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?


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.


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  • 2
    $\begingroup$ Why not just go through a similar derivation that gave you what you cite at the beginning of your question? $\endgroup$ – Aaron Stevens Jun 12 at 4:32
  • $\begingroup$ You say something like proposing $E_n\approx \varepsilon_{0,0}+\alpha \varepsilon_{1,0}+\beta \varepsilon_{0,1}+\alpha \beta \varepsilon_{1,1}$ ?? $\endgroup$ – Ariel Jun 13 at 20:21

You can solve this Hamiltonian exactly. We can write

$$ \hat H = \left(\frac{1}{2m}+\beta\right)p^2+\frac{m\omega^2}{2}\left(x+\frac{\alpha}{m\omega^2}\right)^2-\frac{\alpha^2}{2 m \omega^2} $$

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.

  • $\begingroup$ Thanks for the answer of the exact solution, Jahan. Btw I still want to know how to answer my first question without use that way (I mean, what if there wasn't exact solution?) $\endgroup$ – Ariel Jun 12 at 4:28
  • $\begingroup$ @Ariel Expanding out the exact solution should give you a good hint to the general equation! $\endgroup$ – Jahan Claes Jun 13 at 14:56

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