Exact Perturbation Theory: Solving H0 and Obtaining Precise Energies

[eljose]

Let,s suppose we have a system $$H=H_{0}+\deltaH_{1}$$ where we know how to solve H0 to obtain its eigenfunctions and energies now let,s apply perturbation theory in the form:

$$E_{n}=E^{0}_{n}+<\psi_0|\delta{H_{1}}|\psi_0>$$ but now we have that dH1 is so well behaved that gives us precisely the exact energies to first order in perturbation theory in the sense that $$<\psi_0|\delta{H_{1}}|\psi_0>=E_{n}-E^0_{N}$$ that is that the potential is given in a form that gives the exact energies to first order..but my question is if this would be possible and then what would happen to the rest terms in perturbation theory...thanx.

 

[Dr Transport]

if memory serves me correctly, you can add an $$a x^4$$ term to a harmonic oscillator and solve it exactly along with doing perturbation theory to it to get the same answer.

 

[eljose]

the result could be exact if the follow integral equation for the potential were satisified...
$$E_{n}-E_{n}^0=\int_{-\infty}^{\infty}dx|\psi^0}(n,x)|^{2}V(x)dx$$
no matter how big or weak be the perturbation...