Meaning of adding a term to the Hamiltonian in a quantum harmonic oscillator Let $H$ be the Hamiltonian in a harmonic oscillator,
$$
    H = \sum_{n=0}^{\infty} \hbar \omega \left (n+\frac{1}{2} \right ) |n\rangle \langle n|.
$$
Suppose we introduce the interaction
$$V = \sqrt{2} \hbar \omega
(|0\rangle \langle 1|+|1\rangle \langle 0|
$$
and the new Hamiltonian is $H+V$. How can we understand the physics under this interaction? Is there any classical interpretation?
 A: You can think of the initial Harmonic oscillator with the Hamiltonian $H=\sum_{n}\hbar\omega(n+1/2)|n\rangle\langle n|$ as a free system, that is under time evolution a state prepared in any one of the $|n\rangle$ remains in that state. However, the effect of adding an interaction term of the form that you proposed now allows for transitions among the states $|0\rangle$ and $|1\rangle$.
For example, consider the time evolution of an initial state $|0\rangle$ under the full new Hamiltonian $H'=H+V$.
$|\psi(t)\rangle = e^{-iH't/\hbar}|0\rangle \approx |0\rangle + (-\frac{it}{\hbar})H'|0\rangle$
where I have taken $t$ to be small. We see that $H'|0\rangle=\frac{\hbar\omega}{2}|0\rangle + \sqrt{2}\hbar\omega|1\rangle$ and so
$|\psi(t)\rangle \approx (1 - \frac{it\omega}{2})|0\rangle - it\sqrt{2}\omega|1\rangle$
So, we see that now the system has a finite probability of being in the state $|1\rangle$. This would not be possible when there was only $H$. For now, we just added the interaction term by hand, but such terms usually come into the picture when you couple your original system to another system. For example, if we couple the original Harmonic oscillator to say a laser, one could consider the following interaction.
$V = \sqrt{2}\hbar\omega ( a \otimes |1\rangle\langle0| +  a^{\dagger} \otimes |0\rangle\langle1|)$
Where $a,a^{\dagger}$ are the photo creation and annihilation operators. So, when the system absorbs one photon, the system goes from $|0\rangle$ to $|1\rangle$ and the reverse when it emits a photon.
A: Well the eigenstates would change, but not all of them; since if $H'=H+V$, $H|n \rangle= H'|n \rangle$ for all $n>1$, where $|n \rangle$ is eigenstate of the original QHO with energy: $E_n = \hbar \omega \big(n+\frac{1}{2} \big)$. Now in the case of $n\leq 1$:
$$H'|0 \rangle=E_0|0 \rangle+ \sqrt2 \hbar \omega |1 \rangle$$
$$H'|1 \rangle=E_1|1 \rangle+ \sqrt2 \hbar \omega |0 \rangle$$
Finding the new eigenstates is therefore equivalent to finding the eigenvectors of the following matrix:
$$
\begin{pmatrix}
E_0 &  \sqrt2 \hbar \omega\\
\sqrt2 \hbar \omega & E_1 
\end{pmatrix} = \begin{pmatrix}
\frac{\hbar \omega}{2} &  \sqrt2 \hbar \omega\\
\sqrt2 \hbar \omega & \frac{ \hbar \omega}{2} 
\end{pmatrix}=  \frac{\hbar \omega}{2}\begin{pmatrix}
1 &  \frac{1}{\sqrt2} \\
\frac{1}{\sqrt2} & 3 
\end{pmatrix}
$$
You will find the eigenvalues and associated eigenvectors to  be: $\frac{4-\sqrt6}{4}\hbar\omega \text{ , } \frac{4+\sqrt6}{4}\hbar\omega$ and: $ \frac{-(2+\sqrt6)|0 \rangle +\sqrt2 |1 \rangle}{6\sqrt2 + 2\sqrt{12}} \text{ , } \frac{(\sqrt3 - \sqrt2)|0 \rangle +|1 \rangle}{6-2\sqrt6}$ respectively.
So your new Hamiltonian $H'$ will have eigenbasis:
$$\Big\{ \frac{-(2+\sqrt6)|0 \rangle +\sqrt2 |1 \rangle}{6\sqrt2 + 2\sqrt{12}} 
, \frac{(\sqrt3 - \sqrt2)|0 \rangle +|1 \rangle}{6-2\sqrt6} , |n \rangle\Big\} \text{for all n>1} $$
and the energies are the same for $n>1$ and for $n \leq1$ they are the eigenvalues stated above.
Using this you can solve the S.E. and find your dynamics!
