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.

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.

Time dependent perturbation theory

Time dependent perturbation theory addresses the transition between one eigenstate to another eigenstate when there is a small perturbation with respect to time such that it doesn't change the eigenfunctions.

The Hamiltonian for these problems of perturbation is, $$ H = H_0 + \lambda {H'(t)}$$ where now the perturbing Hamiltonian is time dependent. Solution for $H_0$ is, $$\psi_n(x,t) = \phi_n(x) e^{i\frac{E_n}{\hbar}t} = \phi_n e^{-i\omega_nt}$$ where $$H_0\phi_n = E^0_n\phi_n$$ and $ \omega_n = E_n/\hbar$ and the general state vector is given by, $$ \Psi = \sum_nc_n\psi_n(x,t)$$
In our problem we start with a particular eigenstate and assume that, after switching on the perturbation the state evolves to a general state always expressed as the linear combination of initial eigenstates where the time dependence is accompanied with the coefficient as $$\Psi (x,t) = \sum_n c_n(t) \phi_n(x) e^{i\omega{t}}$$  Because of this assumption, the particle now can be found in any other eigenstate and needn't to be in the same eigenstate forever. As usual, the modulus square of the coefficient gives the probability of finding the particle in any one of eigenstates, except now it changes with perturbation time. Substituting our wave function in general Schrodinger's equation (to find the time dependence of c(t)), $$i\hbar\frac{\partial\Psi}{\partial{t}} = H\Psi$$ gives (taking the inner product with $\psi_k$)$$ i\hbar\dot{c_k} = \lambda\sum_n \langle\phi_k\vert{H'}\vert\phi_n\rangle e^{i(\omega_k-\omega_n)t} f(t) c_n$$ where f(t) is the time-dependent part of the perturbing Hamiltonian H' (in variable separable form).  In the limit $\lambda\rightarrow{0}$ the coefficients are all constant in time and so we seek a solution of the form, $$ c_k(t) = c_k^0 + \lambda{c_k^1(t)} + \lambda^2c_k^2(t) + \cdots$$ with corresponding order of correction in the upper index.  With this expansion we get, $$ i\hbar \dot{c_k}^0 = 0 $$ which means $c_k^0 = constant$ whatever the value of $c_n^0$ is at the beginning of the perturbation remains a constant throughout the perturbation. In specific, if we denote the starting time $t_0$ then $c_n^0(t_0) = \delta_{nl}$ where we assumed the state is in a specific eigenstate at the start.

Similarly, the first order equation gives, $$ i\hbar\dot{c_k}^1 = \sum_n H'_{kn} f(t) c_n^0 e^{i(\omega_k-\omega_n)t}$$ substituting for $c_n^0$ we get, $$ i\hbar\dot{c_k}^1 = H'_{kl}f(t)e^{i(\omega_k-\omega_l)t}$$ finally gives the expression for first order correction as,$$c_k^1(t) = \frac{H'_{kl}}{i\hbar} \int_{t_0}^t f(t) e^{i\omega_{kl}t'} \,dt'$$ where $\omega_{kl} = \omega_k-\omega_l$ From this we can transition probability for each eigenstate by taking modulus square of the above coefficient.

Time Independent Perturbation Theory - Degenerate

Similar to the non-degenerate case, we take our Hamiltonian $$H = H_0 + H'$$ except now, the initial unperturbed Hamiltonian has degenerate states. Assuming it is q- fold degenerate, if the symmetry giving rise to the degeneracy is disturbed, it will produce distinct 'q' non degenerate states. Our motive is to calculate the energy levels of these newly produced distinct energy levels. Let us consider two different initial eigenstates $E_1^0$ and $E_{q+1}^0$ [it is equal to the zeroth order term in the expansion] where $E_1^0$ is q-fold degenerate with states $$E_1^0 = E_2^0 = .... = E_q^0$$. As we have seen in non-degenerate perturbation theory, the coefficients $$c_{jn} = \frac{H'_{jn}}{E_n^0 - E_j^0}\,\,\,j\neq{n}\,\,and\,\, j,n\leq{q}$$ will become infinite for two different energy levels which has the same energy value. This situation can be avoided only if the numerator term is zero for all $j,n \leq {q}$ (Not the diagonal terms because $j\neq{n}$). It is equivalent to diagonalizing the sub-matrix $H'_{jn}\,\,\,\,for j,n\leq{q}$. This can be achieved by transforming to the new set of eigenfunctions from the initial set of eigenfunctions.
Thus, our problem for solving the energy values of the degenerate case reduced to diagonalizing the sub-matrix of H'. The diagonal elements are the change in energy levels of the initial degenerate states due to the perturbation. 
Representing the 'q' new eigenfunctions that diagonalize the $H'_{jn}$ as $\tilde{\psi_n}$ we write, $$ \tilde{\psi_n} = \sum_{m=1}^q a_{nm} \psi_m^0$$ Since, it will diagonalize, $$ \langle\tilde{\psi_n}\vert\tilde{\psi_j}\rangle = H'_{nj}\delta_{nj} $$ So, our new set of basis functions are, $$ Basis \,= \left\{\tilde{\psi_1},\tilde{\psi_2},\cdots\tilde{\psi_q},\psi_{q+1}^0,\cdots\right\}$$The diagonal element of the sub-matrix H' is the first order correction to the energy i.e. $$E'_n = \langle\tilde{\psi_n}\vert\tilde{\psi_n}\rangle = H'_{nn} \,\,n\leq{q}$$
Corresponding eigenfunctions are determined from $$H'\tilde{\psi_n} = E'_n\tilde{\psi_n}$$ substituting for new eigenfunctions in terms of initial wavefunctions, $$H' \sum_{m=1}^q a_{nm}\psi_m^0 = E'_n \sum_{m=1}^q a_{nm} \psi_m^0$$ Taking inner product with $\psi_j^0$ we have, $$\sum_{m=1}^q \left(H'_{jm} - E'_n\delta_{jm}\right)a_nm = 0\,\,\,n,j\leq{q} $$ It could be written simply in matrix form. For non-trivial solution for {a_nm} the determinant of coefficient matrix should vanish gives the secular equation in this case as, $$determinant\,\,\vert{H'_{jm}}-E'_n\mathbb{I}\vert = 0 $$ Using these the new eigenfunctions and eigenvalues are calculated.

Time Independent Perturbation Theory - Non-degenerate

When a given Hamiltonian of a problem is not so much different from the Hamiltonian of an absolutely solvable problem, we use the perturbation theory to find the new eigenfunction from the old  eigenfunctions with the assumption that the perturbing Hamiltonian is small compared to the original one. Let us jump into the mathematics by writing the total Hamiltonian as, $$ H = H_0 +\lambda{H' } \,\,\,\,...(1)$$ where H' is the perturbing Hamiltonian and $\lambda$ is used to denote how small the perturbation is with respect to the original. The eigenfunctions are expressed as, $$ H_0 \psi_n^0 = E_n \psi_n^0 \,\,\,\,...(2)$$ and $$H\psi_n = E_n \psi_n\,\,\,\,...(3)$$ When $\lambda\rightarrow{0}$ the perturbed eigenfunctions should reduce to the original eigenfunctions. So, we expand it in series as, $$\psi_n = \psi_n^0 + \lambda\psi_n^1+\lambda^2\psi_n^0+\cdots \,\,\,...(4)\\ E_n = E_n^0 +\lambda{En^1} + \lambda^2E_n^2+\cdots \,\,\,...(5)$$ where the upper index is used to denote the corresponding order correction on the specific eigenfunction and eigenvalue. Substituting this in (3) and to make this true for arbitrary $\lambda$ values we equate the corresponding terms of lambda in the equation to get, $$ H_0\psi_n^0 = E_n\psi_n^0\,\,\,...(6)\\ H_0\psi_n^1 +H'\psi_n^0 = E_n^1\psi_n^0 + E_n^0\psi_n^1\,\,\,...(7)\\ H_0\psi_n^2 + H'\psi_n^1 = E_n^1\psi_n^1 + E_n^2\psi_n^0+E_n^0\psi_n^2\,\,\,...(8) $$

Rearranging them gives, $$ H_0\psi_n^0 = E_n\psi_n^0\\ \left(H_0 - E_n^0\right)\psi_n^1 = \left(E_n^1-H'\right)\psi_n^0 \\ \left(H_0-E_n^0\right)\psi_n^2 = \left(E_n^1-H'\right)\psi_n^1 + E_n^2\psi_n^0 $$ To avoid $\psi_n^1+ \alpha\psi_n^0$ being a solution with first order energy correction, we choose the correction terms orthogonal to the initial eigenfunctions. $$\langle\psi_n^k\vert\psi_n^0\rangle = 0 \,\,...k>0$$ Assuming all correction terms are in the same Hilbert space of the initial eigenfunctions, the correction terms are expressed as a linear combination. Substituing the linear combination $ \psi_n^k = \sum_j c_{jn} \psi_j^0 $ in the first equation and taking the inner-production with $\psi_l^0$ we get, $$ (E_l^0 - E_n^0) + H'_{ln} = E_n^1\delta_{ln}$$ The first order correction in Energy eigenvalue and eigenfunction is obtained to be, $$ E_n^1 = \langle\psi_n^0\vert{H'}\psi_n^0\rangle \\ \psi_n^1 = \sum_{j\neq{n}} \frac{H'_{jn}}{E_n^0 - E_j^0}\psi_j^0 $$

Similarly solving for second order perturbation,by assuming $\psi_n^2 = \sum_j d_{nj}\psi_j^0$we get, $$ E_n^2 = \sum_{j\neq{n}}\frac{|H'_{nj}|^2}{E_n^0-E_j^0} $$ and $$\psi_n^2 = \sum_{j\neq{n}} \left[\sum_{l\neq{n}}\frac{H'_{jk}H'_{kn}}{\left(E_n^0-E_j^0\right)\left(E_n^0-E_l^0\right)}-\frac{H'_{nn}H'_{jn}}{\left(E_n^0-E_j^0\right)^2}\right]\psi_j^0$$ where $d_{nn} = 0$

Relativistic Field theory - Euler Lagrange equation

To understand some of the advanced theories in monopoles and etc., we will take a look on the relativistic field theory.

Classically, the motion of a particle is understood by solving for the position of the particle as a function of time. Where else the fields are defined over a region in space as $\phi(x_i)\,\,\,i=0,1,2,3$ where $x_0$ is the time component taken together with the space coordinates.These fields give all the information we would like to know about the system. 

In classical method, the Lagrangian is defined as a function of coordinates and its time derivatives. Since, we take here time as one of the components, the new Lagrangian for our fields is a function of both $\phi$ and its derivatives about each of the components. 
We can obtain our new Euler Lagrange equation as,
$$ \delta \int\,L(\partial_\mu\phi,\phi) d\tau = 0 $$ where the L here is called the Lagrangian density and $d\tau = dx_0 \,dx_1 \,dx_2 \,dx_3$
On expansion, $$ \int\,\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\delta(\partial_\mu\phi)+\frac{\partial{L}}{\partial\phi}\delta(\phi)\right) d\tau = 0$$ using $\delta(\partial_\mu\phi) = \partial_\mu(\delta\phi)$ and product rule we get, $$ \int \left(\frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi) }\delta\phi\right) - \frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right)\delta\phi + \frac{\partial{L}}{\partial\phi}\delta{\phi}\right) d\tau = 0 $$
The first term on integrating and applying the condition $\delta\phi$ becomes zero at the end points becomes zero. Taking out the negative sign, $$\int\left(\frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right)-\frac{\partial{L}}{\partial\phi}\right)\delta\phi\,d\tau = 0$$
gives us the final Euler-Lagrange equation for relativistic field theory, $$ \frac{\partial}{\partial{x^\mu}}\left(\frac{\partial{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial{L}}{\partial\phi} = 0 $$
It needn't to be only one function $\phi$ but more functions of $\phi_j$.
Let us consider in specific, the Klein Gordon equation which describes the motion of particles with spin zero. The 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 a scalar field $\phi(x,t)$ 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,$ g_{\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.

This can be derived from the Lagrangian, $$ L = \frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi) - \frac{1}{2}\left(\frac{mc}{\hbar}\right)^2\phi^2 $$ 

Monopoles - 9 - Dirac Monopoles in Quantum Mechanics - Part - 5

Nearly for a month, I searched for the solution of the differential equation we got in the previous post given by, $$ \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL}{dz}\right) + \left[\lambda - \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{2z-z^2}\right] L(z) = 0 $$
Finally I found the solution in a German paper by I.Tamm (which is eventually the same paper mentioned by Dirac).

It took a long time to get the paper and it took even more time to decode it into English (usual google translator buries all the meaning in it), so I took the hard way by translating each and everything as word by word.  Anyways, in the mean time I did learn a lot. 

Our equation looks similar to the general Associated Legendre equation except for the constant term "m", which is replaced with a variable term. 

Away from that, the general theory for solving a second order differential equation with variable coefficients starts from the characteristic equation. 
We will derive for the general differential equation, $$ a(x)y''(x) + b(x)y'(x) + c(x)y(x) = 0 $$ or simply $$ y''(x) + p(x)y'(x) + q(x)y(x) = 0$$ where we divided by a(x) and denote it with new variables. One should note that, a(x) shouldn't have any kind of singularities unless the differential equation itself become meaningless. 

Thus, there are different kinds of singularities and it goes with mathematical literature. Here, we focus only on regular singularities where p(x) or q(x) becomes singular at much slower rate than $\frac{1}{x}$ and $\frac{1}{x^2}$  where we take the singularity at x=0. 

For these regular singularities, it is advised to use the modified power series method known as Frobenius Power series method which we used it for Hermite polynomials, etc. $$ y = x^r \sum_{n=0}^\infty {c_n}x^{n}$$ where $c_0 \neq 0 $
Now, we define, $$ s(x) = xp(x) = \sum_{n=0}^\infty {s_n}\,x^n$$ and $$ t(x) = x^2q(x) = \sum_{n=0}^\infty {t_n}\,x^n$$ 
So, our differential equation becomes, $$ y'' +  \frac{s(x)}{x}y' + \frac{t(x)}{x^2}y = 0 $$ On substitution for y, $$ \sum_{n=0}^\infty(n+r)(n+r-1)c_n\,x^{n+r-2} +  \sum_{n=0}^\infty \frac{s(x)}{x} (n+r) c_n \,x^{n+r-1} + \sum_{n=0}^\infty \frac{t(x)}{x^2}c_n\, x^{n+r} = 0$$
or $$ \sum_{n=0}^\infty\left[(n+r)(n+r-1)+ (n+r)\,s(x) + t(x)\right] c_n\,x^{n+r-2} = 0 $$ Dividing by $x^{r-2}$ we get the equation in powers of $ x^n$ setting x=0 we get, $$ \left[(r)(r-1) + (r) s(0)+t(0)\right]c_0 = 0$$ since $c_0 \neq 0$ we have our indicial equation as, $$ r(r-1) + r s(0) + t(0) = 0 $$ where $$ s(0) = \lim\limits_{x\to{0}}\,s(x) = \lim\limits_{x\to{0}}\,x\,p(x) $$ and $$ t(0) = \lim\limits_{x\to{0}}\,x^2\,q(x) $$
With the same correspondence, our equation has singularities at two points namely, z=0 and z=2 and ofcourse $z=\infty$. Since, our domain lies from 0 to 2 we need to look out for the indicial equation at z=0 and z=2 which is obtained to be,
from our characteristic equation, 
 $r(r-1) + r zp(z) +z^2q(z) = 0$$
where $$ zp(z) = \frac{2(1-z) z}{z(2-z)}\\ z^2 q(z) = z^2\frac{\lambda}{z(2-z)} - z^2 \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{\left(z(2-z)\right)^2}$$
 
at z=0, we get for $r=r_1$ $$ r_1(r_1-1)+r_1 - \left(\frac{m+n}{2}\right)^2 = 0 \,\,\,\,\rightarrow\,\,\,\, r_1 = \pm\frac{m+n}{2}$$
and for z=2, we can change the variable by t = 2-z, 
$$L''(t) - \frac{2(t-1)}{t(2-t)}L'(t) + \left[\frac{\lambda}{t(2-t)} - \frac{\left(m+\frac{n}{2}t\right)^2}{t^2(2-t)^2}\right]L(t) = 0$$  
where use has been made that, $$ \frac{dL}{dx} = - \frac{dL}{dt} \,\,\,\,and\,\,\,\,\frac{d^L}{dx^2}=\frac{d^2L}{dt^2}$$
and substitute for t=0 to get, $ r=r_2$ from, $$ r_2(r_2-1)+r_2-\frac{m^2}{4} = 0 \,\,\,\,\rightarrow\,\,\,\,r_2 = \pm\frac{m}{2}$$

From this, we try for a similar solution we used to derive in associated legendre polynomials by the substitution, $$ ^nP^m_z = 2^{-\frac{S+M}{2}} z^{\frac{S}{2}} (2-z)^{\frac{M}{2}} \,^nV^m(z)$$ where $$ S = |n+m|\,\, and\,\, M=|m|$$ We will derive the resulting equation elaborately on next post.
[You may wonder the difference between regular and essential singularities in simple words - it is just that for a regular singularity, if you consider a plot of a function, you will find finite, continuous values for every point on the curve except for some unique points. Where else in essential singularity, all the nearby points itself tend to infinite or undefined value]

Associated Legendre Polynomial - Part - 1

The General Legendre differential equation as we know, $$ (1-x^2)y''-2xy'+n(n+1)y=0$$ is solved using the frobenius power series method and the solutions are obtained to be the general Legendre polynomials, where suitable normalization constants and boundary conditions are used. 
With this result, we proceed further to completely exploit all the possibile solutions can be obtained from this equation. 
Before that, we will have to use the orthonormalizability property of our solution which is proved as, 
From the generating function, $$ \frac{1}{\sqrt{1-2xz+z^2}}=\sum_n P_n(x) z^n$$ or $$ \frac{1}{1-2xz+z^2} = \sum_n \sum_m z^{n+m} P_n(x) P_m(x) \\ \int\frac{1}{1-2xz+z^2} \,dx = \sum_{m,n} z^{n+m} \int{P_n(x)}P_m(x) \,dx$$ The limits can change maximum from -1 to +1 as in Legendre polynomials and summation is usually implied from 0 to infinity. 
Left hand side term is evaluated to give,$$ \frac{-1}{2z}\int_{(1+z)^2}^{(1-z)^2}\frac{1}{u} du = \frac{-1}{2z} \ln{\frac{(1-z)^2}{(1+z)^2}}\\where\,\, u=1-2xz+z^2 and \,\,du = -2z\,dx$$
and $$ \frac{-1}{2z}\ln{\left(\frac{1-z}{1+z}\right)^2} = \frac{-2}{2z} \ln{\frac{1-z}{1+z}} = \frac{1}{z}\ln{\frac{(1+z)}{(1-z)}}$$
$$\ln{\frac{(1+z)}{(1-z)}} = \ln(1+z) - \ln(1-z)$$
using taylor expansion, 
$$ f(x) = f(a) + f'(a)\, (x-a) + \frac{f''(a)}{2!}\,(x-a)^2+...$$
where the function should be infinitely differentiable at x=a.
So, $$ \ln(1+z) = \ln(1+a) + \frac{1}{1+a}(z-a) - \frac{1}{2!(1+a)^2} (z-a)^2 +...$$
where we can take a=0, $$ \ln(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} +... $$ Similarly, $$ \ln(1-z) = -z - \frac{z^2}{2}  - \frac{z^3}{3} -..$$ This gives, $$ \ln(1+z) - \ln(1-z) = 2 \left[z+\frac{z^3}{3} + \frac{z^5}{5}+...\right] = 2\sum_{n} \frac{z^{2n+1}}{2n+1}$$
Thus we have our integral equal to, $$ \sum_{m,n} z^{n+m} \int_{-1}^{1} P_n(x)\,P_m(x) \,dx = \frac{1}{z} \sum_{n} \frac{z^{2n+1}}{2n+1} = 2  \sum_n \frac{z^{2n}}{2n+1}$$
From this, the left side terms will equal the right side terms only when n=m, so we get the result,
$$ when\,\, n=m\rightarrow\,\, \sum_n z^{2n} \int_{-1}^{+1}P_n(x)\,P_m(x) \,dx = 2\sum_n \frac{z^{2n}}{2n+1}\\ \rightarrow\,\,\,\, \int_{-1}^{1} P_n(x)\,P_m(x)\, dx = \frac{2}{2n+1}$$ $$ when \,\, n\neq{m}\rightarrow\,\, \int_{-1}^{1}P_n(x)\,P_m(x) \,dx = 0 \\ or \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \delta_{nm} \frac{2}{2n+1}$$ which is our orthogonality relation. 
We can also prove the completeness relation - that any arbitrary function can be expanded in terms of the linear combination of these Legendre polynomials. 
$$ f(x) = \sum_n a_n P_n(x)\\ \int_{-1}^{1} f(x)\,P_m(x)\,dx = \sum_n a_n \frac{2}{2n+1}\delta{nm} = \sum_n a_n \int_{-1}^{1}P_n(x)\,P_m(x)\,dx = \frac{2}{2m+1}a_m $$ amd $$ a_m = \frac{2m+1}{2} \int_{-1}^{1} f(x)\,P_m(x)\,dx$$
 Now, we will get back to our differential equation where we substitute our solution and differentiate it "m" times w.r.t. x 
(m < n), $$ (1-x^2)P_n''-2xP_n'+ n(n+1)P_n=0$$
$$ \frac{d^m}{dx^m}\left((1-x^2)\frac{d^2P_n}{dx^2}\right) = (1-x^2) \frac{d^{m+2}P_n}{dx^{m+2}} - m.2x.\frac{d^{m+1}P_n}{dx^{m+1}} - \frac{m(m-1)}{2!}2 \frac{d^mP_n}{dx^m}$$ 
where we made use of the Leibniz formula, $$ \frac{d^m}{dx^m} A(x) \,B(x) = \sum_{k=0}^m \frac{m!}{k!(m-k)!} \frac{d^kA}{dx^k} \frac{d^{n-k}B}{dx^{n-k}}$$
Similarly, $$ -2\frac{d^m}{dx^m}\left[x\frac{dP_n}{dx}\right] = -2x \frac{d^{m+1}P_n}{dx^{m+1}} - 2m\frac{d^mP_n}{dx^m}$$
Which finally gives, $$ \frac{d^m}{dx^m}\left[Legendre\, eqn.\right] \\= (1-x^2)\frac{d^2}{dx^2}\left(\frac{d^mP_n}{dx^m}\right) - 2(m+1)x \frac{d}{dx}\left(\frac{d^mP_n}{dx^m}\right) + \frac{d^mP_n}{dx^m}\left[n(n+1)-m(m+1)\right] = 0 $$
If we assume, $$ \frac{d^mP_n}{dx^m} = V = \frac{W}{(1-x^2)^{\frac{m}{2}}}$$ So, $$ \frac{dV}{dx} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{dW}{dx} +\frac{mxW}{1-x^2}\right]$$ and $$ \frac{d^2V}{dx^2} = \frac{1}{(1-x^2)^{\frac{m}{2}}}\left[ \frac{d^2W}{dx^2} + \frac{2mx}{1-x^2}\frac{dW}{dx}+W\left(\frac{m}{1-x^2}+ \frac{mx^2(m+2)}{(1-x^2)^2}\right)\right]$$ Substituting and doing the necessary manipulation (multiply the whole equation by $(1-x^2)^{\frac{m}{2}}$, We get our associated Legendre equation, $$ (1-x^2)\frac{d^2W}{dx^2} - 2x\frac{dW}{dx} + \left[n(n+1) - \frac{m^2}{1-x^2}\right]W = 0 $$

Eigen values of $L^2$ and $L_z$ operator

The Angular momentum operator is defined by, $$ \vec{L} = \vec{r}\times\vec{p}= -i\hbar\vec{r}\times\nabla $$ In cartesian coordinates we have the usual known coordinates, $$ {L_x} = y{p_z}-z{p_y}\\ {L_y} = zp_x-xp_z\\ L_z = xp_y-yp_x$$ where in quantum mechanics, these terms are replaced by corresponding operators so we get the commutation relations as, $$ [L_x,L_y] = i\hbar{L_z} $$ with other components in cyclic form we have, $$ [L_i,L_j] = i\hbar\epsilon_{ijk}L_k$$ using levi-civita symbol. The components of the angular momentum operators are real, Hermitian since the terms on each coordinate commute with one another. And so, the angular momentum operator itself is a Hermitian operator. 
We can also express it in spherical polar coordinates as given in 
http://scientistech.blogspot.it/2015/11/angular-momentum-operator-in-spherical.html 

From, $$[r_j,p_k] = \delta_{jk}i\hbar$$ since these are operators, Identity operator is intrinsically assumed here. 
Using these we find $L^2$ operator commutes with every other component. 
$$ [L^2, L_x] =  [L_x^2 , L_x] + [L_y^2,L_x] + [L_z^2,L_x] $$
where, $$ [L_x^2,L_x] = 0 $$ $$[L_y^2,L_x] = [L_y,L_x]L_y + L_y[L_y,L_x] = - i\hbar{L_yL_z+L_zL_y}$$
$$ [ L_z^2,L_x] = [L_z,L_x]L_z + L_z[L_z,L_x] = i\hbar{L_yL_z+L_zL_y}$$ Combining all three we get, $$ [L^2, L_x] = 0$$ and similarly for other components. And so, $$ [L^2, \vec{L}]=0$$ 
So, $L^2$ and $L_z$ can be simultaneously diagonalized. It just a convention we use $L_z$. All that matters is that we can simultaneously diagonalize the $L^2$ with one of its components but not with all the three as we know other two are non-commuting with each other. 
Now, we define two new operators as. $$ L_+ = L_x + iL_y \\ L_- = L_x - iL_y$$ where one operator is the adjoint of the other. 
Once you define an operator, you can check for its commutation relation with all other operators, such as, $$[L_+, L_-] = [L_x+iL_y, L_x-iL_y] = 2\hbar{L_z}$$ and $$ [L_z,L_+] = [L_z,L_x+iL_y] = i\hbar{L_y}+ \hbar{L_x} = \hbar{L_+}$$ and $$[L_z, L_-] = -\hbar{L_-}$$ and finally, $$L_+L_- = L_x^2 + L_y^2 +iL_yL_x-iL_xL_y = L_x^2+L_y^2+\hbar{L_z} \\ \rightarrow L^2 = L_+L_- - \hbar{L_z} + L_z^2 $$ using the commutation between the two ladder operators, $$L^2 = L_-L_++ \hbar{L_z} +L_z^2$$

Remember, we have so far not used anything physical in our arguments. It is just that we define an operator as $$ \vec{L} = \vec{r}\times\vec{p} $$ and use the commutation relation $$[r_j,p_k] = i\hbar\delta_{jk}$$ for manipulation. Other than this, it can be or mean anything. It needn't even have any physical correspondence at all. 

Now, we make use of the commutation relation between $L^2,\,and\,L_z$ and seek simultaneous eigenstates[1] for both of these operator (a physical statement from the Heisenberg's principle[2]). 
We start by defining the eigenstate as, $$ L^2|X\rangle = \lambda\hbar^2|X\rangle \\ L_z|X\rangle = m\hbar|X\rangle $$
where $\hbar $ and its square is chosen for dimensional reasons. 
Note: We are just talking about the simultaneous eigenstates and we never checked whether it is the complete set of solution or not?[1].
Now, we check for the action of $L_+$ operator on any eigenstate by making use of the commutation, $$ [L_z,L_+] = L_+\hbar \\ \rightarrow L_zL_+ = L_+L_z + \hbar{L_+}$$ and it gives, $$ L_zL_+|X\rangle = \left(L_+L_z +\hbar{L_+}\right)|X\rangle = \hbar (m+1) L_+|X\rangle$$
similarly, $$L_zL_-|X\rangle = \hbar(m-1)L_-|X\rangle$$ From the above, it is understood that if $L_z$ operator acting on the general $|X\rangle$ gives the eigen value $m\hbar$. But the same operator acting on the wavefunction produced by $L_+ \,\,and\,\,L_-$ operator gives the new eigenvalue which is respectively unit value higher and lower than the old eigenvalue. i.e.These operators works as step operator which makes the eigenfunction to jump from one state to another in fixed unit values. From,$$ L^2 = L_x^2 +L_y^2 +L_z^2$$  we see that the for a fixed "$\lambda$" value there is a maximum and minimum value for "m" since the eigenvalues of Hermitian operators are real and its square is positive quantity $$ \lambda\geq{m^2}$$
Thus, there is a maximum and minimum value for "m" such that, $$ L_+|X_{max}\rangle = 0 \\ L_-|X_{min}\rangle = 0 $$ 
and the corresponding eigenvalue of $L^2$ operator is given by, $$ L^2 = \left(L_-L_++\hbar{L_z} + L_z^2\right)|X_{max}\rangle = \hbar^2m_{max}(m_{max}+1) $$
Similarly for the lowest state , $$ L^2 |X_{min}\rangle= \left(L_+L_- - \hbar{L_z} + L_z^2\right)|X_{min}\rangle = \hbar^2 m_{min}(m_{min}-1)|X_{min}\rangle$$
But $L^2$ operator should have the same eigenvalue and so $$ m_{min}(m_{min}-1) = m_{max}(m_{max}+1) \\ \rightarrow\,\,\, m_{min} = - m_{max}=-l \,\,and\,\, m_{min} = m_{max}+1$$ where the second possibility is excluded from physical reality. So, we do have, "m" varies from -l to +l in integer steps which forces the values of "l" to be the integers and the values of "l" are 0, 1/2, 1, 3/2,2..etc.(either positive or negative).   
Thus we can exploit the properties of the eigenvalues of $L^2$ and $L_z$ operators. 

Check for :
1.Simultaneous eigenstates
2.Heisenberg's Uncertainty principle - general statement

Monopoles - 8 - Dirac Monopoles in Quantum Mechanics - Part - 4

Let us start with the motion of an electron in the field of a magnetic monopole where the usual spherical polar coordinates are described by taking the monopole at the origin. In addition we know that Every wave function describing this system should have a singularity line starting from the origin, should pass through any closed surface.  We use the equation from previous posts (Part-1,2,3)
We consider the same wave function of the type, $$ \psi = \psi_1 e^{i\beta}$$ with corresponding definition on $\beta, \vec{k},\,etc.$
But, now it was introduced two separate things as nodal line and singular line. I couldn't understand it completely, but I just want to proceed with the next step. 
The magnetic field of the monopole is given by, $$ \vec{B} = \frac{q_m}{r^2} \hat{r} $$ and $$ \nabla\times\vec{K} = \frac{e}{\hbar{c}} \vec{B}  $$ On substitution, $$ \nabla\times \vec{K} = \frac{e}{\hbar{c}} \frac{q_m}{r^2}\hat{r} \\~\\ = \frac{e}{\hbar{c}}\frac{n\hbar{c}}{2er^2}\hat{r}\\~\\ = \frac{n}{r^2}\hat{r}$$
Thus, we get the curl of K as radial with magnitude $\frac{n}{2r^2}$ 
So, the solution of K could be worked out by expanding the curl in spherical polar coordinates as, $$ \frac{1}{r^2sin\theta} \left[ \frac{\partial(rsin\theta{k_{\phi}})}{\partial{\theta}} - \frac{\partial(rk_{\theta})}{\partial{\phi}}\right] \hat{r} + \frac{1}{rsin\theta}\left[\frac{\partial(k_r)}{\partial{\phi}} - \frac{\partial(rsin\theta{k_{\phi}})}{\partial{r}}\right] \hat{\theta}+ \\~\\\frac{1}{r}\left[ \frac{\partial(r{k_{\theta}})}{\partial{r}} - \frac{\partial(k_r)}{\partial{\theta}}\right]\hat{\phi} = \frac{n}{2r^2} \hat{r} $$
Equating the components we get a solution as, $$ k_\theta = k_r =  k_0 = 0 \\~\\ k_\phi = \frac{n}{2r} tan\frac{\theta}{2}$$
Then the Schrodinger for non-relativistic electron is given by, $$ \frac{-\hbar^2}{2m} \nabla^2\psi = E\psi$$ 
Applying $$ \psi = \psi_1 e^{i\beta}$$  we get, $$ \nabla\cdot\nabla(\psi_1e^{i\beta}) = \nabla\cdot\left[e^{i\beta}\nabla(\psi_1) + \psi_1 \nabla(e^{i\beta})\right] \\~\\= e^{i\beta} \nabla^2(\psi_1) + \nabla\psi_1\cdot\nabla(e^{i\beta}) + \nabla\cdot\left[\psi_1 \nabla (e^{i\beta})\right]$$ But $$ \nabla(e^{i\beta}) = \frac{\partial(e^{i\beta})}{\partial\vec{r}} = i e^{i\beta} \frac{\partial\beta}{\partial\vec{r}} = ie^{i\beta}\vec{k}$$ (These are 3 vectors - four vectors are separately indicated) Applying this we get, $$ \nabla^2\psi = e^{i\beta}\nabla^2\psi_1 + ie^{i\beta} \nabla\psi_1\cdot \vec{k} + \nabla(\psi_1ie^{i\beta})\cdot\vec{k} + \psi_1ie^{i\beta} \nabla \cdot\vec{k} \\~\\ = e^{i\beta}\nabla^2\psi_1 + ie^{i\beta} \vec{k}\cdot\nabla\psi_1 + ie^{i\beta}\nabla\psi_1\cdot\vec{k} + \psi_1 \nabla(ie^{i\beta})\cdot\vec{k} + \psi_1 ie^{i\beta} \nabla\cdot\vec{k} \\~\\ = e^{i\beta} \nabla^2\psi_1 + ie^{i\beta} \vec{k}\cdot\nabla\psi_1 + ie^{i\beta}\nabla\psi_1\cdot\vec{k} + ie^{i\beta}\psi_1 \nabla\cdot\vec{k} - e^{i\beta}\psi_1 \vec{k}\cdot\vec{k}$$
which finally gives, $$ \nabla^2\psi = e^{i\beta}\left[ \nabla^2\psi_1 + i\vec{k}.\nabla\psi_1 + i \left(\nabla\psi_1\cdot\vec{k} + \psi_1\nabla\cdot\vec{k}\right) - k^2\psi_1\right] $$$$ \nabla^2\psi = e^{i\beta}\left[ \nabla^2 + i\vec{k}.\nabla + i (\nabla\cdot\vec{k}) - k^2\right]\psi_1$$
Now, our initial schrodinger equation can be rewritten as,
$$ \frac{-\hbar^2}{2m}\left[\nabla^2 + i\vec{k}.\nabla + i (\nabla\cdot\vec{k}) - k^2\right]\psi_1 = E\psi_1$$ 
Substituting for the values of k, we will get, 
$$\vec{k^2} = k_\phi^2 = \frac{n^2}{4r^2}tan^2{\theta/2} $$ and 
$$ \vec{k}\cdot\nabla = (\nabla\cdot\vec{k}) = \frac{k_\phi}{rsin\theta}\frac{\partial}{\partial\phi} = \frac{n\,tan\frac{\theta}{2}}{2r^2sin{\theta/2}cos{\theta/2}}\frac{\partial}{\partial\phi} = \frac{n\,sec^2{\theta/2}}{4r^2}\frac{\partial}{\partial{\phi}}$$ 
On substitution, $$ \frac{-\hbar^2}{2m}\left[ \nabla^2 + \frac{2ni}{4r^2} sec^2{\theta/2}\frac{\partial}{\partial{\phi}} - \frac{n^2tan^2{\theta/2}}{4r^2}\right]\psi_1 = E \psi_1 $$
Applying for Laplace operator in polar coordinates and using the regular separation of variables method we finally get (it is just manipulation), for Radial part, $$ \left[ \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} - \frac{\lambda}{r^2}\right]R(r) = \frac{-2mE}{\hbar^2} R(r)$$
and angular part, $$ \left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{sin^2\theta}\frac{\partial^2}{\partial\phi^2} +\frac{ni}{2} sec^2{\theta/2}\frac{\partial}{\partial\phi} - \frac{n^2}{4}tan^2{\theta/2}\right]Y(\theta,\phi) = -\lambda{Y(\theta,\phi)}$$

From here, 
we need to solve two differential equations for which I searched for the solution in various places. I couldn't find the complete solution but just the preview of the starting of the solution by I.Tamm. Only first two pages are free to see and the complete paper costs more money. So, I tried my own to convert the above Angular equation into the usual simple equation of Spherical Harmonics.  

We will start with the Angular part by assuming the solution of type, $$ Y(\theta,\phi) = L(\theta) e^{im\phi} $$ Upon substitution, $$e^{im\phi}\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) - \frac{m^2}{sin^2\theta}-\frac{mn}{2} sec^2{\theta/2} - \frac{n^2}{4}tan^2{\theta/2}\right]L(\theta) = -\lambda{e^{im\phi}}{L(\theta)} $$ With slight alterations, we can again rewrite the above as, $$\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right) - \frac{m^2}{sin^2\theta}- \frac{mn}{1+ cos{\theta}} - \frac{n^2}{4}\frac{1-cos\theta}{1+cos\theta}\right]L(\theta) = -\lambda{L(\theta)} $$ and $$\left[ \frac{1}{sin\theta} \frac{\partial}{\partial\theta}\left(sin\theta\frac{\partial}{\partial\theta}\right)+\lambda - \frac{m^2}{sin^2\theta}- \frac{mn}{1+ cos{\theta}} - \frac{n^2}{4}\frac{1-cos\theta}{1+cos\theta}\right]L(\theta) = 0 $$Now, we will try to convert this into usual equation by replacing with suitable new variable $$ z = 1+cos\theta$$,
So that, $$ \frac{dL}{d\theta} = \frac{dL}{dz} \frac{dz}{d\theta}$$ and the first term becomes $$ \frac{dz}{d\theta} = -sin\theta$$ and $$ \frac{1}{sin\theta}\frac{d}{d\theta}\left(sin\theta\left(-sin\theta\frac{dL}{dz}\right)\right) = \frac{1}{sin\theta}\left[\frac{d}{dz}\left(-sin^2\theta\frac{dL}{dz}\right)\right]\frac{dz}{d\theta}\\~\\ = \frac{d}{dz}\left(sin^2\theta\frac{dL}{dz}\right) = \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL}{dz}\right)$$ where we used the fact that, $$ cos\theta = z - 1\\~\\ sin^2\theta = 2z-z^2$$ and the second term becomes, $$ \left[\lambda - \frac{m^2}{sin^2\theta} -\frac{m}{1+cos\theta} - \frac{n^2}{4} \frac{(1-cos\theta)}{(1+cos\theta)}\right] = \left[\lambda - \frac{m^2}{2z-z^2}-\frac{m}{z}-\frac{n^2}{4}\frac{2-z}{z}\right]\\~\\ = \lambda - \left[\frac{m^2+mn(2-z)+\frac{n^2}{4}(2-z)^2}{z(2-z)}\right]\\~\\ = \lambda - \left[\frac{\left(m+\frac{n}{2}(2-z)\right)^2}{z(2-z)}\right]$$
Finally we get our differential equation as, $$ \frac{d}{dz}\left(\left(2z-z^2\right)\frac{dL(z)}{dz}\right) + \left[\lambda - \frac{\left(m+\frac{n}{2}(2-z)\right)^2}{2z-z^2}\right]L(z) = 0$$
  

Monopoles - 6 - Dirac Monopoles in Quantum Mechanics - Part - 2

From our previous post Monopoles-5-dirac-monopoles-part-1, we have for the wavefunction of the momentum operator, $$ -i\hbar \frac{\partial\psi}{\partial{x}} = -i\hbar\,e^{i\beta}\left[\frac{\partial\psi_1}{\partial{x}} + i\frac{\partial\beta}{\partial{x}}\right] = e^{i\beta} \left(-i\hbar\frac{\partial}{\partial{x}} + \hbar{k_x}\right)\psi_1$$
With all three components we get to know that, if the wavefunction $\psi$ satisfies any wave equation for the operator $\hat{p}$, then $\psi_1$ will satisfy the same wave equation for the operator $\hat{p} + \hbar\vec{k}$ Similarly, for the energy operator, $$ H\psi = i\hbar\frac{\partial\psi}{\partial{t}} = i\hbar\left[e^{i\beta}\frac{\partial\psi_1}{\partial{t}} + \psi_1ie^{i\beta}\frac{\partial\beta}{\partial{t}}\right] \\~\\ = e^{i\beta}\left[i\hbar\frac{\partial}{\partial{t}} - \hbar{k_0}\right]\psi_1$$ where the wave function $\psi_1$ will satisfy the wave equation for the operator, $ E - \hbar{k_0}$ The reason we take our wave function in the above given structure helps us to compare with a similar kind of wave equation, which we will encounter in the Gauge transformation of the wave function of free charge in an Electromagnetic field. 

Let us look at the Schrodinger equation for a particle with mass "m" and charge "e" in an Electromagnetic field described by its potentials $\vec{A}\,,\,\phi$ as, $$ i\hbar\frac{\partial\psi}{\partial{t}} = H\psi = \frac{1}{2m}\left(\vec{p} - \frac{e\vec{A}}{c}\right)^2\psi + e\phi\psi$$
$$ H\psi = \frac{\vec{p}^2}{2m}\psi -\frac{e}{2mc}\left(\vec{p}\cdot\vec{A} + \vec{A}\cdot\vec{p}\right)\psi +\frac{e^2}{2mc^2}\vec{A}^2\psi +e\phi\psi$$

But wait.. What I have done above is wrong!!

The reason is because I treated the operators as some usual terms in multiplication and did the usual product. 

Here is the main point when you deal with operators. 
Never work with operators blindly without any function. You can do all operations with operators only after it acts on any function. 

Let us try it again by acting on a function. 
$$ \left(\vec{p} - \frac{e}{c}\vec{A}\right)^2\psi = \left(\vec{p} - \frac{e}{c}\vec{A}\right)\left(\vec{p\psi}-\frac{e}{c}\vec{A\psi}\right) = \left(\vec{p}^2\psi - \frac{e}{c}\vec{p}\cdot\vec{A\psi}-\frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right)$$
But $$ \left(\vec{p}\cdot\vec{A}\right)\psi = \vec{p}\cdot\vec{A\psi} +\vec{p\psi}\cdot\vec{A} \\\rightarrow\,\,\,\,\,\,\vec{p}\cdot\vec{A\psi} =\left( \vec{p}\cdot\vec{A}\right)\psi - \vec{p\psi}\cdot\vec{A}$$
So, $$\left[\vec{p}^2\psi - \frac{e}{c}\left(\vec{p}\cdot\vec{A}\right)\psi - \frac{e}{c}\vec{A}\cdot\vec{p\psi}-\frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right]$$ Combining the terms and substituting in the Hamiltonian, we get, $$ H\psi = \frac{1}{2m}\left[\vec{p}^2\psi - \left(\frac{e}{c}\vec{p}\cdot\vec{A}\right)\psi - 2 \frac{e}{c}\vec{A}\cdot\vec{p\psi}+\frac{e^2}{c^2}\vec{A}^2\psi\right]$$
or simply, $$H\psi = \frac{\vec{p}^2}{2m}\psi -\frac{e}{2mc}\left(\vec{p}\cdot\vec{A}\right)\psi + \frac{e}{mc}\vec{A}\cdot\vec{p}\psi +\frac{e^2}{2mc^2}\vec{A}^2\psi +e\phi\psi$$
From this, the application of gauge transformation transfers the potential to new values. Accordingly, to maintain the same structure and physical results, the wave function should be transformed into a new wave function. 

It can be derived by substituting the new potentials in the Hamiltonian and comparing it with the old Hamiltonian. 

The Gauge transformation is given by, $$ \vec{A'} = \vec{A}+\nabla\chi \\~\\ V' = V - \frac{1}{c}\frac{\partial\chi}{\partial{t}}$$ The transformation of the wave function is, $$ \psi' = \psi e^{\frac{ie\chi}{\hbar{c}}}$$ 
If I make the initial Potentials zero, then $$\vec{A} = 0 \,\,\,\,\rightarrow\,\,\,\, \vec{A'} = \nabla\chi\\ V = 0 \,\,\,\,\rightarrow\,\,\,\, V' = -\frac{1}{c} \frac{\partial\chi}{\partial{t}}$$ which says that if $\psi$ satisfies the Hamiltonian where there is no Electromagnetic field i.e. in free space, then $\psi'$ will satisfy the Hamiltonian with the Electromagnetic potentials given by the above relations.  

This is exactly similar to the result we derived at the beginning. Comparing with the corresponding equations we get, $$ \beta = \frac{e\chi}{\hbar{c}}$$ and so, $$ \vec{A'} = \nabla\chi = \frac{\hbar{c}}{e}\left[\frac{\partial\beta}{\partial{x}}\hat{x} +\frac{\partial\beta}{\partial{y}}\hat{y} +\frac{\partial\beta}{\partial{z}}\hat{z}\right] = \frac{\hbar{c}}{e}\vec{k} $$ Similarly, $$ V' = \frac{-\hbar{c}}{e} \frac{1}{c} \frac{\partial\beta}{\partial{t}} = -\frac{\hbar{c}}{e}k_0$$ Thus, our initial wave equation now gets a physical meaning with its corresponding wave equation of a particle in an Electromagnetic field described by our Potentials. 

We will try to analyze completely the physical implications imposed by our potentials in the next post.