5
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ Some thoughts: measuring the particle to not be in some region $U\subseteq S$ (where $S$ is the space of possible positions) is the same as measuring it to be in the region $S \setminus U$. So your question isn't noticeably distinct from the case where you measure it in $[x_a,x_b]$ and do find it there. Which is also a good question. $\endgroup$ – jacob1729 Apr 29 at 10:09
  • 2
    $\begingroup$ Possible duplicate: physics.stackexchange.com/questions/232502/… $\endgroup$ – JPattarini Apr 29 at 11:46
  • $\begingroup$ @jacob1729 Could you develop why these two situations are not distinct from each other, please? My point is that in measuring something where the particle is not, you don't interact with it and not interacting with it should not in principle change it's state. $\endgroup$ – Álex De La Calzada Apr 30 at 13:23
  • $\begingroup$ @ÁlexDeLaCalzada before the measurement the wavefunction is non zero almost everywhere, measuring the particle in some region does require you do something with your apparatus in a region that the particle has some amplitude to be found, so I wouldn't call that 'not interacting with it'. $\endgroup$ – jacob1729 Apr 30 at 13:31
5
$\begingroup$

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 $H$.

$\endgroup$
  • $\begingroup$ That's a very good explanation, but I still do not see how measuring something by not interacting with the particle can actually change its state. $\endgroup$ – Álex De La Calzada Apr 30 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.