Finding an exact value for energy in perturbation theory 
Supose a particle of mass $m$ and electric charge $q$, subject to harmonic potential in 1D, is placed in an area with electric field $\vec E = E \hat u_x$. Determine the exact change in its energy spectrum caused by interacting with this field.

I started by writing down the Hamiltonean operator as:
$\hat H = \frac{1}{2m}\hat p_x^2 + \frac{1}{2}m\omega^2 \hat x^2 + qE \hat x$, where the last term represents the perturbation caused by interacting with the electric field. (We can treat it as a perturbation since usually $q \ll 1$.)
Using this, for any state $|n \rangle$, the first order correction to the energy is always going to be zero, since, using the ladder operators: $\hat a_{\pm} = \sqrt{\frac{m\omega}{2\hbar}}(\hat x \pm \frac{i}{m\omega} \hat p)$,
$$\epsilon_1 = \langle n | \hat W |n\rangle = qE\cdot 2\sqrt{\frac{2\hbar}{m\omega}} \langle n | \hat a_+ + \hat a_-|n\rangle = C (\sqrt{n+1}\langle n |n+1 \rangle + \sqrt{n}\langle n |n-1 \rangle) =0,$$
because the states are orthogonal to each other. Therefore, the first order correction is zero. However, this still leads to an approximate answer:
$$E(n) \approx \epsilon_0 (n) + q\epsilon_1 + O(q^2) = \epsilon_0 (n) + O(q^2), $$
so I don't understand how to get to an exact value for the change in energy, specially since in my Quantum Mechanics class we didn't cover higher order corrections.
Is there another way to approach the problem that I'm not seeing?
 A: This isn't intended as a perturbation theory problem.  (It actually can be solved, to all orders in perturbation theory, but that would be unthinkably arduous.)  The actual point is the notice that the potential, including the linear potential due to the electric field, is still a quadratic function of the position,
$$V(x)=\frac{1}{2}m\omega^{2}x^{2}+qEx=\frac{1}{2}m\omega^{2}\left(x-\frac{qE}{m\omega^{2}}
\right)^{2}-\frac{q^{2}E^{2}}{2m\omega^{2}}.$$
The full Hamiltonian therefore represents another harmonic oscillator (centered around a different location), with exactly the same frequency, but a different ground state energy. Thus
$$E_{n}=\left(n+\frac{1}{2}\hbar\omega\right)-\frac{q^{2}E^{2}}{2m\omega^{2}},$$
which does agree with the first-order perturbative expression in the question.
A: If you want an exact solution, you can't use perturbation theory.
Try a change of variables $x\rightarrow y$ so that:
$$ \frac 1 2 m \omega x^2 + qE x = \frac 1 2 m\omega (y-a)^2 - b^2 $$
where $a$ is an offset to the (classical) ground state, and $-b^2$ is a global energy shift.
Classically, that energy shift is stored in the spring, and quantum mechanically, it's in the field that causes $V(x)$.
