2.3. The path integral: from classical to quantum mechanics

The path integral is a very powerful concept for connecting classical and quantum mechanics. Moreover, this formulation renders the connection between quantum mechanics and statistical mechanics very explicit. We shall restrict ourselves here to a discussion of the path integral in quantum mechanics. The reader is advised to consult the excellent book by Feynman and Hibbs (Quantum Mechanics and Path Integrals, McGraw-Hill, 1965) for more details.

The path integral formulation can be derived from the following heuristics, based on the analogy between particles and waves:

• A point particle which moves with momentum ${\bf p}$ at energy $E$ can also be viewed as a wave with a phase $\varphi$ given by $$\varphi = {\bf k}\cdot {\bf r}- \omega t$$ where ${\bf p}= \hslash {\bf k}$ and $E = \hslash \omega$.

• For a single path, these phases are additive, i.e., to find the phase of the entire path, the phases for different segments of the path should be added.

• The probablity to find a particle which at $t=t_0$ was at ${\bf r}_0$, at position ${\bf r}_1$ at time $t=t_1$, is given by the absolute square of the sum of the phase factors $\exp({\rm i}\varphi)$ of all possible paths leading from $({\bf r}_0, t_0)$ to $({\bf r}_1, t_1)$: $$P({\bf r}_0, t_0; {\bf r}_1, t_1) = \left| \sum_{\text{all paths}} e^{{\rm i}\varphi_{\text{path}}} \right|^2.$$ This probability is defined up to a constant which can be fixed by normalisation (i.e. the term within the absolute bars must reduce to a delta-function in ${\bf r}_1-{\bf r}_0$).

These heuristics are the analog of the Huygens principle in wave optics.

To analyse the consequences of these heuristics, we chop the time interval between $t_0$ and $t_1$ into many identical time slices (see Fig. 2.3) and consider one such slice. Within this slice we take the path to be linear.

A possible path running from an initial position $x_\text{i}$ at time $t_\text{i}$ to a final position $x_\text{f}$ at time $t_\text{f}$. The time is divided up into many identical slices.

To simplify the analysis we consider one-dimensional motion. We first consider the contribution of $k x$ to the phase difference. If the particle moves in a time $\Delta t$ over a distance $\Delta x$, we know that its $k$-vector is given by $$k = \frac{mv}{\hslash} = \frac{m \Delta x}{\hslash \Delta t}.$$ The phase change resulting from the displacement of the particle can therefore be given as $$\Delta \varphi = k \Delta x = \frac{m \Delta x^2}{\hslash \Delta t}.$$ We still must add the contribution of $\omega \Delta t$ to the phase. The frequency $\omega$ is related to the energy; we have $$\hslash \omega = \frac{p^2}{2m} + V(x) = \frac{\hslash^2 k^2}{2m} + V(x).$$ Neglecting the potential energy we obtain $$\Delta \varphi = \frac{m \Delta x^2}{\hslash \Delta t} - \frac{\hslash^2 k^2}{2m\hslash} \Delta t = \frac{m \Delta x^2}{2\hslash \Delta t}.$$ The potential also enters through the $\omega \Delta t$ term, to give the result: $$\Delta \varphi = \frac{m \Delta x^2}{2\hslash \Delta t} - \frac{V(x)}{\hslash}\Delta t.$$ For $x$ occurring in the potential we may choose any value between the begin and end point -- the most accurate result is obtained by substituting the average of the value at the beginning and at the end of the time interval.

If we now use the fact that phases are additive, we see that for the entire path the phases are given by $$\varphi = \frac{1}{\hslash} \sum_j \left\{ \frac{m}{2} \left[ \frac{x(t_{j+1}) - x(t_j)}{\Delta t} \right]^2 - \frac{V[x(t_j)] + V[x(t_{j+1})]}{2} \right\} \Delta t.$$ This is nothing but the discrete form of the classical action of the path! Taking the limit $\Delta t \rightarrow 0$ we obtain $$\varphi =\frac{1}{\hslash} \int_{t_0}^{t_1} \left[ \frac{m \dot{x}^2}{2} - V(x) \right] \; dt = \frac{1}{\hslash} \int_{t_0}^{t_1} L(x, \dot{x}) \; dt.$$ We therefore conclude that the probability to go from ${\bf r}_0$ at time $t_0$ to ${\bf r}_1$ at time $t_1$ is given by $$P({\bf r}_0, t_0; {\bf r}_1, t_1) = \left| {\mathcal N} \sum_{\text{all paths}} \exp \left[ \frac{{\rm i}}{\hslash} \int_{t_0}^{t_1} L(x, \dot{x}) \; dt \right] \right|^2$$ where ${\mathcal N}$ is the normalisation factor $${\mathcal N} = \sqrt{\frac{m}{2\pi {\rm i}\Delta t\hslash}}.$$ This now is the path integral formulation of quantum mechanics. Let us spend a moment to study this formulation. First note the large prefactor $1/\hslash$ in front of the exponent. If the phase factor varies when varying the path, this large prefactor will cause the exponential to vary wildly over the unit circle in the complex plane. The joint contribution to the probability will therefore become very small. If on the other hand there is a region in phase space (or 'path space') where the variation of the phase factor with the path is zero or very small, the phase factors will add up to a significant amount. Such regions are those where the action is stationary, that is, we recover the classical paths as those giving the major contribution to the phase factor. For $\hslash \rightarrow 0$ (the classical case), only the stationary paths remain, whereas for small $\hslash$, small fluctuations around these paths are allowed: these are the quantum fluctuations.

You may not yet recognise how this formulation is related to the Schrödinger equation. On the other hand, we may identify the expression within the absolute signs in the last expression for $P$ with a matrix element of the time evolution operator since both have the same meaning: $$\left\langle x_1 \right| \hat{U}(t_1-t_0) \left| x_0 \right\rangle= \sum_{\text{all paths}} {\mathcal N}\exp \left[ \frac{{\rm i}}{\hslash} \int_{t_0}^{t_1} L(x, \dot{x}) \; dt \right].$$ This form of the time evolution operator is sometimes called the propagator. Let us now evaluate this form of the time evolution operator acting for a small time interval $\Delta t$ on the wave function $\psi(x,t)$: $$\begin{gathered} \psi(x_1, t_1) = \int \left\langle x_1 \right|\hat{U}(t_1-t_0) \left| x_0 \right\rangle\left\langle x_0 | \psi \right\rangle dx_0 = \\ {\mathcal N} \int {\rm D} [x(t)] \int_{-\infty}^\infty \exp\left\{ \frac{{\rm i}}{\hslash} \int_{t_0}^{t_1} \left[ m \frac{\dot{x}^2(t)}{2} - V[x(t)]\right] \; dt \right\} \psi(x_0, t_0) \; dx_0 . \end{gathered}$$ The notation $\int {\rm D} [x(t)]$ indicates an integral over all possible paths from $(x_0, t_0)$ to $(x_1, t_1)$. We first approximate the integral over time in the same fashion as above, taking $t_1$ very close to $t_0$, and assuming a linear variation of $x(t)$ from $x_0$ to $x_1$: $$\psi(x_1, t_1) = {\mathcal N} \int_{-\infty}^\infty \exp\left\{ \frac{{\rm i}}{\hslash} \left[ m \frac{(x_1 - x_0)^2}{2\Delta t^2} - \frac{V(x_0)+V(x_1)}{2} \right] \; \Delta t \right\} \psi(x_0, t_0) \; dx_0.$$ A similar argument as used above to single out paths close to stationary ones can be used here to argue that the (imaginary) Gaussian factor will force $x_0$ to be very close to $x_1$. The allowed range for $x_0$ is $$(x_1 - x_0)^2 \ll \frac{\hslash\Delta t}{m}.$$ As $\Delta t$ is taken very small, we may expand the exponent with respect to the $V\Delta t$ term: $$\psi(x_1, t_1) = {\mathcal N}\int_{-\infty}^\infty \exp\left[ \frac{{\rm i}}{\hslash} m \frac{(x_1 - x_0)^2}{2\Delta t} \right] \left[ 1 - \frac{{\rm i}[ V(x_0)+V(x_1)]}{2\hslash} \Delta t\right] \psi(x_0, t_0) \; dx_0 .$$ As $x_0$ is close to $x_1$ we may approximate $\frac{ V(x_0)+V(x_1)}{2\hslash}$ by $V(x_1)/\hslash$. We now change the integration variable from $x_0$ to $u=x_0-x_1$: $$\psi(x_1, t_1) = {\mathcal N}\int_{-\infty}^\infty \exp\left( \frac{{\rm i}}{\hslash} m \frac{u^2}{2\Delta t} \right) \left[ 1 - {\rm i}/\hslash V(x_1) \Delta t\right] \psi(x_1+u,t_0)\; du .$$ As $u$ must be small, we can expand $\psi(x)$ about $x_1$ and obtain $$\psi(x_1, t_1) = {\mathcal N} \int_{-\infty}^\infty \exp\left( \frac{{\rm i}m}{\hslash} \frac{u^2}{2\Delta t} \right) \left[ 1 - \frac{{\rm i}}{\hslash} V(x_1) \Delta t\right] \left[\psi(x_1, t_0)+u \frac{\partial}{\partial x} \psi(x_1,t_0) + \frac{u^2}{2}\frac{\partial^2}{\partial x^2} \psi(x_1,t_0)\right] \; du .$$ Note that the second term in the Taylor expansion of $\psi$ leads to a vanishing integral as the integrand is an antisymmetric function of $u$. All in all, after evaluating the Gaussian integrals, we are left with $$\psi(x_1, t_1) = \psi(x_1, t_0) - \frac{{\rm i}\Delta t}{\hslash} V(x_1) \psi(x_1, t_0) + \frac{{\rm i}\hslash\Delta t}{2m} \frac{\partial ^2}{\partial x^2} \psi(x_1,t_0) .$$

Using $$\frac{\psi(x_1, t_1) - \psi(x_1, t_0)}{\Delta t} \approx \frac{\partial}{\partial t} \psi(x_1, t_1),$$ we obtain the time-dependent Schrödinger equation for a particle moving in one dimension: $${\rm i}\hslash \frac{\partial}{\partial t} \psi(x,t) = \left[ - \frac{\hslash^2}{2m}\frac{\partial^2}{\partial x^2} + V(x) \right] \psi(x,t).$$

You may have found this derivation a bit involved. It certainly is not the easiest way to arrive at the Schrödinger equation, but it has two attractive features;

• Everything was derived from simple heuristics which were based on viewing a particle as a wave and allowing for interference between the waves;

• The formulation shows that the classical path is obtained from quantum mechanics when we let $\hslash \rightarrow 0$.