# How to understand the kernel as a transition amplitude?

Consider the time evolution operator $$U(t_f, t_i)$$ that controls the evolution of a wave function according to $$|\psi(t_f \rangle = U(t_f, t_i) | \psi(t_i) \rangle$$.

As I understand it, the Born rule says that we interpret $$\langle x | \psi(t) \rangle$$ as the complex probability amplitude of measuring the system $$\psi$$ to have position $$x$$ and time $$t$$. Using the concept of the wave function, we write $$\psi(x, t) = \langle x | \psi(t) \rangle$$, and is the square norm of this wave function gives us a pdf.

Thus, it seems reasonable to interpret $$\langle x_f | U(t_f, t_i) | \psi(t_i) \rangle$$ as the probability amplitude of measuring the postion of $$\psi$$ to be $$x_f$$ at time $$t_f$$. Let us say that $$| \psi(t_i) \rangle = | x_i \rangle$$. Then it seems reasonable to interpret $$K(x_f, t_f; x_i; t_i) := \langle x_f | U(t_f, t_i) | x_i \rangle$$ as the "transition amplitude" of measuring a system to have position $$x_f$$ at time $$t_f$$ when we know the system had position $$x_i$$ at time $$t_i$$.

However, I have only seen the term "transition amplitude" used (in Townsend's "A Modern Approach to Quantum Mechanics, Chapter 8, and these notes, Section 1, http://www.blau.itp.unibe.ch/lecturesPI.pdf) used to refer to the notation $$\langle x_f, t_f | x_i, t_i \rangle$$, where the state $$|x_i, t_i \rangle$$ is expressly NOT the time evolution of $$|x_i \rangle$$ but rather the eigenfunction of the time evolved position operator in the Heisenberg picture (thus, $$|x_i, t_i \rangle = \exp\{i \hat{H} t_i\} | x_i \rangle$$).

So, my question (finally) is: why does it seem that the term "transition amplitude" is applied to $$\langle x_f, t_f | x_i, t_i \rangle$$ and NOT (the equivalent) $$K(x_f, t_f; x_i, t_i)$$, when it seems that $$K(x_f, t_f; x_i, t_i)$$ has a very clear understanding as the transition amplitude of a particle starting at position $$x_i$$ at time $$t_i$$ to be measured at position $$x_f$$ at time $$t_f$$, whereas the interpretation of $$\langle x_f, t_f | x_i, t_i \rangle$$ seems much less straightforward?

1. The overlap $${}_H\langle x_f, t_f | x_i, t_i \rangle{}_H$$ is indeed the transition amplitude between the two instantaneous position eigenstates $$| x_i, t_i \rangle{}_H$$ and $$| x_f, t_f \rangle{}_H$$ in the Heisenberg picture.
2. It is also equal to $${}_S\langle x_f | U(t_f, t_i) | x_i \rangle{}_S$$ in the Schrödinger picture.