# What's the meaning of the propagator in QM?

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.

• 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) – ACuriousMind Jan 27 '15 at 12:06
• 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? – user71714 Jan 27 '15 at 12:14
• @ACuriousMind So the propagator is the probability amplitude for a system in one state to be found in another? – Udit Dey Nov 26 '16 at 12:02

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.