The free particle path integral

Intuition

For a free particle the path integral can be done exactly, and it shows how the method works in general. Write every path as the classical straight line plus a deviation that vanishes at both ends. Because the action is quadratic in the path and the straight line makes it stationary, the action splits exactly into the classical action plus the action of the deviation alone; the cross term vanishes. The integral over the deviations does not know where the ends are, so it can depend only on the time: the propagator is a function of $T$ times $e^{iS_{\text{cl}}/\hbar}$ , with $S_{\text{cl}}=m(x_{b}-x_{a})^{2}/2T$ . Doing the sliced Gaussian integrals, or asking that two propagators compose into one, fixes the function: $K=\sqrt{m/2\pi i\hbar T}e^{im(x_{b}-x_{a})^{2}/2\hbar T}$ . It is exactly what the Schrödinger equation gives.

To find how a stretched string sags under its own weight you can split its shape into the smooth average curve and the wiggles about it; the wiggles do not care where the ends are pinned. The free propagator splits the same way into a classical part and a fluctuation part that depends only on the time.

The real part of the free propagator at a fixed time, against $x_{b}-x_{a}$ , in units where $m=\hbar=T=1$ and relative to its size (dashed): . The size is the same everywhere; the phase, the classical action over , turns faster and faster with the distance.

For a free particle of mass $m$ , with $T=t_{b}-t_{a}$ :

\begin{aligned}&K(x_{b},t_{b};x_{a},t_{a})\\ &=\sqrt{\frac{m}{2\pi i\hbar T}}\;\exp\Bigl(\frac{im(x_{b}-x_{a})^{2}}{2\hbar T}\Bigr)\end{aligned}

Properties

The phase is $S_{\text{cl}}/\hbar$ with $S_{\text{cl}}=m(x_{b}-x_{a})^{2}/2T$ , the action of the straight path at constant speed.

The free propagator from the path integral

\begin{aligned}K&=F(T)\,e^{iS_{\text{cl}}/\hbar}\\ S_{\text{cl}}&=\frac{m(x_{b}-x_{a})^{2}}{2T}\\ F(T)&=\sqrt{\frac{m}{2\pi i\hbar T}}\end{aligned}

Split each path into the straight line and a deviation vanishing at the ends. The cross term in the action is the first-order variation about a stationary path, so it vanishes, and the action splits. The deviations’ integral depends on $T$ alone. Composing the short-time propagators, each a Gaussian, fixes it: two Gaussians of times $\varepsilon_{1}$ and $\varepsilon_{2}$ combine into one of time .

Proof steps

x(t)=x_{\text{cl}}(t)+y(t),\qquad y(t_{a})=y(t_{b})=0

Every path is the straight line plus a deviation fixed at the ends.

\begin{aligned}S[x]&=S[x_{\text{cl}}]+\int m\dot{x}_{\text{cl}}\dot{y}\,dt+\int\tfrac12m\dot{y}^{2}\,dt\\ &=S_{\text{cl}}+S[y]\end{aligned}

The middle term is $\delta S$ at the stationary path: it vanishes.

S_{\text{cl}}=\tfrac12m\Bigl(\frac{x_{b}-x_{a}}{T}\Bigr)^{2}T=\frac{m(x_{b}-x_{a})^{2}}{2T}

The straight path at constant speed.

K=e^{iS_{\text{cl}}/\hbar}\int\mathcal{D}y\;e^{iS[y]/\hbar}=F(T)\,e^{iS_{\text{cl}}/\hbar}

The deviations’ integral does not involve the ends.

\int K_{\varepsilon_{2}}(x-y)\,K_{\varepsilon_{1}}(y-z)\,dy=K_{\varepsilon_{1}+\varepsilon_{2}}(x-z)

Each short-time kernel is $K_{\varepsilon}(\Delta)=\sqrt{m/2\pi i\hbar\varepsilon}\;e^{im\Delta^{2}/2\hbar\varepsilon}$ ; the Gaussian integral over adds the times, and repeating it over all slices gives .

Applications

Practice

The Classical Action

A free particle goes straight at constant speed, so its action is $\tfrac12m(\Delta x/T)^{2}$ times $T$ .

S_{\text{cl}}=\frac{m(x_{b}-x_{a})^{2}}{2T}

Try it

With $m=1$ , $x_{b}-x_{a}=4$ and $T=2$ , what is $S_{\text{cl}}$ ?