Feynman's path integral formulation and the classical path

Suppose that we have a quantum-mechanical particle that starts out in the state \(\psi(x)\) at \(t = 0\), that is, 
\[
|\Psi(0)\rangle=\int \psi(x)|x\rangle dx.
\]
And we wish to compute its state \(\phi(x)\) at some other time \(t=T\), i.e.,
\[
|\Psi(T)\rangle=\int \phi(x)|x\rangle dx.
\]
Let
\[
\mathcal{P}(a \to b) = \{ \gamma: [0, T] \to \mathbb{R}^3 : \gamma(0) = a, \gamma(T) = b \}
\]
denote the space of all paths that start at the point \(a\) at time 0 and end at the point \(b\) at time \(T\). 
The probability of finding the particle at a given point \(b\) at time \(T\) is \(\mathcal{A}_\text{total} \in \mathbb C\), the **sum of the amplitudes**, \(\mathcal{A}[\gamma] \in \mathbb C\), of all the paths that end at that point. That is,
\[
\mathcal{A}_\text{total} = \int_{\mathcal{P}(a \to b)} \mathcal{A}[\gamma] \, \mathcal{D}\gamma ,
\]
where \(\mathcal{D}\gamma\) is some sort of mysterious [[measure]] on the space \(\mathcal{P}(a \to b)\). The probability **amplitude of an individual path** is given by the exponential of the action of the path divided by \(\hbar\).
\[
\mathcal{A}[\gamma] = \exp\left(\frac{i}{\hbar}S[\gamma]\right),
\]
where \(S\) is the action of the path. The action is defined as
\[
S[\gamma] = \int_{0}^{T} L(\gamma,\dot{\gamma}) dt
\]
where \(L\) is the [[Lagrangian Mechanics#Jet space approach|Lagrangian]] of the system. If we discretize the process in "finite jumps", all this should be concluded (or, at least, visualized) from [[quantum approach to particle motion- an example]]. I have to develop this...

The quantity \(\mathcal{A}_\text{total}\) is usually denoted by \(K(a,b,T)\) and called the [[propagator]].

For a fixed \(b\), the system will end in state \(|b\rangle\), according to "conditional probability", with amplitude:
\[
\begin{aligned}
\phi(b) &= \int_{\mathbb{R}^3} K(a,b,T) \psi(a) \, da\\
&=\int_{\mathbb{R}^3} \left( \int_{\mathcal{P}(a \to b)} e^{\frac{i}{\hbar} S(\gamma)} \mathcal{D}\gamma \right) \psi(a) \, da.
\end{aligned}
\]
That is, the final state will be 
\[
\Psi(T)=\int \phi(b)|b\rangle db=\int \int_{\mathbb{R}^3} K(a,b,T) \psi(a) \, da|b\rangle db
\]

What Happens in the Classical Limit?

In the **classical limit**, we let Planck’s constant \(\hbar \to 0\). This has a profound effect on the path integral. The amplitude for any given path \(\gamma\) is:
\[\mathcal{A}[\gamma] = e^{\frac{i}{\hbar} S[\gamma]}\]
As \(\hbar\) becomes vanishingly small, even tiny differences in the action \(S[\gamma]\) between adjacent paths lead to **wildly varying phases**. The contributions from most paths interfere destructively and cancel each other out. The only paths that contribute significantly are those for which the phase is **stationary** with respect to small variations in the path. This stationary phase condition is precisely the principle of least action:
\[\delta S = 0 \quad \Rightarrow \quad \textbf{Classical equations of motion}\]
So, only the path that extremizes the action—the **classical path**, \(\gamma_{\text{classical}}\)—and its immediate neighbors contribute constructively. This simplifies the propagator to:
\[K(a,b,T) \approx C_{ab}e^{\frac{i}{\hbar} S[\gamma_\text{classical}]}\]
where \(S[\gamma_\text{classical}]\) is the action evaluated along the classical path from \(a\) to \(b\).

But this still allows for a classical path from *any* starting point \(a\) to *any* endpoint \(b\). How do we recover a single trajectory? The answer lies in the initial state.

Let's consider the full expression for the final wavefunction \(\phi(b)\), using a realistic initial wavepacket \(\psi_{ap_0}(x)\) localized around position \(a\) with average momentum \(p_0\):
\[\phi(b) = \int_{\mathbb{R}^3} K(x,b,T) \psi_{ap_0}(x) \, dx \approx \int_{\mathbb{R}^3} C_{xb}e^{\frac{i}{\hbar} S[\gamma_{\text{classical}}(x,b,T)]} \psi_{ap_0}(x) \, dx\]
The initial wavepacket \[\psi_{ap_0}(x) \sim \text{constant } e^{-\frac{(x-a)^2}{4\sigma^2}} \cdot e^{\frac{i}{\hbar} p_0 \cdot x}\]
also has a phase, corresponding to its momentum: \(e^{\frac{i}{\hbar} p_0 \cdot x}\). The total phase inside the integral is therefore \(\Phi(x, b) = S[\gamma_{\text{classical}}(x,b,T)] + p_0 \cdot x\).

We now apply the **stationary phase approximation a second time**, this time to the integral over the starting positions \(x\). For \(\hbar \to 0\), the quantity:
\[\phi(b) = \int_{\mathbb{R}^3} C_{xb} e^{-\frac{(x-a)^2}{4\sigma^2}}  \left(  e^{\frac{i}{\hbar} (S[\gamma_{\text{classical}}(x,b,T)]+p_0 \cdot x)} \right) \, dx\]
is only appreciable when \(S[\gamma_{\text{classical}}(x,b,T)]+p_0 \cdot x\) is near a extrema, that is, 
\[\frac{\partial S[\gamma_{\text{classical}}]}{\partial x} + p_0 = 0.\]
From classical mechanics, we know that \(\frac{\partial S}{\partial x}\) is the negative of the classical initial momentum, \(-p_i\) (this is Hamilton-Jacobi theory). The condition becomes:
\[-p_i + p_0 = 0 \quad \implies \quad \boldsymbol{p_i = p_0}\]
Finally, we are going to see that if we call \(b_{cl}\) to the end point of the classical trajectory with initial data \((a,p_0)\), **\(\phi(b)\) peaked around \(b_{cl}\)**.
First, observe that if \(x\neq a\) the integrand is small because of the \(e^{-\frac{(x-a)^2}{4\sigma^2}}\) term, so we only have contributions for \(x\approx a\). 
- If \(b=b_{cl}\) then, since \(x\approx a\), the initial momentum for \(\gamma_{cl}\) must be necessarily \(p_0\), so we are in a stationary point for \(S[\gamma_{\text{classical}}(x,b,T)]+p_0 \cdot x\).
- On the contrary, if \(b\neq b_{cl}\), and taking again into account that \(x\approx a\), the initial momentum cannot be \(p_0\), since \(b_{cl}\) is precisely the classical end point obtained with \(a\) and \(p_0\). Then \(x\) is not near an extrema for \(S[\gamma_{\text{classical}}(x,b,T)]+p_0 \cdot x\), and the integral is negligible.

Application: refraction of a laser beam

Let’s apply this path integral framework to the refraction of a laser beam, as seen in the picture where the beam passes from air into a glass block and back into air.

In the Feynman path integral formulation, light (or a photon) travels from one point to another by taking all possible paths. Each path contributes an amplitude, and the total amplitude at the destination determines the probability of detecting the light there.

Now, why do we see the beam? The path integral tells us that the amplitude—and thus the intensity—is highest along the classical path (the refracted path). However, light itself isn’t visible unless it reaches our eyes. In the air and glass, tiny particles (dust or imperfections) scatter some of the light toward us. They act like a global particle detector. Since the intensity is maximized along the classical path due to constructive interference, more light scatters from points along this path, making the beam visible as a sharp, bright line. Away from this path, the amplitude drops off rapidly due to destructive interference, so little scattering occurs, and the beam doesn’t appear diffuse.

In summary, the path integral formulation shows that the laser beam’s path—its refraction at each interface—is the classical path where the action is extremized, leading to constructive interference. This recovers the classical optics result of Snell’s law in the limit \( \hbar \to 0 \). The visibility of the beam comes from scattering along this high-intensity path, revealing the trajectory determined by the path integral’s dominant contribution.

Application: The Classical Path of a Tennis Ball

Imagine throwing the tennis ball from point A to point B. In the path integral picture, the ball could take any path: a straight line, a zig-zag, a loop, or even a path that shoots up into space and back down. Each path has an associated action, a quantity from classical mechanics defined as the integral of the Lagrangian (kinetic energy minus potential energy) over time. The classical path—the parabolic trajectory—is the one that minimizes the action, according to the principle of least action.

In the path integral, paths close to this classical path have similar actions, so their phases (related to the action via the quantum phase factor \( e^{iS/\hbar} \), where \( S \) is the action and \( \hbar \) is the reduced Planck constant) don’t vary much. These paths interfere constructively. Paths far from the classical one, like a loop or zig-zag, have very different actions, so their phases vary rapidly and interfere destructively, canceling out.

When you watch a tennis ball in flight, sunlight (or ambient light) bounces off the ball and reaches your eyes. At each moment, the ball interact with photons coming from the sunlight. This cloud of photons is like a global particle detector. If we were throwing billions of tennis balls, and recording the reflected photons, we would see a sharp trajectory, but like in the case of the laser beam, we would have a "kind of" diffusion around the path, only that infinitesimally small.

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