The propagator

Intuition

Ask the simplest question about motion: if a particle is surely at $x_{a}$ at the time $t_{a}$ , what is the amplitude to find it at $x_{b}$ at the time $t_{b}$ ? The answer is the position matrix element of the evolution operator, $K(x_{b},t_{b};x_{a},t_{a})=\bra{x_{b}}\hat{U}(t_{b},t_{a})\ket{x_{a}}$ , the propagator. It carries any wavefunction forward: the new wavefunction at $x_{b}$ is the old one at every $x_{a}$ , times the amplitude to get from there to $x_{b}$ , added up. Because evolving for a while and then for a while longer is the same as evolving for the whole time, propagators compose: the amplitude from $a$ to $b$ is the sum over every place $c$ the particle might pass at some middle time, of the amplitude from $a$ to $c$ times the amplitude from $c$ to $b$ . That rule, applied again and again, is the seed of Feynman’s sum over paths.

The chance of driving from one city to another through a mountain range adds up the ways through every pass; the amplitude to get from $a$ to $b$ adds up the ways through every point at a middle time — except that amplitudes, unlike chances, can cancel.

Time across and position up. The amplitude to go from $a$ to $b$ is the sum, over every position $x_{c}$ at a middle time (dashed), of the amplitude to reach $x_{c}$ times the amplitude to go on from there to . A few of the two-leg routes are drawn.

For a Hamiltonian that does not depend on time, with $T=t_{b}-t_{a}$ , the propagator is

K(x_{b},t_{b};x_{a},t_{a})=\bra{x_{b}}e^{-i\hat{H}T/\hbar}\ket{x_{a}}

Properties

It propagates wavefunctions: $\psi(x_{b},t_{b})=\int K(x_{b},t_{b};x_{a},t_{a})\,\psi(x_{a},t_{a})\,dx_{a}$ .

Properties of the propagator

\begin{aligned}\psi(x_{b},t_{b})&=\int K(b;a)\,\psi(x_{a},t_{a})\,dx_{a}\\ K(b;a)&=\int K(b;c)\,K(c;a)\,dx_{c}\end{aligned}

Both follow from the evolution operator by inserting the completeness of position states: once between the operator and the old state, and once between the two factors of an evolution split at a middle time. Expanding the evolution operator in stationary states gives the spectral form.

Proof steps

\ket{\psi(t_{b})}=\hat{U}(t_{b},t_{a})\int\ket{x_{a}}\braket{x_{a}|\psi(t_{a})}\,dx_{a}

Completeness of the position states.

\psi(x_{b},t_{b})=\int\bra{x_{b}}\hat{U}\ket{x_{a}}\,\psi(x_{a},t_{a})\,dx_{a}

Take the component along $\ket{x_{b}}$ .

\hat{U}(t_{b},t_{a})=\hat{U}(t_{b},t_{c})\,\hat{U}(t_{c},t_{a})

Evolving to $t_{c}$ and then on to $t_{b}$ is evolving to $t_{b}$ .

K(b;a)=\int\bra{x_{b}}\hat{U}(t_{b},t_{c})\ket{x_{c}}\bra{x_{c}}\hat{U}(t_{c},t_{a})\ket{x_{a}}\,dx_{c}

Insert $\int\ket{x_{c}}\bra{x_{c}}\,dx_{c}$ between the two factors.

K(b;a)=\sum_{n}\psi_{n}(x_{b})\,\psi_{n}^{*}(x_{a})\,e^{-iE_{n}T/\hbar}

$e^{-i\hat{H}T/\hbar}=\sum_{n}e^{-iE_{n}T/\hbar}\ket{n}\bra{n}$ .

Applications

Practice

The Propagator

The propagator is the amplitude to find the particle at $x_{b}$ at $t_{b}$ , given that it was surely at $x_{a}$ at .

K(x_{b},t_{b};x_{a},t_{a})=\bra{x_{b}}\hat{U}(t_{b},t_{a})\ket{x_{a}}

Try it

What is $K(x_{b},t_{b};x_{a},t_{a})$ ?