Perturbation Theory (Non-Degenerate)

7 replies · 4K views · last reply
jhosamelly
Messages
125
Reaction score
0
If I have V(x)=[itex]\frac{1}{2}[/itex]m[itex]\omega^{2}[/itex]x[itex]^{2}[/itex] (1+ [itex]\frac{x^{2}}{L^{2}}[/itex])

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?
 
[itex]H= H_{0} + H_{p}[/itex]

So basically, you have an aditional term, [itex]H_{p} = \frac{1}{2L^{2}}mω^2 x^4[/itex], that perturbates your hamiltonian.
You already know the solution for the harmonic oscillator, [itex]H= H_{0} = \hbarω(n + \frac{1}{2})[/itex], so you just have to find the corrections for the [itex]H_{p}[/itex].

hope i made myself clear ( ;
 
so does this mean my hamiltonian would be [itex]H= \hbarω(n + \frac{1}{2}) + \frac{1}{2L^{2}}mω^2 x^4[/itex] ?
 
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*(Hp)ψ0
 
I actually don't know the wave function.. That's also my prob... if i only know the wave function I'll be able to solve this.
 
Last edited:
Is this the same for an anharmonic oscillator? That is the problem about.
 
No,you use unpertubed harmonic oscillator wave function for calculation.