The Path Integral

If this seems very mysterious, you are not alone. Understanding what is going on here is in some sense equivalent to understanding Quantum Mechanics. I do not understand Quantum Mechanics. Feynman admitted that he never understood Quantum Mechanics. It may be true that nobody can understand Quantum Mechanics in the usual meaning of the word "understand."¹

Intro

Continuing the quantum inquisition, this post aims to explain/derive the Feynman's path integral and describe its utility.

This expression is key to understanding several areas of applied and theoretical mathematics including statistics, field theory, physics, and –notably for us/me/this series– quantum mechanics.

Whereas a standard, definite integral of the form $A = \int_a^b f(x)dx$ can be understood to compute the area under some curve on the range $[a, b]$ by adding up infinitely many infinitesimal rectangular slices of the curve with area $\ell\times w$ where $\ell = f(x)$ and $w = dx$:

and summing up these areas in the limit as we take the width $dx$ to be infinitesimally small, we get the continuous area under the curve. The path integral does something else all together: it allows us to compute quantities such as transition probabilities perturbatively – that is to say, multiple (possibly infinite) areas under several possibilities of curves, rather than single curve.

In our quantum context, the path integral allows us to sum over paths between points:

If we assign to each path a complex number given by some function $\phi : X \rightarrow \mathbb C$, where $X$ is the topological space of paths between points $a, b \in \mathbb R^2$.² $\phi(x)$ can be thought of as a unit vector which we can visualize on the complex plane:

Feynman's path integral finds purchase for us in quantum mechanics by allowing us to measure quantum states. Consider a quantum particle at some place and time: $(x_0, t_0)$. If we wait for some amount of time to elapse and measure the particle again, we must consider all possible trajectories the particle could have taken to reach its destination at ($x_f, t_f$), and our measurement will be the quantum probability of observing the particle at a given location (or, more commonly, a band or window of locations).

To compute this probability, we must sum over all paths:

Specifically, the probability of observing the particle at $(x_f, t_f)$ will be proportional to the modulus of this integral as established by Feynman when he was 23:³

Classical Kinematics

Whereas a classical trajectory can be described with kinematics equations with position as a function of time, e.g. throwing a ball:

The expression describing this function can be straightforwardly derived from $F=ma$, with $a=g$ being a downwards gravitational force:

If we know the state $x$ of the ball/particle at initial time $t_i$, we can precisely compute its state at some future time $x(t_f)$. Crucially, this is not true in quantum mechanics where we're limited to computing a probability of observation of some state in the future.

Double Slit Experiment

Enter the famous Double Slit experiment performed by Thomas Young in 1804,⁴ but more illustratively for electrons in particular by George Thompson,⁵ Davisson and Germer in 1927.⁶ Referring back to the epigraph, this mysterious experiment demonstrates that when slits are placed in the path of an electron, they behave not like particles but instead like waves, causing interference patterns on the measurement backdrop (unless they're explicitly observed). When firing classical particles through the slits at the backdrop, we would expect to see a distribution of measured impacts following a Gaussian like so:

If a quantum particle moved along a unique, definite curve like a baseball, then we could say for certain whether it passes through the left or the right slot. As noted by Heisenberg, only if we observe the path of the electron does it solidify into existence. Prior to/in the absence of observation, the electron behaves strictly probabilistically (or, more or less imprecisely: quantum-ly), perturbing all possible paths through all slits.

As such, after passing many particles through the slits towards the backdrop, we'd expect to see a normal distribution of particle impacts around the center of the backstop. But instead, we see an interference pattern:

Here, the peaks of dense particle clusters and valleys devoid of particle presence. This experiment indicates that quantum particles are somehow probing both slits and "cancelling" themselves out in at the valleys. Note that it is not the case that multiple particles are interfering with one another which can be proved by emitting a single electron at a time, repeatedly.

And this, like the cotton gin was to the civil war, is actually right where we get the path integral.

If we add a third slit, this now allows for quantum trajectories through that slit as well:

And we can inductively generalize this approach, scaling the number and placement of slits horizontally and vertically till we have a mesh of infinitely many many slits in infinitely many bars lying between the location of particle emission $a$ and our backdrop $b$:

In the limit, we find that the particle probes every possible path from $a \rightarrow b$. Thus, we have a physical motivation for the path integral, but what exactly is it doing quantitatively?

Path Function

Recall that we said that $\phi(x)$ is some function which maps a path to a complex number. This is accomplished via:

where $s[x]$ describes the action of the path, and $\hbar$ is Planck's constant.⁷ $s$ can be physically defined as the kinetic energy of the particle $E$ at any moment minus its potential energy $U$ integrated over the trajectory of the given path:

On the complex plane, this quantity will be a vector with angle $s[x]$. The amplitude for a particle to propagate from $x_i \rightarrow x_f$ is given by the product of the path integral with some constant $A$:

and, lastly, when measuring quantum particles, we're really measuring windows rather than individual points (since $p[x_f = x] = 0$ in the limit), so we multiply the complex amplitude by an infinitesimal width:

Computing the Path Integral

If we try to actually compute a path integral, we'll quickly be anti-chuffed because it is impossible to recover a discrete list of infinitely many paths. Recall that an ordinary integral just sums the slices under some curve in the limit:

For path integrals, $\phi$ is a functional which assigns a scalar to a curve. For illustrative purposes, we can discretize our toy curve, and consider what happens if we split a trajectory up into $N$ slices:

Our trajectory is a function which assigns a point in space $x_i$ for each discrete time step $t_i$. Connecting these points, we get a zig-zag shape which approximates our curve. Note that we can move each $x_i$ up or down in order to approximate any path. So, to sum over all trajectories, we just sum over all positions that each $x_i$ can take for all $x$ in our trajectory $f(t)$, so we end up integrating over the entire range of values in the $x, t$ plane which is what's meant by the capital infinitesimal $Dx$:⁸

Recall we also include an $A$ term for getting a sensible factor in the limit because of course we're taking the limit of $N$ intermediate points in our discrete approximation of our path in order to recover a high fidelity, continuous path:

Example

Let's work an example on a free particle. "Free" meaning no forces are acting on it s.t. $F=0$. For a classical particle like a baseball, then $F = ma = 0$ would imply a straight line trajectory from $x_i \rightarrow x_f$, but for a quantum particle, we must compute the action:

So we end up just raising $e$ to the power of $i$ times the kinetic portion of the action divided by Planck's constant for all points. By breaking up the path into discrete time slices, approximating the curve with each piece of the trajectory we're approximating being a straight line, we get:

And we repeat this process for all segments $i \in [1, N-1], j \leftarrow i+1$ to compute $s[x]$:

and, raising $e$ to the power of $\frac{is[x]}{\hbar}$ gives us the weight of this path:

Note that the beastly right hand side has the form of a bunch of Gaussians:

and also that only one term depends on $N$, so in order recover our path integral formula, we know that $A$ must be the inverse of this term:

All together the expression for the path integral of a free particle is:

This is the quantum mechanical amplitude $K$ for a free particle to propagate from from position $x_i \rightarrow x_f$:

Physically, this corresponds to the probability of finding the particle at position $x_f$, or rather, in an infinitesimally thin window of width $dx_f$ centered about $x_f$. To interpret the above amplitude as a probability, we take the absolute value squared:

In other words, the square of $K$ tells us the probability density of finding the particle at any given point:

Taking the modulus squared of our $K_{fi}$ term, we can note that it is comprised of two terms, the square root term and the exponential:

Note that the square of $|e^{bi}| = 1 \; \forall b \in \mathbb R$, so the 2nd term will just be 1, regardless of the mass of the particle, position, or time interval. The square of the modulus of first term will just remove the square root, so we get:

Curiously, this doesn't actually depend on the position at all! This implies that there is a uniform likelihood that the particle can be found anywhere in space... Let's sanity check if/how this makes sense. This stems from the Uncertainty Principle which indirectly states that the more narrowly we restrict a particles position in space, the broader the range of possible velocities it can have:

And furthermore, since we assumed that the particle has a definite initial position of $x_i$, we've effectively unbounded the range of possible initial velocities it might have, therefore it could feasibly shoot off into any other point in space with equal probability when observed later. This is why, in practice, we proscribe windows of initial conditions describing our particle.

Footnotes