
#1
Nov2312, 08:21 AM

P: 126

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? 



#2
Nov2512, 05:34 AM

P: 1

[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 ( ; 



#3
Nov2612, 12:35 AM

P: 126

so does this mean my hamiltonian would be [itex]H= \hbarω(n + \frac{1}{2}) + \frac{1}{2L^{2}}mω^2 x^4 [/itex] ?




#4
Nov2612, 06:58 AM

P: 987

Perturbation Theory (NonDegenerate)
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>=∫ψ_{0}*(H_{p})ψ_{0} 



#5
Nov2612, 07:02 AM

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.




#6
Nov2612, 07:09 AM

P: 987




#7
Nov2612, 07:11 AM

P: 126

Is this the same for an anharmonic oscillator? That is the problem about.




#8
Nov2612, 07:18 AM

P: 987

No,you use unpertubed harmonic oscillator wave function for calculation.



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