# Calculating exact energy levels of perturbed Hamiltonian

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

$$\hat{H}=\frac{p^2}{2m}(1+2m\beta)+\frac{m\omega^2}{2}(x+\frac{2}{m\omega^2}\alpha)^2-\frac{\alpha^2}{m^2\omega^4}$$.

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.

• 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. – ZeroTheHero Jun 17 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.