Harmonic-Oscillator Propagator

For the one-dimensional harmonic oscillator

$$H=\frac{p^2}{2m}+\frac12m\omega^2x^2,$$

the coordinate-space propagator is

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

where $T=t_f-t_i$ and $\sin\omega T\ne0$. At the caustic times $T=n\pi/\omega$, this expression must be interpreted with the correct phase prescription or replaced by the corresponding delta-function limit.

The oscillator propagator is special because the semiclassical expression is exact: a quadratic action makes all fluctuation integrals Gaussian.

Why the Oscillator Is Special

For a general potential, an exact coordinate-space kernel is rarely available. The harmonic oscillator is an exception because both the Hamiltonian and classical action are quadratic. This means:

• the Schrödinger equation is exactly solvable by Hermite functions;
• the path integral is Gaussian;
• the classical action controls the exact phase;
• coherent states remain coherent under time evolution;
• each normal mode of a free quantum field behaves like an oscillator.

The canonical oscillator spectrum and wavefunctions are reviewed in Quantum Harmonic Oscillator.

Spectral Derivation

For a time-independent Hamiltonian, the propagator has the spectral form

$$K(x_f,T;x_i,0) = \sum_{n=0}^{\infty} \psi_n(x_f)\psi_n^*(x_i) e^{-iE_nT/\hbar}.$$

For the harmonic oscillator,

$$E_n=\hbar\omega\left(n+\frac12\right),$$

and

$$\psi_n(x) = \frac{1}{\sqrt{2^n n!}} \left(\frac{1}{\pi\ell^2}\right)^{1/4} H_n\left(\frac{x}{\ell}\right) e^{-x^2/(2\ell^2)}, \qquad \ell=\sqrt{\frac{\hbar}{m\omega}}.$$

Thus

$$\begin{aligned} K_{\rm ho}(x_f,T;x_i,0) &= e^{-i\omega T/2} \sum_{n=0}^{\infty} \psi_n(x_f)\psi_n^*(x_i) e^{-in\omega T}. \end{aligned}$$

The summation is evaluated using Mehler’s formula for Hermite polynomials. With the convergence prescription $e^{-i\omega T}\to e^{-i\omega T-\epsilon}$, the sum gives the closed form above.

The factor $e^{-i\omega T/2}$ is the zero-point-energy phase. Dropping it gives the wrong kernel.

Classical Action Form

The classical oscillator path joining $(x_i,t_i)$ to $(x_f,t_f)$ has action

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

The mixed second derivative is

$$-\frac{\partial^2S_{\rm cl}} {\partial x_f\,\partial x_i} = \frac{m\omega}{\sin\omega T}.$$

Therefore the kernel can be written as

$$K_{\rm ho} = \left[ \frac{1}{2\pi i\hbar} \left( -\frac{\partial^2S_{\rm cl}} {\partial x_f\,\partial x_i} \right) \right]^{1/2} e^{iS_{\rm cl}/\hbar},$$

again with the appropriate phase branch. This is the one-dimensional Van Vleck form, and for a quadratic Hamiltonian it is exact rather than merely semiclassical.

Short-Time Limit

For small $T$,

$$\sin\omega T\approx\omega T, \qquad \cos\omega T\approx1.$$

The oscillator kernel reduces at leading order to the free-particle kernel:

$$K_{\rm ho}(x_f,T;x_i,0) \sim \left( \frac{m}{2\pi i\hbar T} \right)^{1/2} \exp\left[ \frac{im(x_f-x_i)^2}{2\hbar T} \right].$$

The potential contributes at the next orders in $T$. This short-time structure is why the free-particle kernel appears inside time-sliced derivations of path integrals.

Caustics and Phase Issues

The formula above is singular when

$$\sin\omega T=0, \qquad T=\frac{n\pi}{\omega}.$$

These are focal times of the classical oscillator. At such times,

$$x_f=(-1)^n x_i$$

for every classical trajectory that starts at $x_i$, independent of the initial momentum. The kernel becomes a delta-function limit:

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

The phase $e^{-in\pi/2}$ is the caustic, or Maslov, phase in this simple setting. More general caustic and stability-determinant issues belong to the semiclassical pages, especially Semiclassical Propagator and Van Vleck Determinant.

Coherent-State Propagation

Number states evolve by phases:

$$U(T)|n\rangle = e^{-i\omega T(n+1/2)}|n\rangle.$$

A coherent state has the number-state expansion

$$|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}}|n\rangle.$$

Therefore

$$U(T)|\alpha\rangle = e^{-i\omega T/2} |\alpha e^{-i\omega T}\rangle.$$

The packet remains a minimum-uncertainty Gaussian. Its center follows the classical oscillator motion:

$$x_c(T)=\sqrt2\,\ell\,\operatorname{Re}(\alpha e^{-i\omega T}), \qquad p_c(T)=\frac{\sqrt2\hbar}{\ell} \operatorname{Im}(\alpha e^{-i\omega T}).$$

This is a concrete example of exact quantum evolution reproducing classical motion for selected observables and states, without making the state a classical point particle.

QFT Bridge

A free quantum field decomposes into independent oscillator modes after Fourier expansion and choice of boundary conditions. Each mode has a Hamiltonian of the schematic form

$$H_{\mathbf k} = \hbar\omega_{\mathbf k} \left( a_{\mathbf k}^\dagger a_{\mathbf k} +\frac12 \right).$$

The oscillator propagator is therefore the finite-dimensional ancestor of Gaussian functional integrals in free field theory. The field-theory version adds infinitely many modes, regularization, source terms, and relativistic correlation-function conventions.

Common Mistakes

• Using the closed-form expression at $\sin\omega T=0$ without the delta-function and phase prescription.
• Dropping the zero-point phase $e^{-i\omega T/2}$ in the spectral derivation.
• Treating the propagator as a probability density rather than an amplitude.
• Forgetting that the square-root branch changes across caustics.
• Assuming the exact Van Vleck form holds for arbitrary potentials; it is exact here because the Hamiltonian is quadratic.
• Confusing coherent-state propagation with energy-eigenstate stationarity.

Cross-Links

• Propagator Kernel
• Spectral Decomposition of the Propagator
• Propagators in Multiple Dimensions
• Free-Particle Propagator
• Quantum Harmonic Oscillator
• Ladder-Operator Solution
• Coherent States
• Harmonic-Oscillator Propagator Notebook
• Semiclassical Propagator
• Van Vleck Determinant
• Propagator Table

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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 5th ed., World Scientific, 2009.

Exercises

1. Show that the oscillator propagator reduces to the free-particle propagator at leading order as $T\to0$.

Solution

Use

$$\sin\omega T=\omega T+O(T^3), \qquad \cos\omega T=1+O(T^2).$$

Then the prefactor becomes

$$\left[ \frac{m\omega}{2\pi i\hbar\sin\omega T} \right]^{1/2} = \left[ \frac{m}{2\pi i\hbar T} \right]^{1/2} \left[1+O(T^2)\right],$$

and the phase becomes

$$\frac{im\omega}{2\hbar\sin\omega T} \left[ (x_f^2+x_i^2)\cos\omega T -2x_fx_i \right] = \frac{im(x_f-x_i)^2}{2\hbar T} +O(T).$$

2. Verify that the classical action generates the endpoint momenta.

Solution

For

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

the final momentum is

$$p_f=\frac{\partial S_{\rm cl}}{\partial x_f} = \frac{m\omega}{\sin\omega T} (x_f\cos\omega T-x_i).$$

The initial momentum is

$$p_i=-\frac{\partial S_{\rm cl}}{\partial x_i} = \frac{m\omega}{\sin\omega T} (x_f-x_i\cos\omega T).$$

These are exactly the endpoint relations for the classical oscillator trajectory.

3. Use the number-state expansion to derive $U(T)|\alpha\rangle=e^{-i\omega T/2}|\alpha e^{-i\omega T}\rangle$.

Solution

Start from

$$|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}}|n\rangle.$$

Applying $U(T)$ gives

$$U(T)|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n e^{-i\omega T(n+1/2)}}{\sqrt{n!}}|n\rangle.$$

Factor out $e^{-i\omega T/2}$ and combine the remaining phase with $\alpha^n$:

$$U(T)|\alpha\rangle = e^{-i\omega T/2} e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{(\alpha e^{-i\omega T})^n}{\sqrt{n!}}|n\rangle.$$

This is $e^{-i\omega T/2}|\alpha e^{-i\omega T}\rangle$.

4. What is the kernel after one full oscillator period $T=2\pi/\omega$?

Solution

For one full period, $n=2$ in the caustic formula:

$$K_{\rm ho}(x_f,t_i+2\pi/\omega;x_i,t_i) = e^{-i\pi} \delta(x_f-x_i) = -\delta(x_f-x_i).$$

The unitary is $U(2\pi/\omega)=-I$ because every number state picks up the phase $e^{-i2\pi(n+1/2)}=-1$. The global sign has no effect on a single ray but matters in phase-sensitive comparisons.