Propagators in Multiple Dimensions

A multidimensional propagator is not a new kind of quantum object. It is the coordinate-space matrix element of the same time-evolution operator, now taken between points of a configuration space with more than one coordinate:

$$K(mathbf q_f,t_f;mathbf q_i,t_i) = \langle \mathbf q_f\rvert U(t_f,t_i) \lvert \mathbf q_i\rangle.$$

The dimension of $\mathbf q$ is the dimension of configuration space, not necessarily the dimension of physical space. One particle moving in ordinary space has three configuration coordinates. Two distinguishable particles in ordinary space have six. A planar double pendulum has two generalized coordinates even though it is embedded in three-dimensional space.

Three pieces of information must be kept together:

• the Hamiltonian and its operator domain;
• the configuration-space measure;
• the boundary or symmetry conditions imposed on states.

Once those data are fixed, multidimensional kernels obey the same evolution, composition, and initial-value rules as one-dimensional kernels.

Multidimensional Kernels

For Cartesian coordinates on $\mathbb R^d$, use the normalization

$$\langle \mathbf q\vert\mathbf q'\rangle = \delta^{(d)}(\mathbf q-\mathbf q'),$$

and the identity resolution

$$I = \int_{\mathbb R^d}d^d q\, \lvert\mathbf q\rangle\langle\mathbf q\rvert.$$

The kernel evolves a wavefunction according to

$$\psi(\mathbf q_f,t_f) = \int_{\mathbb R^d}d^d q_i\, K(\mathbf q_f,t_f;\mathbf q_i,t_i) \psi(\mathbf q_i,t_i).$$

For a time-independent Hamiltonian, the final variables satisfy

$$i\hbar\frac{\partial}{\partial t_f} K(\mathbf q_f,t_f;\mathbf q_i,t_i) = H_{\mathbf q_f} K(\mathbf q_f,t_f;\mathbf q_i,t_i),$$

with distributional initial condition

$$\lim_{t_f\to t_i^+} K(\mathbf q_f,t_f;\mathbf q_i,t_i) = \delta^{(d)}(\mathbf q_f-\mathbf q_i).$$

At an intermediate time $t$, the composition law becomes

$$\begin{aligned} K(\mathbf q_f,t_f;\mathbf q_i,t_i) &= \int_{\mathbb R^d}d^d q\, K(\mathbf q_f,t_f;\mathbf q,t)\\ &\qquad\times K(\mathbf q,t;\mathbf q_i,t_i). \end{aligned}$$

The integral is over one complete intermediate configuration. For an $N$-particle problem, that generally means integrating all particle coordinates, not one physical-space point.

General coordinate measures

Suppose a coordinate chart carries the measure

$$d\mu(q)=w(q)d^d q.$$

The coordinate basis and its delta distribution must be normalized consistently:

$$I = \int d\mu(q)\, \lvert q\rangle\langle q\rvert,$$ $$\langle q\vert q'\rangle = \delta_\mu(q,q'),$$

where

$$\int d\mu(q)\, \delta_\mu(q,q')f(q) = f(q').$$

In a chart with $w(q)\gt0$, one may write

$$\delta_\mu(q,q') = \frac{\delta^{(d)}(q-q')}{w(q)}$$

as a distribution. Kernel evolution and composition then use $d\mu$, and the equal-time kernel is $\delta_\mu$. The measure is not a decorative factor: changing it changes the identity operator represented by the integral.

Product Systems

Let the Hilbert space split as

$$\mathcal H = \mathcal H_A\otimes\mathcal H_B,$$

and suppose the subsystems do not interact:

$$H = H_A\otimes I_B + I_A\otimes H_B.$$

The two terms commute, so for a time-independent Hamiltonian,

$$U(T) = U_A(T)\otimes U_B(T).$$

In a product coordinate basis,

$$\lvert q_A,q_B\rangle = \lvert q_A\rangle\otimes\lvert q_B\rangle,$$

the kernel factorizes:

$$\begin{aligned} &K_{AB}(q_{A,f},q_{B,f},T; q_{A,i},q_{B,i},0)\\ &\quad= K_A(q_{A,f},T;q_{A,i},0) K_B(q_{B,f},T;q_{B,i},0). \end{aligned}$$

This factorization describes the dynamics, not necessarily the initial state. An entangled initial wavefunction is still evolved by the product kernel when the Hamiltonian is noninteracting. Local unitary evolution preserves its entanglement spectrum even though the full coordinate integral does not factor into two independent wavefunction evolutions.

An interaction term $H_{\mathrm{int}}$ generally prevents

$$U(T)=U_A(T)\otimes U_B(T),$$

and therefore prevents kernel factorization in the subsystem coordinates. The operator structure and entangling consequences are developed in Composite Hamiltonians.

For $N$ distinguishable particles in three dimensions, the configuration coordinate is

$$\mathbf q = (\mathbf r_1,\ldots,\mathbf r_N) \in \mathbb R^{3N}.$$

Identical particles require restriction to the symmetric or antisymmetric subspace. Treating their kernel as an unrestricted distinguishable-particle kernel misses exchange amplitudes; see Symmetric and Antisymmetric Wavefunctions.

Separable Hamiltonians

Product systems are one source of factorization. Another is separation of coordinates within one configuration space. Suppose

$$H = \sum_{a=1}^{s}H_a, \qquad [H_a,H_b]=0,$$

where each $H_a$ acts only on a coordinate block $q_a$. If the operator domain and boundary conditions also separate, then

$$U(T) = \prod_{a=1}^{s}e^{-iH_aT/\hbar},$$

and

$$K(\mathbf q_f,T;\mathbf q_i,0) = \prod_{a=1}^{s} K_a(q_{a,f},T;q_{a,i},0).$$

For example, a Cartesian Hamiltonian

$$H = \sum_{a=1}^{d} \left[ \frac{p_a^2}{2m_a} +V_a(q_a) \right]$$

has a product kernel when each coordinate has an independent domain. The anisotropic oscillator is obtained by taking

$$V_a(q_a) = \frac12m_a\omega_a^2q_a^2,$$

so its multidimensional kernel is a product of the one-dimensional kernels derived in Harmonic-Oscillator Propagator.

Potential separability alone is not sufficient. A boundary that mixes coordinates can destroy the product domain even when $V(\mathbf q)=\sum_aV_a(q_a)$. Conversely, a coupled quadratic Hamiltonian may become separable after a normal-mode transformation. Kernel factorization belongs to coordinates in which the full Hamiltonian and its domain split.

The stationary-state counterpart is Separation of Variables. Its product eigenfunctions and additive energies lead to the same factorization through the spectral decomposition.

Free Particle in Multiple Dimensions

For a free particle on $\mathbb R^d$,

$$H = \frac{\mathbf p^2}{2m}.$$

With

$$\langle\mathbf r\vert\mathbf p\rangle = e^{i\mathbf p\cdot\mathbf r/\hbar},$$

and

$$I = \int\frac{d^dp}{(2\pi\hbar)^d} \lvert\mathbf p\rangle\langle\mathbf p\rvert,$$

the spectral integral is

$$\begin{aligned} K_0^{(d)}(\mathbf r_f,T;\mathbf r_i,0) &= \int\frac{d^dp}{(2\pi\hbar)^d} \exp\left[ \frac{i}{\hbar}\mathbf p\cdot (\mathbf r_f-\mathbf r_i) \right]\\ &\qquad\times \exp\left[ -\frac{iT}{2m\hbar}\mathbf p^2 \right]. \end{aligned}$$

The Cartesian Gaussian integral factorizes into $d$ one-dimensional integrals. For $T\gt0$, with the real-time prescription $T\to T-i0^+$,

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

The phase is the classical free-particle action divided by $\hbar$:

$$S_{\mathrm{cl}} = \frac{m\lvert\mathbf r_f-\mathbf r_i\rvert^2}{2T}.$$

The square-root branch is fixed by the same convergence prescription used in one dimension and by continuity under composition. The compact one-dimensional derivation belongs to Free-Particle Propagator.

Structural checks

The formula passes several checks:

• Dimension: $K_0^{(d)}$ has units of length$^{-d}$, as required by the $d^dr_i$ integration measure.
• Symmetry: it depends only on $\mathbf r_f-\mathbf r_i$ and its Euclidean norm, reflecting translation and rotation invariance.
• Initial value: as $T\to0^+$, it tends to $\delta^{(d)}(\mathbf r_f-\mathbf r_i)$ distributionally.
• Composition: the convolution of kernels for times $T_1$ and $T_2$ gives the kernel for $T_1+T_2$.
• Dynamics: acting with $i\hbar\partial_T+\hbar^2\nabla_f^2/(2m)$ gives zero away from $T=0$.

In three dimensions,

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

The stationary plane-wave and energy-shell structure is reviewed in Free Particle in Three Dimensions.

Different masses

For independent Cartesian coordinates with a positive diagonal mass matrix

$$M = \operatorname{diag}(m_1,\ldots,m_d),$$

the free kernel is

$$\begin{aligned} K_M(\mathbf q_f,T;\mathbf q_i,0) &= \left[ \frac{\det M}{(2\pi i\hbar T)^d} \right]^{1/2}\\ &\quad\times \exp\left[ \frac{i}{2\hbar T} (\mathbf q_f-\mathbf q_i)^{\mathsf T} M (\mathbf q_f-\mathbf q_i) \right]. \end{aligned}$$

This is just the product of one-dimensional free kernels with masses $m_a$.

Example: Center-of-Mass and Relative Motion

Consider two distinguishable particles with a translation-invariant interaction:

$$H = \frac{\mathbf p_1^2}{2m_1} + \frac{\mathbf p_2^2}{2m_2} + V(\mathbf r_1-\mathbf r_2).$$

Define

$$\mathbf R = \frac{m_1\mathbf r_1+m_2\mathbf r_2}{M}, \qquad \mathbf r = \mathbf r_1-\mathbf r_2,$$

with

$$M=m_1+m_2, \qquad \mu=\frac{m_1m_2}{M}.$$

The transformation has unit absolute Jacobian, and the Hamiltonian separates:

$$H = \frac{\mathbf P^2}{2M} + \left[ \frac{\mathbf p^2}{2\mu}+V(\mathbf r) \right].$$

Therefore

$$\begin{aligned} &K(\mathbf R_f,\mathbf r_f,T; \mathbf R_i,\mathbf r_i,0)\\ &\quad= K_{\mathrm{cm}}(\mathbf R_f,T;\mathbf R_i,0) K_{\mathrm{rel}}(\mathbf r_f,T;\mathbf r_i,0). \end{aligned}$$

The center-of-mass factor is a free kernel with mass $M$. All interaction physics lies in the relative kernel with reduced mass $\mu$. This example also shows that a kernel may fail to factor in the original particle coordinates but factor exactly after a physically adapted coordinate transformation.

Central Potentials Preview

For a three-dimensional central potential,

$$H = -\frac{\hbar^2}{2m}\nabla^2 +V(r),$$

rotational invariance organizes the kernel into angular-momentum sectors. Define the reduced radial Hamiltonian on $L^2(\mathbb R_+,dr)$ by

$$h_\ell = -\frac{\hbar^2}{2m}\frac{d^2}{dr^2} + \frac{\hbar^2\ell(\ell+1)}{2mr^2} +V(r),$$

and let

$$K_\ell^{(u)}(r_f,T;r_i,0) = \langle r_f\rvert e^{-ih_\ell T/\hbar} \lvert r_i\rangle$$

be its kernel with respect to the measure $dr$. Then the full kernel is

$$\begin{aligned} K(\mathbf r_f,T;\mathbf r_i,0) &= \frac{1}{r_fr_i} \sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{\ell} Y_{\ell m}(\widehat{\mathbf r}_f)\\ &\qquad\times Y_{\ell m}^*(\widehat{\mathbf r}_i) K_\ell^{(u)}(r_f,T;r_i,0). \end{aligned}$$

Using the spherical-harmonic addition theorem,

$$\begin{aligned} K(\mathbf r_f,T;\mathbf r_i,0) &= \frac{1}{4\pi r_fr_i} \sum_{\ell=0}^{\infty}(2\ell+1)\\ &\qquad\times P_\ell(\cos\gamma) K_\ell^{(u)}(r_f,T;r_i,0), \end{aligned}$$

where

$$\cos\gamma = \widehat{\mathbf r}_f\cdot \widehat{\mathbf r}_i.$$

The factors $1/(r_fr_i)$ convert between the physical radial measure $r^2dr$ and the reduced-wavefunction measure $dr$. The domain of $h_\ell$ fixes the behavior at $r=0$. For regular nonsingular potentials, reduced radial wavefunctions normally obey $u_\ell(0)=0$; singular potentials can require a more careful self-adjoint-domain analysis.

This partial-wave expansion is a structural preview. The canonical radial equation, centrifugal term, and boundary conditions belong to Central Potentials and Radial Schrödinger Equation.

Path Integral Measure Preview

Let

$$T=t_f-t_i, \qquad \epsilon=\frac{T}{N}, \qquad t_j=t_i+j\epsilon.$$

For

$$H = \frac{\mathbf p^2}{2m} +V(\mathbf q)$$

on flat Cartesian configuration space, repeated composition with the short-time kernel gives the time-sliced expression

$$\begin{aligned} K(\mathbf q_f,t_f;\mathbf q_i,t_i) &= \lim_{N\to\infty} \left( \frac{m}{2\pi i\hbar\epsilon} \right)^{Nd/2}\\ &\quad\times \int \prod_{j=1}^{N-1}d^dq_j\, \exp\left[ \frac{i}{\hbar}S_N \right], \end{aligned}$$

where $\mathbf q_0=\mathbf q_i$, $\mathbf q_N=\mathbf q_f$, and one standard discretization is

$$S_N = \sum_{j=0}^{N-1} \epsilon \left[ \frac{m}{2} \left( \frac{\mathbf q_{j+1}-\mathbf q_j}{\epsilon} \right)^2 -V(\mathbf q_j) \right].$$

The formal notation

$$\int\mathcal D\mathbf q(t)$$

hides both the product of $d$-dimensional integrations and the normalization factor with exponent $Nd/2$. The finite-partition limit is the definition used here; $\mathcal D\mathbf q$ is not being treated as an ordinary translation-invariant Lebesgue measure on an infinite-dimensional path space.

Curvilinear and curved configuration spaces

For a metric $g_{ab}(q)$, the natural coordinate volume is

$$d\mu(q) = \sqrt{g(q)}\,d^dq, \qquad g=\det g_{ab}.$$

The corresponding Laplace–Beltrami operator is

$$\Delta_g = \frac{1}{\sqrt g} \partial_a \left( \sqrt g\,g^{ab}\partial_b \right).$$

If the Hamiltonian is fixed as

$$H = -\frac{\hbar^2}{2m}\Delta_g+V,$$

its kernel, measure, and short-time prescription must represent that same operator. A coordinate change generally transforms the integration measure, kinetic term, and short-time prefactor together. Merely inserting a Jacobian into the flat Cartesian formula is not enough. On genuinely curved spaces, discretization and operator-ordering conventions may introduce compensating local curvature terms; the operator and its domain are the unambiguous starting data.

The detailed finite-partition construction belongs to Time Slicing, with normalization and sign choices collected in Path Integral Conventions.

Common Mistakes

• Reading $d$ as the number of physical-space directions when it actually counts configuration coordinates.
• Replacing $\delta^{(d)}$ by a one-dimensional delta function in the initial condition.
• Forgetting the full intermediate configuration-space integration in the composition law.
• Assuming that a separable potential guarantees a product kernel when the boundary conditions mix coordinates.
• Treating kernel factorization as proof that every allowed state is a product state.
• Raising the one-dimensional free prefactor to the $d$th power without keeping a consistent complex branch.
• Using the Cartesian measure in spherical, constrained, or curved coordinates.
• Confusing the reduced radial kernel, which uses $dr$, with the full three-dimensional kernel, which uses $r^2dr\,d\Omega$.
• Applying a distinguishable-particle kernel directly to identical particles without symmetrization or antisymmetrization.
• Writing $\mathcal D\mathbf q$ without specifying the finite-dimensional normalization and endpoint conditions it abbreviates.

Cross-Links

• Propagator Kernel
• Composition Law
• Spectral Decomposition of the Propagator
• Free-Particle Propagator
• Harmonic-Oscillator Propagator
• Propagators and Boundary Conditions
• Causality, Support, and Interpretation in Nonrelativistic QM
• Propagator Table
• Composite Hamiltonians
• Separation of Variables
• Free Particle in Three Dimensions
• Central Potentials
• Time Slicing
• Path Integral Conventions

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.
• C. Grosche and F. Steiner, Handbook of Feynman Path Integrals, Springer, 1998.
• H. Kleinert, Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets, 5th ed., World Scientific, 2009.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.

Exercises

1. Derive the free kernel on $\mathbb R^d$ by factorizing the momentum integral into Cartesian components.

Solution

Write

$$\mathbf p\cdot\Delta\mathbf r = \sum_{a=1}^dp_a\Delta r_a, \qquad \mathbf p^2 = \sum_{a=1}^dp_a^2.$$

Then

$$\begin{aligned} K_0^{(d)} &= \prod_{a=1}^d \int_{-\infty}^{\infty} \frac{dp_a}{2\pi\hbar} \exp\left[ \frac{ip_a\Delta r_a}{\hbar} -\frac{ip_a^2T}{2m\hbar} \right]\\ &= \prod_{a=1}^d \left( \frac{m}{2\pi i\hbar T} \right)^{1/2} \exp\left[ \frac{im(\Delta r_a)^2}{2\hbar T} \right]. \end{aligned}$$

Multiplying the factors gives

$$K_0^{(d)} = \left( \frac{m}{2\pi i\hbar T} \right)^{d/2} \exp\left[ \frac{im\lvert\Delta\mathbf r\rvert^2} {2\hbar T} \right].$$

2. Prove that a noninteracting bipartite Hamiltonian has a product kernel.

Solution

For

$$H = H_A\otimes I_B + I_A\otimes H_B,$$

the two terms commute. Therefore

$$e^{-iHT/\hbar} = e^{-iH_AT/\hbar} \otimes e^{-iH_BT/\hbar}.$$

Taking a matrix element between product coordinate states gives

$$\begin{aligned} K_{AB} &= \langle q_{A,f}\rvert U_A(T) \lvert q_{A,i}\rangle\\ &\quad\times \langle q_{B,f}\rvert U_B(T) \lvert q_{B,i}\rangle\\ &= K_AK_B. \end{aligned}$$

The conclusion concerns the propagator. It does not require the initial state to factorize.

3. Show that the two-particle center-of-mass transformation has unit absolute Jacobian and identify the two kernel factors.

Solution

In one Cartesian direction,

$$\begin{pmatrix} R\\ r \end{pmatrix} = \begin{pmatrix} m_1/M & m_2/M\\ 1 & -1 \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix}.$$

The determinant is

$$-\frac{m_1+m_2}{M}=-1.$$

Thus the absolute Jacobian is one in each Cartesian direction and hence for the full transformation. The kinetic energy becomes

$$\frac{\mathbf P^2}{2M} + \frac{\mathbf p^2}{2\mu}.$$

Because $V$ depends only on $\mathbf r$, the center-of-mass and relative terms commute. The kernel factors into a free center-of-mass kernel of mass $M$ and a relative kernel for

$$H_{\mathrm{rel}} = \frac{\mathbf p^2}{2\mu}+V(\mathbf r).$$

4. Explain the difference between the radial delta distribution for $R(r)$ and for the reduced wavefunction $u(r)=rR(r)$.

Solution

Radial wavefunctions $R(r)$ use the inner product

$$\langle R_1\vert R_2\rangle = \int_0^\infty r^2dr\, R_1^*(r)R_2(r).$$

The identity kernel for this measure is therefore

$$\delta_{r^2dr}(r,r') = \frac{\delta(r-r')}{r^2}.$$

After defining $u=rR$, the inner product becomes

$$\int_0^\infty dr\, u_1^*(r)u_2(r),$$

so the reduced radial identity kernel is simply $\delta(r-r')$. The factors $1/(r_fr_i)$ in the partial-wave expansion convert between these conventions.

5. Determine the normalization power in a time-sliced path integral for $d$ Cartesian coordinates and $N$ time intervals.

Solution

Each short-time kernel contributes

$$\left( \frac{m}{2\pi i\hbar\epsilon} \right)^{d/2}.$$

There are $N$ short-time kernels, so their product contributes

$$\left( \frac{m}{2\pi i\hbar\epsilon} \right)^{Nd/2}.$$

There are only $N-1$ intermediate integrations because the two endpoints are fixed:

$$\prod_{j=1}^{N-1}\int d^dq_j.$$

The normalization and integrations together leave the final kernel with the dimensions appropriate to one $d$-dimensional coordinate-space delta distribution.