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Physicslad78
Nov9-08, 05:19 PM
I have a question..I am trying to solve a differential equation that arises in my research problem. Because the differential equation has no solution in terms of well known functions, I had to construct a series solution for the differential equation which is physical and agrees with the potential energy of the system (potential energy minimum at a point,then wavefunctions maximum at that particular point). Anyhow the eigenvalues of the matrix arising from solving the series are energies but are sometimes complex which makes the solution non physical!! what can be done in this case? Changing the matrix into Hermitian by adding it to the Transpose and dividing by 2 is mathematical I guess and will not produce the correct values of energies..Can anyone help please. Thanks..

N.B: The differential equation is:

(1-x^2)~ \frac{\partial ^2 F(x,y)}{\partial x}+\left(\frac{1-2x^2}{x}\right)~\frac{\partial F(x,y)}{\partial x}-\left[ax^2(1-2y^2)-c\right] F(x,y)=0

where x and y are the variables and a and c are constants

Avodyne
Nov9-08, 11:57 PM
Where did you get your differential equation? Is this supposed to be the time-independent Schrodinger equation corresponding to some hermitian hamiltonian? If so, then it's a theorem that the energy eigenvalues are real, so you must have made a mistake somewhere.

newbee
Nov10-08, 12:45 AM
What is your Hamiltonian?

Physicslad78
Nov10-08, 05:00 AM
well this equation stems from the separation of radial and angular parts and this equation above is a transformed equation of the angular part..Originally it is:

\frac{L^2}{6k^2}\Psi+\frac{w\sqrt{3}}{2}\cos 2 \gamma \sin^2 \theta \Psi=E \Psi

where L is the usual angular momentum (depending on the angles \theta and \phi and there is an extra angle \gamma that arises. I am trying to find the values of the energies (E) where w is just a constant. As you can see the solution is spherical harmonics if w=0!! I did the transformation x=\sin\theta and y=\sin\gamma and got the above equation.

Avodyne
Nov10-08, 02:37 PM
Well, you've assumed no \phi dependence (m=0 in terms of spherical harmonics), but other than than your equation seems correct. However, x is restricted to be between 0 and 1, and your function has to obey some sort of regularity condition at the endpoints, so that may be messing up your numerics.

Physicslad78
Nov10-08, 05:00 PM
Yeah I have assumed m=0 for the first case but will take m=1 later on. I am not sure what the solution could be..U mean i need to add a boundary condition at the endpoints then?

Avodyne
Nov11-08, 02:31 PM
Yes, you need some sort of boundary condition at the end points. Not sure what it should be off hand.

reilly
Nov21-08, 12:39 AM
If gamma is independent of theta, LET Z=COS(THETA). Then the equation becomes
(L^^2 -E)W= g(1-z^^2). Take matrix elements to get l(l+1)W(l) = < l |g(1-z^^2)|W>, where g is constant and |W> is the state vector. The right hand side will be a linear expression in W(l+2), W(l+1),W(l), W(l-2), W(l-1), as in a Clebsch-Gordon expansion. Tough to solve-- probably related to Mathieu functions.
Regards,
Reilly Atkinson

Physicslad78
Nov21-08, 12:27 PM
Thanks very much for all your suggestions..It is an interesting one which reilly suggested. so u mean that the resulting equation could be a Mathieu equation in \gamma ?But what about L^2 ? It will becomes a function of z i suppose. If m was non zero, would it still be as easy as that? I will try doing the transformation and see what happens...Thanks again..

Physicslad78
Nov23-08, 04:46 PM
The differential equation turns out to be the spheroidal wave equation whose solution can be written as a linear combination of Legendre Polynomials which is sensical. The problem is when one tries to find eigenvalues, I form the matrix which contains the term \cos \gamma .This makes eigenvalues unobtainable and expressed in terms of Root in Mathematica...any ideas of how I can circumvent this?:(