I wish to find the exact energy levels of the following perturbed hamiltonian. $$\hat{H}=\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2+\alpha x+\beta p^2.$$

I believe that it can be solved by using the destruction/creation operators, $$\hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^\dagger)$$ $$\hat{p}=\sqrt{\frac{m\omega\hbar}{2}}i(\hat{a}^\dagger-\hat{a})$$ However, as it is rather tedious calculation, I want to use the following method: I complete the square and factorize by $p^2$ $$\hat{H}=\frac{p^2}{2m}(1+2m\beta)+\frac{m\omega^2}{2}(x^2+\frac{2}{m\omega^2}\alpha x)$$.


Now I introduce $\tilde{p}^2=p^2(1+2m\beta)$ and $\tilde{x}=x+\frac{2}{m\omega^2}\alpha$ and the Hamiltonian is thus: $$H=\frac{\tilde{p}^2}{2m}+\frac{m\omega^2}{2}\tilde{x}^2-\frac{\alpha^2}{m^2\omega^4}$$

Thus, the energy levels would be $$E_n=\hbar\omega(n+\frac{1}{2})-\frac{\alpha^2}{m^2\omega^4}$$

My question is, is this method correct? I know it works when there is a perturbation that depends on x, but I am not sure if it works now that the perturbation involves a term with momentum.

  • $\begingroup$ If you try perturbation theory you should see the effect of the small additional term in $p$ quite quickly and verify that it would not agree with your original solution. $\endgroup$ Jun 17 '19 at 8:55

No, your calculation is not correct $-$ you're using a non-canonical transformation in ways which assume that it is canonical.

More specifically, it is correct to say that if your hamiltonian maps into the form $$ H = \frac{1}{2\mu} P^2 + \frac12 \mu\tilde\omega^2 X^2, $$ where $X$ and $P$ are operators such that $[X,P]=i\hbar$, then the eigenvalues of $H$ are of the form $E_n = \hbar\tilde \omega (n+1/2)$, for $n=0,1,2,3,\ldots$.

However, your transformation fails at the second property on that list $-$ you're defining $\tilde x=x+2\alpha/m\omega^2$ and $\tilde p = \sqrt{1+2m\beta} p$, which means that the commutator between the new position and momentum equals $$ [\tilde x, \tilde p] = \sqrt{1+2m\beta} \, i\hbar, $$ i.e., the $\tilde x$ and $\tilde p$ you have defined do not form a canonically conjugate pair.

To fix this, you will need to alter your redefinition of $\tilde x$ (or use a new value of the mass) such that the rephrased problem does fulfill both of the conditions laid out above.

  • $\begingroup$ And if beta were 0 I could do the trick I did, right? $\endgroup$ Jun 17 '19 at 3:44
  • $\begingroup$ Yes, that is correct. $\endgroup$ Jun 17 '19 at 5:02

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