I've been thinking a lot about changes to the harmonic oscillator potential, and I was looking into the problem where
$$V(x) = \frac{1}{2}m\omega ^2 x^2 + C$$
where $C$ is some positive real constant. I was able to convince myself that, in this case, our eigenenergies would take the form $E_n = \hbar \omega (n+\frac{1}{2}) + C$, which is basically our "standard" harmonic oscillator energies shifted by $C$.
What I'm having trouble with are the eigenstates. Since $[H_\text{new}, H_\text{SHO}] =0$, they should share the same set of eigenstates. This means that we can use $|n\rangle$ as are eigenstates.
However, how can this make sense? Suppose we start with
$$H = \frac{p^2}{2m} + \frac{1}{2}m\omega ^2 x^2 + C$$
and our particle is in the state $|0\rangle$. As some time, we snap our fingers that $H$ immediately changes to
$$H = \frac{p^2}{2m} + \frac{1}{2}m\omega ^2 x^2 \, ,$$
our normal harmonic oscillator. If I check the probability of measuring the energy $E_0 = \hbar \omega /2$, I would get
$$|\langle 0|0 \rangle|^2 = 1 \, . $$
This seems weird to me because we originally had an energy of $\hbar \omega /2 + C$ … this would imply that we are measuring in the same basis, but our measurement depends on the Hamiltonian we are using.
My conclusion is that the eigenstates are not the same, but I can't see why!
Best Answer
Your "snap of the fingers" really means that your Hamiltonian is
$$H(t) = \frac{p^2}{2m} + \frac{1}{2}m \omega^2 - \theta(t)C$$
where $\theta(t)$ is the Heaviside step function.
The eigenstates will only be equivalent asymptotically for $t \rightarrow\pm\infty$ since you are now dealing with a time-dependent Hamiltonian.