Central potential QM problem

sachi

this is the last part of a long question.
We first have to serparate the TISE for a central potential into two parts using separation of variables (i.e into and R(r) and Y(theta, phi). Then we take the total differential equation in R(r) and substitute R(r) = f(r)/r and get:

-(hbar^2)/2m * (d^2)f/(dr^2) + (V + c/2m(r^2))f = Ef

where c =l(l+1) where l is the angular momentum quantum no. all the mathematical derivations I can do fine.
The question notes that the equation is exactly the same form as the 1-d TISE with V replaced by V +c/2m(r^2). It then asks if we can conclude that the spectrum is the same as for the 1-d problem.
I'm a bit stumped with this. It seems obvious that if you include another term in the potential of a different form, then the energy levels should change (except for l=0), but presumably they want something better than this. Any suggestions would be greatly appreciated.

Sachi

Physics Monkey

Hi sachi,

Clearly the differential equation is the same, but you know there is more to solving a differential equation than just the equation itself. In general you need to specify things like initial conditions or boundary conditions, domain of the solutions, and so forth. Think about some of these issues and see what progress you can make.

sachi

I think I've essentially nailed the point you were hinting at (i.e we no longer have the boundary condition that psi = 0 outside the box, so we are not going to get the same set of stationary waves corresponding to psi, so the energy levels will be different). There's just one thing that you said that I'm a little bit confused about; surely the differential equation is no longer the same? What if V(r) is not proportional to 1/(r^2). then our modified potential has a completely different form and so does the differential equation?

Sachi

nrqed

You are right that it is obvious that the spectrum will not be the same!! (except for the case l=0, as you noted). The extra term contains 1 over r^é so it chqnges co,pletely the solution (as you said). An example is the infinite 1-D square well vs the infinite spherical well. For l=0, the spectrum is identical with the wavefunctions also identical, i.e. sine functions (with the width of the 1-D well replaced by the radius of the spherical well) but for l not equal to zero, the solutions are the Bessel functions and the energies are given in terms of the zeroes of those functions.

Pat

Physics Monkey

nrqed,

I think they aren't asking the rather dull question of whether two different differential equations have the same solutions. I believe the question is if you already knew the solutions to the 1d problem with potential [tex] V_{eff}(x) [/tex] (not just [tex] V(x) [/tex] ), can you carry those solutions over wholesale. The answer must in general be no. Just for starters, the variable r in the radial equation must always be positive, but you would probably have solved the 1d equation on the whole real line. It isn't hard to imagine situations where this can mess things up.

nrqed

Ah ok. Yes, this makes much more sense as a question! Sorry I did not get it at first.
And I agree with your point completely.

sorry if I was way too naive in my interpretation of the question. Thansk for your comment!

Patrick