# Solving the Schrödinger equation for hydrogen atom

1. Feb 4, 2009

### raul_l

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

Hello. I'd like to solve this: $$-\frac{\hbar^2}{2m}\nabla^2 \Psi(r,\theta,\phi) -U(r) \Psi(r,\theta,\phi) = E\Psi(r,\theta,\phi)$$

2. Relevant equations

3. The attempt at a solution

I can separate the variables, but that's about it.

$$\frac{1}{R(r)} \frac{d}{dr}(r^2 \frac{d}{dr}R(r))+\frac{2mEr^2}{\hbar^2} + \frac{2m\gamma r}{\hbar^2} = C_r$$

$$\frac{1}{F(\phi)} \frac{d^2}{d\phi^2} F(\phi) = C_\phi$$

$$-C_r-\frac{sin\theta}{P(\theta)} \frac{d}{d\theta} (sin\theta \frac{d}{d\theta} P(\theta))=C_\phi$$

The answer should be something like this but I don't know how to get there.
$$\psi_{nlm}(r,\vartheta,\varphi)=\sqrt{{\left(\frac{2}{n a_0} \right)}^3\frac{(n-l-1)!}{2n(n+1)!}}e^{-\rho /2} \rho^{1} L_{n-l-1}^{2l+1}(\rho)\cdot Y_{lm}(\vartheta, \varphi)$$

If somebody could offer me any idea on how to proceed that would be great. I'm not even sure of how to choose the righ boundary conditions.

2. Feb 4, 2009

3. Feb 4, 2009

### nrqed

the derivation is done in almost all quantum mechancis textbooks so you should look on up.
It's a long derivation.

There are two apporaches to this problem: the DE's are standard ones and one can simply look up the solutions. However, if you have to prove the solutions from scratch, you must start with a series expansion. Again, all the details are in QM textbooks

4. Feb 4, 2009

### raul_l

Ok, this looks pretty complicated (not suprisingly).

$$\frac{d^2}{d\phi^2} F(\phi) = C_\phi F(\phi)$$

The solution is in the form $$F(\phi)=Ae^{B\phi}$$

which gives me
$$\frac{d^2}{d\phi^2} Ae^{B\phi} = C_\phi Ae^{B\phi} \Rightarrow -AB^2 e^{B\phi} = C_\phi Ae^{B\phi}$$ and therefore $$C_\phi=-B^2$$

Due to constraints on the wavefunction $$F(\phi)=F(\phi+2\pi)$$

which means that $$Ae^{B\phi}=Ae^{B(\phi+2\pi)}=Ae^{B\phi} e^{2B\pi} \Rightarrow e^{2B\pi}=1 \Rightarrow B=im$$ where m is an integer.

Because of normalization $$\int_{-\infty}^{\infty} F(\phi) d\phi=1$$ and therefore A=1.

I have arrived at $$F(\phi)=e^{im\phi}$$
and $$C_\phi=-m^2$$

which means that $$-C_r-\frac{sin\theta}{P(\theta)} \frac{d}{d\theta} (sin\theta \frac{d}{d\theta} P(\theta))=-m^2$$

Thanks for your help so far. I'll keep on working on this.

5. Feb 5, 2009

### raul_l

I'm half way solving the colatitude equation (I know how to solve differential equations using series expansion).

However, this is a lot of work I think it would probably be a better idea to just look it up in a QM textbook.

I have one question though.
Why is $$C_r=l(l+1)$$ ? I can't find the answer anywhere.

6. Feb 5, 2009

7. Feb 8, 2009

### raul_l

I have a problem with solving the general Legendre equation.

As we know the series solution is in the form
$$\sum a_s x^s$$

I have found the formula for the coefficients
$$a_{s+2}=-a_s \frac{(l-s)(l+s+1)}{(s+1)(s+2)}$$

Now, if $$l=s$$ then $$a_{l+2}$$ and all subsequent coefficients become zero.

In the book it says that the highest coefficient $$a_l$$ is given by the formula $$a_l=\frac{(2l)!}{2^l (l!)^2}$$ but I don't see how.

If I set $$a_0=1$$ I get
$$a_s=\frac{(l-s+2)(l-s+4)...l(l+1)(l+3)...(l+s-1)}{s!}$$

If $$l=s$$ then
$$a_l=\frac{2*4*...*l(l+1)(l+3)...(2l-1)}{l!}$$

I don't know how to proceed. The best I can do is write it like this:
$$a_l=\frac{(2l)!}{1*3*...(l-1)(l+2)(l+4)...2l*l!}$$

which is pretty close to $$\frac{(2l)!}{2^l (l!)^2}$$ but not exactly.

8. Feb 8, 2009

which book?

9. Feb 8, 2009

### raul_l

Mathematical Methods For Physicists by Tai L. Chow

Of course, there is always the possibiliy of a mistake in the book.

10. Feb 8, 2009

### malawi_glenn

11. Feb 8, 2009

### raul_l

At the bottom of page 298.

12. Feb 9, 2009

### raul_l

13. Feb 9, 2009

### malawi_glenn

14. Feb 9, 2009

### raul_l

Yes, as far as I know.

Actually, they use the series expansion method to solve the Laguerre equation. But the Legendre equation is solved differently. I think it's quite clever.

15. Feb 15, 2009

### raul_l

Hello again.

I'm having trouble evaluating $$\frac{(-1)^l}{2^{2l} (l!)^2} \frac{(2l)!(l+m)!}{(l-m)!} \int_{-1}^{1} (x^2-1)^l dx$$

which should be $$\frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$$

My first thought was to use the Newton's binomial formula to evaluate the integral.
$$\int_{-1}^{1} (x^2-1)^l dx = \int_{-1}^{1} \sum_{k=0}^l \frac{l!}{k!(l-k)!} (x^2)^{l-k} (-1)^k = (\sum_{k=0}^l \frac{l!}{k!(l-k)!(2l-2k+1)} x^{2l-2k+1} (-1)^k) _{x=-1}^{x=1}$$

Since $$(2l-2k+1)$$ is always odd $$(x^{2l-2k+1})_{x=-1}^{x=1}$$ always yields 2.

Therefore
$$\frac{(-1)^l}{2^{2l} (l!)^2} \frac{(2l)!(l+m)!}{(l-m)!} \int_{-1}^{1} (x^2-1)^l dx = \frac{(-1)^l}{2^{2l} (l!)^2} \frac{(2l)!(l+m)!}{(l-m)!} \sum_{k=0}^l \frac{l!}{k!(l-k)!(2l-2k+1)} (-1)^k = \frac{(-1)^l}{2^{2l} l!} \frac{(2l)!(l+m)!}{(l-m)!} \sum_{k=0}^l \frac{1}{k!(l-k)!(2l-2k+1)} (-1)^k$$

I don't know how to proceed (or if I've been correct so far).
Btw, here x substitutes $$cos \theta$$ but I don't think that changes anything.