0
$\begingroup$

Yesterday I was solving some exercises, and after solving the time evolution I was asked to find the probability of the system to some state. In specific: $$|\Psi(t)\rangle = \frac{1}{\sqrt{2}}\left(|2\rangle + e^{-i/\hbar}|3\rangle\right) $$ and the state to find the probability is $$ |a\rangle = \frac{1}{\sqrt{3}}\left(|1\rangle + |2\rangle + |3\rangle\right) $$

Now, intuitively I tough that the system with the time evolution given by $|\Psi(t)\rangle$ will never be in $|a\rangle$ because the system evolve only in $|2\rangle$ and $|3\rangle$, but obviously, the propagator $$ \langle a|U_t|\Psi(0)\rangle $$ says another story, being different of zero. So my question is, indeed, what's the meaning of the propagator, or better, of the scalar product $$ \langle a | b \rangle $$ in general.

$\endgroup$
  • 1
    $\begingroup$ Exactly that which you thought - the inner product gives the probability for one state to be found in the other. Your intuition that $\lvert\psi(t)\rangle$ cannot be found in the state $\lvert a \rangle$ is simply wrong. (...and that's what quantum mechanics is all about) $\endgroup$ – ACuriousMind Jan 27 '15 at 12:06
  • $\begingroup$ Yes, I know my intuition is wrong. Maybe I believe I'm beginning to understand what I miss reading to the example of the light polarization states, which are orthogonal, but if you change basis you get a superposition, and hence the probability of being in the two states after a measurement. I'm right here? $\endgroup$ – user71714 Jan 27 '15 at 12:14
  • $\begingroup$ @ACuriousMind So the propagator is the probability amplitude for a system in one state to be found in another? $\endgroup$ – Udit Dey Nov 26 '16 at 12:02
1
$\begingroup$

The point is that your time dependent state can always be written as a linear combination of $\left|a\right \rangle$ and a vector perpendicular to it. The probability is related to the degree to which those vectors overlap.

| cite | improve this answer | |
$\endgroup$

Your Answer

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