Module 5: Path Integral for the Harmonic Oscillator
The Second Exactly Solvable Case — Classical Paths, Fluctuation Determinants, and the Exact Propagator
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In the previous module, we rigorously derived the path integral from the operator formulation of quantum mechanics. We also worked through the free particle as the simplest application. Now we tackle the harmonic oscillator — the second exactly solvable path integral, and arguably the most important one in all of physics.
Why is this calculation so central? Because the harmonic oscillator isn’t just a toy. It’s the prototype for every quadratic action — and in quantum field theory, the free-field path integral is nothing but an infinite collection of harmonic oscillators. The techniques we develop here — decomposing paths into classical + fluctuations, computing fluctuation determinants, using eigenvalue methods — are the same tools that power modern QFT calculations. Master this module, and you’ve mastered the template for essentially all exactly solvable path integrals.
- How to find the classical path satisfying boundary conditions
- How to compute the classical action explicitly
- The crucial decomposition: \(x(t) = x_{\text{cl}}(t) + y(t)\) and why the action factorizes
- How to evaluate the fluctuation path integral via eigenvalue methods
- The exact harmonic oscillator propagator (boxed formula!)
- How to extract the energy spectrum \(E_n = (n + \tfrac{1}{2})\hbar\omega\) from the propagator
- The free-particle limit and the half-period singularity
Textbook references: Feynman & Hibbs, Ch. 3 §3-5; Shankar, Ch. 8 §8.6; Kleinert, Ch. 2; Zinn-Justin, Ch. 2.
1 The Setup
We consider a particle of mass \(m\) in a one-dimensional harmonic potential with angular frequency \(\omega\). The Lagrangian is:
\[ L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}m\omega^2 x^2 \]
This is a quadratic Lagrangian — quadratic in both \(x\) and \(\dot{x}\). This fact will have profound consequences: it means the path integral can be evaluated exactly.
Our goal is to compute the propagator (transition amplitude):
\[ K(x_b, t_b; x_a, t_a) = \int_{x(t_a) = x_a}^{x(t_b) = x_b} \mathcal{D}x(t) \; \exp\!\left[\frac{i}{\hbar} \int_{t_a}^{t_b} L\big(x, \dot{x}\big) \, dt \right] \]
with the boundary conditions:
\[ x(t_a) = x_a, \qquad x(t_b) = x_b \]
We define the total elapsed time:
\[ T \equiv t_b - t_a \]
Think of the path integral as a “quantum election” — every conceivable path from \((x_a, t_a)\) to \((x_b, t_b)\) gets to “vote,” weighted by the phase \(e^{iS/\hbar}\). For the harmonic oscillator, the action is quadratic, which means the “vote counting” reduces to a single Gaussian integral. This is why the harmonic oscillator is exactly solvable — just like any Gaussian integral is exactly computable.
2 The Classical Path
2.1 Setting Up the Equation of Motion
Before doing any quantum calculation, we need to find the classical path — the one that extremizes the action. The Euler-Lagrange equation for our Lagrangian is:
\[ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0 \]
Computing the derivatives:
\[ \frac{\partial L}{\partial \dot{x}} = m\dot{x}, \qquad \frac{\partial L}{\partial x} = -m\omega^2 x \]
So the equation of motion is:
\[ m\ddot{x} + m\omega^2 x = 0 \quad \implies \quad \boxed{\ddot{x} + \omega^2 x = 0} \]
This is the familiar simple harmonic oscillator equation!
2.2 The General Solution
The general solution is:
\[ x(t) = A\cos\omega t + B\sin\omega t \]
where \(A\) and \(B\) are constants determined by boundary conditions.
2.3 Applying Boundary Conditions
We need \(x(t_a) = x_a\) and \(x(t_b) = x_b\). Let’s apply these:
Condition 1: \(x(t_a) = x_a\):
\[ x_a = A\cos\omega t_a + B\sin\omega t_a \tag{1} \]
Condition 2: \(x(t_b) = x_b\):
\[ x_b = A\cos\omega t_b + B\sin\omega t_b \tag{2} \]
From Eq. (1):
\[ A = \frac{x_a - B\sin\omega t_a}{\cos\omega t_a} \]
Substituting into Eq. (2):
\[ x_b = \frac{(x_a - B\sin\omega t_a)\cos\omega t_b}{\cos\omega t_a} + B\sin\omega t_b \]
\[ x_b \cos\omega t_a = x_a \cos\omega t_b - B\sin\omega t_a \cos\omega t_b + B\sin\omega t_b \cos\omega t_a \]
\[ x_b \cos\omega t_a - x_a \cos\omega t_b = B\big(\sin\omega t_b \cos\omega t_a - \sin\omega t_a \cos\omega t_b\big) \]
The expression in parentheses is exactly \(\sin\omega(t_b - t_a) = \sin\omega T\):
\[ B = \frac{x_b \cos\omega t_a - x_a \cos\omega t_b}{\sin\omega T} \]
Similarly:
\[ A = \frac{x_a \sin\omega t_b - x_b \sin\omega t_a}{\sin\omega T} \]
2.4 Writing the Classical Path in a Clean Form
Substituting \(A\) and \(B\) back into \(x(t) = A\cos\omega t + B\sin\omega t\):
\[ x_{\text{cl}}(t) = \frac{(x_a \sin\omega t_b - x_b \sin\omega t_a)\cos\omega t + (x_b \cos\omega t_a - x_a \cos\omega t_b)\sin\omega t}{\sin\omega T} \]
Now, group the \(x_a\) and \(x_b\) terms:
- Coefficient of \(x_a\): \(\sin\omega t_b \cos\omega t - \cos\omega t_b \sin\omega t = \sin\omega(t_b - t)\)
- Coefficient of \(x_b\): \(\cos\omega t_a \sin\omega t - \sin\omega t_a \cos\omega t = \sin\omega(t - t_a)\)
Both identities use the sine subtraction formula \(\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta\).
Therefore, the classical path is:
\[ \boxed{x_{\text{cl}}(t) = \frac{x_a \sin\omega(t_b - t) + x_b \sin\omega(t - t_a)}{\sin\omega T}} \]
Let’s verify the boundary conditions:
- At \(t = t_a\): \(x_{\text{cl}}(t_a) = \frac{x_a \sin\omega T + x_b \cdot 0}{\sin\omega T} = x_a\) ✓
- At \(t = t_b\): \(x_{\text{cl}}(t_b) = \frac{x_a \cdot 0 + x_b \sin\omega T}{\sin\omega T} = x_b\) ✓
Also, for \(\omega \to 0\): using \(\sin\omega\tau \to \omega\tau\), we get \(x_{\text{cl}}(t) \to \frac{x_a(t_b - t) + x_b(t - t_a)}{T}\), which is the straight-line path of a free particle. ✓
3 The Classical Action
Now we compute the action evaluated on the classical path:
\[ S_{\text{cl}} = \int_{t_a}^{t_b} \left[\frac{1}{2}m\dot{x}_{\text{cl}}^2 - \frac{1}{2}m\omega^2 x_{\text{cl}}^2 \right] dt \]
This is a crucial calculation, so we’ll do every step.
3.1 Computing \(\dot{x}_{\text{cl}}\)
Differentiating the classical path:
\[ \dot{x}_{\text{cl}}(t) = \frac{-x_a \omega\cos\omega(t_b - t) + x_b \omega\cos\omega(t - t_a)}{\sin\omega T} \]
\[ = \frac{\omega}{\sin\omega T}\Big[-x_a \cos\omega(t_b - t) + x_b \cos\omega(t - t_a)\Big] \]
3.2 The Integration Trick
Rather than computing \(\dot{x}_{\text{cl}}^2\) and \(x_{\text{cl}}^2\) separately and integrating, there’s a beautiful shortcut. Since \(x_{\text{cl}}\) satisfies the equation of motion \(\ddot{x}_{\text{cl}} + \omega^2 x_{\text{cl}} = 0\), we can write:
\[ \frac{1}{2}m\omega^2 x_{\text{cl}}^2 = -\frac{1}{2}m x_{\text{cl}} \ddot{x}_{\text{cl}} \]
So the Lagrangian on the classical path becomes:
\[ L_{\text{cl}} = \frac{1}{2}m\dot{x}_{\text{cl}}^2 + \frac{1}{2}m x_{\text{cl}} \ddot{x}_{\text{cl}} \]
Now notice that this is a total derivative! Specifically:
\[ \frac{d}{dt}\Big(\frac{1}{2}m x_{\text{cl}} \dot{x}_{\text{cl}}\Big) = \frac{1}{2}m \dot{x}_{\text{cl}}^2 + \frac{1}{2}m x_{\text{cl}} \ddot{x}_{\text{cl}} = L_{\text{cl}} \]
This is a wonderful result. The classical action becomes a boundary term:
\[ S_{\text{cl}} = \int_{t_a}^{t_b} L_{\text{cl}} \, dt = \frac{1}{2}m \Big[x_{\text{cl}}(t) \dot{x}_{\text{cl}}(t)\Big]_{t_a}^{t_b} \]
\[ \boxed{S_{\text{cl}} = \frac{m}{2}\Big[x_b \dot{x}_{\text{cl}}(t_b) - x_a \dot{x}_{\text{cl}}(t_a)\Big]} \]
3.3 Evaluating the Boundary Terms
We need \(\dot{x}_{\text{cl}}\) at the endpoints. From our expression for \(\dot{x}_{\text{cl}}(t)\):
At \(t = t_b\): \(\omega(t_b - t_b) = 0\) and \(\omega(t_b - t_a) = \omega T\):
\[ \dot{x}_{\text{cl}}(t_b) = \frac{\omega}{\sin\omega T}\Big[-x_a \cos 0 + x_b \cos\omega T\Big] = \frac{\omega}{\sin\omega T}\Big[x_b \cos\omega T - x_a\Big] \]
At \(t = t_a\): \(\omega(t_b - t_a) = \omega T\) and \(\omega(t_a - t_a) = 0\):
\[ \dot{x}_{\text{cl}}(t_a) = \frac{\omega}{\sin\omega T}\Big[-x_a \cos\omega T + x_b \cos 0\Big] = \frac{\omega}{\sin\omega T}\Big[x_b - x_a \cos\omega T\Big] \]
3.4 Putting It All Together
\[ S_{\text{cl}} = \frac{m}{2}\left[x_b \cdot \frac{\omega(x_b \cos\omega T - x_a)}{\sin\omega T} - x_a \cdot \frac{\omega(x_b - x_a \cos\omega T)}{\sin\omega T}\right] \]
\[ = \frac{m\omega}{2\sin\omega T}\Big[x_b^2 \cos\omega T - x_a x_b - x_a x_b + x_a^2 \cos\omega T\Big] \]
\[ \boxed{S_{\text{cl}} = \frac{m\omega}{2\sin\omega T}\Big[(x_a^2 + x_b^2)\cos\omega T - 2x_a x_b\Big]} \]
This is one of the most important formulas in path integral quantum mechanics. Memorize its structure!
The classical action has a beautiful physical interpretation. The factor \(m\omega/\sin\omega T\) sets the overall scale and diverges when \(\omega T\) is a multiple of \(\pi\) (we’ll see why shortly). The term \((x_a^2 + x_b^2)\cos\omega T\) describes the “self-coupling” of the endpoints, while \(-2x_a x_b\) describes their mutual coupling. In the free-particle limit (\(\omega \to 0\)), this reduces to \(S_{\text{cl}} = m(x_b - x_a)^2 / 2T\) — the familiar free-particle action.
What physical interpretation can you give to the classical action \(S_{\text{cl}}\) in the limit \(x_a = x_b = x_0\)? What does the action look like?
Setting \(x_a = x_b = x_0\): \[ S_{\text{cl}} = \frac{m\omega x_0^2}{2\sin\omega T}(2\cos\omega T - 2) = \frac{m\omega x_0^2(\cos\omega T - 1)}{\sin\omega T} \] Using the identity \((\cos\theta - 1)/\sin\theta = -\tan(\theta/2)\): \[ S_{\text{cl}} = -m\omega x_0^2 \tan\!\left(\frac{\omega T}{2}\right) \] This is the action for a particle starting and ending at the same point \(x_0\). For \(T = 2\pi/\omega\) (one full period), the tangent vanishes and \(S_{\text{cl}} = 0\) — the particle goes around and comes back, accumulating zero net action. This matches the classical intuition that after a full period, the oscillator returns to its initial state with the same velocity.
4 Decomposition: Classical + Fluctuations
This section contains one of the most important ideas in path integral physics.
4.1 The Decomposition
Any path \(x(t)\) that goes from \(x_a\) at time \(t_a\) to \(x_b\) at time \(t_b\) can be written as:
\[ x(t) = x_{\text{cl}}(t) + y(t) \]
where \(x_{\text{cl}}(t)\) is the classical path we just found, and \(y(t)\) is a fluctuation (deviation) from the classical path. Since \(x(t)\) and \(x_{\text{cl}}(t)\) must both satisfy the same boundary conditions, the fluctuation must vanish at both endpoints:
\[ \boxed{y(t_a) = 0, \qquad y(t_b) = 0} \]
This is a Dirichlet boundary condition on the fluctuations.
4.2 The Action Splits Exactly
Let’s compute \(S[x_{\text{cl}} + y]\) by substituting into the action. The Lagrangian is:
\[ L = \frac{1}{2}m(\dot{x}_{\text{cl}} + \dot{y})^2 - \frac{1}{2}m\omega^2(x_{\text{cl}} + y)^2 \]
Expanding:
\[ L = \underbrace{\frac{1}{2}m\dot{x}_{\text{cl}}^2 - \frac{1}{2}m\omega^2 x_{\text{cl}}^2}_{L_{\text{cl}}} + \underbrace{m\dot{x}_{\text{cl}}\dot{y} - m\omega^2 x_{\text{cl}} y}_{\text{cross terms}} + \underbrace{\frac{1}{2}m\dot{y}^2 - \frac{1}{2}m\omega^2 y^2}_{L_{\text{fluct}}} \]
The action therefore has three pieces:
\[ S[x_{\text{cl}} + y] = S_{\text{cl}} + S_{\text{cross}} + S_2[y] \]
where:
\[ S_{\text{cross}} = \int_{t_a}^{t_b} \big(m\dot{x}_{\text{cl}}\dot{y} - m\omega^2 x_{\text{cl}} y\big) \, dt \]
\[ S_2[y] = \int_{t_a}^{t_b} \frac{1}{2}m\big(\dot{y}^2 - \omega^2 y^2\big) \, dt \]
4.3 The Cross Terms Vanish!
This is the key step. Let’s show that \(S_{\text{cross}} = 0\). Integrate the \(\dot{x}_{\text{cl}}\dot{y}\) term by parts:
\[ \int_{t_a}^{t_b} m\dot{x}_{\text{cl}}\dot{y} \, dt = \Big[m\dot{x}_{\text{cl}} y\Big]_{t_a}^{t_b} - \int_{t_a}^{t_b} m\ddot{x}_{\text{cl}} y \, dt \]
The boundary term vanishes because \(y(t_a) = y(t_b) = 0\). So:
\[ S_{\text{cross}} = -\int_{t_a}^{t_b} m\ddot{x}_{\text{cl}} y \, dt - \int_{t_a}^{t_b} m\omega^2 x_{\text{cl}} y \, dt \]
\[ = -\int_{t_a}^{t_b} m(\ddot{x}_{\text{cl}} + \omega^2 x_{\text{cl}}) y \, dt = 0 \]
The integrand vanishes identically because \(x_{\text{cl}}\) satisfies the equation of motion \(\ddot{x}_{\text{cl}} + \omega^2 x_{\text{cl}} = 0\)!
4.4 The Exact Result
The action splits cleanly into two parts:
\[ \boxed{S[x_{\text{cl}} + y] = S_{\text{cl}} + S_2[y]} \]
with the fluctuation action:
\[ S_2[y] = \frac{m}{2}\int_{t_a}^{t_b}\Big(\dot{y}^2 - \omega^2 y^2\Big) \, dt \]
There are no cubic, quartic, or higher-order terms — because the original action was at most quadratic in \(x\). This exact splitting is the hallmark of quadratic actions.
Why does the path integral split into \(e^{iS_{\text{cl}}/\hbar} \times (\text{fluctuation factor})\)? Is this always true?
This happens because the action is at most quadratic in \(x\). For a quadratic action, expanding \(x = x_{\text{cl}} + y\) gives: (1) a piece that depends only on the endpoints (the classical action \(S_{\text{cl}}\)), (2) cross terms that vanish because \(x_{\text{cl}}\) satisfies the equation of motion, and (3) a fluctuation piece \(S_2[y]\) that is independent of the endpoints.
For non-quadratic actions (e.g., with a \(\lambda x^4\) anharmonic term), the decomposition \(x = x_{\text{cl}} + y\) would produce cubic terms like \(\sim x_{\text{cl}} y^2\) and quartic terms \(\sim y^3, y^4\) in the fluctuation action. These “interaction terms” couple the fluctuations to the classical path and to each other, so the path integral does not factorize. In that case, one must resort to perturbation theory or numerical methods.
5 The Key Insight: Factorization
5.1 The Propagator Factorizes
Since \(S[x] = S_{\text{cl}} + S_2[y]\) and the measure \(\mathcal{D}x\) becomes \(\mathcal{D}y\) (shifting by the fixed classical path doesn’t change the measure), we have:
\[ K(x_b, t_b; x_a, t_a) = \int \mathcal{D}y \; \exp\!\left[\frac{i}{\hbar}\big(S_{\text{cl}} + S_2[y]\big)\right] \]
The classical action is just a number (it depends on \(x_a, x_b, T\), but not on \(y\)), so it comes outside the integral:
\[ \boxed{K(x_b, t_b; x_a, t_a) = e^{iS_{\text{cl}}/\hbar} \times \underbrace{\int \mathcal{D}y \; e^{iS_2[y]/\hbar}}_{F(T)}} \]
This is a profound factorization:
- The classical phase \(e^{iS_{\text{cl}}/\hbar}\) carries all the information about the endpoints \(x_a, x_b\).
- The fluctuation integral \(F(T)\) is the path integral over all fluctuations \(y(t)\) with Dirichlet boundary conditions \(y(t_a) = y(t_b) = 0\). It depends only on \(T\) (and on \(m, \omega, \hbar\)), not on the endpoints!
The fluctuation integral is Gaussian — it’s the functional analog of \(\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\pi/a}\). Computing it is our next task.
The factorization \(K = e^{iS_{\text{cl}}/\hbar} \times F(T)\) is the universal template for quadratic actions. To find \(K\), you need to:
- Find the classical path \(x_{\text{cl}}(t)\) satisfying the boundary conditions
- Compute the classical action \(S_{\text{cl}}\)
- Compute the fluctuation prefactor \(F(T)\)
This template works for the free particle, the harmonic oscillator, a particle in a linear potential, a charged particle in a uniform magnetic field, and many more.
6 Computing the Fluctuation Determinant
This is the most technically involved part of the calculation. We need to evaluate the Gaussian path integral:
\[ F(T) = \int \mathcal{D}y \; \exp\!\left[\frac{im}{2\hbar}\int_{t_a}^{t_b}\big(\dot{y}^2 - \omega^2 y^2\big) \, dt\right] \]
over all paths \(y(t)\) satisfying \(y(t_a) = y(t_b) = 0\).
6.1 Step 1: Eigenfunction Expansion
Since \(y(t)\) vanishes at both endpoints, we can expand it in a Fourier sine series. Let’s shift the time variable so that \(t_a = 0\) and \(t_b = T\) (this is just a relabeling and doesn’t affect the physics). Then:
\[ y(t) = \sum_{n=1}^{\infty} a_n \sin\!\left(\frac{n\pi t}{T}\right) \]
Each basis function \(\sin(n\pi t/T)\) automatically satisfies the Dirichlet boundary conditions \(y(0) = y(T) = 0\).
The coefficients \(\{a_n\}\) are the new integration variables — summing over all paths \(y(t)\) is equivalent to integrating over all possible values of the coefficients \(a_1, a_2, a_3, \ldots\)
6.2 Step 2: Evaluate the Fluctuation Action
Let’s compute \(S_2[y]\) in terms of the \(a_n\).
The \(\dot{y}^2\) term:
\[ \dot{y}(t) = \sum_{n=1}^{\infty} a_n \frac{n\pi}{T} \cos\!\left(\frac{n\pi t}{T}\right) \]
\[ \int_0^T \dot{y}^2 \, dt = \sum_{n,m} a_n a_m \frac{n\pi}{T} \frac{m\pi}{T} \int_0^T \cos\!\left(\frac{n\pi t}{T}\right)\cos\!\left(\frac{m\pi t}{T}\right) dt \]
The orthogonality of cosines gives us:
\[ \int_0^T \cos\!\left(\frac{n\pi t}{T}\right)\cos\!\left(\frac{m\pi t}{T}\right) dt = \frac{T}{2}\delta_{nm} \]
Therefore:
\[ \int_0^T \dot{y}^2 \, dt = \sum_{n=1}^{\infty} a_n^2 \frac{n^2\pi^2}{T^2} \cdot \frac{T}{2} = \frac{1}{2}\sum_{n=1}^{\infty} a_n^2 \frac{n^2\pi^2}{T} \]
The \(y^2\) term:
\[ \int_0^T y^2 \, dt = \sum_{n,m} a_n a_m \int_0^T \sin\!\left(\frac{n\pi t}{T}\right)\sin\!\left(\frac{m\pi t}{T}\right) dt = \sum_{n=1}^{\infty} a_n^2 \frac{T}{2} \]
where we used the orthogonality \(\int_0^T \sin(n\pi t/T)\sin(m\pi t/T) \, dt = (T/2)\delta_{nm}\).
Combining:
\[ S_2[y] = \frac{m}{2}\left[\frac{1}{2}\sum_n a_n^2 \frac{n^2\pi^2}{T} - \omega^2 \sum_n a_n^2 \frac{T}{2}\right] \]
\[ = \frac{m}{4}\sum_{n=1}^{\infty} a_n^2 \left(\frac{n^2\pi^2}{T} - \omega^2 T\right) \]
\[ \boxed{S_2[y] = \frac{m}{4}\sum_{n=1}^{\infty} \lambda_n \, a_n^2} \]
where we have defined the eigenvalues:
\[ \lambda_n = \frac{n^2\pi^2}{T} - \omega^2 T = \frac{T}{1}\left(\frac{n^2\pi^2}{T^2} - \omega^2\right) \]
The eigenvalues \(\lambda_n\) are the eigenvalues of the differential operator \(\hat{O} = -d^2/dt^2 - \omega^2\) acting on functions with Dirichlet boundary conditions. The eigenfunctions are \(\sin(n\pi t/T)\) with eigenvalues \((n\pi/T)^2 - \omega^2\). We’ll explore this operator interpretation much more deeply in Module 7 on Gaussian integrals and functional determinants.
6.3 Step 3: The Path Integral as a Product of Ordinary Integrals
The path integral measure becomes:
\[ \mathcal{D}y = \mathcal{N} \prod_{n=1}^{\infty} da_n \]
where \(\mathcal{N}\) is a normalization constant (which we’ll track carefully). The path integral factorizes into a product of independent Gaussian integrals:
\[ F(T) = \mathcal{N} \prod_{n=1}^{\infty} \int_{-\infty}^{\infty} da_n \; \exp\!\left[\frac{im}{4\hbar}\lambda_n a_n^2\right] \]
Each integral is a Fresnel integral of the form \(\int_{-\infty}^{\infty} e^{i\alpha x^2} dx = \sqrt{\pi / (-i\alpha)} = \sqrt{i\pi/\alpha}\):
\[ \int_{-\infty}^{\infty} da_n \; \exp\!\left[\frac{im\lambda_n}{4\hbar} a_n^2\right] = \sqrt{\frac{4\pi i \hbar}{m\lambda_n}} \]
So:
\[ F(T) = \mathcal{N} \prod_{n=1}^{\infty} \sqrt{\frac{4\pi i\hbar}{m\lambda_n}} \]
6.4 Step 4: Ratio with the Free Particle
The overall normalization \(\mathcal{N}\) is tricky, and it’s much cleaner to compute \(F(T)\) as a ratio with the free-particle case (\(\omega = 0\)). For \(\omega = 0\), the eigenvalues are \(\lambda_n^{(0)} = n^2\pi^2/T\), and we know the free-particle propagator is:
\[ K_{\text{free}} = \sqrt{\frac{m}{2\pi i\hbar T}} \exp\!\left[\frac{im(x_b - x_a)^2}{2\hbar T}\right] \]
So the free-particle fluctuation prefactor is \(F_{\text{free}} = \sqrt{m/2\pi i\hbar T}\).
The ratio of the harmonic oscillator fluctuation factor to the free-particle one involves:
\[ \frac{F(T)}{F_{\text{free}}(T)} = \prod_{n=1}^{\infty} \sqrt{\frac{\lambda_n^{(0)}}{\lambda_n}} = \prod_{n=1}^{\infty} \sqrt{\frac{n^2\pi^2/T}{n^2\pi^2/T - \omega^2 T}} \]
\[ = \prod_{n=1}^{\infty} \frac{1}{\sqrt{1 - \omega^2 T^2/(n^2\pi^2)}} \]
So:
\[ \left(\frac{F(T)}{F_{\text{free}}(T)}\right)^2 = \prod_{n=1}^{\infty} \frac{1}{1 - \omega^2 T^2/(n^2\pi^2)} \]
6.5 Step 5: The Infinite Product Identity
Now we use the famous Euler infinite product formula for the sine function:
\[ \sin z = z \prod_{n=1}^{\infty}\left(1 - \frac{z^2}{n^2\pi^2}\right) \]
Setting \(z = \omega T\):
\[ \prod_{n=1}^{\infty}\left(1 - \frac{\omega^2 T^2}{n^2\pi^2}\right) = \frac{\sin\omega T}{\omega T} \]
Therefore:
\[ \left(\frac{F(T)}{F_{\text{free}}(T)}\right)^2 = \frac{\omega T}{\sin\omega T} \]
\[ \frac{F(T)}{F_{\text{free}}(T)} = \sqrt{\frac{\omega T}{\sin\omega T}} \]
And since \(F_{\text{free}} = \sqrt{m/2\pi i\hbar T}\):
\[ F(T) = \sqrt{\frac{m}{2\pi i\hbar T}} \cdot \sqrt{\frac{\omega T}{\sin\omega T}} = \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega T}} \]
\[ \boxed{F(T) = \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega T}}} \]
The fluctuation prefactor tells you “how many paths contribute effectively.” When \(\sin\omega T\) is small (near \(T = n\pi/\omega\)), the prefactor diverges — this means the paths are “focusing” toward a single configuration, producing a delta-function-like propagator. When \(\sin\omega T\) is large, many paths contribute with similar phases, producing a broad, smooth propagator.
7 The Final Propagator
7.1 Assembling the Pieces
We now combine the classical phase and the fluctuation prefactor:
\[ K(x_b, t_b; x_a, t_a) = F(T) \cdot \exp\!\left[\frac{i}{\hbar}S_{\text{cl}}\right] \]
Substituting our results:
\[ \boxed{K(x_b, t_b; x_a, t_a) = \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega T}} \; \exp\!\left\{\frac{im\omega}{2\hbar\sin\omega T}\Big[(x_a^2 + x_b^2)\cos\omega T - 2x_a x_b\Big]\right\}} \]
This is the exact propagator for the quantum harmonic oscillator, first derived by Feynman. It is one of the crown jewels of path integral quantum mechanics.
Let’s appreciate what we’ve accomplished: starting from \(\int \mathcal{D}x \, e^{iS[x]/\hbar}\), a sum over infinitely many paths, we’ve obtained a closed-form expression. This was possible because the action is quadratic.
Textbook reference: Feynman & Hibbs, §3-5, Eq. (3-59); Shankar, Ch. 8, Eq. (8.6.14).
7.2 Structure of the Propagator
| Component | Expression | Physical meaning |
|---|---|---|
| Prefactor | \(\sqrt{m\omega/2\pi i\hbar \sin\omega T}\) | Fluctuation determinant; “how many paths contribute” |
| Phase: diagonal | \((x_a^2 + x_b^2)\cos\omega T\) | Self-coupling of the endpoints |
| Phase: off-diagonal | \(-2x_a x_b\) | Mutual coupling between endpoints |
| Overall scale | \(m\omega/2\hbar\sin\omega T\) | Sets the “sharpness” of the propagator |
What happens at \(T = \pi/\omega\) (half-period)? The propagator diverges! What does this mean physically?
At \(T = \pi/\omega\), we have \(\sin\omega T = \sin\pi = 0\). But the propagator doesn’t just blow up randomly — it becomes a delta function: \(K \to \delta(x_b + x_a)\).
To see this, note that as \(\sin\omega T \to 0\) and \(\cos\omega T \to -1\), the exponent becomes: \[ \frac{im\omega}{2\hbar\sin\omega T}\Big[-(x_a^2 + x_b^2) - 2x_a x_b\Big] = \frac{-im\omega}{2\hbar\sin\omega T}(x_a + x_b)^2 \]
This is a Gaussian in \((x_a + x_b)\) with width \(\to 0\), which is a representation of \(\delta(x_b + x_a)\).
Physical meaning: After a half-period, the particle arrives at the mirror-image point \(x_b = -x_a\). This is the classical focusing effect — all classical paths starting from \(x_a\) converge at \(-x_a\) after a half-period, regardless of their initial velocity. The quantum propagator reflects this classical focusing by becoming a delta function.
Similarly, at \(T = 2\pi/\omega\) (full period), \(K \to \delta(x_b - x_a)\): the particle returns to its starting point.
8 Recovering the Energy Spectrum
The propagator contains the complete information about the energy spectrum. Let’s extract it.
8.1 Spectral Decomposition of the Propagator
From operator quantum mechanics, we know that the propagator can be written as:
\[ K(x_b, t_b; x_a, t_a) = \sum_{n=0}^{\infty} \psi_n(x_b) \psi_n^*(x_a) \, e^{-iE_n T/\hbar} \]
where \(\psi_n(x)\) and \(E_n\) are the energy eigenfunctions and eigenvalues of the harmonic oscillator Hamiltonian.
If we can expand our exact propagator in this form, we can read off both the energy levels and the wavefunctions. This is a spectacular consistency check — and it also demonstrates that the path integral and operator formulations are completely equivalent.
8.2 The Mehler Formula
The mathematical identity that connects our propagator to the spectral sum is the Mehler formula (also known as Mehler’s Hermite formula). Let’s derive the expansion directly.
First, rewrite the exponent in a more symmetric form. Define the dimensionless variables \(\xi_a = \sqrt{m\omega/\hbar}\, x_a\) and \(\xi_b = \sqrt{m\omega/\hbar}\, x_b\), and set \(q = e^{-i\omega T}\). Then:
\[ \cos\omega T = \frac{q + q^{-1}}{2} = \frac{1+q^2}{2q}, \qquad \sin\omega T = \frac{q^{-1} - q}{2i} = \frac{1-q^2}{2iq} \]
After some algebra, the exponent becomes:
\[ \frac{im\omega}{2\hbar\sin\omega T}\Big[(\xi_a^2 + \xi_b^2)\cos\omega T - 2\xi_a\xi_b\Big] \cdot \frac{\hbar}{m\omega} \]
Working through the algebra carefully, the propagator can be written as:
\[ K = \sqrt{\frac{m\omega}{\pi\hbar}} \cdot \frac{1}{\sqrt{1 - q^2}} \exp\!\left[-\frac{\xi_a^2 + \xi_b^2}{2}\cdot\frac{1+q^2}{1-q^2} + \frac{2\xi_a \xi_b q}{1-q^2}\right] \]
Now expand using the generating function for Hermite polynomials. The Mehler formula states:
\[ \frac{1}{\sqrt{1 - q^2}} \exp\!\left[-\frac{(\xi_a^2 + \xi_b^2)q^2 - 2\xi_a\xi_b q}{1-q^2}\right] = \sum_{n=0}^{\infty} \frac{q^n}{2^n n!} H_n(\xi_a) H_n(\xi_b) \]
where \(H_n\) are the Hermite polynomials.
8.3 Reading Off the Spectrum
Noting that \(q^n = e^{-in\omega T}\), the exponential prefactor contributes \(e^{-(\xi_a^2 + \xi_b^2)/2} \cdot e^{-i\omega T/2}\) (from careful treatment of the \(1/(1-q^2)\) and the prefactor). The full expansion becomes:
\[ K = \sum_{n=0}^{\infty} \left[\frac{1}{\sqrt{2^n n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x_b^2/2\hbar} H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}x_b\right)\right] \] \[ \times \left[\frac{1}{\sqrt{2^n n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x_a^2/2\hbar} H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}x_a\right)\right] e^{-i(n+1/2)\omega T} \]
Comparing with \(K = \sum_n \psi_n(x_b)\psi_n^*(x_a) e^{-iE_n T/\hbar}\), we read off:
8.3.1 The Energy Levels
\[ \boxed{E_n = \left(n + \frac{1}{2}\right)\hbar\omega, \qquad n = 0, 1, 2, \ldots} \]
The familiar harmonic oscillator spectrum, complete with the zero-point energy \(E_0 = \hbar\omega/2\), emerges naturally from the path integral!
8.3.2 The Wavefunctions
\[ \boxed{\psi_n(x) = \frac{1}{\sqrt{2^n n!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp\!\left(-\frac{m\omega x^2}{2\hbar}\right) H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}\, x\right)} \]
These are exactly the harmonic oscillator wavefunctions obtained from the operator (ladder operator) approach!
The path integral propagator \(K\) is not just a transition amplitude — it’s a generating function for the entire energy spectrum and all wavefunctions. By computing a single object (the propagator), you get all the eigenvalues and eigenfunctions simultaneously. This is the power of the path integral approach.
9 Connection to the QHO Course
If you’ve studied the Quantum Harmonic Oscillator course on this site, you’ll recognize that the results we just derived — \(E_n = (n+\frac{1}{2})\hbar\omega\) and the Hermite-Gaussian wavefunctions — are exactly what was obtained there using ladder operators (the algebraic method) and by solving the Schrödinger equation directly.
What’s remarkable is that the path integral approach gives us these results from a completely different starting point. We never once wrote down the Schrödinger equation. We never used operators \(\hat{a}\) and \(\hat{a}^\dagger\). We simply computed a sum over all paths — and the energy levels and wavefunctions emerged automatically.
| Approach | Starting Point | How \(E_n\) is Obtained |
|---|---|---|
| Ladder operators | \([\hat{a}, \hat{a}^\dagger] = 1\) | Algebraic: count rungs on the ladder |
| Schrödinger equation | \(\hat{H}\psi = E\psi\) | Solve ODE, demand normalizability |
| Path integral | \(K = \int \mathcal{D}x \, e^{iS/\hbar}\) | Compute \(K\), expand as spectral sum |
All three approaches are equivalent. The path integral doesn’t give you “different physics” — it gives you a different perspective on the same physics, one that generalizes beautifully to field theory.
10 Worked Examples
10.1 Worked Example 1: The Free-Particle Limit (\(\omega \to 0\))
Problem: Show that the harmonic oscillator propagator reduces to the free-particle propagator when \(\omega \to 0\).
Solution: We need to take the limit \(\omega \to 0\) carefully in:
\[ K_{\text{HO}} = \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega T}} \; \exp\!\left\{\frac{im\omega}{2\hbar\sin\omega T}\Big[(x_a^2 + x_b^2)\cos\omega T - 2x_a x_b\Big]\right\} \]
Step 1: Expand the trigonometric functions.
For small \(\omega\):
\[ \sin\omega T \approx \omega T - \frac{(\omega T)^3}{6} + \cdots \approx \omega T \]
\[ \cos\omega T \approx 1 - \frac{(\omega T)^2}{2} + \cdots \approx 1 \]
Step 2: Expand the prefactor.
\[ \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega T}} \approx \sqrt{\frac{m\omega}{2\pi i\hbar \cdot \omega T}} = \sqrt{\frac{m}{2\pi i\hbar T}} \]
This is exactly the free-particle prefactor! ✓
Step 3: Expand the exponent.
\[ \frac{m\omega}{2\hbar\sin\omega T}\Big[(x_a^2 + x_b^2)\cos\omega T - 2x_a x_b\Big] \]
\[ \approx \frac{m\omega}{2\hbar \cdot \omega T}\Big[(x_a^2 + x_b^2) \cdot 1 - 2x_a x_b\Big] = \frac{m}{2\hbar T}(x_a - x_b)^2 \]
using \((x_a^2 + x_b^2 - 2x_a x_b) = (x_a - x_b)^2\).
Step 4: Assemble.
\[ K_{\text{HO}} \xrightarrow{\omega \to 0} \sqrt{\frac{m}{2\pi i\hbar T}} \exp\!\left[\frac{im(x_b - x_a)^2}{2\hbar T}\right] = K_{\text{free}} \quad \checkmark \]
The harmonic oscillator propagator smoothly reduces to the free-particle propagator as the restoring force vanishes.
10.2 Worked Example 2: Extracting the Ground State from \(T \to \infty\)
Problem: Extract the ground-state energy and wavefunction from the harmonic oscillator propagator by taking the limit \(T \to \infty\) (with appropriate \(i\epsilon\) prescription).
Solution: The spectral decomposition tells us:
\[ K = \sum_{n=0}^{\infty} \psi_n(x_b)\psi_n^*(x_a)\, e^{-iE_n T/\hbar} \]
For large \(T\), all terms oscillate wildly and the sum doesn’t converge. The trick is to give \(T\) a small negative imaginary part: \(T \to T(1 - i\epsilon)\) with \(\epsilon > 0\) small. This is equivalent to a small Wick rotation. Then:
\[ e^{-iE_n T/\hbar} \to e^{-iE_n T/\hbar} \cdot e^{-E_n \epsilon T/\hbar} \]
The real exponential \(e^{-E_n \epsilon T/\hbar}\) suppresses all terms with \(E_n > E_0\). As \(T \to \infty\), only the ground state (\(n = 0\)) survives:
\[ K \xrightarrow{T \to \infty} \psi_0(x_b)\psi_0^*(x_a)\, e^{-iE_0 T/\hbar} \]
Step 1: Take \(T \to \infty\) in the propagator.
With \(T \to \infty (1 - i\epsilon)\), we have \(\omega T \to \infty\) with a negative imaginary part. Therefore:
\[ \sin\omega T = \frac{e^{i\omega T} - e^{-i\omega T}}{2i} \to \frac{-e^{-i\omega T}}{2i} = \frac{e^{-i\omega T}}{2i} \cdot (-1) \]
Wait — let’s be more careful. With \(T \to T(1-i\epsilon)\):
\[ e^{i\omega T} \to e^{i\omega T} \cdot e^{\omega\epsilon T} \to \infty \] \[ e^{-i\omega T} \to e^{-i\omega T} \cdot e^{-\omega\epsilon T} \to 0 \]
So \(\sin\omega T \approx e^{i\omega T}/2i\) and \(\cos\omega T \approx e^{i\omega T}/2\).
Step 2: Evaluate the prefactor.
\[ \sqrt{\frac{m\omega}{2\pi i\hbar \sin\omega T}} \approx \sqrt{\frac{m\omega}{2\pi i\hbar \cdot e^{i\omega T}/2i}} = \sqrt{\frac{m\omega}{\pi\hbar \cdot e^{i\omega T}}} = \sqrt{\frac{m\omega}{\pi\hbar}} \, e^{-i\omega T/2} \]
Step 3: Evaluate the exponent.
\[ \frac{im\omega}{2\hbar\sin\omega T}\Big[(x_a^2 + x_b^2)\cos\omega T - 2x_a x_b\Big] \]
\[ \approx \frac{im\omega}{2\hbar} \cdot \frac{2i}{e^{i\omega T}} \left[(x_a^2 + x_b^2)\frac{e^{i\omega T}}{2} - 2x_a x_b\right] \]
\[ = \frac{im\omega}{2\hbar} \cdot 2i \left[\frac{x_a^2 + x_b^2}{2} - 2x_a x_b \, e^{-i\omega T}\right] \]
The term \(e^{-i\omega T} \to 0\) as \(T \to \infty\) (with the \(i\epsilon\) prescription), so:
\[ \text{exponent} \to \frac{im\omega}{2\hbar} \cdot 2i \cdot \frac{x_a^2 + x_b^2}{2} = -\frac{m\omega}{2\hbar}(x_a^2 + x_b^2) \]
(since \(i \cdot i = -1\)).
Step 4: Assemble.
\[ K \xrightarrow{T \to \infty} \sqrt{\frac{m\omega}{\pi\hbar}} \, e^{-i\omega T/2} \exp\!\left[-\frac{m\omega}{2\hbar}(x_a^2 + x_b^2)\right] \]
\[ = \left[\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x_b^2/2\hbar}\right] \left[\left(\frac{m\omega}{\pi\hbar}\right)^{1/4} e^{-m\omega x_a^2/2\hbar}\right] e^{-i\omega T/2} \]
Step 5: Read off the results.
Comparing with \(K \to \psi_0(x_b)\psi_0^*(x_a) e^{-iE_0 T/\hbar}\):
\[ \boxed{\psi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp\!\left(-\frac{m\omega x^2}{2\hbar}\right)} \]
\[ \boxed{E_0 = \frac{1}{2}\hbar\omega} \]
The ground-state Gaussian wavefunction and the zero-point energy emerge directly from the large-\(T\) behavior of the propagator. The zero-point energy \(\hbar\omega/2\) is not put in by hand — it comes from the fluctuation determinant (the \(\sqrt{\sin\omega T}\) in the denominator).
11 Summary
11.1 Key Formulas at a Glance
| Quantity | Formula |
|---|---|
| Classical path | \(x_{\text{cl}}(t) = \dfrac{x_a \sin\omega(t_b - t) + x_b \sin\omega(t - t_a)}{\sin\omega T}\) |
| Classical action | \(S_{\text{cl}} = \dfrac{m\omega}{2\sin\omega T}\Big[(x_a^2 + x_b^2)\cos\omega T - 2x_a x_b\Big]\) |
| Action decomposition | \(S[x_{\text{cl}} + y] = S_{\text{cl}} + S_2[y]\), with \(S_2 = \dfrac{m}{2}\displaystyle\int(\dot{y}^2 - \omega^2 y^2)\,dt\) |
| Fluctuation prefactor | \(F(T) = \sqrt{\dfrac{m\omega}{2\pi i\hbar \sin\omega T}}\) |
| Full propagator | \(K = \sqrt{\dfrac{m\omega}{2\pi i\hbar \sin\omega T}} \exp\!\left\{\dfrac{im\omega}{2\hbar\sin\omega T}\Big[(x_a^2+x_b^2)\cos\omega T - 2x_ax_b\Big]\right\}\) |
| Energy levels | \(E_n = (n + \tfrac{1}{2})\hbar\omega\) |
| Ground state wavefunction | \(\psi_0(x) = (m\omega/\pi\hbar)^{1/4}\, e^{-m\omega x^2/2\hbar}\) |
11.2 Free Particle vs. Harmonic Oscillator — Side by Side
| Feature | Free particle (\(\omega = 0\)) | Harmonic oscillator |
|---|---|---|
| Classical path | Straight line: \(x_{\text{cl}}(t) = x_a + \dfrac{x_b-x_a}{T}(t-t_a)\) | \(\dfrac{x_a \sin\omega(t_b-t) + x_b\sin\omega(t-t_a)}{\sin\omega T}\) |
| Classical action | \(\dfrac{m(x_b-x_a)^2}{2T}\) | \(\dfrac{m\omega}{2\sin\omega T}\Big[(x_a^2+x_b^2)\cos\omega T - 2x_ax_b\Big]\) |
| Prefactor | \(\sqrt{\dfrac{m}{2\pi i\hbar T}}\) | \(\sqrt{\dfrac{m\omega}{2\pi i\hbar \sin\omega T}}\) |
| Energy spectrum | \(E_k = \dfrac{\hbar^2 k^2}{2m}\) (continuous) | \(E_n = (n+\tfrac{1}{2})\hbar\omega\) (discrete) |
| Ground state | Plane wave (delocalized) | Gaussian \(\psi_0 \propto e^{-m\omega x^2/2\hbar}\) |
| Singularities in \(K\) | None | At \(T = n\pi/\omega\) (caustics) |
| Replacement rule | \(T \to \sin\omega T / \omega\) | — |
The replacement rule is striking: in every formula, you can go from the free particle to the harmonic oscillator by replacing \(T \to (\sin\omega T)/\omega\). This works for the prefactor and (with appropriate modifications for \(\cos\omega T\)) for the exponent.
Before moving on, make sure you can:
- Derive the classical path \(x_{\text{cl}}(t)\) from the equation of motion with boundary conditions
- Compute the classical action using the boundary-term trick \(S_{\text{cl}} = \frac{m}{2}[x_b\dot{x}_{\text{cl}}(t_b) - x_a\dot{x}_{\text{cl}}(t_a)]\)
- Explain why the cross terms \(S_{\text{cross}}\) vanish (it’s because \(x_{\text{cl}}\) satisfies the EOM)
- State why the factorization \(K = e^{iS_{\text{cl}}/\hbar} \times F(T)\) is exact for quadratic actions but not for anharmonic ones
- Reproduce the eigenvalue computation: \(\lambda_n = (n\pi/T)^2 - \omega^2\), and the use of the Euler product to get \(\sin\omega T / \omega T\)
- Write down the exact propagator from memory
- Take the \(\omega \to 0\) limit and recover the free-particle propagator
- Explain the physical meaning of the singularities at \(T = n\pi/\omega\)
12 Practice Problems
Problem 1 (Warm-up) ⭐
Verify that the harmonic oscillator propagator satisfies the Schrödinger equation:
\[ i\hbar \frac{\partial K}{\partial t_b} = -\frac{\hbar^2}{2m}\frac{\partial^2 K}{\partial x_b^2} + \frac{1}{2}m\omega^2 x_b^2 \, K \]
Hint: Differentiate the exact propagator with respect to \(t_b\) and \(x_b\). This is an algebraic exercise — tedious but important.
Problem 2 (Classical action verification) ⭐⭐
Compute the classical action \(S_{\text{cl}}\) by directly integrating \(\int_{t_a}^{t_b} L_{\text{cl}} \, dt\) without using the boundary-term trick. That is, compute \(\int \dot{x}_{\text{cl}}^2 \, dt\) and \(\int x_{\text{cl}}^2 \, dt\) separately using the explicit classical path. Verify that you get the same answer.
Hint: You’ll need the integrals \(\int_0^T \sin^2(\alpha t + \beta)\, dt\) and \(\int_0^T \cos^2(\alpha t + \beta) \, dt\).
Problem 3 (Full-period propagator) ⭐⭐
Show that at \(T = 2\pi/\omega\) (one full classical period), the propagator becomes:
\[ K\big(x_b, x_a; T = 2\pi/\omega\big) = \delta(x_b - x_a) \]
Hint: Take the limit \(\omega T \to 2\pi\) carefully. You’ll find \(\cos\omega T \to 1\), \(\sin\omega T \to 0\), and the exponent becomes a representation of the delta function.
What does this result mean physically?
Problem 4 (Propagator composition) ⭐⭐⭐
Verify the composition property explicitly: compute
\[ \int_{-\infty}^{\infty} K(x_b, T_2; x, 0) \, K(x, 0; x_a, -T_1) \, dx \]
and show that the result equals \(K(x_b, T_1 + T_2; x_a, 0)\). You’ll need to do a Gaussian integral.
Hint: Collect the exponent into a quadratic form in \(x\) and complete the square.
Problem 5 (The driven harmonic oscillator) ⭐⭐⭐
Consider a harmonic oscillator with an additional driving force \(f(t)\). The Lagrangian is:
\[ L = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}m\omega^2 x^2 + f(t)\,x \]
Find the classical path \(x_{\text{cl}}(t)\) by solving the driven equation of motion \(\ddot{x} + \omega^2 x = f(t)/m\) with boundary conditions \(x(t_a) = x_a\), \(x(t_b) = x_b\). Express your answer using the Green’s function of the harmonic oscillator.
Show that the action still splits as \(S = S_{\text{cl}} + S_2[y]\) and that the fluctuation prefactor \(F(T)\) is the same as for the undriven case.
Why is the fluctuation prefactor independent of the driving force?
Problem 6 (Quarter-period propagator) ⭐⭐⭐⭐
At \(T = \pi/(2\omega)\) (quarter period), show that the propagator simplifies to:
\[ K\big(x_b, x_a; T = \pi/2\omega\big) = \sqrt{\frac{m\omega}{2\pi i\hbar}} \exp\!\left[\frac{im\omega}{\hbar}\left(-x_a x_b\right)\right] \]
Show that this propagator, when applied to a Gaussian wavepacket centered at \(x_0\) with width \(\sigma\), produces the wavefunction at time \(T = \pi/(2\omega)\). What does this operation correspond to in phase space?
Hint: A quarter-period evolution of the harmonic oscillator is essentially a Fourier transform. This connection is deep — the harmonic oscillator generates continuous fractional Fourier transforms!