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.

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]

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. 

Monopoles - 5 - Dirac Monopoles in Quantum Mechanics - Part - 1

We know, Magnetic vector potential plays the crucial part in the Hamiltonian of an Electromagnetic system where the Hamiltonian formulation is the basis definition of transformation of equations from classical to Quantum.   
And Experimental results like Aharonov-Bohm effect make it look like the magnetic vector potential is inevitable in Quantum mechanics. Together, they state the importance of vector potential

But, if we consider the possibility of monopoles, then we may have to reject the concept of vector potential and need to redefine our Hamiltonian with new potentials, which in turn may result into contradictions with unexpected results. 

To prevent this, we reject at the beginning itself the possibility of an isolated magnetic monopole in our Universe, so that everyone can live in their happy little world with conventional equations.
But, Dirac first took a completely different approach with his new mathematical treatment along with vector potential and proposed a new magnetic monopole that can even co-exist with the vector potential. i.e.without any change in our old potential formalism. 
Moreover, the vector potential itself allows for a such particle to exist in nature without any violations. 

Let us look at that approach  from his 1931 paper..
First we introduce the wave function in the usual form as, $$ \vert\psi\rangle = Ae^{i\gamma} $$ where $\gamma$ is the function of x,y,z,t and also we take A - amplitude as the function of position and time since we take the general case. 
Now, the indeterminacy in this wave function can be regarded as the possible addition of any constant to the phase. That is equivalent to $$\psi = A e^{i\gamma + \chi} = Ae^{i\gamma} e^{\chi}$$ where $\chi$ is some constant. It doesn't change anything physical about the wave function, because we know that from superposition, the wave function will not change from the possible multiplication of any arbitrary real or complex number. 
So, you can never determine any wave function up to an arbitary constant which can be chosen arbitrarily to normalize the wave function. 

From this fact, the wave function cannot have a definite phase at all the points but can have only definite difference. All it does matter is the difference in $\gamma$ value. But we are not sure whether this difference is unique for any two arbitrary points, as there are many paths from going from one point to another. We are not even sure whether the closed integral of this phase change will vanish or not.  

Away from this, the definitions of our phase change in any sense should not give rise to ambiguity in the applications of the theory.       
First it is seen that the concept of phase doesn't change anything in the density function as, $$ \langle\psi\vert\psi\rangle = A^2 e^{i\gamma} e^{-i\gamma} = A^2 $$ which is a pure real number that doesn't depend on the value of phase (as the phase vanishes by its complex conjugate). 

But, it is no longer the case if we take two different wavefunctions.
$$ \langle\psi_m\vert\psi_n\rangle = \langle\psi_m\vert[c_1\psi_1 + c_2\psi_2 + ...+ c_m\psi_m + ...]\rangle = c_m $$
where I expanded $\psi_n$ in terms of the eigen functions $\psi_m$. 

We know from the Quantum Mechanics postulates that, $\vert{c_m}\vert^2$ gives the probability of $\psi_m$ state in $\psi_n $ state which termed by dirac as, probability of agreement of the two states. 

The limits of the integral in the bra-ket notation needn't to be from $(-\infty,\infty)$.
I think, this may be the only disadvantage in Dirac notation. The limits are not represented explicitly.

Since the integral does depend on the end points, even though the wave functions don't have any definite phase, they should have definite phase difference between two points (because the integral is a number).

The same physical argument gives us that, the change in phase round the closed path should be zero. 

Let us look at it like this by saying, 

the phase of the wave function $\psi_m$ is $\gamma_m$ and for the second $\psi_n$ is $\gamma_n$ So, the phase of $ \langle\psi_m\vert\psi_n\rangle $ is $e^{i(\gamma_n - \gamma_m)}$

When we go along the path from one point to another, there is a corresponding change in phase for each wave function respectively $\chi_m\,,\,\chi_n$. 

And so, the new phase difference at this point is $$ e^{i(\gamma'_n -\gamma'_m)} = e^{i[(\gamma_n +\chi_n)-(\gamma_m+\chi_m)]}$$ For different set of two points, the phase difference can have definite values. But, we know that from the physical fact that, if we come again at the same initial point, we should have the same probability and so the same integral value on closed path integral, i.e. $$ e^{i(\gamma'_n-\gamma'_m)} = e^{i(\gamma_n -\gamma_m)} $$ or $$ \gamma_n + \chi_n -\gamma_m-\chi_m = \gamma_n - \gamma_m \\ \rightarrow \chi_n = \chi_m $$ 
Which says that the change in phase of $\psi_m$ and $\psi_n$ should be the same and opposite around a closed path. 

Since it is a general result, it can be stated as in Dirac's paper, 
The change in phase of a wave function round any closed curve must be the same for all the wave functions. 

The change in phase doesn't talk anything about the nature of wave function or concerned with any specific system. So, the change in phase should be something a property of the dynamical system or the force field in which it moves. 

For the mathematical treatment, it was expressed the wave function as, $$ \psi = \psi_1 e^{i\beta}$$ $\psi_1$ being the usual wave function with definite phase and the uncertainty in phase is put in the factor $ e^{i\beta}$ where $beta$ is the same as $\chi$ we used in the previous. This $\beta$ having definite values at each point is not a function of x,y,z,t. (because different paths at the same point will possibly give different values of phase change). But it has definite derivatives at each point (x,y,z,t). 

We represent its derivatives as, $$ k_x =\frac{\partial\beta}{\partial{x}} \\ k_y =\frac{\partial\beta}{\partial{y}}\\k_z =\frac{\partial\beta}{\partial{z}}\\k_0 =\frac{\partial\beta}{\partial{t}}$$ In general these derivatives needn't be integrable following the condition $$ \frac{\partial\beta}{\partial{y}\partial{x}} = \frac{\partial\beta}{\partial{x}\partial{y}}$$ 

Now, using Stokes' theorem, we try to calculated the change in phase around a closed path as, $$ \oint \vec{K}\cdot\vec{dl} = \int (\nabla\times\vec{K})\cdot\vec{dS} $$ 
where the length and area element is considered in four dimensions, since K has four components. 

Here is one essential point we shouldn't assume, that is all the wave functions should have the same phase factor $e^{i\beta}$ because of the fact that all wave functions have same phase difference along a closed path. 

The reason is because only the change in phase depend on the curl of K vector. We can still change the components by the gradient of any scalar function, so that same phase difference can be obtained to be the same.
   
We can start from here in the next post.    
      

Monopoles - 4 - Thomson dipole

Let us start with a new problem that involves an imaginary magnetic monopole, producing magnetic field that follows inverse square law similar to the electrostatic case. 
As a consequence, we can exploit some more information about this system.
A combination of electric monopole and a magnetic monpole is known to be Thomson dipole. 
Here, we assume the magnetic field of the magnetic charge is equal to , $$ \vec{B} = \frac{q_m}{r^2}\hat{r}$$ Just from the Lorentz force, it is known that Magnetic force can never do work on a electric charge since it is always perpendicular to the velocity. Similarly, the electric force can never do work on magnetic charge. 
Mathematically, work done by the Magnetic field (of the magnetic charge) on electric charge is, $$ dW = \int \vec{F_{em}}\cdot\vec{dl} = q_e \int(\frac{\vec{v}}{c}\times\vec{B})\cdot\vec{dl} \\~\\ = q_e \int (\vec{v}\times\vec{B})\cdot \vec{v}\, dt = 0 $$
Similarly, work done by Electric field on Magnetic charge is always zero, $$dW = \int\vec{F_{me}} \cdot\vec{dl} = \frac{-\vec{v}}{c}\times\vec{E}\cdot\vec{v}\, dt = 0 $$ 
From this fact, we can be sure that the particles will not gain any kinetic energy from Work-Energy theorem. So, the velocity of the particles should be a constant. 

Other than this, you can also determine the basic constants of motion you can find in any mechanics problem i.e.Total Energy, total linear and angular momentum. 

Except now, the definition of momentum is upgraded to Electromagnetic momentum which has wider applicability than the former definition.    

There are  some astonishing results about this system because of the above fact. 
First, if you solve the system completely you will find out that the motion of the particle will lie only along the surface of cone centered about an axis passing through the direction of a constant vector quantity i.e. $$\vec{L'} = (\vec{r}\times{m\vec{v}}) - q_eq_m\hat{r} $$. 

where the spherical coordinates are measured from where the z-axis lie along the direction of vector Q.

And the Second which is very peculiar that, 
this system has some intrinsic angular momentum stored in its fields which is independent of the distance between the charges. This in fact happens to be the basis for the idea of quantization of the Electric and Magnetic charge in Quantum mechanics - one of the famous ideas of Dirac.

It is derived as follow, 
The magnetic field, $$ \vec{B} = \frac{q_m}{r^3}\vec{r}$$ and the electric field is placed at a distance "d" from this origin. So,
the Electric field, $$ \vec{E} = \frac{q_e}{r'^3} \vec{r'} $$ 
Using vector addition rule, $$ \vec{r'} = \vec{r} + \vec{d} $$ where $\vec{d}$ is directed from electric charge to magnetic charge. 
Henceforth, $$ \vec{E} = \frac{q_e(\vec{r}+\vec{d})}{(r^2 + d^2+ 2\,r\,d\, cos\theta)^{3/2}} $$ Then linear momentum density stored in the fields is calculated as,  $$ \vec{P} = \frac{1}{4\pi{c}}\vec{E}\times\vec{B}= \frac{q_eq_m}{4\pi{c}}\frac{((\vec{r}+\vec{d})\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}\\~\\ = \frac{q_eq_m}{4\pi{c}}\frac{(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$ 
Now, finding the angular momentum density stored in the fields,
$$ \vec{l} = \vec{r}\times\vec{P} = \frac{q_eq_m}{4\pi{c}}\frac{\vec{r}\times(\vec{d}\times\vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$

Using the vector relation, $$ \vec{r}\times(\vec{d}\times\vec{r}) = \vec{d} (\vec{r}\cdot\vec{r}) - \vec{r} (\vec{r}\cdot\vec{d}) $$
$$ \vec{l} = \frac{q_eq_m}{4\pi{c}}\frac{(r^2 \vec{d} - r\,d\,cos\theta\, \vec{r})}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}}$$

To get the total angular momentum, we integrate this all over the space using spherical coordinate system where we assume $r\,cos\theta$ lie along the direction $\vec{d}$ and we can split the vector into its components as $\vec{r} = r cos\theta \hat{d} + \,\,components\,\, perpendicular\,\, to\,\, the \,\,direction\,\,\hat{d}$$ 
so that  $$ \vec{L} = \frac{q_eq_m}{4\pi{c}}\int_{space}\frac{(r^2 \,d - r^2\,d\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} d\tau $$
The other perpendicular components will integrate to the value zero. $$ \vec{L} = \frac{q_eq_md}{4\pi{c}}\int_{space}\frac{(r^2  - r^2\,cos^2\theta)}{r^3 (r^2+d^2+2rdcos\theta)^{3/2}} r^2 sin\theta \,d\theta \,d\phi\, dr $$
On integration we will get, $$ \vec{L} = \frac{q_eq_m}{c} $$ 

When we go to quantum mechanics, we have studied the angular momentum is quantized in terms $$ L = n\frac{\hbar}{2} $$
Comparing the results we get, $$ L = \frac{q_eq_m}{c} = n\frac{\hbar}{2} $$
Therefore, $$ \frac{2q_eq_m}{\hbar{c}} = Integer $$ 

which is the exact result obtained by Dirac. From this condition, even if one magnetic charge exists in nature, it would imply the quantization of all the electric charges in the Universe. 

Monopoles - 3 - Consequence of Duality transformation

The major consequence of this duality transformation is that, you can never say it for sure that whether any charged particle in Nature has only electric charge or magnetic charge or both. 
For example, let us say there are 3 planets separated very far from each other  with 3 different kind of aliens, where the laws of physics are the same and so the Maxwell's equations are equally applicable anywhere in these 3 planets. 

They will predict exactly the same results for any Electromagnetic phenomena in their universe using the common Maxwell's equations. But they needn't to have the same form. If they vary their definition of Electric and magnetic fields according to duality transformation, there is no way finding which one is true. (It is not correct use the word "true" - after all their definition are different but they will conclude the same results).


It is a possibility for the first one (let us say humans are the first type of aliens) to describe any EM phenomena with our usual definition of Electric and Magnetic field where $q^m = 0$ magnetic monopole charge is zero, and the second planet is our inverse where they define only the magnetic charge with no electric charge by choosing $q^e = 0$
Unlike it so happens that, the third Planet define their electron with both electric and magnetic charge!

So, let us leave the first two planets and go to the third planet, where we will try to get some intrinsic physical understanding of their definitions.  

In this planet, we will first take two positive electric charges with respect to our conventional Maxwell's equations $Q_1$ and $Q_2$. 
The electrostatic repulsion force is given by coulomb's law as,
$$\vec{F_{21}}=\frac{Q_1Q_2}{r^2}\hat{r_{12}}$$
But, if the aliens define it in a such a way that it has both the electric and magnetic charge as, $$ Q_1 = q_1^e + q_1^m $$ and $$ Q_2 = q_2^e + q_2^m $$
Then the two charges as we study in Electrostatics and Magneto statics (no changing Electric or Magnetic fields), the force on charge 2 due to charge 1 is given by, $$ F_{Q_2Q_1} = q_2^e \left[\vec{E} + \frac{(\vec{v}\times\vec{B})}{c}\right] + q_2^m \left[ \vec{B} - \frac{(\vec{v}\times\vec{E})}{c}\right] $$

Since we are working with static conditions, $\vec{v} = 0$.
So, $$F_{Q_2Q_1} = q_2^e\vec{E} + q_2^m\vec{B}$$ where E and B due to $Q_1$ is given by, 
$$ \vec{E} = \frac{q_1^e}{r^2}\hat{r_{12}}$$ and $$ \vec{B} = \frac{q_1^m}{r^2} \hat{r_{12}}$$ because, now we just consider the problem as the combination of Electric and Magnetic charge placed closed together at the same point.

The Net force, $$ F_{Q_2Q_1} = q_2^e\frac{q_1^e}{r^2}\hat{r_{12}} + q_2^m \frac{q_1^m}{r^2}\hat{r_{12}}$$
or simply, $$ F_{21} = \frac{(q_2^eq_1^e+q_2^mq_1^m)}{r^2}\hat{r_{12}} $$ Since, they both direct along the same direction, we will just see some Net force acting as a repulsion force. 

So, you will always see the same result independent of the assigned electric or magnetic charge to the charged particle. Physically observable results are invariant under different definitions. 

Then, how do we decide the truth? what we really mean by a magnetic monopole?

The definition Magnetic monopole is,
Given the usual conventional Maxwell's equations, where there is only Electric charge, we haven't found any particle in Nature with pure magnetic charge i.e.the particle that transformed with $\frac{\pi}{2}$ angle in duality transformation equations.  

But, still it doesn't answer whether I have an Electromagnetic charge or pure electric charge in my hand!! 

If I say, I have pure electric charge and expect to find in Nature a new particle with pure magnetic charge, then I can also expect for another particle with both Electric and Magnetic charge (i.e. Electromagnetic charge).

After all there is no any kind of specification about the quantization or anything about the charge in Maxwell's equations. It can just assume any arbitrary value of unlike the reality where it can have only discrete values (also in energy, angular momentum, etc.).

Thus, the role of Quantum Mechanics is inevitable when you talk about any subatomic particle in reality. It is the reason why, subsequent development about Magnetic monopoles were first made by Paul Dirac with his new concept of Dirac string.

Monopoles - 2 - Duality Transformation

We can derive, how an arbitrary vector transforms under the rotation of coordinate system. For example, if a point in x-y plane is given by P(x,y). The same point in a rotated coordinate frame (conventionally we take - anticlockwise as a positive angle with respect to x axis i.e. angle $\alpha$). The new coordinates can be denoted as P(x',y').

If we want to go from one system to another, the relation between old and new coordinates axes is imminent. It can be easily verified these relations, x' = x cos$\alpha$ (projection of old x-axis on new x'-axis) + y sin$\alpha$ (projection of old y-axis with new x'-axis)  and y' = x cos(90+$\alpha$) ( projection of old x-axis on new y'-axis) + y cos$\alpha$ ( projection of old y-axis on new y'-axis).

Simply they are written as, $$ x' = x cos\alpha + y sin\alpha \\ y' = -xsin\alpha + y cos\alpha $$
where x,y are measured in same units. In a similar way, Electric field and Magnetic field is the only thing you need to know, when you are dealing with Electrodynamics, which is analogues to our usual coordinate system. 

Instead of any point, Any EM phenomena can be pointed in a plane as a point where we can put Electric field on the x-axis and Magnetic field on the y-axis. 

Once we made this analogy, all the equations and condition we derive for coordinate axes can be transferred here with careful analysis. Now, we just need  the above coordinate rotation property where we change variables (x,y) $\rightarrow$ (E,B)
E,B should be in the same units. So, we prefer Gaussian system. In SI we just need to use "cB" instead of B where "c" is the speed of light used for pure dimensional reasons.

In our new system, the rotation of coordinate system is given by, $$ E' = E cos\alpha +  B sin\alpha \\ B' = -E sin\alpha + B cos\alpha $$ with  the corresponding transformation of charge densities $$ \rho_e' = \rho_e cos\alpha + rho_m sin\alpha \\ \rho_m' = -\rho_e sin\alpha + \rho_m cos\alpha $$ 
This transformation specifically known as Duality transformation. 

As in the previous case, Rotation of coordinate system doesn't change any physical fact about the location of the point, Maxwell's equations are invariant under this duality transformation. We can check it as follows, 
Maxwell's equations before transformation,


$$\nabla\cdot \vec{E} = 4\pi{\rho_e} \\ \nabla\cdot\vec{B} = 4\pi\rho_m \\ \nabla\times \vec{E} = -\frac{4\pi}{c}\vec{J_m}- \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \frac{4\pi}{c}\vec{J_e}+\frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}}$$

After the transformation, 

(1)
$$ \nabla\cdot\vec{E'} = (\nabla\cdot\vec{E}) cos\alpha + (\nabla\cdot\vec{B}) sin\alpha = 4\pi\rho_e cos\alpha + 4\pi\rho_m sin\alpha = 4\pi\rho'_e $$  

(2)$$ \nabla\cdot\vec{B'} = (\nabla\cdot\vec{B}) cos\alpha - (\nabla\cdot\vec{E}) sin\alpha = 4\pi\rho_m cos\alpha - 4\pi\rho_e sin\alpha = 4\pi\rho'_m $$ 

(3) $$ \nabla \times \vec{E'} = (\nabla\times\vec{E}) cos\alpha + (\nabla\times\vec{B}) sin\alpha \\~\\= - \frac{4\pi}{c} \vec {J_m} cos\alpha + \frac{4\pi}{c} \vec{J_e} sin\alpha - \frac{1}{c} \frac{\partial{\vec{B}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}} sin\alpha \\~\\= - \frac{4\pi}{c} \vec{J_m'} - \frac{1}{c}\frac{\partial\vec{B'}}{\partial{t}} $$

(4)    
$$\nabla \times \vec{B'} = (\nabla\times\vec{B}) cos\alpha - (\nabla\times\vec{E}) sin\alpha \\~\\= \frac{4\pi}{c} \vec {J_e} cos\alpha + \frac{4\pi}{c} \vec{J_m} sin\alpha + \frac{1}{c} \frac{\partial{\vec{E}}}{\partial{t}} cos\alpha + \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} sin\alpha \\~\\=  \frac{4\pi}{c} \vec{J_e'} + \frac{1}{c}\frac{\partial\vec{E'}}{\partial{t}}$$ 

Also, the Lorentz force, 
(5)
$$ F' = q_e' \left[ \vec{E'} + \frac{(\vec{v}\times\vec{B'})}{c}\right] + q_m' \left[\vec{B'} - \frac{(\vec{v}\times\vec{E'})}{c} \right] \\~\\ = (q_e cos\alpha + q_m sin\alpha) \left[\vec{E} cos\alpha + \vec{B} sin\alpha + \frac{(\vec{v} \times \vec{B}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{E}) sin\alpha}{c}\right] +\\~\\ (q_m cos\alpha - q_e sin\alpha ) \left[ \vec{B} cos\alpha - \vec{E} sin\alpha - \frac{(\vec{v}\times\vec{E}) cos\alpha}{c} - \frac{(\vec{v}\times\vec{B}) sin\alpha}{c}\right] $$

After doing the arithmetic manipulations (8 terms will cancel out), the remaining 8 terms are, $$ F' = q_e\vec{E} cos^2\alpha + q_m\vec{B} sin^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c} cos^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} sin^2 \alpha +\\~\\ q_e \vec{E} sin^2\alpha + q_m \vec{B} cos^2\alpha + q_e \frac{(\vec{v}\times\vec{B})}{c}sin^2\alpha - q_m \frac{(\vec{v}\times\vec{E})}{c} cos^2 \alpha \\~\\ = q_e \left[\vec{E} + \frac{(\vec{v}\times\vec{B})}{c}\right] + q_m \left[ \vec{B} - \frac{(\vec{v}\times\vec{E})}{c}\right] = F $$

Thus, we proved all the four Maxwell's equations with the Lorentz force is invariant under duality transformation. 

We should note that, the angle $\alpha$ can vary arbitrarily. There is no any restriction to the values of $\alpha$ in the classical sense.

We will see the consequence in the next post. 

Monopoles -1 - Introduction

I just want to start from the basics where the idea of mono poles come into play in Classical Electrodynamics. We can straightly start from Maxwell's equations given by, $$ \nabla\cdot \vec{E} = \frac{\rho_e}{\epsilon_0} \\ \nabla\cdot\vec{B} = 0 \\ \nabla\times \vec{E} = \frac{-\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \mu_0\vec{J_e}+\mu_0\epsilon_0\frac{\partial{\vec{E}}}{\partial{t}} $$ with conventional notation of charge and current density. 
These equations will transform into a symmetrical set of equations in vacuum where there is no charge or current as, $$ \nabla\cdot\vec{E} = 0 \\ \nabla\cdot\vec{B} = 0 \\ \nabla\times \vec{E} = \frac{-\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \mu_e\epsilon_0 \frac{\partial{\vec{E}}}{\partial{t}} $$
We don't need to put much attention towards the constant factors that shows on the front. But, these are just a matter of unit system. If you take Gaussian system, all these complexities will disappear where E and B will be measured in same units. 

From this symmetry, it will arise a question whether we can prevail this symmetry even when charges and currents are present. 

In a pure mathematical perspective, Maxwell's equations are symmetrical when it is introduced magnetic charges and currents. The new Maxwell's equations are given by, 
$$\nabla\cdot \vec{E} = \frac{\rho_e}{\epsilon_0} \\ \nabla\cdot\vec{B} = \mu_0\rho_m \\ \nabla\times \vec{E} = -\mu_0\vec{J_m}- \frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \mu_0\vec{J_e}+\mu_0\epsilon_0\frac{\partial{\vec{E}}}{\partial{t}}$$

From this symmetry, mathematically we can never differentiate Electric fields from Magnetic fields. 
The equations are invariant when you make a transformation such that $$ \vec{E} \rightarrow \vec{B} \\ \vec{B} \rightarrow -\mu_0\epsilon_0\vec{E} $$

It shows that, if there is an alternate universe where the Electric and Magnetic fields are related to the Electric and Magnetic fields in our universe in such a way as above transformation relation then both observers will explain the same result of Physics from their Maxwell's equations. 

From this fact, it can be deduced that the laws of Nature (from Maxwell's equations) does allow any stable particle with pure magnetic charge or the particle with both Electric and Magnetic charge. It will not affect our Mathematical formalism in anyway.

For the first time, I believed in its existence, but when I did learn this next line - I really got confused. 

It was mentioned that, from Duality transformation between Electric and Magnetic fields, we can never say whether an electron has electric charge or magnetic charge or both. It is just a convention, not a condition that electron should have electric charge. 

So, which transformation I should use? 
What should I choose - whether electric or magnetic or both for electrons? (to make the Maxwell's equations look more symmetric). 
Why should I expect and search specifically for magnetic mono poles with pure magnetic charge? How do I confirm?

To answer these, probably I should do elaborately on the scale transformation with the analogue of coordinate transformation rules. 

For any future reference, in a way to make the equations in much simpler form, we will use Gaussian system of Units instead of SI units. 


$$\nabla\cdot \vec{E} = 4\pi{\rho_e} \\ \nabla\cdot\vec{B} = 4\pi\rho_m \\ \nabla\times \vec{E} = -\frac{4\pi}{c}\vec{J_m}- \frac{1}{c}\frac{\partial{\vec{B}}}{\partial{t}} \\ \nabla \times \vec{B} = \frac{4\pi}{c}\vec{J_e}+\frac{1}{c}\frac{\partial{\vec{E}}}{\partial{t}}$$

Dipole Moment

     It is very important to understand the mathematical beauty of the terms occurring in the Multipole expansion. Since higher order terms vanish faster than the monopole, dipole terms, they are important only when the need of the accuracy is high. 
     As a consequence, in most of the common problems, dipole terms plays the most significant role after the monopole term . 
    
    The dipole term in the expansion is , 
$$ V(\vec{r})_{dipole} = \frac{1}{4\pi\epsilon_0} \frac{1}{r^2} \int_{V'} r'cos\theta' \rho(\vec{r'}) \,dV'  ...\ldots eq.(1)$$
To make the integrand a vector quantity, we know that
 $\hat{r} \cdot \vec{r'} = r' cos\theta' $
and so the dipole term becomes, $$ V(\vec{r})_{dipole} = \frac{\hat{r}}{4\pi\epsilon_0 r^2}\cdot \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV'$$ Terms they depend only on r' is separated and called as dipole moment, $$ \vec{p} = \int_{V'} \vec{r'} \rho(\vec{r'}) \,dV'         ...\ldots eq.(2)$$
and the dipole potential becomes, $$ V_{dipole}(\vec{r}) = \frac {\vec{p}\cdot \hat{r}}{4\pi\epsilon_0 r^2}           ...\ldots eq.(3)$$ From eq.(2) it is known that the dipole moment defined only in terms of r' and it implies that, 'dipole moment of a volume charge depends only on it distribution'.
For point charges, $$ \vec{p} = \sum_{i=1}^N q_i \vec{r'_i}$$ and for a physical dipole it becomes, $ \vec{p} = q(\vec{r'_+} - \vec{r'_-}) = q\vec{d}$ where d is the vector from -q to +q.
    But we should remember that dipole moment doesn't mean there should be only two charges. Our definition is general for any charge distribution. It so happens the physical dipole has similar kind of representation. It is always possible to ask for the dipole moment of any number of charges e.g. three charges in a triangle.
    Some properties of the dipole moment are, Change in coordinate system usually changes the dipole moment except when the total charge is zero. And if we place a physical dipole in a uniform Electric field E, it will experiences a torque and if it is non-uniform it will experience an additional force other than the torque given by, $  \vec{F} = (\vec{p}\cdot\nabla)\vec{E} $
   Each problem will give more insight. 
Similar kind of multipole expansion for a vector potential reveals that the dipole term for a vector potential as,
$$ \vec{A_{dipole}(\vec{r})} = \frac{\mu_0}{4\pi} \frac {\vec{m}\times\hat{r}}{r^2} $$ where $\vec{m}$ is the magnetic dipole moment $$ \vec{m} = I \int \,d{\vec{a}} = I\vec{a} $$
'a' is the area enclosed by the loop and 'I' is the current. Magnetic dipole moment is always independent of the coordinate system since they don't play any role. 
   As we did in Electric dipole, certain properties for a magnetic dipole are obtained and are,
In a uniform field, the net force on any loop is zero. In a non-uniform magnetic field, an infinitesimal loop of dipole moment 'm' will experience a force, $$ \vec{F} = \nabla (\vec{m}\cdot\vec{B})$$ Thus, the concept of polarization explains the newer ideas namely, bound charges and bound currents. 

Multipole Expansion and its symmetry with common dipole

       Concept of dipole - You may ask why do we need to study about these dipoles, quadruples or etc. ?

      All of the objects around us are neutral in Nature but we know that everything is made of positive and negative charges. That is it... 

      A system with total charge zero but having separate positive and negative charge is what we need to study. You can think of quadrupole, octopole, etc. Simplest of those system is a dipole.

But the real mathematical definition of dipole is not made from this physical fact. It is a completely mathematical abstraction derived from the so called Multipole expansion. 

Note: To get a feel, analyze the following sentence,
A combination of three or more or any volume distribution of charges possess dipole moment from its definition.

All our definitions are more general. It has nothing to do with dipole or quadrupole that we used to imagine in our lower classes. This derivation is applicable for all 1/r potential.  

     For general discussion, let us consider the potential for an arbitrary volume charge distribution $\rho(r')$ at a point r from the origin and r' is the distance to the source from origin. 
Coulomb potential $$ V(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int_{V'} \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|} \,dV' \ldots.. eq.(1)$$  but we know that from the vector addition,$$|\vec{r}-\vec{r'}|=R=(r^2-2\vec{r}\cdotp\vec{r'}+r'^2)^{\frac{1}{2}} = r \left(1-2\frac{\hat{r} \cdotp\vec{r'}}{r}+\frac{r'^2}{r^2}\right)^{\frac{1}{2}}$$
Hence $$ \frac{1}{R} = \frac{1}{r} \left[\frac{1}{(1-2\frac{\hat{r}\cdotp\vec{r'}}{r}+\frac{r'^2}{r^2})^{\frac{1}{2}}}\right].... eq.(2)$$
From special functions, the generating function of Legendre polynomials are given by,
$$\frac{1}{(1-2xz+z^2)^\frac{1}{2}} = \sum_{n=0}^\infty P_n(x) z^n \ldots.. eq.(3) \\~\\ \\~\\  for -1\leq x \leq1  and   |z|<1$$ 
In eq.(2) we know that $\hat{r}\cdotp\vec{r'} = r' cos\theta$ where $-1 \leq cos\theta \leq 1$ and we assumed here that source charges are near to origin and the potential is comparatively calculated very far from the origin and so, $r>>r' or \frac{r'}{r}<< 1$.

      Thus comparing eq(2) and (3) we get the general expansion of 1/R in terms of legendre polynomial as, $$\frac{1}{R} = \frac{1}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \ldots.. eq.(4)$$
Substituting this at our primary eq(1) we get, $$ V(r) = \frac{1}{4\pi {\epsilon}_0} \int_{V'} \frac{\rho(\vec{r'})}{r} \sum_{n=0}^\infty P_n(cos\theta) (\frac{r'}{r})^n \,dV' \ldots... eq.(5)$$which is the general expansion for multipole expansion.  

      But what else this expansion tells us..? 

      I think it could be examined by looking at the first few terms of the expansion of eq.(5).. $$ V(\vec{r})= \frac{1}{4\pi\epsilon_0} \left[ \frac{1}{r} \int_{V'} \rho(\vec{r'}) \,dV' + \frac{1}{r^2} \int_{V'} r' cos\theta' \rho(\vec{r'}) \,dV' + \\~\\ \frac{1}{r^3} \int_{V'} r'^2 \left(\frac{3}{2}cos^2\theta'-\frac{1}{2}\right) \rho(\vec{r'}) \,dV' + \ldots.. \right] \ldots.. eq.(6)$$

   The first term is just the familiar Electric monopole term from the definition. Where else the second term is the dipole and the third is the quadrupole term and so on. The higher order terms are useful for better approximations. 

  [We know it is a dipole and the quadrupole term because it is possible to arrive at the same resultant potential by separately taking the case of a single dipole or quadruple system]. 

  But wait.. It is quite mysterious because in the whole derivation I haven't used any distinct physical input about dipoles or quadrupoles or octopoles, etc. Then how it comes into play automatically? What is the physical significance and from where it arise into this picture?

 I think the reason maybe as simple as follows, 
 1 = 1 + [1 - 1] + [2 - 2] + [4 - 4] + ... 
The reason for the non existance of tripole term in the expansion is due to its lack of physical symmetry in its distribution [because dipole moments follows vector addition and it plays the role as the integrand].

It looks there is a much more mathematical connection than anticipated with the physical structure of these multi poles in Nature.