Time slicing

Intuition

The propagator over a long time is hard to find; over a very short time $\varepsilon$ it is easy, because then the kinetic and potential parts of the Hamiltonian can be applied one after the other, the error being of order $\varepsilon^{2}$ , which vanishes faster than $\varepsilon$ . The kinetic part is diagonal in momentum and the potential part in position, so inserting momentum states turns the short-time propagator into an integral over one momentum — a Gaussian integral that can be done exactly. What comes out is a plain number: a normalising factor times $e^{i\varepsilon L/\hbar}$ , where $L=\tfrac12m(\Delta x/\varepsilon)^{2}-V$ is the classical Lagrangian of a straight step from $x$ to $x+\Delta x$ . Split a finite time into $N$ such steps and compose them: the propagator becomes $N$ short-time factors, integrated over the $N-1$ positions in between.

A film is a stack of still frames: each frame is simple, and motion is what you get by running them quickly one after another. Time slicing replaces a hard evolution by many easy steps.

A time interval cut into six slices, dashed, with one position chosen at each cut: a broken path from $a$ to $b$ . The time-sliced propagator integrates over the position at every cut, so it sums over all such broken paths.

For $\hat{H}=\hat{p}^{2}/2m+V(\hat{x})$ and a short time , with :

\begin{aligned}&\bra{x'}e^{-i\varepsilon\hat{H}/\hbar}\ket{x}\\ &\quad\approx\sqrt{\frac{m}{2\pi i\hbar\varepsilon}}\,\exp\Bigl\{\frac{i\varepsilon}{\hbar}\Bigl[\frac{m}{2}\Bigl(\frac{\Delta}{\varepsilon}\Bigr)^{2}-V(x)\Bigr]\Bigr\}\end{aligned}

Properties

The error is of order $\varepsilon^{2}$ in each step, so $N=T/\varepsilon$ steps together err by order $\varepsilon$ , which vanishes as $N\to\infty$ .

The short-time propagator

\bra{x'}e^{-i\varepsilon\hat{H}/\hbar}\ket{x}=\sqrt{\frac{m}{2\pi i\hbar\varepsilon}}\;e^{i\varepsilon L/\hbar}\bigl(1+O(\varepsilon)\bigr)

Split the short evolution into a kinetic and a potential factor, at the cost of order $\varepsilon^{2}$ . Insert momentum states: the potential factor gives a number at $x$ , the kinetic factor a phase in $p$ , and the overlaps plane waves. The remaining integral over $p$ is a Gaussian; completing the square gives the normalising factor and the kinetic part of the Lagrangian.

Proof steps

e^{-i\varepsilon(\hat{T}+\hat{V})/\hbar}=e^{-i\varepsilon\hat{T}/\hbar}\,e^{-i\varepsilon\hat{V}/\hbar}+O(\varepsilon^{2})

The two exponents differ from one by terms of order $\varepsilon^{2}$ in their commutator (stated).

\begin{aligned}&\bra{x'}e^{-i\varepsilon\hat{H}/\hbar}\ket{x}\\ &\approx\int\frac{dp}{2\pi\hbar}\,e^{ip\Delta/\hbar}\,e^{-i\varepsilon p^{2}/2m\hbar}\,e^{-i\varepsilon V(x)/\hbar}\end{aligned}

Insert $\int\ket{p}\bra{p}dp$ , with $\braket{x'|p}=e^{ipx'/\hbar}/\sqrt{2\pi\hbar}$ .

\frac{p\Delta}{\hbar}-\frac{\varepsilon p^{2}}{2m\hbar}=-\frac{\varepsilon}{2m\hbar}\Bigl(p-\frac{m\Delta}{\varepsilon}\Bigr)^{2}+\frac{m\Delta^{2}}{2\hbar\varepsilon}

Complete the square in $p$ .

\int\frac{dp}{2\pi\hbar}\,e^{-i\varepsilon(p-p_{0})^{2}/2m\hbar}=\sqrt{\frac{m}{2\pi i\hbar\varepsilon}}

The Gaussian integral $\int e^{-a u^{2}}du=\sqrt{\pi/a}$ with .

\bra{x'}e^{-i\varepsilon\hat{H}/\hbar}\ket{x}\approx\sqrt{\frac{m}{2\pi i\hbar\varepsilon}}\,e^{\frac{i\varepsilon}{\hbar}\bigl[\frac{m}{2}(\Delta/\varepsilon)^{2}-V(x)\bigr]}

The exponent is $i\varepsilon L/\hbar$ for the straight step.

Applications

Practice

Slices

A time $T$ is cut into $N$ slices of length $\varepsilon=T/N$ ; each slice is short enough to handle exactly.

\varepsilon=\frac{T}{N}

Try it

A time $T=2$ is cut into $N=8$ slices. What is $\varepsilon$ ?