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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 $c_n$?

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    $\begingroup$ Just to check: you really mean $\lambda q^2$ not some other power like $q^4$? $\endgroup$
    – jacob1729
    Jan 3, 2020 at 19:10
  • $\begingroup$ Yes it's $\lambda q^2$. $\endgroup$ Jan 3, 2020 at 19:12
  • $\begingroup$ As written this question appears off-topic. Do you have a conceptual question? We don't provide solutions to standard exercises. $\endgroup$
    – Bill N
    Jan 3, 2020 at 20:47

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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.

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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.

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