2-D Harmonic Osc. with Perturbation

proto0
Messages
1
Reaction score
0
2-D Harmonic Oscillator with Perturbation

Homework Statement



A 2-D harmonic oscillator has an energy Ensubxnsuby and wavefunctions \phinsubxnsuby

The first excited states are 2-fold degenerate E_{01}=E_{10}=2\hbar\omega

What are the energies and wavefunctions if we add the perturbation

H'=gwyp_{x}?



Homework Equations



En=\hbar\omega(nx + ny + 1)

\phisubnxny=\phisubnx(x)\phisubny(y)

H=g\omegayPsubx

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
 

Attachments

  • hw question.JPG
    hw question.JPG
    33.1 KB · Views: 447
Last edited:
Physics news on Phys.org
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_{01} it will look much better, e.g.
The first excited states are 2-fold degenerate E_{01} = E_{10} = 2 \hbar \omega (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, H'=H_{0}+K'xy

I'm suposed to solve the secular equation right? By calculating the H_{ij}, 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, H'=H_{0}+K'xy

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


Thanks

Yes, the perturbation H'=H_{0}+K'xy couples states of different levels, remember from c your class you should have seen <n'|x|n> ~ c_{1}\delta_{n+1,n} + c_{2}\delta_{n-1,n}. Since your perturbation is separable, you should be able to get it quickly.
 
Got it! Thanks a lot for your help! :-p
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top