Path Integral for Propagators: Time-Slicing Derivation

1. The Core Idea: A Sum Over Histories

Before we dive into the mathematical derivation, let's build some physical intuition. The central idea of the path integral is that a quantum particle, in moving from an initial point to a final point, does not follow a single, well-defined trajectory. Instead, it simultaneously explores every possible path that connects the two points. The total probability amplitude to get from start to finish is found by "summing" the contributions from all these paths.

The following video provides an excellent conceptual introduction to this "sum over histories" idea, grounding it in the famous double-slit experiment.

As the video explains, the total amplitude is a sum of complex numbers (or "arrows"), one for each path. The phase of each arrow is determined by a quantity you're very familiar with from classical mechanics: the action, $S$.

2. The Derivation: From Operators to Paths

Our goal is to derive this "sum over paths" formula starting from the standard operator formalism of quantum mechanics. The object we want to calculate is the propagator, or time-evolution kernel, which is the amplitude for a particle at position $x_i$ at time $t_i$ to be found at position $x_f$ at time $t_f$.

In the Schrödinger picture, you know that the time evolution of a state is governed by the Hamiltonian operator $\hat{H}$: $|\psi (t_f)⟩=e^{-i\hat{H}(t_f-t_i)}|\psi (t_i)⟩$. The propagator $K$ is simply the matrix element of this time evolution operator in the position basis:

Knowing the propagator is equivalent to solving the system, because you can find the wavefunction at any later time from the initial wavefunction $\psi (x,t_i)$ via:

Now, how do we evaluate $K$? The exponential of the Hamiltonian operator is generally very difficult to compute. Feynman's ingenious trick was to break the total time interval $T=t_f-t_i$ into a large number of infinitesimal slices.

The following lecture from MIT OpenCourseWare provides a step-by-step chalkboard derivation of the entire procedure. We will walk through its key segments together.

Let's summarize the key steps you just saw.

1. Time-Slicing: We divide the total time interval $T=t_f-t_i$ into $N$ small steps of duration $\Delta t=T/N$. The time evolution operator becomes a product of $N$ operators for these small steps:

   $$e^{-i\hat{H}T}=\underset{N\text{ times}}{\underbrace{e^{-i\hat{H}\Delta t}{e}^{-i\hat{H}\Delta t}\cdots e^{-i\hat{H}\Delta t}}}$$

2. Inserting Complete Sets: We insert a resolution of the identity, $1=\int dx|x⟩⟨x|$, between each of these factors. This transforms the propagator into:

   $$K=\int d{x}_1\cdots d{x}_{N-1}⟨x_f|e^{-i\hat{H}\Delta t}|x_{N-1}⟩⟨x_{N-1}|\cdots |x_1⟩⟨x_1|e^{-i\hat{H}\Delta t}|x_i⟩$$

   where we have labeled $x_f=x_N$ and $x_i=x_0$.

The problem is now reduced to evaluating the propagator for an infinitesimal time step, $⟨x_{j+1}|e^{-i\hat{H}\Delta t}|x_j⟩$.

Evaluating the Infinitesimal Propagator

This is the most crucial part of the derivation. For a Hamiltonian of the form $\hat{H}=\frac{{\hat{p}}^2}{2m}+V(\hat{q})$, the kinetic and potential energy operators do not commute. Therefore, we cannot simply write $e^{-i(\hat{T}+\hat{V})\Delta t}=e^{-i\hat{T}\Delta t}{e}^{-i\hat{V}\Delta t}$.

However, for a very small time step $\Delta t$, the Baker-Campbell-Hausdorff (BCH) formula tells us that we can make this separation, with corrections that are of order $(\Delta t{)}^2$, which will vanish as we take the limit $\Delta t\to 0$.

The next segment of the video shows exactly how to use this approximation, insert a complete set of momentum states, perform the resulting integrals, and magically arrive at an expression involving the classical Lagrangian.

Let's review the result. After inserting momentum states and performing the (Gaussian) integral over momentum, the infinitesimal propagator becomes:

The term in the exponent is precisely $\Delta t$ times the discretized classical Lagrangian $L=T-V$!

Assembling the Full Path Integral

Now we just need to put all the pieces back together and take the continuum limit. We have a product of $N$ such terms, which means the exponentials combine into an exponential of a sum.

In the limit $N\to \infty$ and $\Delta t\to 0$:

• The sum over the discrete time steps becomes a time integral: $$\sum _{j=0}^{N-1}\Delta tL(x_j,{\dot{x}}_j)⟶{\int }_{t_i}^{t_f}dtL(x(t),\dot{x}(t))\equiv S[x(t)]$$ This is the classical action.
• The product of integrals over all intermediate positions $d{x}_1\cdots d{x}_{N-1}$, along with the normalization factors, becomes the path integral measure, denoted $\mathcal{D}x(t)$.

Putting it all together, we arrive at the Feynman path integral representation for the propagator:

(We have restored $\hslash$ for clarity). This remarkable formula states that the quantum amplitude is a sum over all possible paths from the initial to the final state, where each path is weighted by a phase factor proportional to the classical action evaluated along that path.

3. A Note on Rigor

There are a couple of important technical points worth noting, which a mathematically rigorous treatment would need to address. For our purposes, it is sufficient to be aware of them.

1. Convergence: The integral we derived involves the complex exponential $e^{iS/\hslash }$, which oscillates and does not decay at infinity, making the integral's convergence problematic. This is typically handled by performing a Wick rotation, where we analytically continue time to the imaginary axis: $t\to -i\tau$. This changes the weighting factor to $e^{-S_E/\hslash }$, where $S_E$ is the Euclidean action. This factor is a real exponential that decays, making the integrals well-defined. At the end of the calculation, one rotates back to real time.

2. Hamiltonian vs. Lagrangian Form: We derived the Lagrangian form $\int \mathcal{D}x{e}^{i{S}_L[x]}$. If we had not performed the momentum integrals at each step, we would have arrived at the Hamiltonian path integral, which is an integral over both position and momentum paths in phase space:

   $$K=\int \mathcal{D}x(t)\mathcal{D}p(t)\exp \left[i\int dt(p\dot{x}-H(p,x))\right]$$

   The two forms are equivalent. The Lagrangian form is often more convenient for calculations and for making the connection to classical action explicit. You can find detailed derivations of both forms in these excellent notes.

Conclusion

In this lesson, we have constructed a bridge from the operator formalism of quantum mechanics to the path integral formulation. We started with the familiar definition of the propagator and, through the methodical process of time-slicing and inserting complete sets of states, derived Feynman's "sum over histories."

Key Takeaways:

• The quantum mechanical propagator can be expressed as a path integral.
• The derivation relies on the time-slicing method: breaking the total time evolution into infinitesimal steps and integrating over all intermediate states.
• For each infinitesimal step, the Baker-Campbell-Hausdorff approximation allows us to separate the kinetic and potential energy operators.
• The final expression is an integral over the space of all possible paths connecting the initial and final points. Each path is weighted by a phase factor $e^{iS[x]/\hslash }$, where $S[x]$ is the classical action of that path.

Preview of the Next Lesson:
We have successfully reformulated single-particle quantum mechanics. The true power of this method will become apparent when we apply it to fields. In our next lesson, we will extend the path integral formalism to scalar field theory to define the generating functional for Green's functions. This generating functional will be our master object, from which we can derive all the correlation functions (and thus S-matrix elements) of our theory in a remarkably elegant way.