Wave function. Measurement of the absence

Question

Imagine we have a particle in an eigenstate of a Hamiltonian, as time passes it will remain in that state.

We suppose in this question that the position can take a continuum of values.

If we measure the position of the particle at $x_0$ it's wave function will collapse and the new wave function $\psi(x,t_0) = \delta(x-x_0)$ which will evolve in time as a superposition of eigenstates of the Hamiltonian.

Now if instead of measuring the position of the particle, which is initially in an eigenstate of the Hamiltonian, we measured if the particle is in a given range $x\in[x_a, x_b]$ at $t_0$, where the wave function is non-zero in this range, and with $[x_a,x_b]$ different to the whole range of $x$, and we found that the particle is not there. Does the particle continue to be in the same eigenstate of the Hamiltonian? Because now we know for sure that the wave function at $t_0$ was zero at that region, should we then take another wave function that meets this requirement? I guess it would be pretty naïve to just take the wave function of the eigenstate of the Hamiltonian we had originally and make it zero through the range $[x_a, x_b]$ and normalize again and express it as a superposition of the eigenstates of the Hamiltonian to study it's time evolution.

Thank you for your answers!

Answer 1

Does the particle continue to be in the same eigenstate of the Hamiltonian?

No. You've performed a binary measurement, i.e. the question "is the particle in the interval $[x_a,x_b]$?", with answers "yes" and "no" corresponding to the projection operators $$ \Pi_1 = \int_{x_a}^{x_b} |x\rangle\langle x | \,\mathrm dx $$ and $$ \Pi_0 = \mathbb I - \Pi_1 = \int_{-\infty}^{x_a} |x\rangle\langle x | \,\mathrm dx + \int_{x_b}^\infty |x\rangle\langle x | \,\mathrm dx. $$

If the particle starts off in the eigenstate $|\psi_n\rangle$ of some hamiltonian $H$, and then you perform that measurement and get a negative answer, then the state of the system will evolve to $$ |\psi_n\rangle \mapsto \frac{1}{N}\Pi_0|\psi_n\rangle = \frac{1}{||\Pi_0|\psi_n\rangle||}\Pi_0|\psi_n\rangle = \frac{1}{\sqrt{\langle \psi_n|\Pi_0|\psi_n\rangle}}\Pi_0|\psi_n\rangle $$ (with the last equality using the fact that $\Pi_0^2 = \Pi_0$). The particle will then evolve according to the previous hamiltonian $H$ ─ probably with some important time evolution, since $\Pi_0|\psi_n\rangle$ is likely to be far from being an eigenstate of $H$.