Particle in XY Plane Homework: Separating Variables

[gfd43tg]

Homework Statement

Homework Equations

The Attempt at a Solution

a) I know that I am allowed to separate variables, such that $\psi (r, \theta) = R(r) \Theta(\theta)$, but the part with the exponential is yet to be seen. So I put it into the Hamiltonian,
$$H = - \frac {\hbar^{2}}{2 \mu} \Big ( \frac {1}{R} \frac {\partial^{2} R}{\partial r^{2}} + \frac {1}{R} \frac {1}{r} \frac {\partial R}{\partial r} + \frac {1}{r^{2}} \frac {1}{\Theta} \frac {\partial^{2} \Theta}{\partial \theta^{2}} \Big ) + V $$

Since V is a function of r, it is not a constant, so I can't just call every term with derivatives as constants, can I? This is what I would normally be able to do, particularly if V = 0. So how do I proceed from here?

 

[mfb]

You can take another angular derivative, then all the non-trivial R terms will go away.
The problem statement is misleading, there are solutions that cannot be written in that way, e.g. superpositions of states with different m.

 

[gfd43tg]

What is the other angular derivative you are talking about? I just see the one in my equation.

 

[mfb]

"Another" as in "one more". You can calculate $\frac{dH}{d\theta}$.

 

[gfd43tg]

Wouldn't that give some sort of triple derivative on the right side of my equation for the third term in the parenthesis?

 

[mfb]

Sure.

 

[vela]

It might help you to write out the complete equation.
$$-\frac{\hbar^2}{2\mu}\left(\frac{R''}{R} + \frac{R'}{rR} + \frac{\Theta'}{r^2\Theta}\right) + V = E.$$ Multiply by $r^2$ and you can separate the terms into those depending on $r$ and those that depend only on $\theta$.

 

[gfd43tg]

Okay, so I can separate the equation

$$ -(H-V)r^{2} \frac {2 \mu}{\hbar^{2}} - \frac {r^{2}}{R} R'' - \frac {r}{R} R' = \frac {\Theta''}{\Theta} $$
But how can I solve this?

 

[mfb]

The left side is independent of theta, which means the whole equation has to have the same value for all values of theta. You can set the left side equal to some constant (it will depend on r but not on theta). This is the same as setting the derivative (with respect to theta) to zero.