# Concerning spherical Bessel and Neumann functions

1. Oct 3, 2014

### moriheru

When transforming the Schrodinger equation into sphericall coordinates one usually substitutes
psi(r,theta,phi) into the equation and ends up with something like this:

-h(bar)^2/2m* d^2/dr^2*[rR(r)]+[V(r)+(l(l+1)*h(bar)^2)/2mr^2]*[rR(r)]=E[r R(r)]

Question 1: How do I replace the Rnl(r) with rho?
Question 2: How do I get to Neumann functions and spherical Bessel?

Sorry for the top equation! Thanks for any help.

2. Oct 3, 2014

### Simon Bridge

Q1. depends on what rho is.
Q2. You explicitly state the potential function and solve the equation.

3. Oct 3, 2014

### moriheru

On Q1
One substitutes rho for (k*r) where k is 2mE^1/2/h(bar). Yet where is the kr does one reform the equation to get kr?
On Q2
By reforming the equation with rho one gets a new differential equation and the solution is a combination of bessel and neumann. The question is how do I get there

Last edited: Oct 3, 2014
4. Oct 3, 2014

### Simon Bridge

You get the kr from the solution, not the DE. You may need $k=\frac{2m}{\hbar}\sqrt{E-V}$ instead ;)
To get a solution, you need an explicit form for V(r).

See, for example, the atomic energy levels or a spherical infinite square well. (Look them up.)
If you suspect that the solution is a combination of specific functions, you can use their form as your proposed solution.

5. Oct 3, 2014

### moriheru

The reformed equation looks like this

d^2R(rho)/d(rho)^2+2/(rho)*dR(rho)/d(rho)+[1-l(l+1)/rho^2] R(rho)=0

and the solution is said to be:

R(rho)=Aj(rho)+Bn(rho) A and B are constants

and j(rho) are the spherical bessel functions

6. Oct 3, 2014

### Simon Bridge

That would be: $$\frac{d^2}{d\rho^2} R(\rho) +\frac{2}{\rho}\frac{d}{d\rho} R(\rho) + \left[\frac{1-l(l+1)}{\rho^2}\right] R(\rho) = 0$$

Notice how there is no $V$ in the reformed equation? Where did it go?
Notice how the form of the DE? Look familiar?

However:

7. Oct 3, 2014

### moriheru

I think I have got it Rnl(r) where nl is the subscript and l the angular quantum number gamma max If I remember correctly, has the index n and one can remove the n and so Rnl(r) becomes Rl(kr)??

8. Oct 3, 2014

### moriheru

The potential is assumed to be zero, to keep things simple.

9. Oct 3, 2014

### Simon Bridge

So you are finding the wavefunction for a free particle in spherical coordinates?

10. Oct 3, 2014

### moriheru

I am trying to solve for the radial part of the spherical Schrodinger equation (BTW thanks for all replies this is a hellish question!).

11. Oct 3, 2014

### Staff: Mentor

Do you mean: "How do I rewrite the SchrÃ¶dinger equation so it uses rho instead of r?" or "How do I solve the rewritten SchrÃ¶dinger equation?"

12. Oct 3, 2014

### moriheru

Actually both but first of all the first. But I think the first can be answerd by looking at the index of the R?!

13. Oct 3, 2014

### moriheru

I have just writen the deriviation so things may be clearer:

inserting the spherical laplacian

Epsi(r)=-h(bar)^2/2m(1/rd^2/dr^2r-(1/h(bar)^2r^2)L^2

-h(bar)^2/2m*1/r*d^2/dr^2*rpsi(r)+1/2mr^2L^2psi(r)+V(r)psi(r)=E *psi(r)

substituting psi(r,theta,phi)
gives the above...

14. Oct 3, 2014

### king vitamin

You need to be more clear - it seems that you're trying to find the solutions of the Schrodinger equation in a region with a zero or constant potential, but people are having to read your mind.

In your first post, you said that you have:

$$-\frac{\hbar^2}{2m}\frac{d^2}{dr^2}\left( rR(r) \right) + \left( \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} \right)rR(r) = ErR(r) \\ \Rightarrow -\frac{\hbar^2}{2mE}\frac{d^2}{dr^2}R(r) - \frac{\hbar^2}{mEr}\frac{d}{dr}R(r) + \left( \frac{\hbar^2}{2mE}\frac{l(l+1)}{r^2} - 1 \right)R(r) = 0$$

where I've carried out the derivative, divided by Er, I'm only considering V=0 in this problem (since I'm pretty sure this is what you want). Now you want to substitute $\rho = kr = (\sqrt{2mE}/\hbar)r$. This is straight-forward for the last part in the parenthesis, so I'm guessing you're just having trouble with the derivative terms. I'll give you a big hint by reminding you of how changing coordinates changes derivatives (this is basically the product rule):

$$\frac{d}{dr} = \frac{d\rho}{dr} \frac{d}{d\rho} = k\frac{d}{d\rho}$$

Using this for the derivative terms should give you what you're looking for.

15. Oct 4, 2014

### moriheru

Yes it works thanks!!!