Earth in the Universe

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. 


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

Quantum Mechanics - Postulates (Part -1) - Wave function, Hermitian Operators

I am not going to give the Postulates as it is in the books or anything, but I just want to postulate and speak its' mathematical importance, in a way I understood. 

From the Classical Physics of Lagrangian and Hamiltonian, we know that any system [it can be single particle or multi particle or anything] can be associated with a function so called Lagrangian or Hamiltonian, such that all the information about the system can be extracted from this function using the corresponding Equations of motion. 
Mathematically, we assume that the Lagrangian or Hamiltonian function contains all the necessary information we need to describe the system completely. 

In the same way, here we assume that "Every Quantum Mechanical System is completely described by an arbitrary State Vector or a Wave function $ \vert{\psi(t)}\rangle $ , read as "ket - psi" is an element of complex linear vector space called Hilbert Space. The State vector contains all the information about the system and it changes only with time.  

The state vector is an abstract concept and you can never measure this state vector or imagine it in a physical manner. 

From the concept of Vector space, we assume that it is always possible to define a set of vectors which are linearly independent and forms the basis for the Vector Space.  

Note: You needn't to panic on hearing the term Vector Space. Your Euclidean space follows the rules of Vector space. Whenever you get in trouble understanding vector space, you can always make a comparison with your 3 dimensional Euclidean space. 

The set of basis vectors needn't to be unique, but it is always possible to represent any vector in the Vector space as a linear combination of these basis vectors. 


As a consequence, you can imagine this arbitrary state vector as the linear combination of all the basis vectors.
We don't know what are these basis vectors, since there is many possible ways of choosing a set of basis vectors from different possible sets. Let us consider this as a general linear combination.  

It is represented as, $$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle = A_1(t) \vert{\phi_1}\rangle + A_2(t) \vert{\phi_2}\rangle + \ldots.....eq.(1)$$

All the basis vectors are ket vectors, after all left side should be equal to right side. And we can always make the time dependence of ket vector to come into the coefficients.  

We already said that, these are in abstract Hilbert space, so we cannot measure anything about them. 
To measure anything, we need to make the projection of this abstract quantities in the known space where we could describe the wave function completely. 

To measure the projection, we make the dot product of desired known parameter with this abstract Wave function. 
So that, the wave function and all its basis vectors are now described using our desired known parameter. 

For example, if the desired known parameter is position, then all the Wave function and its basis vectors will be projected into position space (where position is the parameter). And so, the new projected wave function is called "Position Space Wave function".

$$ \vert{\psi(t)}\rangle = \sum_b A_b(t) \vert{\phi_b}\rangle $$

Dotted with x to give the projection in Position space, 

$$ \langle{x}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{x}\vert{\phi_b}\rangle   \,\, \ldots...eq.(2)$$

Now, the new projection of Wave function in Position space, i.e. Position Space wave function is,
$$ \psi(x,t) = \sum_b A_b(t) \phi_b(x) $$

Where $\langle{x}\vert{\psi(t)}\rangle = \psi(x,t)$ and $ \langle{x}\vert{\phi}\rangle = \phi(x)$ 

If we choose momentum as the desired known parameter, then we can dot momentum with the general wave function. It will result into, $$ \langle{p}\vert{\psi(t)}\rangle = \sum_b A_b(t) \langle{p}\vert{\phi_b}\rangle \,\,\ldots...eq.(3)$$ 
And the new wave function is called Momentum Space Wave function, $$ \psi(p,t) = \sum_b A_b(t) \phi_b(p) $$

That is all we can do with the first Postulate. 

The second Postulate is stated as, "Each dynamical variable that relates to the motion of the particle can be associated with a linear operator". 

An operator is called to be linear if it satisfies the condition, $$ \hat{Q}(c_1\psi_1 + c_2\psi_2) = c_1 \hat{Q}\psi_1 + c_2 \hat{Q}\psi_2 \,\,\,....\ldots.eq.(4)$$

With Each operator, it can be associated a linear eigen value equation such that $$ \hat{Q} \psi_i = \lambda_i \psi_i \,\,\,\,\ldots..eq.(5)$$
where $\psi_i $ is called the eigen state and 
$\lambda_i$ is called the eigen value. 

A linear operator is also an abstract concept, which is represented using a matrix. A linear operator is determined by how it acts on the basis vectors because any vector can be expanded as the linear combination of these basis vectors. 

If we know how an operator acts on the basis, then it gives us everything we need to know about the operator on that Vector Space. 

Let me represent the basis vectors as $$\vert{e_1}\rangle, \vert{e_2}\rangle, \ldots...$$

Therefore, $$ \hat{Q}\vert{\psi}\rangle = \hat{Q} \vert{e_1}\rangle + \hat{Q} \vert{e_2}\rangle + ... \,\,\ldots...eq.(6)$$


If we represent the linear operators with the matrix, knowing the matrix elements is knowing the operator itself. 

Let me take a basis vector $\vert{e_i}\rangle$ in the Hilbert Space. To understand how a linear operator works on this basis vector, we operate it on this basis and it will result some new vector. 

For example, you can consider the rotation of the coordinates as an operation that acts on the basis vectors. 

Due to linear property, $\hat{Q}\vert{e_i}\rangle$ - the new vector itself can be written again as a linear combination of the basis vectors, represented as $$ Q\vert{e_i}\rangle = \sum_k Q_{kj}\vert{e_k}\rangle $$

You can compare it with the coordinate transformation rules.   

Now, the third postulate says that, "Any observable in Quantum Mechanics is a linear Hermitian operator on the Hilbert space, where the eigenvalues are the only possible results of a precise measurement of that observable. 
Definition of a Hermitian operator:
$$ \int \psi_i^* (\hat{Q} \psi_i)\, dx = \int (\hat{Q}\psi_i)^* \psi_i \,dx \,\,\,\ldots...eq.(7)$$

Eq.(7) which gives a special property on expanding with eigenvalue eq.(5) as follows,
$$ \int \psi^* (\lambda_i \psi_i)\,dx =  \int (\lambda_i \psi_i)^* \psi_i\,dx $$
which gives, $$ \lambda_i \int \psi_i^*\psi_i \,dx = \lambda_i^* \int \psi_i^* \psi_i \,dx $$

$$(\lambda_i - \lambda_i^*) \int \psi_i^*\psi_i \,dx = 0 $$

But $\int\psi_i^*\psi_i \,dx = \int {|\psi_i|}^2 \,dx > 0 $ and it is equal to zero only when $|\psi_i| = 0 $ where wave function itself vanishes and that is not a desirable solution. 

So, the only solution is $$ \lambda_i - \lambda_i^* = 0 $$ or $$ \lambda_i = \lambda_i^* $$ It is only possible when $"\lambda_i"$ is a real number. This is a characteristic result of any Hermitian operator, which states that "The eigenvalues of an Hermitian Operator is always a real number". 
This is the reason why, eigenvalues of an operator is the only possible results on a precise measurement, because measurement should give a real number. 

There are much more things to talk about an operator and important relations like Completeness, Orthogonality, Hermiticity of an operator and etc. It should be dealt separately. 

New solution to Einstein's puzzle - "Who owns the Zebra?" or "Who owns the fish?"

I was just solving, the so called Einstein puzzle. I don't know whether it is Einstein's or not but it is a good puzzle. 

As given in Wikipedia, it was called "Who owns the Zebra? - Puzzle".

The puzzle was given as follows, 

1. There are five houses.
2. The Englishman lives in the red house.
3. The Spaniard owns the dog.
4. Coffee is drunk in the green house.
5. The Ukrainian drinks tea.
6. The green house is immediately to the right of the ivory house.
7. The Old Gold smoker owns snails.
8. Kools are smoked in the yellow house.
9. Milk is drunk in the middle house.
10. The Norwegian lives in the first house.
11. The man who smokes Chesterfields lives in the house next to the man with the fox.
12. Kools are smoked in the house next to the house where the horse is kept.
13. The Lucky Strike smoker drinks orange juice.
14. The Japanese smokes Parliaments.
15. The Norwegian lives next to the blue house.

Now, who drinks water? Who owns the zebra?
In the interest of clarity, it must be added that each of the five houses is painted a different color, and their inhabitants are of different national extractions, own different pets, drink different beverages and smoke different brands of American cigarets [sic]. One other thing: in statement 6, right means your right.
                                     — Life International, December 17, 1962

[Give some time to this puzzle and its previous solution and then scroll down to the end of this post for my new solution]

I don't know exactly, but I tried this for some 2 days. When I was travelling back to my home, I made some "hit and trial" between two choices to solve the puzzle. 
Surprisingly I got the answer with the given specifications. 
With too much joy, after reaching my home I just googled to check the answer.

Here is where I got the amazement. 

My answer is Different!!!

But it looks like it follows each and every rule. I was sure, my answer satisfies all the given conditions. 

When I checked the solutions, I found out the turning point. 

In the 11th, 12th and the 15th statement, it was used the word "next".
In the solution, "next" was used in the sense "to the right or the left". 

But I have been concluding that the fifth house and the first house could be thought as next to each other.

And my conclusion is not completely wrong because it was said in the first statement itself, there are only five houses considered.
So, I counted as the first house as the next house from the fifth one. 
Moreover it can be situated in a circular way.

Thus I end up with a new answer where the Norwegian owns the Zebra and drinks the water. 

Definitely, I believe it is a new solution. 

Away from this, there are other views such as, 
There needn't be Zebra and water. The solution can be some other things since it was never said that Zebra should be there in one of the houses. And there are some discussion on the words used in the puzzle and the way it was presented, which you can find it in the wikipedia.   

If those discussions are valid based on the words, then the above conclusion is also perfectly valid since it was never said that the houses should not be in a circular order. 


Thus I can justify my solution. 

The solution is 

For complete explanation - click Who owns the zebra solution