Propagator Kernel

The propagator kernel is the coordinate-space matrix element of the time-evolution operator:

$$K(x_f,t_f;x_i,t_i) = \langle x_f\rvert U(t_f,t_i)\lvert x_i\rangle.$$

It is an amplitude density between delta-normalized positions, with units of inverse length on the line. It becomes a physical position wavefunction only after integration against a normalizable initial state; see transition amplitudes.

Wavefunction Evolution

Given an initial wavefunction $\psi(x_i,t_i)$, the later wavefunction is

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

For other configuration spaces, the integration domain and measure must match the Hilbert-space inner product.

Initial Condition

At equal times, the propagator reduces to a delta distribution:

$$K(x_f,t_i;x_i,t_i) =\delta(x_f-x_i).$$

With zero elapsed time, the identity kernel leaves every initial wavefunction unchanged.

Differential Equation

As a function of the final variables, the kernel satisfies the time-dependent Schrödinger equation:

$$i\hbar\frac{\partial}{\partial t_f} K(x_f,t_f;x_i,t_i) = H_{x_f} K(x_f,t_f;x_i,t_i),$$

with the delta-function initial condition above. Here $H_{x_f}$ means that the coordinate-space Hamiltonian acts on the final coordinate.

Composition Law

Time-evolution operators compose:

$$U(t_b,t_a) =U(t_b,t_c)U(t_c,t_a).$$

In coordinate representation this becomes

$$K(x_b,t_b;x_a,t_a) = \int dx_c\, K(x_b,t_b;x_c,t_c) K(x_c,t_c;x_a,t_a).$$

This identity is the Composition Law and is the direct route to path-integral time slicing.

Spectral Form

For a time-independent Hamiltonian with discrete eigenstates,

$$K(x_f,t;x_i,0) = \sum_n \psi_n(x_f)\psi_n^*(x_i) e^{-iE_nt/\hbar}.$$

Continuous spectra add integrals over continuum labels. Spectral Decomposition of the Propagator develops the discrete, continuous, and mixed cases and makes clear how bound states and scattering states contribute to propagation.

Free-particle example

For $H=p^2/(2m)$ on the line and $T>0$,

$$K_0(x_f,T;x_i,0)=\sqrt{\frac{m}{2\pi i\hbar T}} \exp\!\left[\frac{im(x_f-x_i)^2}{2\hbar T}\right].$$

The square root has phase $e^{-i\pi/4}$, fixed by the damped momentum integral and the delta-function limit. The exponent is $iS_{\mathrm{cl}}/\hbar$, with $S_{\mathrm{cl}}=m(x_f-x_i)^2/(2T)$. The free-particle propagator derives this expression and checks composition. Its constant magnitude as a function of $x_f-x_i$ is another reason not to interpret $|K_0|^2$ as a normalized probability density.

Relation to Path Integrals

The path integral begins by applying the composition law many times, inserting intermediate positions, and taking a formal continuum limit. The result is a sum over histories weighted by $e^{iS/\hbar}$.

Relation to Green Functions

Propagators evolve states in time. Green functions often solve inhomogeneous differential equations or resolvent equations with specified boundary conditions. The two are related, but not identical; boundary conditions and time-ordering conventions matter.

Common Mistakes

• Treating $K$ as a probability instead of an amplitude.
• Forgetting the integration over the initial coordinate when evolving a wavefunction.
• Ignoring the measure and boundary conditions of the configuration space.
• Confusing the propagator kernel with an energy-domain Green function.
• Forgetting that the equal-time limit is a distribution.
• Quoting the free-particle square root without its convergence and branch prescription.
• Using the infinite-line kernel when walls, periodicity, or another configuration-space boundary condition changes the kernel.

References

• R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Show that the composition law for $K$ follows from inserting a position resolution of identity at the intermediate time.

Solution

Start from

$$K(x_b,t_b;x_a,t_a) = \langle x_b\rvert U(t_b,t_c)U(t_c,t_a)\lvert x_a\rangle.$$

Insert

$$I=\int dx_c\,\lvert x_c\rangle\langle x_c\rvert.$$

This gives

$$K(x_b,t_b;x_a,t_a) = \int dx_c\, K(x_b,t_b;x_c,t_c) K(x_c,t_c;x_a,t_a).$$

2. Use dimensional analysis to verify that the one-dimensional free kernel has units of inverse length.

Solution

Because $\hbar T/m$ has units of length squared, the prefactor $\sqrt{m/(\hbar T)}$ has units of inverse length. The exponential is dimensionless. This is also required by $\psi(x_f,T)=\int dx_i\,K\psi(x_i,0)$.

3. For the free kernel, apply stationary phase to $\int dp\,e^{i[p(x_f-x_i)-p^2T/(2m)]/\hbar}$ and identify the stationary momentum.

Solution

The phase is $\Phi(p)=p(x_f-x_i)-p^2T/(2m)$. Solving $d\Phi/dp=0$ gives

$$p_{\mathrm{cl}}=\frac{m(x_f-x_i)}{T},$$

the momentum of the classical straight path joining the endpoints in time $T$.

4. Derive the three-dimensional free kernel by factorizing the Cartesian momentum integrals.

Solution

Since $p^2=p_x^2+p_y^2+p_z^2$ and the measure factorizes, the integral is the product of three one-dimensional kernels. Therefore

$$K_0(\mathbf x_f,T;\mathbf x_i,0) = \left(\frac{m}{2\pi i\hbar T}\right)^{3/2} \exp\!\left[ \frac{im\lvert\mathbf x_f-\mathbf x_i\rvert^2}{2\hbar T} \right].$$