What path integrals are for

Intuition

The path integral gives the same answers as the Schrödinger equation, so what is it for? First, symmetry: the action is a single number, and a symmetry of the action is visible at a glance, which is why relativistic field theories are written this way. Second, perturbation theory: expanding $e^{iS/\hbar}$ in the interaction gives Feynman diagrams. Third, the classical limit and semiclassical methods, from the last lessons. Fourth, and most practically, imaginary time. Replace $t$ by $-i\tau$ : the phase $e^{iS/\hbar}$ becomes a positive weight $e^{-S_{E}/\hbar}$ , with $S_{E}=\int(\tfrac12m\dot{x}^{2}+V)d\tau$ , and the path integral becomes a sum of the kind statistical physics and computers handle well. At long imaginary times every excited state dies away faster than the ground state, so the decay of the imaginary-time propagator measures the ground-state energy; and a closed path of imaginary length $\hbar/kT$ gives a system in thermal equilibrium. Lattice calculations of the proton’s mass run on exactly this.

A mountain range seen by day is a confusion of ridges; seen as a relief map under low evening light, its shape stands out. Rotating time to imaginary values is like that change of light: the same landscape of paths, but with weights that can be summed by chance sampling.

The logarithm of an oscillator’s imaginary-time trace, $\ln\sum_{n}e^{-E_{n}\tau/\hbar}=-\ln\bigl(2\sinh(\omega\tau/2)\bigr)$ , against . At long times it runs parallel to the dashed line of the ground state alone: its slope is .

In imaginary time, $t=-i\tau$ , the propagator becomes a sum over paths with real positive weights:

\begin{aligned}K_{E}(x_{b},x_{a};\tau)&=\bra{x_{b}}e^{-\hat{H}\tau/\hbar}\ket{x_{a}}=\int\mathcal{D}x\;e^{-S_{E}[x]/\hbar}\\ S_{E}&=\int\Bigl(\tfrac12m\dot{x}^{2}+V\Bigr)\,d\tau\end{aligned}

Properties

$K_{E}=\sum_{n}\psi_{n}(x_{b})\psi_{n}^{*}(x_{a})\,e^{-E_{n}\tau/\hbar}$ : at long the ground state dominates, and .

The ground-state energy from imaginary time

E_{0}=-\lim_{\tau\to\infty}\frac{\hbar}{\tau}\,\ln K_{E}(x,x;\tau)

Continue the evolution operator to imaginary time: its phases become decaying exponentials, one for each level. Every excited level decays faster than the ground level, so at long times only the ground term is left, and the rate of decay of the propagator is the ground-state energy. The action continues to a positive one, so the same propagator is a sum of positive weights over paths.

Proof steps

t=-i\tau:\qquad e^{-i\hat{H}t/\hbar}=e^{-\hat{H}\tau/\hbar}

The evolution operator continued to imaginary time.

\begin{aligned}iS&=i\int\Bigl(\tfrac12m\Bigl(\frac{dx}{dt}\Bigr)^{2}-V\Bigr)dt\\ &=-\int\Bigl(\tfrac12m\Bigl(\frac{dx}{d\tau}\Bigr)^{2}+V\Bigr)d\tau=-S_{E}\end{aligned}

With $dt=-i\,d\tau$ and $\frac{dx}{dt}=i\frac{dx}{d\tau}$ .

K_{E}(x,x;\tau)=\sum_{n}|\psi_{n}(x)|^{2}\,e^{-E_{n}\tau/\hbar}

The spectral form with real exponentials.

K_{E}(x,x;\tau)\approx|\psi_{0}(x)|^{2}\,e^{-E_{0}\tau/\hbar}\qquad(\tau\to\infty)

Each excited term decays faster by $e^{-(E_{n}-E_{0})\tau/\hbar}$ .

-\frac{\hbar}{\tau}\ln K_{E}\to E_{0}-\frac{\hbar}{\tau}\ln|\psi_{0}(x)|^{2}\to E_{0}

The last term vanishes as $\tau\to\infty$ , wherever $\psi_{0}(x)\neq0$ .

Applications

Practice

Imaginary Time

With $t=-i\tau$ the phases become decaying exponentials; at long $\tau$ the ground state decays slowest and alone survives.

K_{E}\approx|\psi_{0}|^{2}\,e^{-E_{0}\tau/\hbar}

Try it

With $\hbar=1$ , an imaginary-time propagator falls by the factor $e^{-2}$ between $\tau=10$ and $\tau=14$ , long after the excited states have died. What is $E_{0}$ ?