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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?

Qmechanic
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MathMath
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    Possible duplicate: Normalization of the path integral – Qmechanic Jul 05 '21 at 18:15
  • @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? – MathMath Jul 05 '21 at 18:21
  • 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. – Youran Jul 05 '21 at 18:26
  • @Youran got it! Thanks! – MathMath Jul 05 '21 at 18:27
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    @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: $\int dy | U(y,t;x,t_0) |^2 = \int dy \langle x | U^{\dagger}(t-t_0) | y \rangle \langle y | U(t-t_0) | x \rangle = \langle x | x \rangle = \delta(0)$. – Zack Jul 05 '21 at 18:31
  • @Zack so what is the correct formula? If this is a probability density function, it must be normalizable in some way. – MathMath Jul 05 '21 at 18:34
  • I am reading Qmechanic's answer. It is obscure and takes some time to understand. However, I do not agree with your last equality $\langle x|x \rangle=\delta(0)$. I think as $|x\rangle$ is a normalized state, $\langle x|x \rangle$ should equals 1. – Youran Jul 05 '21 at 18:40
  • Yes. If I understood it correctly, the answer on the other post discusses why the posted formulas are wrong, but the correct formula is not mentioned. – MathMath Jul 05 '21 at 18:42
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    @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 $\langle y | x \rangle = \delta(y-x)$. – Zack Jul 05 '21 at 18:44
  • For me (and I believe for many physicists), space although sometimes written as a continuous variable is a very dense lattice (maybe the $a$ is Plank length?). So there should not be any difference between discrete and continuous cases. – Youran Jul 05 '21 at 18:47
  • @Zack In other words, you did not normalize your initial state to 1, rather, to $\delta(0)$. Maybe the equation should be expressed as $\int dx_f U^2=\langle x,t_0|x,t_0\rangle$. – Youran Jul 05 '21 at 18:51
  • Metaphysics aside, the difference is that an integral is not just a sum. You can roughly think of $dy$ in the integrand as proportional to the lattice constant $a$, and as $a \rightarrow 0$ you must also divide by $a$ in other normalizations so that everything remains finite. But I would prefer not to further debate an elementary formula which can be found in any introductory or advanced quantum mechanics textbook. – Zack Jul 05 '21 at 18:51
  • Qmechnic's answer, in a nutshell, is that the correct normalization is $\lim_{t\to 0}U(x_f,x_i,t)=\delta(x_f-x_i)$. Indeed this is the normalization used in computation. – Youran Jul 06 '21 at 04:53
  • However, for a wavefunction $\psi(x)$ at $t=0$, the wavefunction at $t=t$ is $\phi(x_f)=\int dx_i U(x_f,x_i,t)\psi(x_i)$. We know that the probability is conserved in QM, so $\int dx \psi^2(x)=\int dx \phi^2(x)$. – Youran Jul 06 '21 at 04:55
  • Now suppose $\psi(x_i)=\delta(x_i-x_0)$, then $\phi(x_f)=U(x_f,x_0,t)$. So $\int dx_f U(x_f,x_0,t)^2 = \int dx_i \delta(x_i-x_0)^2 = \delta(0)$. Here shows up a $\delta(0)$ because I did not normalize the initial state $\psi(x_i)$. – Youran Jul 06 '21 at 04:57
  • For a concrete example, consider a free particle. Somehow you know that $U(x_f,x_i,t)\propto \exp((x_f-x_i)^2/2it)$. Then by $\lim_{t\to 0}U(x_f,x_i,t)=\delta(x_f-x_i)$, you can fix the normalization the same as Wikipedia. And you can persuade yourself that $\int dx U^2$ indeed equals $\delta(0)$. There will be an infinity in this step, you can use either discrete lattice or periodic boundary condition to bypass it. – Youran Jul 06 '21 at 05:00
  • I had once the same question when learning QM II (that is, maybe, four years ago). But this is no longer a problem for me now. – Youran Jul 06 '21 at 05:04

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