Harmonic-Oscillator Path Integral

The harmonic oscillator is the second canonical exact path integral after the free particle. Its action is quadratic, so the path integral is Gaussian after the path is decomposed into a classical solution plus fluctuations. The result is the exact coordinate-space kernel

$$\begin{aligned} K_{\rm ho}(x_f,T;x_i,0) &= \left[ \frac{m\omega}{2\pi i\hbar\sin\omega T} \right]^{1/2} \\ &\quad\times \exp\left\{ \frac{im\omega}{2\hbar\sin\omega T} \left[ (x_f^2+x_i^2)\cos\omega T -2x_fx_i \right] \right\}, \end{aligned}$$

for $\sin\omega T\neq0$, with the usual phase prescription across caustics. The Harmonic-Oscillator Propagator page treats the same kernel from the spectral and Van Vleck viewpoints. This page makes the path-integral Gaussian structure explicit.

Quadratic Actions

For the one-dimensional harmonic oscillator,

$$L=\frac{m}{2}\dot x^2-\frac{m\omega^2}{2}x^2,$$

so

$$S[x] = \int_0^T dt\, \left[ \frac{m}{2}\dot x^2 - \frac{m\omega^2}{2}x^2 \right].$$

The fixed-endpoint path integral is

$$K_{\rm ho}(x_f,T;x_i,0) = \int_{x(0)=x_i}^{x(T)=x_f} \mathcal D x(t)\, e^{iS[x]/\hbar}.$$

The important feature is not merely solvability of the Schrödinger equation. It is the quadratic dependence of $S[x]$ on the path. After discretization, the path integral is an ordinary multidimensional Gaussian integral over intermediate positions. The continuum notation packages the same Gaussian determinant into a compact form.

Classical Path

Write each path as

$$x(t)=x_{\rm cl}(t)+\eta(t),$$

where $x_{\rm cl}$ satisfies the endpoint conditions

$$x_{\rm cl}(0)=x_i, \qquad x_{\rm cl}(T)=x_f,$$

and the fluctuation obeys Dirichlet boundary conditions

$$\eta(0)=\eta(T)=0.$$

For $\sin\omega T\neq0$, the classical solution is

$$x_{\rm cl}(t) = \frac{ x_i\sin\omega(T-t)+x_f\sin\omega t } {\sin\omega T}.$$

It solves

$$\ddot x_{\rm cl}+\omega^2 x_{\rm cl}=0.$$

At the focal times $T=n\pi/\omega$, the endpoint problem is singular: many initial momenta focus to the same endpoint relation. The closed-form kernel must then be replaced by its delta-function caustic limit.

Splitting the Action

Substitute $x=x_{\rm cl}+\eta$ into the action:

$$S[x_{\rm cl}+\eta] = S[x_{\rm cl}] + S[\eta] + S_{\rm cross}.$$

The cross term is

$$S_{\rm cross} = m\int_0^T dt\, \left( \dot x_{\rm cl}\dot\eta - \omega^2x_{\rm cl}\eta \right).$$

Integrating by parts gives

$$S_{\rm cross} = m\dot x_{\rm cl}\eta\bigg\rvert_0^T - m\int_0^T dt\, \left( \ddot x_{\rm cl} + \omega^2x_{\rm cl} \right)\eta.$$

The boundary term vanishes because $\eta(0)=\eta(T)=0$, and the integral vanishes because $x_{\rm cl}$ obeys the classical equation of motion. Hence

$$S[x_{\rm cl}+\eta] = S_{\rm cl}+S[\eta].$$

The path integral factorizes:

$$K_{\rm ho}(x_f,T;x_i,0) = e^{iS_{\rm cl}/\hbar} F_\omega(T),$$

where

$$F_\omega(T) = \int_{\eta(0)=0}^{\eta(T)=0} \mathcal D\eta\, e^{iS[\eta]/\hbar}.$$

All endpoint dependence is in $S_{\rm cl}$; the fluctuation factor depends only on $T$ and $\omega$.

Classical Action

For the classical solution above, the action is

$$S_{\rm cl}(x_f,T;x_i,0) = \frac{m\omega}{2\sin\omega T} \left[ (x_f^2+x_i^2)\cos\omega T - 2x_fx_i \right].$$

One efficient derivation uses the identity

$$\frac{d}{dt}(x_{\rm cl}\dot x_{\rm cl}) = \dot x_{\rm cl}^2 + x_{\rm cl}\ddot x_{\rm cl}.$$

Since $\ddot x_{\rm cl}=-\omega^2x_{\rm cl}$,

$$\dot x_{\rm cl}^2-\omega^2x_{\rm cl}^2 = \frac{d}{dt}(x_{\rm cl}\dot x_{\rm cl}).$$

Therefore

$$S_{\rm cl} = \frac{m}{2} \left[ x_{\rm cl}\dot x_{\rm cl} \right]_0^T.$$

Substituting the endpoint velocities from the classical path gives the expression for $S_{\rm cl}$.

Fluctuation Determinant

The fluctuation action is

$$S[\eta] = \frac{m}{2} \int_0^T dt\, \left( \dot\eta^2-\omega^2\eta^2 \right).$$

With Dirichlet boundary conditions, integration by parts gives

$$S[\eta] = \frac{m}{2} \int_0^T dt\, \eta(t) \left[ -\frac{d^2}{dt^2}-\omega^2 \right] \eta(t).$$

Expand the fluctuation in orthonormal Dirichlet modes:

$$\eta(t) = \sum_{n=1}^{\infty}q_n \sqrt{\frac{2}{T}} \sin\frac{n\pi t}{T}.$$

Then

$$S[\eta] = \frac{m}{2} \sum_{n=1}^{\infty} \left[ \left(\frac{n\pi}{T}\right)^2-\omega^2 \right]q_n^2.$$

The fluctuation integral is a product of Gaussian integrals. Its absolute normalization is fixed by matching to the free-particle path integral when $\omega\to0$. Formally,

$$\frac{F_\omega(T)}{F_0(T)} = \left[ \frac{ \det(-d^2/dt^2) }{ \det(-d^2/dt^2-\omega^2) } \right]^{1/2}_{\rm D},$$

where the subscript ${\rm D}$ reminds us that Dirichlet boundary conditions are used.

Using the eigenvalues above,

$$\frac{F_\omega(T)}{F_0(T)} = \left[ \prod_{n=1}^{\infty} \frac{(n\pi/T)^2} {(n\pi/T)^2-\omega^2} \right]^{1/2}.$$

Euler’s product

$$\frac{\sin z}{z} = \prod_{n=1}^{\infty} \left( 1-\frac{z^2}{n^2\pi^2} \right)$$

gives

$$\frac{F_\omega(T)}{F_0(T)} = \left[ \frac{\omega T}{\sin\omega T} \right]^{1/2}.$$

Since the free fluctuation factor is

$$F_0(T) = \left( \frac{m}{2\pi i\hbar T} \right)^{1/2},$$

the oscillator fluctuation factor is

$$F_\omega(T) = \left( \frac{m\omega}{2\pi i\hbar\sin\omega T} \right)^{1/2}.$$

This determinant formula is formal but extremely useful. A careful definition requires a discretization, regulator, or spectral prescription; the ratio is the stable object.

Exact Kernel

Combining the classical action and fluctuation factor gives

$$\begin{aligned} K_{\rm ho}(x_f,T;x_i,0) &= \left[ \frac{m\omega}{2\pi i\hbar\sin\omega T} \right]^{1/2} \\ &\quad\times \exp\left\{ \frac{i}{\hbar} \frac{m\omega}{2\sin\omega T} \left[ (x_f^2+x_i^2)\cos\omega T - 2x_fx_i \right] \right\}. \end{aligned}$$

This is exact because the action is quadratic. No stationary-phase approximation has been made. In a nonquadratic potential, the same split $x=x_{\rm cl}+\eta$ produces interactions among fluctuations, and the Gaussian determinant is only the leading semiclassical piece.

Caustics and Phase Branches

The determinant factor is singular when

$$\sin\omega T=0.$$

At these times, the operator $-d^2/dt^2-\omega^2$ has a zero mode under Dirichlet boundary conditions. The classical endpoint problem is also singular. For

$$T=\frac{n\pi}{\omega},$$

the kernel is the distributional limit

$$K_{\rm ho}(x_f,n\pi/\omega;x_i,0) = e^{-in\pi/2} \delta\left(x_f-(-1)^n x_i\right).$$

The phase jumps are the one-dimensional Maslov phases. They are the exact counterpart of caustic phases in semiclassical propagation.

Relation to Oscillator Modes

The fluctuation calculation already has the structure of a normal-mode decomposition. Each mode coefficient $q_n$ contributes a Gaussian integral with eigenvalue

$$\lambda_n = \left(\frac{n\pi}{T}\right)^2-\omega^2.$$

The full determinant is the product of all mode contributions. This is the same structural fact that appears in free fields: after Fourier expansion, a free field becomes a collection of independent harmonic oscillators, and the Gaussian path integral becomes a product of mode determinants.

The oscillator also explains why zero modes and negative modes matter. When an eigenvalue crosses zero, the determinant vanishes and the naive square root is no longer a regular prefactor. In semiclassical and field-theoretic path integrals, zero modes signal collective coordinates or gauge redundancies; negative modes often signal instability or tunneling saddle structure.

Source and Correlator View

Adding a source,

$$S_J[x]=S[x]+\int dt\,J(t)x(t),$$

keeps the oscillator path integral Gaussian. The result is a source functional whose derivatives generate the oscillator time-ordered correlators. In ground-state form,

$$\mathcal Z_0[J] = \exp\left[ -\frac{1}{2\hbar^2} \int dt\,dt'\, J(t)G_F(t,t')J(t') \right],$$

where

$$G_F(t,t') = \frac{\hbar}{2m\omega} e^{-i\omega|t-t'|}.$$

This is the same Gaussian structure seen in the fluctuation determinant: quadratic actions produce exact Gaussian generating functionals. The details of source conventions are handled in Sources and Generating Functionals in QM, and the correlator interpretation is handled in Correlation Functions in Path Integrals.

QFT Bridge

A free scalar field in a box decomposes into normal modes:

$$\phi(\mathbf x,t) = \sum_{\mathbf k} q_{\mathbf k}(t)u_{\mathbf k}(\mathbf x),$$

schematically. The free-field action becomes a sum of oscillator actions,

$$S[\phi] = \sum_{\mathbf k} \int dt\, \left[ \frac12\dot q_{\mathbf k}^2 - \frac12\omega_{\mathbf k}^2q_{\mathbf k}^2 \right],$$

up to normalization and reality constraints. Thus the harmonic-oscillator path integral is the finite-dimensional ancestor of the free-field Gaussian path integral.

In QFT, the determinant becomes a functional determinant over infinitely many spacetime modes, and the source functional encodes field correlation functions. Regularization and renormalization are then essential, not optional technicalities. The oscillator gives the algebraic pattern, while field theory supplies the continuum and locality issues.

Common Mistakes

• Forgetting that the fluctuation $\eta(t)$ has zero endpoints.
• Keeping a cross term after the classical equation of motion and boundary conditions have already made it vanish.
• Treating the determinant as an ordinary finite product without specifying a regulator or ratio.
• Missing the $\omega T/\sin\omega T$ determinant factor.
• Using the closed form at $\sin\omega T=0$ instead of the caustic distribution.
• Interpreting the exactness of the oscillator as evidence that stationary phase is exact for generic potentials.
• Confusing the oscillator’s time-ordered source correlator with a retarded response function.

Cross-Links

• Time Slicing
• Action and Phase
• Free-Particle Path Integral
• Harmonic-Oscillator Propagator
• Sources and Generating Functionals in QM
• Correlation Functions in Path Integrals
• Path Integrals from QM to QFT
• Quantum Harmonic Oscillator
• Harmonic Oscillator to Fields

References

• R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965.
• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.
• H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 5th ed., World Scientific, 2009.
• J. Zinn-Justin, Path Integrals in Quantum Mechanics, Oxford University Press, 2005.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014.

Exercises

1. Verify the classical path formula.

Solution

The proposed path is

$$x_{\rm cl}(t) = \frac{ x_i\sin\omega(T-t)+x_f\sin\omega t } {\sin\omega T}.$$

At $t=0$,

$$x_{\rm cl}(0) = \frac{x_i\sin\omega T}{\sin\omega T} = x_i.$$

At $t=T$,

$$x_{\rm cl}(T) = \frac{x_f\sin\omega T}{\sin\omega T} = x_f.$$

Each sine term solves $\ddot x+\omega^2x=0$, so their linear combination solves the oscillator equation.

2. Show that the cross term vanishes in the action split.

Solution

The cross term is

$$S_{\rm cross} = m\int_0^Tdt\, \left( \dot x_{\rm cl}\dot\eta-\omega^2x_{\rm cl}\eta \right).$$

Integrating the first term by parts gives

$$S_{\rm cross} = m\dot x_{\rm cl}\eta\bigg\rvert_0^T - m\int_0^Tdt\, \left( \ddot x_{\rm cl}+\omega^2x_{\rm cl} \right)\eta.$$

The boundary term is zero because $\eta(0)=\eta(T)=0$. The remaining integral is zero because $x_{\rm cl}$ satisfies $\ddot x_{\rm cl}+\omega^2x_{\rm cl}=0$.

3. Use Euler’s product to derive the determinant ratio.

Solution

The ratio is

$$\frac{F_\omega(T)}{F_0(T)} = \left[ \prod_{n=1}^{\infty} \frac{(n\pi/T)^2} {(n\pi/T)^2-\omega^2} \right]^{1/2}.$$

Factor each denominator:

$$\frac{(n\pi/T)^2-\omega^2}{(n\pi/T)^2} = 1-\frac{\omega^2T^2}{n^2\pi^2}.$$

Euler’s product gives

$$\prod_{n=1}^{\infty} \left( 1-\frac{\omega^2T^2}{n^2\pi^2} \right) = \frac{\sin\omega T}{\omega T}.$$

Taking the reciprocal square root yields

$$\frac{F_\omega(T)}{F_0(T)} = \left[ \frac{\omega T}{\sin\omega T} \right]^{1/2}.$$

4. Check the free-particle limit of the oscillator kernel.

Solution

As $\omega\to0$,

$$\sin\omega T\sim\omega T, \qquad \cos\omega T\sim1.$$

The prefactor becomes

$$\left[ \frac{m\omega}{2\pi i\hbar\sin\omega T} \right]^{1/2} \longrightarrow \left[ \frac{m}{2\pi i\hbar T} \right]^{1/2}.$$

The classical action becomes

$$\frac{m\omega}{2\sin\omega T} \left[ (x_f^2+x_i^2)\cos\omega T-2x_fx_i \right] \longrightarrow \frac{m(x_f-x_i)^2}{2T}.$$

Thus the kernel reduces to the free-particle kernel.

5. Why is the harmonic-oscillator path integral exact while a generic potential is not?

Solution

The oscillator action is quadratic in $x(t)$. After writing $x=x_{\rm cl}+\eta$, the fluctuation action is purely quadratic in $\eta$, so the fluctuation integral is Gaussian and can be evaluated by a determinant.

For a generic potential,

$$V(x_{\rm cl}+\eta) = V(x_{\rm cl}) + V'(x_{\rm cl})\eta + \frac12V''(x_{\rm cl})\eta^2 + \frac{1}{3!}V'''(x_{\rm cl})\eta^3 +\cdots.$$

The terms cubic and higher in $\eta$ make the fluctuation integral non-Gaussian. The determinant then gives only the Gaussian approximation around a classical path, not the full exact answer.