Path Integrals for Free Particles and Harmonic Oscillators

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Hello! Welcome back to our exploration of the path integral formulation of quantum mechanics.

In our last lesson, we took a significant conceptual leap. Starting from the familiar ground of the Schrödinger picture, we derived the Feynman path integral, expressing the quantum propagator as a sum over all possible histories:

K (x_{f}, t_{f}; x_{i}, t_{i}) = \int D [x (t)] e^{i S [x (t)] /ℏ}

Today, we will move from this beautiful formalism to concrete calculation. Our goal is to evaluate this "integral over all paths" for two of the most fundamental systems in physics: the free particle and the quantum harmonic oscillator. These are two of the very few models for which the path integral can be solved analytically, and they provide an invaluable window into how the formalism works in practice. The key to their solubility lies in the fact that their Lagrangians are quadratic functions of position and velocity, which, as we will see, turns the path integral into a set of solvable Gaussian integrals.

1. The Free Particle Propagator

Let's begin with the simplest case: a free particle of mass $m$ . Its Lagrangian is purely kinetic, $L = \frac{1}{2} m x ˙^{2}$ . As we saw in the derivation, evaluating the path integral involves discretizing time into $N$ steps of duration $ϵ$ , integrating over all intermediate positions $x_{1}, \dots, x_{N - 1}$ , and then taking the limit as $N \to \infty$ .

This process boils down to calculating a sequence of Gaussian integrals. The following video provides an excellent and clear walkthrough of this exact calculation.

Feynman's (almost) impossible integral

The YouTube channel 'Physics with Elliot' offers a very clear, step-by-step calculation of the free particle path integral. This segment starts from the discretized action and meticulously carries out the integration.

Please watch the segment from 16:56 to 22:12. Focus on how the discretized action for a free particle is constructed and how the resulting series of Gaussian integrals is evaluated iteratively.

After performing the $N - 1$ integrals and taking the continuum limit, we arrive at the final expression for the free particle propagator:

K (x_{f}, t_{f}; x_{i}, t_{i}) = \frac{m}{2 π i ℏ T} exp [\frac{im ( x _{f} - x _{i} ) ^{2}}{2ℏ T}]

where $T = t_{f} - t_{i}$ .

Take a close look at the structure of this result. It consists of two parts:

A prefactor $\frac{m}{2 π i ℏ T}$ , which arose from the normalization of the Gaussian integrals.
An exponential phase factor. The term in the exponent is precisely $i S_{c l} /ℏ$ , where $S_{c l}$ is the classical action for a free particle moving in a straight line from $(x_{i}, t_{i})$ to $(x_{f}, t_{f})$ .

This structure—a fluctuation prefactor multiplied by the exponentiated classical action—is a general feature of path integrals for systems with quadratic Lagrangians.

Test your understanding!

Verify that the phase of the free particle propagator, $\frac{m ( x _{f} - x _{i} ) ^{2}}{2 T}$ , is indeed the classical action for a free particle.

Show answer

For a free particle, the classical path is a straight line, so the velocity is constant: $x ˙_{c l} (t) = \frac{x _{f} - x _{i}}{T}$ . The Lagrangian is $L = \frac{1}{2} m x ˙^{2}$ . The classical action is the integral of the Lagrangian along this classical path:

S_{c l} = \int_{t_{i}}^{t_{f}} L (x_{c l}, x ˙_{c l}) d t = \int_{t_{i}}^{t_{f}} \frac{1}{2} m (\frac{x _{f} - x _{i}}{T})^{2} d t

Since the integrand is constant, the integral is simply the integrand multiplied by the time duration $T$ :

S_{c l} = \frac{1}{2} m (\frac{x _{f} - x _{i}}{T})^{2} T = \frac{m ( x _{f} - x _{i} ) ^{2}}{2 T}

This confirms that the phase of the propagator is determined by the classical action.

2. The Quantum Harmonic Oscillator

Now let's tackle a more complex and illuminating case: the quantum harmonic oscillator (QHO), with Lagrangian $L = \frac{1}{2} m x ˙^{2} - \frac{1}{2} m ω^{2} x^{2}$ .

Path Integral for Simple Harmonic Oscillator — A visual summary of the problem: evaluating the path integral for the harmonic oscillator by summing over all paths, each weighted by the exponential of its action.

While we could, in principle, follow the same iterative integration procedure as for the free particle, the $x^{2}$ term in the potential makes the integrals significantly more coupled and complicated. A more powerful and elegant approach is to exploit the quadratic nature of the action directly.

A General Method for Quadratic Lagrangians

The key insight is to decompose any arbitrary path $q (t)$ into the unique classical path $q_{c} (t)$ that satisfies the boundary conditions, plus a deviation or "fluctuation" path $y (t)$ :

q (t) = q_{c} (t) + y (t)

Since $q (t)$ and $q_{c} (t)$ must both start at $q_{0}$ and end at $q_{f}$ , the fluctuation path must start and end at zero: $y (t_{0}) = y (t_{f}) = 0$ .

When we substitute this decomposition into a quadratic action, a crucial simplification occurs. The action separates into a sum of the classical action and an action for the fluctuations, with no cross-terms.

S [q (t)] = S [q_{c} (t)] + S [y (t)]

(The cross-term vanishes precisely because $q_{c}$ is an extremum of the action, meaning the first variation $δ S$ around $q_{c}$ is zero).

The propagator then factorizes beautifully:

K (q_{f}, q_{0}; t_{f}, t_{0}) = \int D [q_{c} + y] e^{i S [q_{c} + y] /ℏ} = e^{i S_{c} /ℏ} \int D [y] e^{i S [y] /ℏ}

The problem is now split in two:

Calculate the classical action $S_{c}$ for the harmonic oscillator.
Calculate the remaining path integral over the fluctuations $y (t)$ . This integral, $\int D [y] e^{i S [y] /ℏ}$ , depends only on the time interval $T = t_{f} - t_{0}$ , not the endpoints, and is often called the "fluctuation determinant".

The following reading provides a masterful derivation of a general formula for the propagator of any system with a quadratic Lagrangian.

Path integrals in quantum theory

These notes by Marcos Mariño derive a general and powerful formula for the propagator of any system with a quadratic Lagrangian. This is a more advanced but highly efficient method compared to direct integration.

Please read section 1.6, 'Gaussian theories' (pages 15-18). Focus on understanding the derivation leading to the main result, equation (1.149). This equation expresses the full propagator in terms of the classical action and a function f(t; t0) which is a solution to the classical equation of motion.

The remarkable result you just studied is:

K (q_{f}, q_{0}; t_{f}, t_{0}) = \frac{m}{2 π i ℏ f ( t _{f} ; t _{0} )} e^{i S_{c} (q_{f}, q_{0}; t_{f}, t_{0}) /ℏ}

Here, $f (t; t_{0})$ is the solution to the homogeneous classical equation of motion $m f ¨ + c (t) f = 0$ with initial conditions $f (t_{0}; t_{0}) = 0$ and $f ˙ (t_{0}; t_{0}) = 1$ .

Now, let's apply this powerful formula to solve our specific problem: the harmonic oscillator.

Path integrals in quantum theory

Having established the general formula, we can now apply it to the harmonic oscillator. This example in Mariño's notes walks through the final steps.

Please read 'Example 1.5. The harmonic oscillator' (pages 18-19). This section calculates the function f(t; t0) for the QHO. The classical action S_c itself is not derived in this section, but it is a standard result from classical mechanics. For completeness, you might also quickly look at 'Example 1.4. The free particle redux' on page 18 to see how this same method reproduces the free particle result we found earlier.

By finding the specific $f (t_{f}; t_{0})$ for the harmonic oscillator and combining it with the known classical action for the QHO, we arrive at the final propagator:

K (q_{f}, q_{0}; T) = \frac{mω}{2 π i ℏ sin ( ω T )} exp {\frac{imω}{2ℏ sin ( ω T )} [(q_{f}^{2} + q_{0}^{2}) cos (ω T) - 2 q_{f} q_{0}]}

This is a celebrated result in quantum mechanics, obtained here through the full machinery of the path integral. Notice that in the limit $ω \to 0$ , this expression correctly reduces to the free particle propagator (using $sin (θ) \approx θ$ and $cos (θ) \approx 1 - θ^{2} /2$ for small $θ$ ).

Conclusion

Today we have put the path integral formalism into practice, deriving the exact propagators for the free particle and the quantum harmonic oscillator. This was possible because their quadratic Lagrangians lead to Gaussian path integrals, which are analytically solvable.

Key Takeaways:

The free particle propagator was calculated by discretizing the path integral and performing a series of Gaussian integrals.
For general quadratic Lagrangians (including the QHO), a more powerful method involves separating the path into a classical trajectory and quantum fluctuations.
The classical path determines the phase of the propagator via the classical action, $e^{i S_{c l} /ℏ}$ .
The quantum fluctuations around the classical path determine the overall amplitude prefactor, which can be found by solving a related classical equation of motion.

Preview of the Next Lesson:
The free particle and QHO are special cases. For most systems, the Lagrangian is not quadratic (e.g., due to anharmonic potentials), and the path integral cannot be solved exactly. In the next lesson, we will learn how to approximate the path integral for these more general systems using the stationary phase approximation. This method will provide a profound justification for the Principle of Least Action and give us a clear picture of how classical mechanics emerges from quantum mechanics in the limit where the action is large compared to $ℏ$ .