Solving the Schrödinger equation for hydrogen atom

[raul_l]

Homework Statement

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

Homework Equations

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.

 

[malawi_glenn]

You might get some help here: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html

You need to know a bit about differential equations, start with what values on the constants are allowed, start with the radial equation for instance.

Standard solutions to some differential equations (you will encounter them)
http://en.wikipedia.org/wiki/Spherical_harmonic
http://en.wikipedia.org/wiki/Laguerre_polynomial#Generalized_Laguerre_polynomials

http://en.wikipedia.org/wiki/Hydrogen_atom

 

[nrqed]

Homework Statement

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

Homework Equations

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.

the derivation is done in almost all quantum mechanics 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

 

[raul_l]

Ok, this looks pretty complicated (not suprisingly).

I'll start with the azumuthal equation.

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

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.

 

[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.

 

[malawi_glenn]

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydcol.html#c1

 

[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.

 

[malawi_glenn]

which book?

 

[raul_l]

Mathematical Methods For Physicists by Tai L. Chow

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

 

[malawi_glenn]

which page?

http://www.des.upatras.gr/physics/sfetsos/Math-Methods- Physics, T.L.Chow.pdf

that book is "famous" for its many tyops.. =/

 

[raul_l]

At the bottom of page 298.

 

[raul_l]

I found a site that should answer pretty much all my questions.
http://www.physics.drexel.edu/~tim/open/hydrofin/hyd.html

What's interesting is that it doesn't use series expansion to solve the equations.

 

[malawi_glenn]

I found a site that should answer pretty much all my questions.
http://www.physics.drexel.edu/~tim/open/hydrofin/hyd.html

What's interesting is that it doesn't use series expansion to solve the equations.

they just use standard differential equations right?

 

[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.

 

[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.