# Perturbation Theory (Non-Degenerate)

by jhosamelly
Tags: nondegenerate, perturbation, theory
 PF Gold P: 126 If I have V(x)=$\frac{1}{2}$m$\omega^{2}$x$^{2}$ (1+ $\frac{x^{2}}{L^{2}}$) How do I start to solve for the hamiltonian Ho, the ground state wave function ?? Calculate for the energy of the quantum ground state using first order perturbation theory?
 P: 1 $H= H_{0} + H_{p}$ So basically, you have an aditional term, $H_{p} = \frac{1}{2L^{2}}mω^2 x^4$, that perturbates your hamiltonian. You already know the solution for the harmonic oscillator, $H= H_{0} = \hbarω(n + \frac{1}{2})$, so you just have to find the corrections for the $H_{p}$. hope i made myself clear ( ;
 PF Gold P: 126 so does this mean my hamiltonian would be $H= \hbarω(n + \frac{1}{2}) + \frac{1}{2L^{2}}mω^2 x^4$ ?
P: 928

## Perturbation Theory (Non-Degenerate)

Don't you know the ground state wave function of unperturbed oscillator.you can see them elsewhere and then just evaluate(with normalized eigenfunctions)
<E>=∫ψ0*(Hp0
 PF Gold P: 126 I actually dont know the wave function.. That's also my prob... if i only know the wave function I'll be able to solve this.
 P: 928
 PF Gold P: 126 Is this the same for an anharmonic oscillator? That is the problem about.
 P: 928 No,you use unpertubed harmonic oscillator wave function for calculation.

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