What can be done when eigenvalues of a Non Hermitian Hamiltonian are complex?

[Physicslad78]

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]

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]

What is your Hamiltonian?

 

[Physicslad78]

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]

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]

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]

Yes, you need some sort of boundary condition at the end points. Not sure what it should be off hand.

 

[reilly]

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]

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]

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?:(