Perturbing the Harmonic oscillator Assume that a quantum harmonic oscillator is described by the Hamiltonian,
\begin{equation}
H=H_0+\lambda q^2
\end{equation}
where,
\begin{equation}
H_0=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2
\end{equation}
If $|\Psi_0⟩$ is the ground state of $H$, it can always be written as a linear combination of the eigenstates of $H_0$ ({$|n⟩$}), 
\begin{equation}
|\Psi_0⟩=\sum_{n}c_n|n⟩
\end{equation}
What are the values of $c_n$?
 A: Since this is a homework type problem, I'll try to not give the entire solution, but this is how I'd do it:
It should not have escaped your attention that both $H$ and $H_0$ are very similar in their forms, in that both of them have the same dependence on $p$ and $q$. The difference, you'll notice, is only in the pre-factor of $q$. 
Show that you can write $H$ as a harmonic oscillator Hamiltonian, but with a different frequency (call it $\omega^\prime$) which depends on $\lambda$. (As a test, you could show that when $\lambda=0$, $\omega = \omega^\prime$.)
Now, since you have written $H$ as a harmonic oscillator with frequency $\omega^\prime$, you can immediate write out its normalised ground state wavefunction, $\Psi_0(x)$. All that's left is to decompose this wavefunction in terms of the energy eigenbasis of $H_0$ (let's call these functions $\psi_n^0(x)$), which you can do by using the orthogonality property of these eigenfunctions, i.e.
$$c_n = \int_{-\infty}^\infty \Psi_0^*(x) \psi_n^0(x)\text{d} x.$$
I admit this isn't very elegant as you'd need the general form of the Harmonic Oscillator wavefunctions (which I doubt many people could remember offhand without looking at a book), but it's not obvious to me how else this could be done. I'm actually curious to know if there's a more elegant way to do this.
A: If you are completely sure that your perturbed Hamiltonian is correctly written you don't need to do anything fancy, just consider
$$
V(q) = (\frac{1}{2}mw^2 + \lambda)q,
$$
solve this new problem, and try to related how this ground state compare to the standard case.
