# Show that the radial component of the wave function satisfies the radial equation

1. Nov 14, 2011

### vst98

1. The problem statement, all variables and given/known data
Show that radial components of the continuum electron wave function

${\left[\frac{-\hbar }{2m}\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{\hbar ^2l(l+1)}{2m r^2}-\frac{Z e^2}{r}\right]R=E R}$

where:
E=(k*hbar)2/2m
R=Rkl(r)

2. Relevant equations
so, the radial component R is given as:

${R_{\text{kl}}(r)=\frac{C_{\text{kl}}}{(2l+1)!}(2{kr})^l\text{Exp}[-\text{ikr}]*F(i/k+l+1,2l+1;2\text{ikr})}$

3. The attempt at a solution
First I have rewritten R as:

Rkl(r)= C rl Exp[-ikr] F[i/k + l + 1,2l+1;2ikr]

where in C is everything that does not depend on r.

Great, I was thinking this will be pretty straightforward to show.

My reasoning was to insert R in the above radial equation, take those derivatives and at the end I will be able the factorize R out, so that I'm left with something like
HR=AR where A is the energy.

It did not work out like that
After taking the derivatives I'm left with a mess and I don't see how to factorize R out.
I mean, derivatives of Exp[-ikr] give me back the same function (multiplied by constant) so I can factorize that out but rl and F give me problems after derivation.
Also I was looking at the properties of these confluent hypergeometric functions F
and tried to adjust them back to original function but without success.

Can someone say whether I'm on the right track trying to solve the problem like that or
have I completely missed the point ?

2. Nov 14, 2011

### dextercioby

Interesting problem. I would first try to relabel the confluent hypergeom function as u(r) and plug the solution into the original ODE. Then I'd be led to a second order ODE for u(r) which should have exactly the same solution with the confluent hypergeom function I'm given.

3. Nov 15, 2011

### vst98

I'm not sure if I understood you well, but this is what I have tried:

Now R is given as:

Rkl(r)= C rl Exp[-ikr] u[r]

And a second order ODE that u[r] obeys is:

$z\frac{d^{2}u}{dz^{^{2}}}+(b-z)\frac{du}{dz}-au=0$

where u is u[r]~F[a,b,z]=F[i/k+l+1,2l+1,2ikr]

So, I should insert R the into radial equation and with the help of this ODE for u[r] somehow simplify the result, right ?
Ok, I focused here only on the result I get when acting on R with this part of the radial equation
$\frac{1}{r^{2}}\frac{\partial }{\partial r}(r^{2}\frac{\partial }{\partial r})$

the result after acting on R with this part is:
${c e^{-i k r} r^{-2+l} \left(\left(l+l^2-2 i k l r-k r (2 i+k r)\right) u[r]+r \left(2 (1+l-i k r) u'[r]+r u''[r]\right)\right)}$

Now, I have tried to bring this expression(or a part of it) to form of ODE for u[r].
But when I adjust the coefficient that multiplies u''[r] so that I have 2ikr*u''[r] , which is required by the ODE for u[r] then the coefficient for u'[r] is not in a proper form anymore.
And the rest of the equation is also of no help since these are the only u''[r] and u'[r] terms.
So I'm stuck again :)

Last edited: Nov 15, 2011