In QM, the following object: $$U(x_{f},t_{f}; x_{i},t_{i}) = \langle x_{f},t_{f}|x_{i},t_{i}\rangle$$ is called propagator. Its interpretation is that it is the transition amplitude from a particle to go from $x_{i}$ at $t_{i}$ to $x_{f}$ at a later time $t_{f}$.

Question: As a transition amplitude, it should be a probability density function, right? But what does it mean in terms of the total probability? I mean, does it mean that $$\int dy |U(y,t; x,t_{0})|^{2} = 1$$ where the integral is over $y$ while $x$ and $t_{0}$ are not considered as variables but as known points? Is that correct?

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    $\begingroup$ Possible duplicate: Normalization of the path integral $\endgroup$
    – Qmechanic
    Jul 5, 2021 at 18:15
  • $\begingroup$ @Qmechanic I've read the answers in the linked post. You mention that my formula is not correct. But I didn't get what is the correct normalization. If is not this or any other, why is this a probability amplitude at all? $\endgroup$
    – MathMath
    Jul 5, 2021 at 18:21
  • $\begingroup$ Delete the integral of $\textrm{d}t$ then the normalization is correct. The equation means that for a particle at $x$ at $t_0$, for any time instance $t$, the probability of finding it in the whole space is 1. $\endgroup$
    – Youran
    Jul 5, 2021 at 18:26
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    $\begingroup$ @Youran your comment is unfortunately incorrect, as explained in Qmechanic's link. It would be correct if position eigenstates $| x \rangle$ had discrete (ie Kronecker delta) normalization, but it is not correct as written above. You can see this immediately from a simple calculation: $\int dy | U(y,t;x,t_0) |^2 = \int dy \langle x | U^{\dagger}(t-t_0) | y \rangle \langle y | U(t-t_0) | x \rangle = \langle x | x \rangle = \delta(0)$. $\endgroup$
    – Zack
    Jul 5, 2021 at 18:31
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    $\begingroup$ @Youran Position eigenstates form a continuous spectrum, and cannot be normalized to 1. Explicitly, $\langle x | y \rangle = \delta(x-y)$. You can find such a formula in any quantum mechanics textbook. But as a sanity check, consider how the resolution of the identity would work with your normalization: $1 |x \rangle = \int dy |y \rangle \langle y | x \rangle$ does not make sense unless $\langle y | x \rangle = \delta(y-x)$. $\endgroup$
    – Zack
    Jul 5, 2021 at 18:44