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Tuesday, 5 April 2016

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]

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