# First-order correction to energy in perturbed harmonic oscillator [closed]

I know, from the perturbation theory, that, if I have the hamiltonian $$\hat H = \hat H_0 + \lambda \hat W$$ where $$\hat H_0$$ is the unperturbed hamiltonian of which I know its eigenvectors and eigenvalues, and $$W$$ is the perturbation. Then the energy of the perturbed hamiltonian, corrected to the first order, is given by $$E_n\approx\varepsilon_0+\lambda \varepsilon_1 \tag{1}$$ where $$\varepsilon_0$$ is the nth eigenvalue of $$H_0$$ (i.e. $$H_0|\varphi_n\rangle =\varepsilon_0 |\varphi_n\rangle$$) and $$\varepsilon _1=\langle \varphi_n|\hat W|\varphi_n\rangle$$.

My question comes from a specific problem in which the hamiltonian given to me is a peculiar "anharmonic" oscillator $$\hat H = \frac{p^2}{2m}+\frac{m\omega^2}{2}x^2+\alpha x + \beta p^2$$ In this case I don't have just one parameter $$\lambda$$ but two ($$\alpha$$ and $$\beta$$). What should the expressión (1) be here?

Thanks

PS:In addittion I would like to know how to solve this problem in exact way, or at least, if this is possible. I suppose it could work with a change of variables.

## closed as off-topic by Aaron Stevens, Kyle Kanos, Jon Custer, GiorgioP, ZeroTheHeroJun 17 at 16:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Kyle Kanos, Jon Custer, GiorgioP, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

• Why not just go through a similar derivation that gave you what you cite at the beginning of your question? – Aaron Stevens Jun 12 at 4:32
• You say something like proposing $E_n\approx \varepsilon_{0,0}+\alpha \varepsilon_{1,0}+\beta \varepsilon_{0,1}+\alpha \beta \varepsilon_{1,1}$ ?? – Ariel Jun 13 at 20:21

$$\hat H = \left(\frac{1}{2m}+\beta\right)p^2+\frac{m\omega^2}{2}\left(x+\frac{\alpha}{m\omega^2}\right)^2-\frac{\alpha^2}{2 m \omega^2}$$
Define $$\bar{m}$$ to satisfy $$\frac{1}{2\bar m}=\frac{1}{2m}+\beta$$, define $$\bar\omega$$ to satisfy $$m\omega^2=\bar m\bar \omega^2$$, and define $$\bar{x}=x+\frac{\alpha}{m\omega^2}$$. Then the Hamiltonian becomes
$$\hat H = \frac{p^2}{2\bar m}+\frac{\bar m\bar\omega^2}{2}\bar{x}^2-\frac{\alpha^2}{2 m \omega^2}$$ which is just a SHO Hamiltonian plus a constant energy offset. So, you should be able to find the energy levels for this Hamiltonian, and write them out in terms of $$m,\omega, \alpha, \beta$$. Expanding the result to first order in $$\alpha,\beta$$ will immediately answer your first question.