# Differential equation, quantum physics

1. Aug 4, 2007

### _Andreas

1. The problem statement, all variables and given/known data

Solve for u(r) in the region r>R (i.e r goes towards infinity). Show that psi(r)=(A/r)e^-alpha*r

2. Relevant equations

$$\frac{\(d^2u(r)}{dr^2}=\alpha^2u(r)$$

$$\(u(r)=\psi(r)*r$$

3. The attempt at a solution

$$\frac{\(d^2u(r)}{dr^2}=\alpha^2u(r)\Longleftrightarrow\frac{\(d^2\psi(r)*r}{dr^2}-\alpha^2\psi(r)*r=0\Longleftrightarrow\frac{\(d^2\psi(r)}{dr^2}+\frac{\(2}{r}\psi(r)-\alpha^2\psi(r)=0\Longrightarrow\mbox{characteristical equation}\Longrightarrow$$

(z^2) + (2/r)z - alpha^2 = 0

which has the roots z = (-1/r) +/- ((1/r^2) + alpha^2 )^(1/2)

This doesn't seem correct. What have I done wrong? (BTW: sorry for not using LATEX everywhere; I didn't have enough time)

2. Aug 4, 2007

### Gokul43201

Staff Emeritus
Wouldn't it be simpler to just solve for the DE in u(r) first?

3. Aug 5, 2007

### dextercioby

Sides, characteristic equations cannot be used for ODE-s z with variable coefficients.

4. Aug 5, 2007

### _Andreas

Yes, but I cannot see how to get the 1/r term. I only seem to get Ae^-alpha*r (after the boundary condition has been imposed).

5. Aug 5, 2007

### _Andreas

This was actually something I was thinking of. Makes sense.

6. Aug 5, 2007

### Gokul43201

Staff Emeritus
Is this a solution for u(r) or psi(r)?

7. Aug 5, 2007

### _Andreas

The former I think. Now I didn't substitute psi(r)*r for u(r) as I did in the OP.

Last edited: Aug 5, 2007
8. Aug 5, 2007

### nrqed

That's right. So you found the solution for u(r) with the approprioate boundary conditions. To get psi, just divide u(r) by r!! $\psi(r) = \frac{u(r)}{r}$ and you are done.

9. Aug 5, 2007

### _Andreas

It's that simple?! I thought I had to somehow include the r term in the derivation (d^2)/dr^2 Well, thanks then!