2-D Harmonic Osc. with Perturbation

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2-D Harmonic Oscillator with Perturbation

Homework Statement



A 2-D harmonic oscillator has an energy Ensubxnsuby and wavefunctions [tex]\phi[/tex]nsubxnsuby

The first excited states are 2-fold degenerate E[tex]_{01}[/tex]=E[tex]_{10}[/tex]=2[tex]\hbar[/tex][tex]\omega[/tex]

What are the energies and wavefunctions if we add the perturbation

H'=gwyp[tex]_{x}[/tex]?



Homework Equations



En=[tex]\hbar[/tex][tex]\omega[/tex](nx + ny + 1)

[tex]\phi[/tex]subnxny=[tex]\phi[/tex]subnx(x)[tex]\phi[/tex]subny(y)

H=g[tex]\omega[/tex]yPsubx

The Attempt at a Solution



I'm unsure exactly how to approach this problem, and I think it's a combination of not knowing exactly how to work with this perturbation and never having dealt with 2-D Harm. Osc. I suppose for the energies just a simple degenerate calculation of the three matrix elements of H' (where I come to trouble working with the ypsubx) and then plugging in for the two fold degeneracy equation. The wavefunctions would be a similar sum over H'mn m!=n. Also I know the wavefunctions (just nx and ny of 1-D 0 and 1 states multiplied together), just don't know where to go from here. Any help would be appreciated. THanks
 

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Firstly, welcome to PF proto0.

Secondly, a little hint to make your post more legible: I see that you use LaTeX to produce Greek letters, but the subscripts are terrible, if you don't mind me saying so. If you put the entire thing between the TeX tags (and replace the word "sub" with an underscore (_) everywhere, as you did in E[tex]_{01}[/tex] it will look much better, e.g.
The first excited states are 2-fold degenerate [tex]E_{01} = E_{10} = 2 \hbar \omega[/tex] (note how that looks better, and if you click it you'll see how I made it).

Having said that, I'll await the attachment to be approved and answer your question when I get back :smile:
(Excuse me for not really answering your question right now, but I have to go now and don't want to rush into posting something.)
 
Hi, I'm having the same question, but with a perturbation like K'xy

Beeing H' the perturbated hamiltonian, [tex]H'=H_{0}+K'xy[/tex]

I'm suposed to solve the secular equation right? By calculating the [tex]H_{ij}[/tex], is that the way?


Thanks
 
Last edited:
carapauzinho said:
Hi, I'm having the same question, but with a perturbation like K'xy

Beeing H' the perturbated hamiltonian, [tex]H'=H_{0}+K'xy[/tex]

I'm suposed to solve the secular equation right? By calculating the [tex]H_{ij}[/tex], is that the way?


Thanks

Yes, the perturbation [itex]H'=H_{0}+K'xy[/itex] couples states of different levels, remember from c your class you should have seen [itex]<n'|x|n> ~ c_{1}\delta_{n+1,n} + c_{2}\delta_{n-1,n}[/itex]. Since your perturbation is separable, you should be able to get it quickly.
 
Got it! Thanks a lot for your help! :-p