= Solution
Introduce a bookkeeping parameter $\lambda$ in $H=H_0+\lambda H_1$ and expand $|\phi_n\rangle=|\psi_n\rangle+\lambda|\eta_n\rangle+O(\lambda^2)$ and $\mathcal E_n=E_n+\lambda\Delta E_n+O(\lambda^2)$. Fix <phase> and normalization at this order by $\langle\psi_n|\eta_n\rangle=0$. The first-order Schrödinger equation is
$$
(H_0-E_n)|\eta_n\rangle=-(H_1-\Delta E_n)|\psi_n\rangle.
$$
<angular momentum> projection onto $\langle\psi_n|$ gives $\Delta E_n=\langle\psi_n|H_1|\psi_n\rangle$. <angular momentum> projection onto a different <eigenstate> gives
$$
\langle\psi_j|\eta_n\rangle=\frac{\langle\psi_j|H_1|\psi_n\rangle}{E_n-E_j},\qquad j\ne n.
$$
Completeness therefore yields the <first-order nondegenerate perturbation theory> result
$$
\boxed{|\phi_n\rangle=|\psi_n\rangle+\sum_{j\ne n}
\frac{\langle\psi_j|H_1|\psi_n\rangle}{E_n-E_j}|\psi_j\rangle+O(H_1^2).}
$$
Nonzero <energy> denominators are essential. The displayed state has unit norm through first order; exact normalization would change terms of second order. A continuum, if present, contributes its corresponding <integral> as well as any discrete sum.
For the oscillator in a static <electric field>, electrostatic <potential energy> is $H_1=-q\epsilon x$ and $\omega=\sqrt{k/m}$. <parity> gives $\langle0|x|0\rangle=0$, and the supplied Gaussian <wavefunctions> give
$$
x_{\rm zp}\equiv\langle1|x|0\rangle
=\sqrt{2\alpha}\int x^2|\psi_0(x)|^2dx
=\frac1{\sqrt{2\alpha}}=\sqrt{\frac{\hbar}{2m\omega}}.
$$
Since $E_0-E_1=-\hbar\omega$, the two-state first-order <ground state> is
$$
|\phi_0\rangle=|0\rangle+\eta|1\rangle+O(\epsilon^2),\qquad
\eta=\frac{q\epsilon x_{\rm zp}}{\hbar\omega}.
$$
Thus $\langle x\rangle=2\eta x_{\rm zp}+O(\epsilon^2)$, and the induced electric dipole is
$$
\boxed{d=q\langle x\rangle=\frac{q^2\epsilon}{m\omega^2}=\frac{q^2}{k}\epsilon
\quad\text{to first order}.}
$$
It is aligned with the field for either sign of $q$. The <harmonic-oscillator electric polarizability> is $q^2/k$. The <position operator> connects $|0\rangle$ only to $|1\rangle$, so the requested two-state calculation contains the complete first-order dipole response, rather than dropping other first-order terms.
Complete the square to find the exact <electric-field displacement of a quantum harmonic oscillator>:
$$
H=\frac{p^2}{2m}+\frac12k(x-x_0)^2-\frac{q^2\epsilon^2}{2k},\qquad x_0=\frac{q\epsilon}{k}.
$$
The exact normalized ground <wavefunction> is $\psi_0(x-x_0)$, with <energy> $\hbar\omega/2-q^2\epsilon^2/(2k)$. Its mean displacement is $x_0$, so
$$
\boxed{d_{\rm exact}=q x_0=\frac{q^2\epsilon}{k}.}
$$
\b[The first-order dipole equals the exact dipole], because the restoring force is exactly linear and translating a harmonic well does not change its width or <frequency>. This agreement of one expectation value does not make the truncated <wavefunction> exact. The <Hellmann–Feynman theorem> provides an independent check: $d=-\partial E_0/\partial\epsilon$.
For <perturbation theory> to approximate the state, $|\eta|\ll1$, or $|x_0|\ll2x_{\rm zp}$. The exact translated <ground state> has the original-oscillator expansion
$$
|0;x_0\rangle=e^{-\eta^2/2}\sum_{n=0}^{\infty}\frac{\eta^n}{\sqrt{n!}}|n\rangle.
$$
It is a <coherent state> with mean occupation $\eta^2$. Large fields require many oscillator levels, whereas the first-order state retains only two. The <normalized two-state truncation of a displaced oscillator> instead gives
$$
|\phi_{01}\rangle=\frac{|0\rangle+\eta|1\rangle}{\sqrt{1+\eta^2}},\qquad
\langle x\rangle_{01}=\frac{x_0}{1+\eta^2},
$$
which incorrectly tends towards zero for large $|\eta|$ instead of following the displaced well. In <position> space the truncated polynomial times Gaussian also develops a node absent from the exact <ground state>. Taking a linearized formula outside its small-state-correction regime is therefore invalid even though its dipole coefficient happens to be exact.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2015/paper-1-displaced-oscillator.png]
{title=Exact translated ground state compared with its normalized two-state truncation at weak and strong displacement}
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