Quantum Harmonic Oscillator - Operator Method - 3

From our previous post harmonic-oscillator-operator-method-2,
we got the differential equation to solve for the ground state of the harmonic oscillator problem given by, $$\frac{1}{\sqrt{2}}\left(\rho +\frac{\partial}{\partial\rho}\right)u_0 = 0$$ with corresponding scale factors for non-dimensionalization. 
Solving this we get, $$u_0 = Ae^{\frac{-\rho^2}{2}}$$ with the normalizing constant as, $$A = \frac{\sqrt{\alpha}}{\pi^{\frac{1}{4}}}$$ All other excited states are obtained from the ground state using creation operator as, $$u_n = A_n a{u_0}$$ and Energy eigenvalues are obtained as $$E_n = \hbar\omega\left(n+\frac{1}{2}\right)$$ Now, we define an operator named as Number operator and defined as $$\hat{N} = aa^\dagger \\~\\ \hat{H} = \hbar\omega\left(\hat{N} + \frac{1}{2}\right) \\~\\ \hat{N}u_n = nu_n \,\,\,\,\,\,where \,\,\,n=0,1,2,..$$ from the analogy $$E_n = \hbar\omega\left(n+\frac{1}{2}\right)\,\,\,\, n = 0,1,2,..$$ Then we can work out for the normalizing constants of all eigen functions as follows, 
Let us start with defining, $$a{u_n} = c_n u_{n+1} \\~\\ a^\dagger{u_n} = d_n u_{n-1}$$ Now, we need to normalize this new eigenfunction $u_{n-1} \,\,or\,\,u_{n+1}$ from the fact that $u_n$ is normalized. So, we get, $$\langle{u_{n-1}}\vert{u_{n-1}}\rangle = 1 = \langle \frac{a^\dagger{u_n}}{d_n}\vert\frac{a^\dagger{u_n}}{d_n}\rangle \\ \rightarrow |d_n|^2 = \langle{aa^\dagger{u_n}}\vert{u_n}\rangle \\~\\ = \langle\hat{N}u_n\vert{u_n}\rangle = ||u_n||^2 n = n$$  
where we know $a,a^\dagger$ are adjoint to each other and $\hat{N}$ is hermitian $$\left(aa^\dagger\right)^\dagger = aa^\dagger$$ and Remember $$\langle {au}\vert{au}\rangle = \langle {a^\dagger{a}u}\vert{u}\rangle = \langle{u}\vert{a^\dagger{a}u}\rangle \\~\\\neq \langle{aa^\dagger{u}}\vert{u}\rangle \neq \langle{aua^\dagger}\vert{u}\rangle$$ Similarly, $$\langle{u_{n+1}}\vert{u_{n+1}}\rangle = 1 = \langle\frac{a{u_n}}{c_n}\vert\frac{a{u_n}}{c_n}\rangle \\ \rightarrow \,\,\,\, |c_n|^2 = \langle{u_n}\vert{a^\dagger{a}}u_n\rangle$$ where $$a^\dagger {a}= \hat{N} + 1$$ and $$|c_n|^2 = \langle{u_n}\vert\left(\hat{N}+1\right)u_n\rangle = (n+1)u_n\\\rightarrow \,\,\,\, c_n = \sqrt{(n+1)}$$ Then, we get $$u_n = \left[\prod_{n=0}^{n-1} \frac{1}{\sqrt{n+1}}\right](a^n) u_0 = \frac{1}{\sqrt{n!}}(a^n)u_0$$  In general, $$u_n = \frac{A}{\sqrt{n!2^n}} \left(\rho - \frac{\partial}{\partial\rho}\right)^nu_0 \propto H_n(\rho)e^{-\frac{\rho^2}{2}}$$ where $H_n(\rho) $ is the Hermite polynomials. And thus we get the general solution of a Simple Linear Harmonic Oscillator as, $$u_n(\rho) = \sqrt{\frac{\alpha}{\sqrt{\pi}2^nn!}} H_n(\rho) e^{-\frac{\rho^2}{2}}$$  

Quantum Harmonic Oscillator - Operator Method - 2

We have from Harmonic oscillator operator method - 1, $$a = \sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p \\~\\ a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p\\~\\ [a,a^\dagger] = -1\\~\\ aa^\dagger = \frac{1}{\hbar\omega}\left[H - \frac{\hbar\omega}{2}\right] \\~\\ a^\dagger{a}u = \lambda{u} \\~\\ a^\dagger{a} (au) = (\lambda+1) (au) \\~\\ a^\dagger{a} (a^\dagger{u}) = (\lambda - 1)(a^\dagger{u})$$
Now, we can derive the result of Hamiltonian operator on this eigen vector, $$H (a^\dagger{u}) = \hbar\omega \left(a^\dagger{a} - \frac{1}{2}\right)(a^\dagger{u}) = \hbar\omega \left( a^\dagger{a}a^\dagger{u} - \frac{a^\dagger{u}}{2}\right) \\~\\ = \hbar\omega\left[(\lambda -1) - \frac{1}{2} \right] (a^\dagger{u})$$
Using the relation, $$Hu = \hbar\omega (a^\dagger{a} - \frac{1}{2})u = \hbar\omega\left[\lambda-\frac{1}{2}\right]u = Eu$$
With its correspondence we can rewrite,
$$H (a^\dagger{u}) = (E - \hbar\omega)(a^\dagger{u})$$
Similarly, $$H (au) = (E+\hbar\omega) (au)$$  
or the general proof, $$H (a^\dagger{u}) = Ha^\dagger{u} - a^\dagger{H}u + a^\dagger{H}u = [H,a^\dagger]u + a^\dagger{H}u \\~\\= [-\hbar\omega{a^\dagger} + E {a^\dagger}]u = (E - \hbar\omega)(a^\dagger{u})$$
where, $$[H, a^\dagger] = \hbar\omega[{aa^\dagger}, a^\dagger] +\hbar\omega[\frac{1}{2}, {a^\dagger}] \\~\\=\hbar\omega \left([ a,a^\dagger ] a^\dagger + a [ a^\dagger, a^\dagger] + [1/2, a^\dagger]\right) \\~\\ =\hbar\omega\left([a,a^\dagger]a^\dagger +0\right) = -\hbar\omega{a^\dagger}$$
As we can see, the repeated application of $"a^\dagger"$ continuously on any eigen function will give negative values for Energy. But, we have assumed the harmonic oscillator has positive energy, which restricts the negative energy states. 
So, we demand a wave function such that it gets annihilated when operated by operator $"a^\dagger"$ on its lower state. Let us say, the lowest ground state is $u_0$, then we have, $$a^\dagger{u_0} = 0$$ or $$\sqrt{\frac{m\omega}{2\hbar}}x{u_0} + \sqrt{\frac{1}{2m\omega\hbar}} i p{u_0} = 0$$ Non-dimensionalization gives, $$\frac{1}{\sqrt{2}}(\rho +\frac{\partial}{\partial{\rho}})u_0 = 0$$ where $\rho = \alpha{x} $ and $ \alpha = \sqrt{\frac{m\omega}{\hbar}}$ 

By solving this, the ground state wave function can be obtained. Once we get the ground state wave function, using creation operator we can derive all other eigen states. 

Quantum Harmonic Oscillator - Operator Method - 1

We have seen, how to obtain the formal solution of Harmonic oscillator using power series method. Now, we will follow somewhat a completely different kind of new approach named operator method. 

We will start with defining two operators in dimensionless form with position and momentum operator as follows, $$a = \sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p$$ and $$a^\dagger = \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p$$ The only physical fact we use here is the commutation relation between x and p with its corresponding representation in position basis, $$[x,p] = i\hbar$$ because, $$[x,p]\psi = xp\psi - px\psi = -i\hbar \left[x\frac{\partial\psi}{\partial{x}} - \frac{\partial{(x\psi)}}{\partial{x}}\right] \\~\\ = \left[x\frac{\partial\psi}{\partial{x}} - x\frac {\partial\psi}{\partial{x}} - \psi\frac{\partial{x}}{\partial{x}}\right] \\~\\ = i\hbar\psi$$
Using this result, we will compute the following relations, 
$$[a,a^\dagger] = \left[\sqrt{\frac{m\omega}{2\hbar}} x - \sqrt{\frac{1}{2m\omega\hbar}} i p, \sqrt{\frac{m\omega}{2\hbar}} x + \sqrt{\frac{1}{2m\omega\hbar}} i p\right] \\~\\ = \frac{1}{2\hbar}[x,ip] + \frac{1}{2\hbar} [-ip,x]$$
Using, $$[x,x] = [p,p] = 0$$ and $$[x,p] = i\hbar = -[p,x]$$
$$[a,a^\dagger] = \frac{1}{2\hbar}{-2\hbar} = -1$$ $$\rightarrow\,\,\,\, [a,a^\dagger] = -1$$

Similarly, the operation $aa^\dagger$ can be calculated as, $$aa^\dagger = \frac{m\omega}{2\hbar}x^2 + \frac{i}{2\hbar}xp - \frac{i}{2\hbar}px + \frac{1}{2m\omega\hbar}p^2 = \frac{m\omega}{2\hbar}x^2 + \frac{1}{2m\omega\hbar}p^2 -\frac{1}{2}$$ Then, $$aa^\dagger = \frac{1}{\hbar\omega} [ H - \frac{\hbar\omega}{2} ]$$ or $$H = \hbar\omega \left[ aa^\dagger + \frac{1}{2} \right] = \hbar\omega \left[a^\dagger{a} - \frac{1}{2}\right]$$  

Now, if we assume "u" is an eigen function of $aa^\dagger$ operator with eigen value $\lambda$ then we have, $$a^\dagger{a}u = \lambda{u}$$ then we get, $$a^\dagger{a}(au) = (1+aa^\dagger)(au) = au + aa^\dagger{a}u \\~\\= au + a\lambda{u} = (\lambda+1)u$$
Similarly, $$a^\dagger{a}(a^\dagger{u}) = (\lambda - 1) (a^\dagger{u})$$  
What these results mean is that, if "u" is an eigen function of "$a^\dagger{a}$" operation, then there exists another eigen function "au" with "$\lambda + 1 $" eigen value and "$a^\dagger{u}$"  eigen function with "$\lambda -1 $" eigen value. 
We will continue 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. 


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