SE equation with a strong potential

[eljose]

let be the SE with two potentials V and V_0 with N>>>>1 a big number..

i\hbar\frac{d\psi}{dt}=-\frac{\hbar^{2}}{2m}D^{2}\psi+(V+NV_{0})\psi

then my question is how could we solve it approximately..thanks...

 

[lalbatros]

If you know a solution for V_0, then change t as tN and x² as Nx² and V as V/N. Then perform a first order development with V/N as pertubation.

Note that V0 probably depends on x, and therefore you need to manage the change of variables x² -> Nx² in the potential term too.
If V0 as a dependence like VO(x/xref), then xref² has simply to be replaced by N xref². Should be simple.

 

[eljose]

thanks..i manage in a very similar way described by you:
first i divide all equation by N e=1/N tehn we would have:

ie\hbar\frac{d\psi}{dt}=-\frac{e\hbar^{2}}{2m}D^{2}\psi+(eV+V_{0})\psi

after that i define the Hamiltonian H_{0}=-\frac{e\hbar^{2}}{2m}D^{2}\psi+V_{0}\psi to solve this i use the WKB approach as e<<<1

then after that i treat V as a perturbation and solve it to first and second order...

But what is this good for?..let,s suppose we have a Lagrangian of the form L0+V with the potential then we could add a term NV0 in the Feynman Path-integral, to obtain the K0 propagator we use the development of Taylor of S near its classical solution in the form:

S[\phi]=S[\phi_{c}]+(1/2)\delta^{2}S[\phi_{c}]\phi^{2}+...

then we evaluate this functional integral to calculte K0,for the rest we use perturbation theory to calculate the corrections to first and second order...

 

[Dr Transport]

Solve for the NV_0 and then use perturbation theory for the other potential...

 

[reilly]

Seems to me that more info is required. WKB is great, but not always valid. Can one solve with either potential; maybe one could solve exacly with both -- two square wells. Do you have a specific problem in mind? Are you talking bound states or scattering, or perhaps both? Given the magnitudes involved, will first order perturbation theory work? (One solvable case is a 1/r potential, with a very large angular momentum, with n=L*(L+1) so the effective potential is (-)q*q/r + n/(r*r), a good test case.

Regards,
Reilly Atkinson



Regards,
Reilly Atkinson