Putting it all together: The Deuteron

[ognik]

I would like to take a 'real' ODE where I have some intuition of what it represents, and take it through to eigenvalues and vectors. I chose a highly simplified model of Deuteron from earlier in the text. (later I might try the real thing)

Given $ -\frac{\bar{h}^2}{2M}\nabla^2\psi +V\psi=E\psi , V=V_0 \: for\: 0\le r<a, 0 $ outside

I easily rearranged this to $ \nabla^2\psi+k_1^2\psi=0 $

At this stage the text substitutes $u(r)=r\psi(r)$ and gets the radial wave equation $ \d{^2{u}}{{r}^2}+k_1^2u=0 $

1) Despite the tempting similarity, I can't put $\nabla^2= \pd{^2{}}{{r}^2}=\d{^2{}}{{r}^2} \:for\: \psi(r) $, I know I have to transform the eqtn to spherical cords.

Pls ignore next 2 points for a while, I found a separation of variables approach that I think is what I need.

2) Laplaces eqtn in spherical form is a 3 term eqtn in $ r, \theta, \phi $. If $\psi $ was independent of both angles, I could drop the last 2 terms; probably the book does something like this - I need some help on this please?

3) Then I would be left with $ \pd{}{r}(r^2 \pd{\psi}{r}) $. Although looking promising, when I substitute $\psi=\frac{u(r)}{r}$ and do the differentiations, I do not end up with $ \d{^2{u}}{{r}^2}$ - and the last term would anyway be u/r instead of just u. Clearly I'm missing something?

 

[ognik]

I would like to take a 'real' ODE where I have some intuition of what it represents, and take it through to eigenvalues and vectors. I chose a highly simplified model of Deuteron from earlier in the text. (later I might try the real thing)

Given $ -\frac{\bar{h}^2}{2M}\nabla^2\psi +V\psi=E\psi , V=V_0 \: for\: 0\le r<a, 0 $ outside

I easily rearranged this to $ \nabla^2\psi+k_1^2\psi=0 $

At this stage the text substitutes $u(r)=r\psi(r)$ and gets the radial wave equation $ \d{^2{u}}{{r}^2}+k_1^2u=0 $

OK, I need some help now please, because ...

I tried a trial solution of $ \psi =R(r)T(\theta)P(\phi)$
then $\partial\phi/\partial r=R'TP; \partial\phi/\partial \theta=RT'P; \partial\phi/\partial \phi=RTP'; $
Worked this through to get: $ \nabla^2\psi=\frac{1}{R}\d{}{r}(r^2R') +\frac{1}{Tsin\theta } \d{}{\theta}(sin\theta T') + \frac{P''}{Psin^2 \theta } = 0 $

Part A:
1) This is not going to end with $\d{^2{u}}{{r}^2} +k_1 ^2u=0 $, so maybe my idea of using some sort of symmetry to drop the 2 angle terms was right after all, but I am not sure of a valid argument to do that?
- To start with, it's a square well, but we are using spherical cords., so this would make more sense to me if we were thinking of the potential as a sphere?
- The only argument I can think of is geometric, V is given as dependent on r so the V equipotentials by inspection are the set of concentric spheres, but I can't see how $\psi$ is independent of the angles?

2) Undaunted, I tried again to work just the 1st term in r, using $u=r\psi$, but got $u''=2\psi' +r\psi''$ a hint for the correct approach here please?
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Part B:
Out of interest if I continued my separation approach above, (we haven't done PDEs yet), I gather I could set each term = a constant, Ex: $ \frac{1}{R}\d{}{r}(r^2R')=k_r $

1) Why is that valid, to me it looks like assuming the particular solution = constant?

2) I've no doubt the choice of constant comes from experience etc., what 3 constants are appropriate and why please?