# Perturbing the Harmonic oscillator [closed]

Assume that a quantum harmonic oscillator is described by the Hamiltonian, $$$$H=H_0+\lambda q^2$$$$ where, $$$$H_0=\frac{p^2}{2m}+\frac{1}{2}m\omega^2q^2$$$$ If $$|\Psi_0⟩$$ is the ground state of $$H$$, it can always be written as a linear combination of the eigenstates of $$H_0$$ ({$$|n⟩$$}), $$$$|\Psi_0⟩=\sum_{n}c_n|n⟩$$$$ What are the values of $$c_n$$?

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

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.

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.