t_1}^{t_2}dt\,L(\dot{x}_c)\,\,\tag{2.5}\]
Using (2.1)
\[\frac{\partial^2L}{\partial\dot{x}\partial\dot{x}}\vert_{\dot{x}_c} = mass = const.\,\tag{2.6}\]
and using integral by parts
\[\int_{t_1}^{t_2}dt\,\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c} \dot{y} = \left[\left.\frac{\partial{L}}{\partial{\dot{x}}}\right\vert_{\dot{x}_c}y(t)\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}dt\, \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right)y \]
for the lagrangian
\[ \frac{d}{dt}\left(\left.\frac{\partial{L}}{\partial\dot{x}}\right\vert_{\dot{x}_c}\right) = m\dot{x}_c = 0 \]
Which results finally,
\[S = S_c + \frac{m}{2} \int_{t_1}^{t_2} dt \,\dot{y}^2\,\tag{2.7}\]
and
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}S_c} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.8}\]
where it is changed the integration variable to $dy(t)$ using (2.2)
The classical action is calculated to be
\[S_c = \frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}\,\tag{2.9}\]
which finally yields,
\[K(x_2,t_2;x_1,t_1)= e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}} \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.10}\]
The integral limit points out the fact that the deviation from the classical path at the end points is zero.
The integral over $y(t)$ is independent of $x_1$ and $x_2$. Its value depends only on $t_1$ and $t_2$ , since the entire problem is time translation invariant,
\[A(t_2-t_1) = \int_{y(t_1)=0}^{y(t_2)=0}dy(t)\,e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,\frac{m}{2}\dot{y}^2}\,\tag{2.11}\]
and
\[K(x_2,t_2;x_1,t_1)= A(t_2-t_1)e^{\frac{i}{\hbar}\frac{m}{2}\frac{(x_2-x_1)^2}{t_2-t_1}}\,\tag{2.12}\]
To determine A(t) for $t_1=0$ making use of the group property (1.23) and (1.19),
\[ \delta(x_2-x_1) = K(x_2,t;x_1,t) = \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K(x,0;x_1,t)\] \[\delta(x_2-x_1) =  \int_{-\infty}^{\infty} dx K(x_2,t;x,0)K^*(x_1,t;x,0) \,\tag{2.13}\]
Substituting for K and using (2.12),
\[K(x_2,t;x,0)= A(t)e^{\frac{i}{\hbar}S_c(x_2,t;x,0)}\]
\[K^*(x_1,t;x,0)= A^*(t)e^{\frac{-i}{\hbar}S_c(x_1,t;x,0)}\]
and
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} dx \,\left\vert{A(t)}\right\vert^2 e^{\frac{i}{\hbar}\left(S_c(x_2,t;x,0)-S_c(x_1,t;x,0)\right)}\,\tag{2.14}\]
When $x_2 = x_1+\Delta{x}$ the argument of exponential can be modified with Taylor expansion as,
\[ S_c(x_2,t;x,0)-S_c(x_1,t;x,0) = \frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} \Delta{x_1}\]
Considering $\Delta{x_1}\rightarrow\,0$ all higher order terms are neglected.
Substituting for $S_c$,
\[\frac{\partial{S_c(x_1,t;x,0)}}{\partial
{x_1}} = \frac{\partial(\frac{m\,(x_1-x)^2}{2\,t})}{\partial{x_1}} = \frac{m\,(x_1-x)}{t} = \gamma(x) \,\tag{2.15}\]
Since $\gamma(x)$ is linear function of $x$, its derivative $\frac{d\gamma}{dx}$ is independent of $x$. Using this information to write,
\[\delta(x_2-x_1) = \int_{-\infty}^{\infty} d\gamma\,\left\vert\frac{dx}{d\gamma}\right\vert \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)\left(x_2-x_1\right)}\,\tag{2.16}\]
From Fourier transform,
\[\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\beta(x_2-x_1)} \,d\beta = \delta(x_2-x_1)\,\tag{2.17}\]
for $\beta = \frac{\gamma}{\hbar}$, 
\[\frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} e^{\frac{i}{\hbar}\gamma(x_2-x_1)} \,{d\gamma} = \delta(x_2-x_1)\,\tag{2.18}\]
Thus, multiplying and dividing by the factor $2\pi\hbar$ in (2.16),
\[ \delta(x_2-x_1) = \int_{-\infty}^{\infty} \frac{d\gamma}{2\pi\hbar} \,|A(t)|^2 e^{\frac{i}{\hbar}\gamma(x)(x_2-x_1)}\frac{2\pi\hbar}{\left\vert\frac{d\gamma}{dx}\right\vert}\,\tag{2.19}\]
Comparing both the left and right side, 
\[\delta(x_2-x_1) = \delta(x_2-x_1)\frac{2\pi\hbar|A(t)|^2}{\left\vert\frac{d\gamma}{dx}\right\vert}\]
so that,
\[|A(t)|^2 = \frac{1}{2\pi\hbar}\left\vert\frac{d\gamma}{dx}\right\vert = \frac{1}{2\pi\hbar}\left\vert\frac{-m}{t}\right\vert =  \frac{1}{2\pi\hbar}\left\vert\frac{\partial^2S_c(x_1,t;x,0)}{\partial{x}\partial{x_1}}\right\vert\,\tag{2.20}\]
The phase can be chosen such that,
\[A(t) = \sqrt{\frac{m}{2i\pi\hbar{t}}}\]
Thus, it is determined the propagator for a free particle as,
\[K(x_2,t_2;x_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[S_c = \frac{m\,(x_2-x_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.21}\]
In general for three dimensions,
\[K(r_2,t_2;r_1,t_1) =\sqrt{\frac{m}{2i\pi\hbar{(t_2-t_1)}}}e^{\frac{i}{\hbar}\left[\frac{m\,(r_2-r_1)^2}{2\,(t_2-t_1)}\right]}\,\tag{2.22}\]
For fixed $x_1=0\,\,,t_1=0$ using (3.5) the propagator for a free particle should reduce to the Schr\"{o}dinger wave function for a free particle. Thus, the probability amplitude of the free particle from the propagator is,
\[\psi(x,t) = K(x,t,0,0) = \sqrt{\frac{m}{2i\pi\hbar{t}}}e^{\frac{i}{\hbar}\frac{m\,x^2}{2\,t}}\,\tag{2.23}\]
If it is considered a specific point $(x_0,t_0)$, then the classical momentum at this point is $$p_0 = mv_0 = m\frac{x_0}{t_0}$$ with energy $$E = \frac{mv_0^2}{2}= \frac{mx_0^2}{2t_0^2}$$ The change in phase in the vicinity of $(x_0,t_0)$ is then using again Taylor expansion,
\[
\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} exp\frac{i\,m}{\hbar\,2}\left[\frac{x_0^2}{t_0}+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{x}}\right\vert_{(x_0,t_0)}(x-x_0)+\left.\frac{\partial{\frac{x^2}{t}}}{\partial{t}}\right\vert_{(x_0,t_0)}(t-t_0)+...\right]
\]
The bracketed term reduces to (neglecting higher order terms),
\[\left[\frac{x_0^2}{t_0}+ \frac{2x_0}{t_0}(x-x_0) -\frac{x_0^2}{t_0^2}(t-t_0)\right]\] Simplifying with arithmetic, 
\[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,exp{\frac{i}{\hbar}\left[m\frac{x_0}{t_0}x - \frac{m}{2}{\frac{x_0^2}{t_0^2}t}\right]}\]
Thus, the wave function varies in the immediate vicinity of $(x_0,t_0)$ according to \[\psi(x,t) = \sqrt{\frac{m}{2i\pi\hbar{t}}} \,e^{\frac{i}{\hbar}\left[p_0x - E_0t\right]}\,\tag{2.24}\]
which is the well-known Einstein-de Broglie relation, according to which a particle with momentum $p$ and energy $E$ is assigned a wave function with wave length and wave number respectively $\lambda = \frac{h}{p}\,\,\rightarrow\,\, k = \frac{2\pi}{\lambda}$ Similarly with frequency and angular frequency $\nu = \frac{E}{h}\,\,\rightarrow\,\,\omega = {2\pi\nu}$
So,
\[ e^{i\left(kx- \omega{t}\right)}= e^{i\left(\frac{2\pi}{\lambda}x- {2\pi\nu}t\right)} = e^{\frac{i}{\hbar}(px-Et)} \tag{2.25}\]
which is equivalent to (2.24).
 
Schr\"{o}dinger’s time evolution equation

So far, everything is discussed in position basis. Whereas it needs to be in momentum or energy basis for explicit derivation of Schr\"{o}dinger’s time evolution equation. This is achieved using the group property of our propagators,
\[K(x,t;p,0) = \int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,K(x',0;p,0)\,\tag{2.26}\]
For fixed initial momentum $p$ at time $t$,
\[K(x,t;p,0) = \chi_{p,0}(x,t)=\int_{-\infty}^{\infty} dx'\,K(x,t;x',0) \,\chi_{p,0}(x',0)\,\tag{2.27}\]
Let us take this as an ansatz for the transformation amplitude,
\[\chi_{p,0}(x,0) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}xp}\,\tag{2.28}\]
Substituting this in (2.27) with the corresponding substitution for $K$, 
\[\chi_{p,0}(x,t) =\int_{-\infty}^{\infty}\sqrt{\frac{m}{2i\pi\hbar{(t)}}}e^{\frac{i}{\hbar}\frac{m\,(x-x')^2}{2\,t}} \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{i}{\hbar}x'p} \,dx'\tag{2.29}\]
As $x'$ is the integrating variable, the x' terms are combined together and simplified using arithmetic, 
\[\chi_{p,0}(x,t) = e^{\frac{i}{\hbar}\left(xp-\frac{p^2}{2m}t\right)}\sqrt{\frac{m}{2i\pi\hbar{t}}}\sqrt{\frac{1}{2\pi\hbar}} \int_{-\infty}^{\infty}e^{\frac{i}{\hbar}\frac{m}{2\,t}\left[x'-\left(x-\frac{pt}{m}\right)\right]^2}\,\tag{2.30}\]
changing the variable to $u = \left[x'-\left(x-\frac{pt}{m}\right)\right]$ and making use of Gaussian Integral, 
\[\chi_{p,0}(x,t) = \sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[xp-\frac{p^2}{2m}t\right]}\,\tag{2.31}\]
In three dimensions,
\[\chi_{p,0}(x,t) = \left(\frac{1}{2\pi\hbar}\right)^{\frac{3}{2}} e^{\frac{i}{\hbar}\left[r.p-\frac{p^2}{2m}t\right]}\tag{2.32}\]
Again using the group property with momentum arguments,
\begin{align*}
K(p_2,t;p_1,0) &=\int_{-\infty}^{\infty} dx'\,K(p_2,t;x,t) \,K(x,t;p_1,0) \\ &= \int_{-\infty}^{\infty} dx\,K(p_2,0;x,0) \,\chi_{p_1,0}(x,t) \\  & = \int_{-\infty}^{\infty} dx\,\chi^*_{p_2,0}(x,0) \,\chi_{p_1,0}(x,t) \\ &= \int_{-\infty}^{\infty}dx\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{-i}{\hbar}\left[p_2x\right]}\sqrt{\frac{1}{2\pi\hbar}} e^{\frac{i}{\hbar}\left[p_1x-\frac{p_1^2}{2m}t\right]}
\end{align*}
Using Fourier transform,$$ \frac{1}{2\pi\hbar}\int_{-\infty}^{\infty} dx \,e^{\frac{-i}{\hbar}x(p_2-p_1)} = \delta(p_2-p_1)$$ which results,
\[K(p_2,t;p_1,0) = \delta(p_2-p_1) \,e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}}\,\tag{2.33}\]
From this, it can be shown that $K(p_2,t;p_1,0)$ satisfies Schr\"{o}dinger equation,
\[i\hbar\frac{\partial{K(p_2,t;p_1,0)}}{\partial{t}} = \delta(p_2-p_1)\frac{p_1^2}{2m}e^{\frac{-i}{\hbar}\frac{p_1^2{t}}{2m}} = \frac{p_2^2}{2m} K(p_2,t;p_1,0) \,\tag{2.34}\]
Thus, it is shown the equivalence between Schr\"{o}dinger formulation and path integral formulation.
It can also be checked that,
\[i\hbar\frac{\partial}{\partial{t}}\chi_{p,0}(x,t) = \frac{p^2}{2m}\chi_{p,0}(x,t)\,\tag{2.35}\]
and
\[i\hbar\frac{\partial}{\partial{t}}K(x,t;x',0) = \frac{-\hbar^2}{2m}K(x,t;x',0)\,\tag{2.36}\]
Reference: Classical and Quantum Dynamics - W.Dittrich, M.Reuter

Path Integral formulation - Part - 2 - Quantum Paths

Things become complicated when it is tried to introduce the concept of paths in the domain of Quantum Mechanics. Though the position in Quantum Mechanics is completely a measurable quantity, the position in consecutive time intervals is not a determinate one.

In the sense, even if it observed a perfect value by making a position measurement on a quantum mechanical system at time $t_1$, there is no way of predicting, what would be the result of a position measurement at time $t_2$. All one can talk about is the average value of the position of the particle [known as the expectation value of position operator].

Nevertheless the wave function in quantum mechanics is absolutely defined in terms of probability. Because of this, even for a single particle there is a non-zero probability of the particle to be found at any point in the entire three dimensional space.

The particle can be found anywhere in the universe in any two consecutive position measurements. This indeterminacy makes it extremely difficult to apply the concept of paths for a particle in the Quantum World. 

To put forth the idea, it is to be started with a simple definition and expanded in terms of probability arguments.

For instance, a classical path in position space is defined as the consecutive value of the position of the particle over a time interval. If a particle is found at position $x_1$ at time $t_1$ and found at a later time in position $x_2$ at time $t_2$ then it is described as, the particle travels from the position $x_1$ at time $t_1$ to position $x_2$ at time $t_2$ in the specific path determined by the extremum principle of action. This same classical path by incorporating the concept of Probability can be restated as,
the path of the classical particle is the one where the quantum wave function reduces to Dirac delta function at every point.

Similar to the above, first it is started with the restated definition of well known classical concepts in terms of probability argument and then the concepts are extrapolated to the Quantum domain.
This way of extrapolation of the 'concept of classical paths' to the quantum particles was first done by Richard Feynman in 1948. The idea is basically described in the simplest form as,

the quantum particles can follow any path as well as every path in the three dimensional Euclidean space.
This gives rise to an infinite number of possible paths for a quantum particle even if the particle wants to go the most nearest point. And the mathematical model consists of two types of strategies for the summation procedure of these paths in the form of integrals. To demonstrate, let us consider the case of a free particle going from region one to region two, where it is restricted with an infinite wall with only two slits for the particle to cross between the regions [Figure (1)].
 

 

It is known for sure that, to reach the point $(r_2,t_2)$ from $(r_1,t_1)$, the particle can only take either of the two slits. But just from making a single observation at $(r_2,t_2)$, one cannot say anything about the initial point from where the particle has started its path. The particle could have started from anywhere within the left side region of the wall.

From the figure(1), it could be from $(r_1,t_1)$ or $(r_1^*,t_1^*)$ or from some other point. But the essential point is that, a particle found at position $r_2$ at time $t_2$ will never tell anything about the initial position $r_1$ from where it started its motion.

This lack of information leads to the first of the two strategies that needs to be addressed in the general theory.
The first one is,

In the general motion of a Quantum particle in the three dimensional space, if the particle is observed at a specific position [It is possible to take out the wall by considering infinite number of slits instead of two], then the particle is said to have come from every possible initial point in the entire three dimensional space.
And each of those paths contribute to the net probability amplitude.
So, it is introduced an integral such that it is carried over every possible initial position of the particle.

The second strategy is,

Once the initial position and the final position of the particle is fixed, the next difficulty comes from the fact that the particle can now follow any path as well as every path between those two points.

In the figure(1), once the initial position is fixed, the particle can take any one of the infinite possible paths via slit 1 or slit 2 or through any one of the infinite number of slits.
To account for this new fact, it is introduced a second integral within the first integral to take into account the contribution from every possible paths.

Technically, using the definition of wave function, $$ \psi_1(r_1, t_1) $$ is the probability amplitude of finding the particle at the position ${r_1}$ at time $t_1$ and $$ \psi_2({r_2},t_2)$$ is the probability amplitude of finding the particle at ${r_2}$ at time $t_2$. Using the first integral,  one would like to obtain $\psi_2({r_2},t_2)$ from $\psi_1({r_1},t_1)$ by defining a correlating function called transition amplitude $$ K (r_2,t_2,r_1, t_1)$$ which is the probability amplitude for finding a particle at $({r_2},t_2)$ when it was initially found at $({r_1},t_1)$.

Hence forth, the fundamental dynamical equation is stated as, \[\psi_2({r_2},t_2) = \int_{space} K({r_2},t_2;{r_1},t_1) \psi({r_1},t_1) d^3{r_1} \,\,\,\tag{1.14}\]  
The only unknown term in this equation is $K({r_2},t_2;{r_1},t_1)$. It is also called the Feynman propagator. Once the explicit form of this propagator is known, it is then possible to determine how it controls the dynamical development of the Schr\"{o}dinger wave function.

To obtain the Propagator, one can make use of the second integral where it is summed over all possible paths between the two points A$({r_1},t_1)$ and B$({r_2},t_2)$. So, it is taken a general point C between A and B and considered the motion of the particle along this point. i.e. Path from A to B via C. The probability amplitude for this path A-C-B is denoted as $\phi_{BA}(C)$ and the propagator is obtained by integrating through all possible A-C-B paths, \[K(B,A) = \int_{all\,possible\, paths} dC \,\phi_{BA}(C) \,\,\,\tag{1.15}\]  

Determining this integral for $({r_1}, t_1) \rightarrow ({r_2},t_2)$ in general consists of an infinite number of possible paths with their corresponding probability amplitudes. And obviously there is no fundamental physical principle that determines this amplitude $\phi_{BA}(C)$ directly.

This difficulty was first overcame by Dirac, who postulated that each path contributes same amount of probability amplitude to the final result with different phase factors given by,  \[\phi_{BA}(C) = e^{\frac{i}{\hbar}S(C)}\,\,\,\tag{1.16}\] where S is the classical action integral.
Thus one finally obtains the formula for the Feynman propagator as, \[ K({r_2},t_2; {r_1},t_1) = \int_{{r}(t_1)={r_1}}^{{r}(t_2)={r_2}} d{r(t)}\, e^{\frac{i}{\hbar}\int_{t_1}^{t_2}dt\,L({r}(t), \dot{{r}}(t),t)}\,\tag{1.17}\]
It is easily seen that, in the classical limit when $\hbar\rightarrow\,0$ or $S/\hbar>>1$ the exponential factor oscillates rapidly for all regions except where "S" remains stationary. Thus, the corresponding amplitudes of rapidly oscillating factor will be washed out by destructive interference and the major contribution will come from stationary "S" value which occurs for the classical action. Thus consistent with the correspondence principle, classical results are obtained as a limiting case of the quantum theory. 

In addition, if it is considered $K(x_2,t_2,x_1,t_1)$  (for simplicity it is considered in one dimension) , and $(x_1,t_1)$ kept fixed, one obtains \[K(x,t;x_1,t_1) = K_{(x_1,t_1)}(x,t)\,\,\,\tag{1.18}\] which is a function of $x$ and $t$ alone.

From (1.14) for $t=t_1$ as, $$ \psi(x,t)=\int_{-\infty}^{\infty} dx_1 K(x,t;x_1,t_1) \psi(x_1,t_1)$$ and
$$\psi(x,t)=\int_{-\infty}^{\infty} dx_1 K(x,t;x_1,t) \psi(x_1,t)$$
And from the definition of Dirac delta function, our new function reduces as, \[K(x,t;x_1,t) = K_{(x_1,t_1)}(x,t)\vert_{t=t_1} = \delta(x-x_1)\,\tag{1.19}\]

It is seen from the above results that the propagator function $K_{(x_1,t_1)}(x,t)$,
at the initial point $(x_1, t_1)$ reduces to Dirac delta function where the particle was found certainly without the amplitude being smeared out. And at later times, the propagator is just the probability amplitude for finding the particle at a variable point $(x,t)$.

Comparing the above properties with the postulates of Quantum Mechanics, it is understood that the propagator function is exactly what it is meant by the Schr\"{o}dinger wave function. 
Thus, we have found an intrinsic way to determine the general schrodinger wave function from the propagator function.  

Another interesting property of the propagator is that, from the definition (1.14), \[\psi(x_3,t_3)=\int_{-\infty}^{\infty} dx_2 K(x_3,t_3;x_2,t_2) \psi(x_2,t_2)\,\tag{1.20}\] and \[\psi(x_2,t_2)=\int_{-\infty}^{\infty} dx_1 K(x_2,t_2;x_1,t_1) \psi(x_1,t_1)\,\tag{1.21}\] and \[\psi(x_3,t_3)=\int_{-\infty}^{\infty} dx_1 K(x_3,t_3;x_1,t_1) \psi(x_1,t_1)\,\tag{1.22}\] substituting (1.21) in (1.20) and equating with (1.22) one obtains an important group property, \begin{align*}
& \int_{-\infty}^{\infty} dx_1 K(x_3,t_3;x_1,t_1) \psi(x_1,t_1) \\ & = \int_{-\infty}^{\infty} dx_2 K(x_3,t_3;x_2,t_2) \int_{-\infty}^{\infty} dx_1 K(x_2,t_2;x_1,t_1) \psi(x_1,t_1)
\end{align*}
which gives,  \[K(x_3,t_3;x_1,t_1) = \int_{-\infty}^{\infty} dx_2 K(x_3,t_3;x_2,t_2)K(x_2,t_2;x_1,t_1)\,\tag{1.23}\] In general, for $f(x_f,t_f)$ and $i(x_i,t_i)$ one can write \[K(f;i) = \int_{-\infty}^{\infty} dx_{N-1}.....\int_{-\infty}^{\infty}dx_1 K(f;N-1)\,K(N-1;N-2)\,...K(2;1)K(1;i)\,\tag{1.24}\] It is to be noted that, the intermediate times are not integrated over.
 
Reference: Classical and Quantum Dynamics - W.Dittrich, M.Reuter

Path Integral formulation - Part - 1 - Euler Lagrange equations from Variational Principle

The study of Classical Mechanics usually starts with the three laws of motion given by Sir Isaac Newton. One knew from these three laws that, if it is possible to completely determine the net force acting on a particle at each instant, the motion of the particle is uniquely determined by the solution of the second order differential equation

\[
{\vec{F}=m\vec{a}}\tag{1.1}
\]

(where it should be provided with the necessary initial conditions).
These laws work so fine that all the celestial and terrestrial motions are perfectly described using them.

But, it continues to be a matter of curiosity to ask questions such as "why the particle follows Newton's laws of motion?"
For example, In the general motion of a classical particle,
"why doesn't the particle take any other path, other than the prescribed path given by Newton's laws of motion?"
Instead of directly going from an initial point A to some final point B, "why doesn't it take an infinite path all around the universe and come to point B ?".

These questions asked by some of the greatest Physicists led to the so called Action principle.
It states that in any classical motion, there is a quantity defined as the "action" which is always either a maximum or a minimum.
Though it may seem an abstract concept at the first glance, it has a pure logical reasoning.
For example, if it is observed the particle to follow a path with some finite value of action other than the extremum value, the question then arises, "why this path and not the nearby path with a different value of action".

This question applies symmetrically to everywhere, which affects all other paths nearby to the original considered path. As a consequence it is inferred that, the resultant path can never be an ordinary one like others but it should have some unique features that always distinguishes it from others. And thus it is arrived at the extremum paths, the only one that follows these unique features.

The significance of the extremum paths can be illustrated from a famous quote by Euler, (the English translation)

"For since the fabric of the Universe is most perfect and the work of a most wise creator, nothing at all takes place in the Universe in which some rule of maximum or minimum does not appear"

So far, it is simplified the fundamental question from "why not all other paths" to "why the extremum path". But it is left with a new quantity named "Action" that needs to be defined and quantified in a mathematical form.

For this, it is begun with the Variational principle that states that, the motion of a system from time $t_1$ to time $t_2$ will be such that the line integral "S" called action or action integral has stationary value for the actual path of the motion.  And Action "S" is defined as
\[
S = \int_{t_1}^{t_2} L \,dt \tag{1.2}
\]
The function $L$ which will be later called Lagrangian, is a function of position and velocity  and it can also be an explicit function of time,

\[ L  = L(x(t),\dot{x}(t),t) \tag{1.3}\]
 
It is normally used $x(t), \dot{x}(t)$ to denote the generalized position and generalized velocity. 

By the term "stationary value" for a line integral, it is meant that the integral along the given path has the same value to within first-order infinitesimals as that along all neighbouring paths. The notion of a stationary value for a line integral thus corresponds in ordinary function theory to the vanishing of the first derivative.
The statement is equivalent to saying that the action should be an extremum for the actual path. Out of all possible paths by which the system point in configuration space could travel from its position at time $t_1$ to its position at time $t_2$, it will actually travel along the path where the action integral is an extremum or has a stationary line integral.

Euler Lagrange Equations:
 
From the definition, the action integral gives some definite value for every possible path, out of which the maximum value will correspond to the actual path. So, the general path is taken as,

\[x(t,\alpha) = x(t,0) + \alpha\eta(t)\tag{1.4}\]

where $\eta(t)$ is an arbitrary function defined to create an arbitrary variation to the initial path. It is well behaved, continuous in first and second derivatives and non-singular between $t_1$ and $t_2$.

Therefore, any path $x(t,\alpha)$can be expressed from $x(t,0)$ as the variation due to $\eta(t)$.
This variation is chosen such that it should vanish at the end points $t_1$ and $t_2$. i.e.
$\eta(t) = \frac{\partial{x}}{\partial{\alpha}}$ and
\[\eta(t_1)=\eta(t_2) = 0\]

The action for the general path is,
\[S(\alpha) = \int_{t_1}^{t_2} L \,dt = \int_{t_1}^{t_2}L(x(t,\alpha),\dot{x}(t,\alpha),t)\,dt \tag{1.5}\]
The condition for this quantity to have a stationary value is that,
\[ \left.\frac{\partial{S}}{\partial\alpha} \right\vert_{\alpha=0} = 0\,\,\,\tag{1.6}\]
Substituting for S and differentiating under the integral sign,
\[\frac{\partial{S}}{\partial\alpha} = \int_{t_1}^{t_2} \left(\frac{\partial{L}}{\partial{x}}\frac{\partial{x}}{\partial\alpha} + \frac{\partial{L}}{\partial{\dot{x}}} \frac{\partial{\dot{x}}}{\partial\alpha}  \right)\,dt \,\,\,\tag{1.7}\]
(time is an independent parameter). 

The last term (using integral by parts),
\[\int_{t_1}^{t_2} \frac{\partial{L}}{\partial{\dot{x}}} \frac{\partial{\dot{x}}}{\partial\alpha} \,dt = \int_{t_1}^{t_2} \frac{\partial{L}}{\partial{\dot{x}}} \frac{\partial^2{x}}{\partial\alpha\partial{t}} \,dt \\=\left.\left(\frac{\partial{L}}{\partial{\dot{x}}} \frac{\partial{x}}{\partial\alpha}\right)\right\vert_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{\partial{x}}{\partial{\alpha}} \frac{d}{dt}\left(\frac{\partial{L}}{\partial\dot{x}}\right) \,dt \,\,\,\tag{1.8}\]
Using the end point conditions of $\eta(t)$,
\[ \frac{\partial{S}}{\partial\alpha} = \int_{t_1}^{t_2} \left(\frac{\partial{L}}{\partial{x}}-\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{x}}}\right)\right)\frac{\partial{x}}{\partial\alpha}\,dt\,\tag{1.9}\]
Using the arbitrariness of $\eta(t)$, one can choose it as well as a positive quantity through out the domain ($t_1$,$t_2$). But, the condition must hold true for all types of $\eta(t)$ which implies that the other integrand should always be zero. Thus it gives,
\[ \frac{\partial{L}}{\partial{x}}-\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{x}}}\right) = 0 \,\,\tag{1.10}\] which is known as the Euler-Lagrange equation. Using variational differentials notation,
\[\frac{\partial{x}}{\partial\alpha} d\alpha = \delta{x} \,\tag{1..11}\]
\[\frac{\partial{S}}{\partial\alpha} d\alpha = \delta{S} \,\tag{1.12}\]
\[\delta{S} = \int_{t_1}^{t_2} \left(\frac{\partial{L}}{\partial{x}}-\frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{x}}}\right)\right)\,dx\,dt = 0 \,\tag{1.13}\]
Thus, for every problem in classical mechanics there exists a function called Lagrangian from which it can be extracted everything that needs to be known about the system using Euler-Lagrange equations. It is completely equivalent to Newton's laws of motion except for the fact that there is no need to deal with the force explicitly. Instead, together with the concept of path and action, the new Lagrangian plays the most fundamental role.

Therefore in all the classical problems, determining this Lagrangian for a specific problem is the only task one needs to accomplish.

General Solution of Klein Gordon Equation

Let us discuss something about the general solution of the Klein Gordon equation which will be later useful when we adopt for the field formulation of our theory.  
Assuming our solution as a general waveform solution in the form,$$ \phi(x) = \frac{1}{(2\pi)^{2}}\int \,d^4k \tilde{\phi}(k) \,e^{-(ikx = ik_{\mu}x^{\mu})}$$ where 4 vector notation is generally implied unless it is specified. 
We substitute this general form in our equation to obtain the conditions to be satisfied from the KG equation. It gives, $$ \frac{1}{(2\pi)^{2}}\int \,d^4k \,e^{-ikx}\left(-k^2+m^2\right)\tilde{\phi}(k) $$ [Normalization factor $\sqrt{2\pi}$ for four integrations]. 

From the above equation, the solution should be of the form such that, if $k^2=m^2$, $\tilde{\phi}(k)$ can take any arbitrary value and if $k^2\neq {m^2}$ then $\tilde{\phi}(k)$ should be zero, 
therefore it should be in the form, $$\tilde{\phi}(k) = \delta(k^2-m^2)\,\tilde{f}(k)$$ 
Thus, our general wave solution cannot take any arbitrary form but should always satisfy the energy momentum relation - which is imposed by the dirac delta function. It is equivalent to saying that the energy can take only positive and negative values of $\omega_k$ given by $\delta(k^2-m^2) = \delta(k_0^2-|\vec{k}|^2-m^2) = \delta(k_0^2-\omega_k^2)$ where $\omega_k^2=|\vec{k}|^2+m^2$ 
Using the identity, $$\delta(f(x)) = \sum_{k}\frac{\delta(x-x_k)}{f'(x_k)}$$, we can rewrite our solution as,
$$\phi(x) = \frac{1}{(2\pi)^{2}}\int \,d^4k \left[e^{-ikx}\delta(k_0-\omega_k)\tilde{f}(k) + e^{-ikx}\delta(k_0+\omega_k)\tilde{f}(k)\right]$$ Now, integrating the zeroth component of momentum vector and using the fact that the integration is symmetric under $+\vec{k}$ and $-\vec{k}$ we can arrive at the form, $$\phi(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{-ikx}\frac{\tilde{f}(\omega_k,\vec{k})}{\sqrt{4\pi\omega_k}}+ e^{ikx}\frac{\tilde{f}(-\omega_k,-\vec{k})}{\sqrt{4\pi\omega_k}}\right]$$
Now we identify the terms other than exponential as $ a(k), b^*(k)$ just for a general convenience. $$\phi(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{-ikx} a(k) + e^{ikx} b^*(k)\right]\vert_{k_0=\omega_k} = \phi_+ +\phi_-$$ 
The complex conjugate is, $$\phi^*(x) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3k}{\sqrt{2\omega_k}} \left[e^{ikx} a^*(k) + e^{-ikx} b(k)\right]\vert_{k_0=\omega_k} $$ 

For real fields, one can show that, $a(k) = b(k)$ and $a^*(k) = b^*(k)$

Similarly, the momentum space terms can be written as, $$a(p) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3x}{\sqrt{2\omega_p}} \left[\omega_p \phi(x) +i\partial_0\phi(x)\right]e^{ipx}\vert_{p_0=\omega_p} $$ and $$b^*(p) = \frac{1}{(2\pi)^{3/2}}\int \,\frac{d^3x}{\sqrt{2\omega_p}} \left[\omega_p \phi(x) - i\partial_0\phi(x)\right]e^{ipx}\vert_{p_0=\omega_p}$$ 
Hint: One can prove from RHS to LHS by integrating with respect to $d^3x$, making use of the definition of Dirac delta function, integrating again with respect to $d^3k$ and finally using the fact that $\omega_k=\omega_{-k}=\omega_p=\omega_{-p}$ - the energy is independent of the direction of the momentum vector. 
It is always a good practice to use different symbols to discuss it in a general form of the derivation. 
Finally, one can obtain $a^*(p), b(p)$ by taking the complex conjugate. 

Dirac Equation

The essential fact about the Klein Gordon equation is that it is a non-linear equation. But, equations are consistent with Lorentz transformation when they are linear in space and time. Dirac came forward with a new way of resolving the problem of this non-linearity by factorizing the relativistic energy momentum relation into two linear equations. The equation is given by the format, 
$$ i\frac{\partial\psi}{\partial{t}} = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right)\psi $$
But, still it should satisfy the energy momentum relation $$ E^2 \psi = \left(m^2+|\vec{p}|^2\right)\psi $$
that is,
$$ E (E\psi) = i \frac{\partial}{\partial{t}}\left(\frac{i\partial\psi}{\partial{t}}\right) \\ = \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \left(-i \vec{\alpha}\cdot\vec{p}+\beta{m}\right) \psi $$ [the differentials commute]
In summation convention,
$$\frac{\partial^2\psi}{\partial{t^2}} = \left[\alpha_i\alpha_j\partial_i\partial_j + i(\alpha_i\beta+\beta\alpha_i)m\partial_i - \beta^2m^2\right]\psi $$
[It is assumed $\beta$ commutes with differentials]

The above can be evaluated separately from Klein Gordon equation as, $$\frac{\partial^2\psi}{\partial{t^2}} = \left[\vec{\nabla}^2 - m^2\right]\psi $$

Comparing both of them, we can see that $\alpha_i,\beta$ shouldn't commute and so they cannot be numbers. Dirac proposed the idea of matrices instead of numbers. This leads to the following properties that should be satisfied by $\alpha_i$ and $\beta$ as, 

$$\{\alpha_i,\alpha_j\} = 2 \delta_{ij} \\ \{\alpha_i,\beta\} = 0 \\ \alpha_i^2=\beta^2 =1$$
$$Tr[\alpha_i^2{X}\, (or)\, \beta^2{X}=\lambda^2{X}=IX]$$ gives us the eigenvalues to be $\pm 1$ 

$$Tr[\alpha_i] \\ = Tr[\alpha_i{\beta^2}] = Tr[\beta\alpha_i{\beta}] (using \,the \,property\, Tr(AB)=Tr(BA)) \\ = Tr[-\beta\alpha_i\beta] (using \,anti-commutation \,relation)$$ 

Doesn't matter whether we start with $\alpha_i$ or $\beta$, this proves the trace of the matrices $\alpha_i,\beta$ should be zero. 
From the above two properties, dimension of the matrix should be even. 

Starting with the dimension N=2, there are no sufficient number of independent matrices to describe our four matrices (since pauli matrices span the space completely).

Thus, we are left with the next possibility of N=4 by constructing our matrices as, 

$$ \alpha_i = \left(\begin{matrix}0 & {\sigma_i}\\ {\sigma_i}&0\end{matrix}\right) \\ \beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0 & \mathbb{-I}\end{matrix}\right) $$

To make it simple, we introduce the symbol $\gamma$ and defined everything in terms of new Dirac $\gamma$ matrices. And, the Dirac equation is finally given in the simplest form as, 

$$\gamma^0 =\beta = \left(\begin{matrix}\mathbb{I} & 0 \\ 0& \mathbb{-I}\end{matrix}\right) \\ \gamma^i = \beta\alpha_i = \left(\begin{matrix}0 &{\sigma_i}\\ {-\sigma_i}&0\end{matrix}\right)$$

with the properties,

$$\gamma^0 = (\gamma^0)^\dagger \\ (\gamma^0)^2 = \mathbb{I} \\ \gamma^i = -(\gamma^i)^\dagger \\ (\gamma^i)^2=-\mathbb{I} \\ \{\gamma^\mu,\gamma^\nu\} = 2g_{\mu\nu} $$

Using this new representation, the Dirac equation is written as, $$\left(i\gamma^\mu\partial_\mu - m \right)\psi = 0 $$

In slashed notation, any Lorentz vector is $\slashed{x} = \gamma^\mu{x}_\mu$ and so, Dirac equation is, $$ \left(i\slashed{\partial}-m\right)\psi = 0$$ 

Klein Gordon Equation from Correspondence Principle in Relativistic domain

The Klein Gordon equation is developed from the general relativistic energy momentum relation by the substitution of corresponding operators as, $$ E = i\hbar\frac{\partial}{\partial{t}}$$ $$ \vec{p} = -i\hbar\nabla$$
acting on $\phi(x,t)$ to give, 
$$ -\hbar^2\frac{\partial^2\phi}{\partial{t^2}} = -\hbar^2c^2\nabla^2\phi +m^2c^4\phi$$ $\rightarrow$$$\hbar^2c^2\left[\nabla^2\phi-\frac{1}{c^2}\frac{\partial^2\phi}{\partial{t^2}} \right]= m^2c^4\phi$$ $\rightarrow$$$\left[\Box+\frac{m^2c^2}{\hbar^2}\right]\phi(x_i)=0$$
where the d'alembertian operator is defined here as $$ \Box =  \frac{1}{c^2}\partial_t^2 - \nabla^2$$

We know that, $$p^\mu{p_\mu} = -\hbar^2 \partial^\mu\partial_\mu = -\hbar^2\Box$$ where the minkowski metric is given by,$ \eta_{\mu\nu}$ = diagonal(1,-1,-1,-1), and thus we have,
$$ \left(\partial^\mu\partial_\mu + \frac{m^2c^2}{\hbar^2}\right)\phi = 0 $$ which is known as Klein Gordon Equation.
To make life simple, we adopt to the convention,
where we equate the planck's constant and the speed of light equal to 1 (dimensionless number). The consequences are, the dimension of Length and Time are the same and the dimension of Mass is just the inverse of Length or Time. In addition, Energy and Momentum is measured in the same unit as of the Mass. 

And, contravariant vectors are $ A^{\mu}= (A_0, \vec{A})$ and corresponding covariant transformation is $ A_{\mu}= \eta_{\mu\nu}A^{\nu} = (A_0, -\vec{A})$ where the differential is defined in reverse way 
$ \partial_\mu = \left(\partial_0,\vec{\nabla}\right) $
[$\partial_0 = \partial_t$] and 
$ \partial^{\mu}= \eta^{\mu\nu}\partial_{\nu} = ({\partial}_0,-\vec{\nabla})$
With these substitution we get our KG equation as, $$\left(\Box + m^2\right)\phi = 0 $$ where $p^\mu{p}_\mu = m^2 \\ \,\, E^2 = \omega^2 = m^2 + {|\vec{p}|^2} $
Since $m^2$ and d'Alembertian operator is invariant under Lorentz transformation, if the function $\phi$ satisfies the condition $ \phi'(x') = \phi(x)$ then the whole KG equation is invariant under Lorentz transformation. 
But, there are some problems with this equation in the basic definition of $\phi(x)$. First of all it cannot be the wave function of the particle as it is in the case of Non-relativistic Schr\"{o}dinger equation.

The problem arises because of the probability statements. To understand its significance, let us just assume that $\phi(x)$ is the usual wave function. 

Then, it says that the probability of finding the particle at a position x is the same as of the finding the particle in $x'$ position in some other reference frame [Lorentz invariance condition]. It implies the wave function should behave like a scalar quantity and independent of direction, which is not true in general for spin half particles. The properties of spin half particles depends on from which direction it is measured and change with respect to different orientations of the reference frame. 

And the second and the most important problem is that the probability density definition changes completely and it allows for the weird possibility of the negative probability. 
The continuity equation with the time component becomes, $$\partial_0\rho+\vec{\nabla}\cdot\vec{J} = 0 = \partial_\mu{J}^\mu$$ where $ \rho = \frac{1}{2} \left[\phi^*(\partial_0\phi) - (\partial_0\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $ 
and \newline
$\vec{J} = \frac{-1}{2}\left[\phi^*(\nabla\phi) - (\nabla\phi)^*\phi\right] = \frac{1}{2} \phi^*\overleftrightarrow{\partial_0}\phi $
where $\rho$ is the equivalent probability density as defined in non-relativistic case can now have positive as well as negative values, that is not compatible with the definition of probability - you can check it for monochromatic wave of the for $\phi = Ae^{\pm{ikx}}$. Thus, either we will have to abandon KG equation or should find an alternative way of description for its definition.

Fourier series and Fourier transform

The concept of Fourier series is based on the orthogonality of sine and cosine functions. So, it is a perfect thing to start from these properties.
A sine function with time period $2\pi$ radians is given by $sin(t)$ and the integral of the function over its time period is zero. In general, for an arbitrary sine function with Time period $T_0$, the function is given by $sin (\frac{2n\pi{t}}{T_0})$, where the time integral over this new time period is zero. Everything applies to "cosine" function as much as the same.

In functional form, $$sin\left(\frac{2n\pi}{T_0}(t+T_0)\right) = sin(\frac{2n\pi{t}}{T_0})$$ and $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2m\pi{t}}{T_0}) = 0  \\ for \,\,\,\,n,m > 0 $$
From our trigonometric identities, we can write $$ sinA cosB = \frac{sin(A+B)+ sin(A-B)}{2}$$ Using this in our case, we will get (n+m) and (n-m) where n and m are integers. This gives, $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$
Similarly, with the other identity $$ sinA sinB = cos (A-B) - cos(A+B) \\ cosA cosB = cos(A+B) + cos(A-B)$$ we can prove $$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$ except now, there arise a problem for the case n = m as we will get an extra non-zero term $$\int_{t_0}^{t_0+T_0} \,dt \frac{cos(\frac{2(n-m)\pi{t}}{T_0})}{2} = \int_{t_0}^{t_0+T_0} \frac{1}{2}\,dt = \frac{T_0}{2}$$
Finally we get the following identities,
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = 0 $$ and
$$\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) sin(\frac{2m\pi{t}}{T_0}) = \int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) cos(\frac{2m\pi{t}}{T_0}) = \delta_{nm}\frac{T_0}{2} $$
This is the fundamental orthogonality relation we require for the sine and cosine function to serve as the basis vectors.
Now, the Fourier series is defined as, $$f(t) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n cos(\frac{2n\pi{t}}{T_0}) + \sum_{n=1}^\infty b_n sin(\frac{2n\pi{t}}{T_0})$$
The only unknown terms in this series are the Fourier coefficients $a_0, a_n, b_n$ which can be determined using our orthogonality relations as, $$ a_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt cos(\frac{2n\pi{t}}{T_0}) f(t) \,\,\,\, n = 0,1,2..\\ b_n = \frac{2}{T_0}\int_{t_0}^{t_0+T_0} \,dt sin(\frac{2n\pi{t}}{T_0}) f(t) $$
Once we are able to write a function in terms of sine and cosine, we proceed to write it in terms of exponential functions using Euler's formula. And we can contract the series in a simple form as, $$ f(t) = \sum_{n=-\infty}^\infty c_n e^{i\frac{2n\pi{t}}{T_0}}$$
Now, the only unknown in this series is $c_n$ and it can also be determined using the orthogonality relation of exponentials $$ \int_{t_0}^{t_0+T_0}\,dt\,e^{i\frac{2n\pi{t}}{T_0}}e^{-i\frac{2m\pi{t}}{T_0}} = \delta_{nm}T_0 $$ which gives finally, $$ c_m = \frac{1}{T_0} \int_{t_0}^{t_0+T_0} f(t) e^{\frac{i2n\pi{t}}{T_0}} \,dt $$

Fourier Transform:

Substituting for $ \omega_n = \frac{2n\pi}{T_0}$ we get $$ f(t)  = \sum_{n= -\infty}^\infty c_n e^{i\omega_nt}$$ and $$ c_m = \frac{1}{T_0} \int_{-\frac{T_0}{2}}^{\frac{T_0}{2}} f(t) e^{-i\omega_nt}\,dt$$ (from the substitution $t_0 = -\frac{T_0}{2}$)

It is easily seen from the above equation that, it fails for $T_0 = \infty$ For a function with infinite period, the coefficient cannot be calculated from above equation.

Since, many of the real function used in Physics has this structure (Time period infinity), it is necessary to look our for an alternative to deal with these functions. This is where the integral comes about to play the significant role and known as fourier transform.

For very large time period, the change in $\omega$ - the step size becomes negligible i.e. $$\omega_{n+1} -\omega_n = \frac{2\pi}{T_0} \\ \rightarrow \,\,\,\, \delta\omega = d\omega\\ \frac{1}{T_0} = \frac{d\omega}{2\pi}$$ and there is no need for the index 'n'.

Combining everything together, $T_0\rightarrow\infty$ and substituting for $c_n$ with some dummy variable, $$ f(t) = \sum_{n= -\infty}^\infty\frac{d\omega}{2\pi} \left[\int_{-\infty}^{\infty}f(\tau) e^{-i\omega_n\tau}\,d\tau\right]e^{i\omega_n{t}}$$ As "n" changes $d\omega$ is very small and the sum can be replaced by Integral,

$$ f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega_nt}\,d\omega \left[\int_{-\infty}^\infty f{\tau} e^{-i\omega_n\tau}\,d\tau\right] $$ the bracketed term depends only on $\omega$. Thus, we have finally arrived at our fourier and inverse fourier transform. It doesn't matter which one you call the fourier transform and which one is the inverse one. By splitting $\frac{1}{2\pi}$ factor in two different ways, the final equations are, $$f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}\,dt \,e^{-i\omega{t}}\,f(t) \,\,\,\,\\or\,\,\,\,\\ f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty}\,d\omega e^{i\omega{t}}F(\omega) \\ F(\omega) = \int_{-\infty}^{\infty}\,dt e^{-i\omega{t}}\,f(t)$$ or you can put the $2\pi$ factor in the other way. 

Volume of a Hypersphere in N - dimension

    To find the number of micro states in a statistical system for N number of particles in a fixed volume and energy within the range E and $E+\delta{E}$, we require the knowledge on how to find the volume for N dimensional hypersphere in momentum space.

In the simplest way, the problem is to find the volume of an N-dimensional hypersphere. If we try to find the general formula by mathematical induction, using the equation in Cartesian form, it will produce a complex equation.

    Instead there is a clever trick to simplify the procedure based on the concept of infinity. The equation of a sphere in Cartesian form is $$ x_1^2+x_2^2+x_3^2+...+x_N^2 = r^2$$ where 'r' is the radius of the sphere.

    In general, volume of an n-dimensional sphere should be proportional to $r^N$ and its surface is proportional to $r^{N-1}$. So, $$V \propto r^N\\ S \propto r^{N-1} $$ or $$ S = A r^{N-1}$$ where A is the proportionality constant that is equal to the surface area of a unit sphere. If we could find somehow this value of A in n-dimension, it is then a simple task to proceed to the volume. The special characteristic of the surface of a sphere is that, once we know the surface, the volume is just a simple line integral of the surface value along 'r'.

$$ dV = S dr  = A r^{N-1} dr $$

The generalization of the equation of sphere in Cartesian form is a straight task. To connect all these things, we use the property of the infinity given as,

    Integral of a function over the infinite volume is the same independent of whether the integral is done with Cartesian cubic volume element (cubic volume extending to infinity) or spherical volume element (spherical volume extending to infinity), the result is the same.  This is not a completely new one because we use this same fact to find the integral of $$\int_{-\infty}^{\infty}e^{-\alpha{x^2}}\,dx = \sqrt{\frac{\pi}{\alpha}}$$

    Using the same integral we have, $$ I = \int_{-\infty}^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} \,dx_1\,dx_2\,...\,dx_N = \pi^{N/2} $$ Now,the same function integrated over spherical volume element gives, $$ I = \int_0^{\infty} e^{-{x_1^2+x_2^2+...+x_N^2}} A r^{N-1}dr = \int_0^{\infty} A e^{-{r^2}} r^{N-1} dr $$ Substituting $y = r^2 $ we have $$I  = \int_0^{\infty} A e^{-y} y^{\frac{N-1}{2}} \,\frac{1}{2y^{1/2}}\,dy = \int_0^{\infty} A \frac{1}{2} e^{-y} y^{\frac{N}{2}-1}dy = \frac{1}{2} A \Gamma(N/2)$$ then we have, by equating the value of I in two ways, $$ A = \frac{2\pi^{N/2}}{\Gamma(N/2)}$$ Now, the volume is evaluated to be $$V  =  \int _0^R \,S \,dr  = \int_0^R \,A r^{N-1}\,dr \\ V =  \int_0^R\,\frac{2\pi^{N/2}}{\Gamma(N/2)} r^{N-1} \, dr = \frac{\pi^{N/2}}{\Gamma(N/2)} \frac{R^{N}}{N/2} = \frac{\pi^{N/2} R^N}{\Gamma(\frac{N}{2}+1)}$$ where we used the fact  $$\frac{N}{2} \Gamma(\frac{N}{2}) = \Gamma(\frac{N}{2}+1)$$

Thus we can determine the volume and so the surface area of an n-dimensional hypersphere in a simple form.