# Exact perturbation

1. Oct 11, 2005

### 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.

2. Oct 11, 2005

### 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.

3. Oct 12, 2005

### 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...

Last edited: Oct 12, 2005
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