# Solving Schrodinger Equation in Two Dimension, r and theta

1. Sep 20, 2009

### gatztopher

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

It's a two-part problem, the first part was deriving a Schrodinger equation from when x = r cos(theta) and y = r sin(theta)

I got:
$$-\frac{\hbar^2}{2m}[\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}]\Psi(r,\theta)+V(r)\Psi(r,\theta)=E\Psi(r,\theta)$$

And now I have to solve it by dividing the wave equation into
$$\Psi(r,\theta)=R(r)\Theta(\theta)$$

2. Relevant equations

This (7-B):
http://bcs.wiley.com/he-bcs/Books?a...&assetId=17333&resourceId=1342&newwindow=true
was my guide for the derivation of the Schrodinger equation

3. The attempt at a solution

I've tried making two Schrodinger equations such that
$$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}R(r)+V(r)R(r)=ER(r)$$
$$-\frac{\hbar^2}{2m}\frac{1}{r^2}\frac{\partial^2}{\partial\theta^2}\Theta(\theta)+V(r)\Theta(\theta)=E\Theta(\theta)$$

And I figure the solutions to be R=C1e^ik1r, with k1=sqrt(2m(E-V))/hbar and Theta=C2e^ik2r where k2=sqrt(2m(E-V))r/hbar

My problem: my solution for Theta has an r in it, which I equate to an r dependence, which will screw up the whole partial derivation of Psi=Theta*R right??? And also, V(r) has an r in it, which I also equate to an r dependence, which similarly botches the whole separation, no? I feel so close and yet so far! How can I solve this?

Last edited: Sep 20, 2009
2. Sep 20, 2009

### gabbagabbahey

You'd better post the original problem, because it looks like you've done the first part incorrectly.

I'll help you with the second part after the first part is corrected.

3. Sep 20, 2009

### gatztopher

The original problem states, quote: Work out the Schrodinger equation in polar coordinates r and theta, with x=rcos(theta) and y=rsin(theta), for a potential that depends on r.

First I got,
dx=cos(theta)dr-rsin(theta)dtheta
dy=sin(theta)dr+rcos(theta)dtheta

Then, I solved for dr and dtheta
dr=sin(theta)dy+cos(theta)dx
dtheta=(1/r)(cos(theta)dy-sin(theta)dx)

that I solved so
$$\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial\theta}=cos\theta\frac{\partial}{\partial r}-(1/r)sin\theta\frac{\partial}{\partial\theta}$$

$$\frac{\partial}{\partial y}=\frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial y}\frac{\partial}{\partial\theta}=sin\theta\frac{\partial}{\partial r}+(1/r)cos\theta\frac{\partial}{\partial\theta}$$

I plugged this into delta squared
$$\Delta^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$$

which became (after a little algebra)
$$\Delta^2=\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}$$

which is what's in my original equation
$$-\frac{\hbar^2}{2m}[\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta ^2}]\Psi(r,\theta)+V(r)\Psi(r,\theta)=E\Psi(r,\theta)$$

Mostly I did it all following this (7-B)
http://bcs.wiley.com/he-bcs/Books?a...&assetId=17333&resourceId=1342&newwindow=true

Last edited: Sep 20, 2009
4. Sep 20, 2009

### gatztopher

5. Sep 20, 2009

### gabbagabbahey

Okay, so far so good!

Not quite, you need to be careful when taking the derivatives:

\begin{aligned}\Delta^2 & =\left(\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)^2+\left(\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial\theta}\right)^2\\ & =\left[\cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial}{\partial r}\right)-\cos\theta\frac{\partial}{\partial r}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\left(\cos\theta\frac{\partial}{\partial r}\right)+\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)\right]\\ & \;\;\;\;+\left[\sin\theta\frac{\partial}{\partial r}\left(\sin\theta\frac{\partial}{\partial r}\right)+\sin\theta\frac{\partial}{\partial r}\left(\frac{\cos\theta}{r}\frac{\partial}{\partial\theta}\right)+\frac{\cos\theta}{r}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial r}\right)+\frac{\cos\theta}{r}\frac{\partial}{\partial\theta}\left(\frac{\cos\theta}{r}\frac{\partial}{\partial\theta}\right)\right]\end{aligned}

Terms like $\cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial}{\partial r}\right)$ and $\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\left(\cos\theta\frac{\partial}{\partial r}\right)$ are easily simplified.

For example,

$$\cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial}{\partial r}\right)=\cos^2\theta\frac{\partial^2}{\partial r^2}$$

But, terms like $\cos\theta\frac{\partial}{\partial r}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)$ and $\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)$ will require the product rule!

For example,

$$\cos\theta\frac{\partial}{\partial r}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\right)=-\frac{\sin\theta\cos\theta}{r^2}\frac{\partial}{\partial\theta}+\frac{\sin\theta\cos\theta}{r}\frac{\partial^}{\partial r \partial\theta}\right)$$

Keeping this in mind, you should find a slightly different result.

6. Sep 20, 2009

### gatztopher

Right! The product rule!

Thank you!

7. Sep 20, 2009

### gatztopher

It was simpler than that even - the results of the product rule canceled out - I simply was moving cos's and sin's left of the partial derivatives with complete absence of mind! :P

8. Sep 20, 2009

### gabbagabbahey

Okay, so what's your new Schroedinger equation? What do you get when you plug $\Psi(r,\theta)=R(r)\Theta(\theta)$ into it?

9. Sep 20, 2009

### gatztopher

I got
$$-\frac{\hbar^2}{2m}[\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial\theta ^2}]\Psi(r,\theta)+V(r)\Psi(r,\theta)=E\Psi(r,\theta)$$

Which is still quite difficult to work but fortunately a little birdie informed me that
$$L=-i\hbar(\vec{r}\times\nabla)=-i\hbar\frac{\partial}{\partial\theta}$$

which goes that
$$L\Psi(r,\theta)=-i\hbar\frac{\partial}{\partial\theta}\Psi(r,\theta)=m\hbar\Psi(r,\theta)$$

so it ends up
$$\Psi(r,\theta)=e^{im\theta}R(r)$$

which gives a normalized
$$\Theta(\theta)=\frac{1}{\sqrt{2}}e^{im\theta}$$

And this makes a new Schrodinger...
$$-\frac{\hbar^2}{2m}[\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}-\frac{m^2}{r^2}]R(r)+V(r)R(r)=ER(r)$$

And now how to derive R(r)...

Last edited: Sep 20, 2009
10. Sep 21, 2009

### gabbagabbahey

Good!

Unfortunately, I don't think the "little birdie's" comment is all that appropriate here....instead of using it, I suggest you just substitute $\psi(r,\theta)=R(r)\Theta(\theta)$ into your Schroedinger equation, and work out the solution from scratch...upon substitution, you should get:

$$\frac{-\hbar^2}{2m}\left[R''(r)\Theta(\theta)+\frac{R'(r)}{r}\Theta(\theta)+\frac{R(r)}{r^2}\Theta''(\theta)\right]+V(r)R(r)\Theta(\theta)=ER(r)\Theta(\theta)$$

The first thing you should notice is that every term involves both $r$ and $\theta$; so it is difficult to see how to solve it. But, if you multiply both sides of the equation by $$\frac{r^2}{R(r)\Theta(\theta)}$$, what happens?