# QM - the Hydrogen atom

1. Apr 28, 2008

### maria clara

the Hamiltonian for the hydrogen atom appears as

H= (Pr)^2/2m+(L^2)/2mr^2-e^2/r

The Schroedinger equation is

((Pr)^2/2m+(L^2)/2mr^2-Ze^2/r+|E|)PHI=0

|E| because we're looking for bound states.

m - reduced mass.

{(-h/2m)[(1/r)(d^2/dr^2)r]+h^2l(l+1)/2mr^2-e^2/r+|E|}R(r)=0

(h denotes h-bar)

we change the the dependent variable to u=rR(r)

and we get

(-d^2/dr^2+l(l+1)/r^2-(2m/h^2)(e^2/r)+2m|E|/h^2)u=0

I don't understand how exactly the last equation is obtained. By the change u=rR, shouldn't we get the factor 1/r inside the brackets?

Last edited: Apr 28, 2008
2. Apr 28, 2008

### IHateMayonnaise

Maria,

In your first equation (the "Radial Equation"), you have 1/r in the first term - why not multiply both sides (or, just the left side, since the right side is just equal to zero) by r? This way, you get rid of the r in the first term, and when you distribute out the R(r) you can then make the substitution for U(r)=rR(r) (The "Radial Function"). At this point, most authors make the following substitution

$$V_{eff}=\frac{l(l+1)\hbar^2}{2mr^2}-\frac{e^2}{4\pi\epsilon_0 r}$$

which simplifies things a bit. Of course, you'll have to substitute this back in eventually, however it may make it easier to see things temporarily. Additionally, following this it becomes convenient to make the following substitution

$$\rho=\frac{\sqrt{8m|E|}r}{\hbar}$$

The math that follows this is really quite messy, however none of it is particularly difficult.

3. Apr 29, 2008

thank you(: