Radial side

Granted that $\Delta=l(l+1)$ , the radial equation becomes simply:

$$\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)-\frac{2mr^2}{\hbar^2}\left(V(r)-E\right)R=l(l+1)R$$ (4.1)

When finding solutions to such equations, it is generally in our best interest to isolate the leading derivatives. We can do so with the substitution:

$$u(r)=rR(r)\rightarrow R=\frac{u}{r}, \frac{dR}{dr}=\frac{r\frac{du}{dr}-u}{r^2} \ni \frac{d}{dr}(r^2\frac{dR}{dr}) = r\frac{d^2u}{dr^2}$$

$$\therefore -\frac{\hbar^2}{2m}\frac{d^2u}{dr^2}+\left(V-E+\frac{\hbar^2 l(l+1)}{2mr^2}\right)u =$$ 0

$$\blacktriangledown V = -\frac{e^2}{4\pi\epsilon_or} , \zeta = \frac{\sqrt{-2mE}}{\hbar} \blacktriangleright$$

$$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2}+\left(-\frac{e^2}{4\pi\epsilon_or}+\frac{\zeta^2 \hbar^2}{2m}+\frac{\hbar^2 l(l+1)}{2mr^2}\right)u =$$ 0

$$\frac{1}{\zeta^2}\frac{d^2u}{dr^2} = \left(1-\frac{me^2}{2\pi\epsilon_o\hbar^2\zeta^2r}+\frac{l(l+1)}{\zeta^2r^2}\right)u$$

$$\blacktriangledown \rho = \zeta r , \rho_o = \frac{me^2}{2 \pi \epsilon_o \hbar^2 \zeta} \blacktriangleright \frac{d^2u}{d\rho^2} = \left(1 - \frac{\rho_o}{\rho} + \frac{l(l+1)}{\rho^2}\right)u$$ (4.2)

Here we have considered the bound states of the electron (of course) where $E<0$ .

Typically, one looks and supposes asymptotic solutions. For $\rho \rightarrow \infty$ ,

$$\frac{d^2u}{d\rho^2}=u \longrightarrow u(\rho)=\alpha e^{-\rho} + \beta e^{\rho}$$ (4.3)

But of course, finiteness requires that $\beta=0$ As $\rho \rightarrow 0$ , the $\rho^{-2}$ term dominates $\ni$

$$\frac{d^2u}{d\rho^2}=\frac{l(l+1)u}{\rho^2}$$ (4.4)

The solution to this equation is interesting to derive. We suppose the existence of some function $z(\rho)$ , and rewrite the equation thereby:

$$\frac{du}{d\rho} = \frac{du}{dz}\frac{dz}{d\rho}$$

$$\frac{d^2u}{d\rho^2} = \frac{d}{d\rho}\left(\frac{d}{dz}u\frac{dz}{d\rho}\right)$$

$$= \frac{dz}{d\rho}\frac{d}{dz}\left(\frac{du}{dz}\right)+\frac{du}{dz}\frac{d^2z}{d\rho^2}$$

$$= \frac{d^2u}{dz^2}\left(\frac{dz}{d\rho}\right)^2 + \frac{du}{dz}\frac{d^2z}{d\rho^2}$$

$$\therefore \frac{d^2u}{d\rho^2} = \frac{l(l+1)u}{\rho^2} \blacktriangleright$$

$$= \frac{d^2u}{dz^2}\left(\frac{dz}{d\rho}\right)^2 + \frac{du}{dz}\frac{d^2z}{d\rho^2} -\frac{l(l+1)u}{\rho^2}$$ 0 (4.5)

The trick here is to assume that there exist a $z(\rho)$ such that the previous equation is a differential equation with constant coefficients, i.e.:

$$\frac{\left(\frac{dz}{d\rho}\right)^2}{\frac{l(l+1)}{\rho^2}} = 1$$

$$\frac{dz}{d\rho} = \frac{\sqrt{l(l+1)}}{\rho}$$

$$z=\sqrt{l(l+1)}\ln(\rho) = \ln(p^{\sqrt{l(l+1)}})$$ (4.6)

Our characteristic equation follows from:

$$\frac{l(l+1)}{\rho^2}u'' - \frac{\sqrt{l(l+1)}}{\rho^2}u' -\frac{l(l+1)}{\rho^2}=0$$ (4.7)

It is easy to show the resulting characteristic is $(1\pm(2l+1))/(2\sqrt{l(l+1)})$ with solutions:

$$u = \gamma exp(\frac{2(l+1)z}{2\sqrt{l(l+1)}})+\beth exp(\frac{-2lz}{2\sqrt{l(l+1)}})$$

$$= \gamma exp(\frac{2(l+1)\sqrt{l(l+1)}ln\rho}{2\sqrt{l(l+1)}}) +\beth exp(\frac{-2l\sqrt{l(l+1)}ln\rho}{2\sqrt{l(l+1)}})$$

$$= \gamma exp(ln\rho^{(l+1)})+\beth exp(ln\rho^{-l})$$

$$= \gamma \rho^{l+1} + \beth \rho^{-l}$$ (4.8)

Since this solution contains the zero point, $\beth=0$ , and we can write our asymptotic general solution in total as $u(\rho)=\rho^{l+1}e^{-\rho}v(\rho)$ , where $v(\rho)$ is some unknown function which properly completes the radial function and is to be found. Reformulating the equation thus far,

$$u(\rho) = \rho^{l+1}e^{-\rho}v(\rho)$$

$$\frac{du}{d\rho} = \rho^{l}e^{-\rho}\left((l+1-\rho)v + \rho \frac{dv}{dp}\right)$$

$$\frac{d^2u}{d\rho^2} = \rho^le^{-\rho}\left(\left(-2l-2+\rho+\frac{l(l+1)}{\rho}\right)v +2(l+1-\rho)\frac{dv}{d\rho} + \rho\frac{d^2v}{d\rho^2}\right)$$

With the radial equation as,

$$\frac{d^2u}{d\rho^2}=\left(1-\frac{\rho_o}{\rho}+\frac{l(l+1)}{\rho^2}\right)u$$ (4.9)

We can plug in the values we found above to reformulate the equation in terms of v. Via algebra which can be easily done in mind, we find:

$$\rho\frac{d^2v}{d\rho^2}+2(l+1-\rho)\frac{dv}{d\rho}+(\rho_o-2(l+1))v=0$$ (4.10)

Following typical derivations (though we will part with it eventually), we assume a solution of a series solution form, $v(\rho)=\sum^{\infty}_{j=0}a_jp^j$ . Done so, the prior equation becomes,

$$\sum^{\infty}j(j+1)a_{j+1}p^j + 2(l+1)\sum^{\infty}_{j=0}a_{j+1}p^j -2\sum^{\infty}_{j=0}ja_jp^j$$

$$+\left(p_o-2(l+1)\right)\sum^{\infty}_{j=0}a_jp^j = 0 \longrightarrow$$

$$j(j+1)a_{j+1}+2(l+1)(j+1)a_{j+1}-2ja_j + \left(p_o - 2(l+1)\right)a_j =$$ 0

$$\therefore a_{j+1} = \left(\frac{2(j+l+1)-\rho_o}{(j+1)(j+2l+2)}\right)a_j$$

$$\blacktriangledown \lim_{j\rightarrow \infty}\left(\frac{2(j+l+1)-\rho_o}{(j+1)(j+2l+2)}\right)a_j \approx \frac{2^j}{j!}\gamma$$

$$\blacktriangleright \mu(\rho) = \sum^{\infty}_{j=0}\frac{2^j}{j!}p^j = \epsilon e^{2p} \rightarrow \infty @ \rho \rightarrow \infty$$ (4.11)

Obviously the series must terminate dynamically, i.e.

$$\exists j_{max} \ni a_{j_{max}+1}=0 \rightarrow 2(j_{max}+l+1)-\rho_o=0 \blacktriangledown n = j_{max} + l + 1 \blacktriangleright \rho_o=2n$$

Our new recursion equation is:

$$a_{j+1}=\frac{2(j+l+1 -2n)}{(j+1)(j+2l+2)}a_j$$ (4.12)

We consider a few sample results of $v^{l}_{n}$ . E.g.

$$v^0_2 = a_o-a_ox$$

$$v^1_3 = a_o-\frac{a_o}{2}x$$

$$v^o_3 = a_o-2a_ox+\frac{2a_o}{3}x^2$$

$$v^l_4 = a_o-a_ox + \frac{2a_o}{10}$$

We can go on ad infinitum, but we seek an analytical solution to this equation. The asymptotically suggested form gives us a starting point for its completing function. We also suspect that the solution might follow the onion-derivative similar to that of the angular solution. With these educated guesses, we start as follows: We assume the solution has a kernel $w=e^{-x}x^q$

$$w'=-e^{-x}x^q + qe^{-x}x^{q-1}$$

$$w''=e^{-x}x^q -qe^{-x}x^{q-1}+q(q-1)e^{-x}x^{q-2}-qe^{-x}x^{q-1}$$

$$@ q=2 w''=2e^{-x}-4e^{-x}x+e^{-x}x^2,$$

$$e^x w''(q=2) = 2 -4x + x^2$$ (4.13)

We had plugged in $p=2$ as a simple test of low level results. Take the derivative of the latter equation and find, $2x-4$ a la $v^3_1 = a_o -\frac{a_o}{2}x$ . One could follow through this examination ad infinitum, but we already know the ending to this story so we assume immediately that we have happened upon the correct solution form, i.e.

$$v^p_{q-p}(x)=\daleth \left(\frac{d}{dx}\right)^p\left(e^x\left(\frac{d}{dx}\right)^qe^{-x}x^q\right)$$ (4.14)

When $\daleth = (-1)^p$ , the above equation is the associated Laguerre Polynomial, $L^p_{q-p}$ . We can bring this all home as follows. We consider our $v^3_1$ example. In that case, for $n=3$ , $l=1$ , we can write the the v completing function with coefficients in terms of the recursion variables. Using $v(\rho)=L^{2l+1}_{n-l-1}(2\rho)$ , which, due to the multiplication by two matches our $v^3_1$ example,

Now $R=\frac{u(\rho)}{r}$ , where:

$$u(\rho)=\rho^{l+1}e^{-\rho}L^{2l+1}_{n-l-1}(2\rho)$$ (4.15)

But with $\rho_o=2n$ , and

$$\rho_o=\frac{me^2}{2 \pi \epsilon_o \hbar^2 \kappa}$$

$$\kappa = \frac{me^2}{4\pi n \epsilon_o \hbar^2}$$ (4.16)

Let

$$a=\frac{4\pi \epsilon_o \hbar^2}{me^2}$$

which is the Bohr Radius we had previously derived. Then we write $\kappa = \frac{1}{an}$ . Thus $\rho = \kappa r = \frac{r}{an}$ . Ergo

$$R(r)=\frac{N}{an}\left(\frac{r}{an}\right)^l \exp(-\frac{r}{na}) L^{2l+1}_{n-l-1}\left(\frac{2r}{na}\right)$$ (4.17)

The normalization factor N is determined as follows.