Free-Particle Propagator

The one-dimensional free-particle propagator is

$$K_0(x_f,t_f;x_i,t_i) = \left[ \frac{m}{2\pi i\hbar (t_f-t_i)} \right]^{1/2} \exp\left[ \frac{im(x_f-x_i)^2}{2\hbar(t_f-t_i)} \right],$$

for $t_f\ne t_i$, with the square-root phase chosen by the usual real-time evolution prescription. It is the coordinate-space kernel of

$$U_0(t_f,t_i) = \exp\left[ -\frac{i}{\hbar}\frac{p^2}{2m}(t_f-t_i) \right].$$

This page derives the kernel, checks its normalization and composition law, and explains why it is the basic short-time building block for path integrals.

Momentum-Space Derivation

Let

$$T=t_f-t_i, \qquad \Delta x=x_f-x_i.$$

Insert the momentum resolution of identity into

$$K_0(x_f,t_f;x_i,t_i) = \langle x_f|U_0(T)|x_i\rangle.$$

With

$$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar},$$

one obtains

$$\begin{aligned} K_0(x_f,t_f;x_i,t_i) &= \int_{-\infty}^{\infty} \frac{dp}{2\pi\hbar}\, \exp\left[ \frac{ip\Delta x}{\hbar} - \frac{ip^2T}{2m\hbar} \right]. \end{aligned}$$

This is a Fourier transform of the free-particle energy phase $e^{-ip^2T/(2m\hbar)}$.

Gaussian Integral Evaluation

Complete the square:

$$\frac{p\Delta x}{\hbar} - \frac{p^2T}{2m\hbar} = - \frac{T}{2m\hbar} \left( p-\frac{m\Delta x}{T} \right)^2 + \frac{m(\Delta x)^2}{2\hbar T}.$$

Including the factor of $i$ in the exponent,

$$K_0 = e^{im(\Delta x)^2/(2\hbar T)} \int_{-\infty}^{\infty} \frac{dp}{2\pi\hbar}\, \exp\left[ -\frac{iT}{2m\hbar} \left( p-\frac{m\Delta x}{T} \right)^2 \right].$$

The remaining Fresnel integral is defined by a convergence prescription, for example $T\to T-i0^+$ for $T\gt0$. The result is

$$K_0 = \left( \frac{m}{2\pi i\hbar T} \right)^{1/2} \exp\left[ \frac{im(\Delta x)^2}{2\hbar T} \right].$$

The prefactor is not optional: it enforces the delta-function initial condition and unitarity.

Delta-Function Initial Condition

At equal times, the propagator must become the identity kernel:

$$\lim_{T\to0} K_0(x_f,t_i+T;x_i,t_i) = \delta(x_f-x_i)$$

in the distributional sense. This does not mean the expression has an ordinary pointwise limit. Instead, for a sufficiently nice test wavefunction,

$$\lim_{T\to0} \int_{-\infty}^{\infty}dx_i\, K_0(x_f,t_i+T;x_i,t_i)\psi(x_i) = \psi(x_f).$$

The rapidly oscillating phase away from $x_i=x_f$ is what produces the delta distribution.

Schrödinger Equation Check

The kernel satisfies the free Schrödinger equation in the final coordinate:

$$i\hbar\frac{\partial K_0}{\partial t_f} = -\frac{\hbar^2}{2m} \frac{\partial^2K_0}{\partial x_f^2}.$$

Together with the delta-function initial condition, this characterizes the free-particle kernel. The same expression also satisfies the adjoint equation in the initial variables, as required by unitarity.

Composition Law

For $t_i\lt t\lt t_f$, the kernel composes as

$$K_0(x_f,t_f;x_i,t_i) = \int_{-\infty}^{\infty}dx\, K_0(x_f,t_f;x,t) K_0(x,t;x_i,t_i).$$

The cleanest verification uses the momentum representation. Each short kernel is an integral over a momentum phase; the intermediate position integral gives a delta distribution equating the two momenta. The result is the single momentum integral for the total time $t_f-t_i$.

This composition law is the starting point for the time-sliced construction in From Propagators to Path Integrals.

Classical Action Form

The classical path from $(x_i,t_i)$ to $(x_f,t_f)$ is a straight line with velocity

$$v_{\rm cl} = \frac{x_f-x_i}{t_f-t_i}.$$

The classical action is

$$S_{\rm cl} = \int_{t_i}^{t_f} \frac12m v_{\rm cl}^2\,dt = \frac{m(x_f-x_i)^2}{2(t_f-t_i)}.$$

Therefore

$$K_0(x_f,t_f;x_i,t_i) = \left[ \frac{1}{2\pi i\hbar} \left( -\frac{\partial^2 S_{\rm cl}} {\partial x_f\,\partial x_i} \right) \right]^{1/2} e^{iS_{\rm cl}/\hbar}.$$

For the free particle this semiclassical-looking expression is exact because the action is quadratic.

Wave-Packet Evolution

The propagator evolves an initial wavefunction by

$$\psi(x_f,t_f) = \int_{-\infty}^{\infty}dx_i\, K_0(x_f,t_f;x_i,t_i)\psi(x_i,t_i).$$

For an initial Gaussian packet

$$\psi(x,0) = \frac{1}{(2\pi\sigma_0^2)^{1/4}} \exp\left[ -\frac{(x-x_0)^2}{4\sigma_0^2} + \frac{ip_0(x-x_0)}{\hbar} \right],$$

the probability density remains Gaussian. Its center follows the classical trajectory

$$x_c(t)=x_0+\frac{p_0}{m}t,$$

while its width becomes

$$\sigma_t = \sigma_0 \left[ 1+ \left( \frac{\hbar t}{2m\sigma_0^2} \right)^2 \right]^{1/2}.$$

The packet spreads because the free-particle dispersion relation $\omega=\hbar k^2/(2m)$ makes different momentum components accumulate different phases. For the canonical wave-packet discussion, see Gaussian Wave Packets and Wave-Packet Spreading.

Path-Integral Role

For a short time step $\epsilon$, the free-particle kernel is

$$K_0(x_{j+1},t_j+\epsilon;x_j,t_j) = \left( \frac{m}{2\pi i\hbar\epsilon} \right)^{1/2} \exp\left[ \frac{im(x_{j+1}-x_j)^2}{2\hbar\epsilon} \right].$$

When a potential is present, a short-time kinetic-potential splitting gives the approximate kernel

$$K(x_{j+1},t_j+\epsilon;x_j,t_j) \approx K_0(x_{j+1},t_j+\epsilon;x_j,t_j) e^{-i\epsilon V(x_j)/\hbar}.$$

Multiplying many such short-time kernels and integrating over intermediate positions is the formal route to the real-time path integral.

Common Mistakes

• Treating $K_0$ as a probability density. It is an amplitude and is not normalizable as a function of $x_f$ for fixed $x_i$.
• Dropping the phase of the square-root prefactor.
• Forgetting that the equal-time limit is a distribution, not an ordinary function.
• Reusing the free kernel in a box, on a half-line, or on a ring without enforcing the correct boundary conditions.
• Confusing the free-particle propagator with a relativistic or QFT propagator.
• Ignoring Fourier-transform conventions when deriving the prefactor.

Cross-Links

• Propagator Kernel
• Spectral Decomposition of the Propagator
• Propagators in Multiple Dimensions
• Propagators and Boundary Conditions
• Causality, Support, and Interpretation in Nonrelativistic QM
• Free Particle
• Gaussian Wave Packets
• Wave-Packet Spreading
• Trotter Product Formula
• From Propagators to Path Integrals
• Free-Particle Propagator Notebook
• Fourier Transform
• 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. Starting from the momentum integral for $K_0$, complete the square and recover the stated prefactor.

Solution

Start with

$$K_0 = \int_{-\infty}^{\infty} \frac{dp}{2\pi\hbar} \exp\left[ \frac{ip\Delta x}{\hbar} - \frac{ip^2T}{2m\hbar} \right].$$

Completing the square gives

$$\frac{ip\Delta x}{\hbar} - \frac{ip^2T}{2m\hbar} = \frac{im(\Delta x)^2}{2\hbar T} - \frac{iT}{2m\hbar} \left( p-\frac{m\Delta x}{T} \right)^2.$$

The shifted Fresnel integral is

$$\int_{-\infty}^{\infty} \frac{dq}{2\pi\hbar} \exp\left( -\frac{iTq^2}{2m\hbar} \right) = \left( \frac{m}{2\pi i\hbar T} \right)^{1/2},$$

with the standard convergence prescription. Combining factors gives the result.

2. Verify the free Schrödinger equation for the kernel by differentiating with respect to $T$ and $\Delta x$.

Solution

Write

$$K_0=C T^{-1/2} \exp\left[ \frac{im(\Delta x)^2}{2\hbar T} \right].$$

Then

$$\frac{\partial K_0}{\partial T} = \left[ -\frac{1}{2T} - \frac{im(\Delta x)^2}{2\hbar T^2} \right]K_0,$$

so

$$i\hbar\frac{\partial K_0}{\partial T} = \left[ -\frac{i\hbar}{2T} + \frac{m(\Delta x)^2}{2T^2} \right]K_0.$$

Also,

$$\frac{\partial^2K_0}{\partial x_f^2} = \left[ \frac{im}{\hbar T} - \frac{m^2(\Delta x)^2}{\hbar^2T^2} \right]K_0.$$

Multiplying by $-\hbar^2/(2m)$ gives the same expression as $i\hbar\,\partial_TK_0$.

3. Use the momentum representation to verify the composition law for the free-particle propagator.

Solution

Write each kernel as a momentum integral:

$$K_0(x_f,t_f;x,t) = \int\frac{dp}{2\pi\hbar} e^{ip(x_f-x)/\hbar} e^{-ip^2(t_f-t)/(2m\hbar)}$$

and similarly for $K_0(x,t;x_i,t_i)$. The intermediate integral over $x$ gives

$$\int dx\,e^{i(q-p)x/\hbar} = 2\pi\hbar\,\delta(q-p).$$

The delta distribution sets $q=p$, leaving

$$\int\frac{dp}{2\pi\hbar} e^{ip(x_f-x_i)/\hbar} e^{-ip^2(t_f-t_i)/(2m\hbar)},$$

which is the single free kernel for the total time.

4. Why does the free-particle kernel not decay at large $|x_f-x_i|$ for fixed nonzero $T$?

Solution

The magnitude of the real-time free kernel is set by the prefactor:

$$\left|K_0\right| = \left( \frac{m}{2\pi\hbar |T|} \right)^{1/2}.$$

The dependence on $x_f-x_i$ is purely in the phase. This is not a probability distribution over endpoints. Physical probabilities arise after convolving the kernel with a normalizable initial wavefunction and then taking the modulus squared.