Solving Schrodinger Equation in Two Dimension, r and theta

[gatztopher]

Homework Statement

It's a two-part problem, the first part was deriving a Schrödinger 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
$$\Psi(r,\theta)=R(r)\Theta(\theta)$$

Homework 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 Schrödinger equation

The Attempt at a Solution

I've tried making two Schrödinger 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?

 

[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.

 

[gatztopher]

The original problem states, quote: Work out the Schrödinger 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

Thank you for your help!

 

[gatztopher]

Found a solution here: http://galileo.phys.virginia.edu/classes/751.mf1i.fall02/OrbitalEgnfns.htm

Must have gotten something wrong in my derivation of the Schrödinger equation... though what I still don't know...

 

[gabbagabbahey]

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

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

Okay, so far so good!

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)
$$\Delta^2=\frac{\partial^2}{\partial r^2}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}$$

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.

 

[gatztopher]

Right! The product rule!

Thank you!

 

[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

 

[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?

 

[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 Schrödinger...
$$-\frac{\hbar^2}{2m}[\frac{d^2}{dr^2}+\frac{1}{r}\frac{d}{dr}-\frac{m^2}{r^2}]R(r)+V(r)R(r)=ER(r)$$

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

 

[gabbagabbahey]

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

Good!

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

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?