All right, i did manage to did it down eventually by writing
g=e^{-(K+\delta K)}(e^{i\psi/2},-i e^{-i\psi/2})^T
and solving for \delta K and \psi to first order.
The resulting differential equations for \delta K and \psi show that the first correction to K is second order and \psi comes out as...
Thanks, DrDu. That does seem to be a good idea. And your explanation explains Bill_K's suggestion.
If i label the unperturbed solutions as |g_1> and |g_2> then the vectors
g_{\pm}=e^{-K}|g_1> \pm e^{K}|g_2>
are proportional to \sigma_z eigenstates and are therefore the perturbation is...
Yes DrDu, I would be very surprised if it turns out to be incorrect...
the solution after treating it to first order, claim the authors, is
g=const (e^{i\psi/2}, -ie^{-i\psi/2})^T e^{-K}
where
\psi(r)=-\int_r^{\infty}e^{2K(r)-2K(r')} \frac{2}{\lambda}\left(\epsilon...
i'm sorry but i don't understand how that helps... the perturbation is ε+\frac{\mu\hbar^2}{2mr^2} and that term as I see it has the same eigenvalue/ expectation value for both g_+ = (1,1)^T (1+i) and g_- = (1,-1)^T (1-i). Did I misunderstand you?
Hi,
I have an equation of the form
(-i \lambda \frac{d}{dr}\sigma_z+\Delta(r)\sigma_x) g =(\epsilon + \frac{\mu \hbar^2}{2mr^2}) g
where \sigma refers to the Pauli matrices, g is a two component complex vector and the term on the right hand side of the equation is small compared to the other...