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.