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A classical path joining prescribed endpoints is a stationary point of the action, satisfying the Euler-Lagrange equation and the endpoint boundary conditions. Expanding the path integral about each such path separates its classical phase from its quadratic fluctuation determinant. There may be multiple classical paths or none within a chosen real saddle class; the semiclassical propagator must respect that saddle structure rather than assuming a unique trajectory for every potential.

Write the Hamiltonian operator as $\hat{H}={\hat{p}}^2/(2m)+V(\hat{q})$, with $[\hat{q},\hat{p}]=i$, and define the quantum-mechanical propagator by $U(\beta ,\alpha ;T)=⟨\beta |e^{-iT\hat{H}}|\alpha ⟩$. Use $N$ intervals of length $\epsilon =T/N$, endpoint coordinates $q_0=\alpha ,q_N=\beta$, and the Trotter product formula. Inserting the completeness relations of the position eigenstates between the short-time factors givesNormalize momentum eigenstates by $⟨q|p⟩=e^{ipq}/\sqrt{2\pi }$. The Gaussian integral over the intermediate momentum isThe square-root branch is fixed by continuation from $\mathrm{Im}\epsilon <0$. Consequently the precise time-sliced configuration-space path integral isThis formula defines the notation $\int \mathcal{D}q{e}^{iS[q]}$; it is not an unnormalized infinite product of ordinary integrals. For real time take the boundary value from $T-i\delta$, $\delta >0$, or the corresponding oscillatory-distribution limit. For a self-adjoint, bounded-below Hamiltonian operator under the usual product-formula hypotheses, the operator limit supplies the definition even where the coordinate kernel is a distribution. The discrete exponent approaches the action with kinetic term $m{\dot{q}}^2/2$, and the prefactor is part of the functional measure. A different consistent slicing gives the same kernel for this separated kinetic-plus-potential Hamiltonian.

For the quantum harmonic oscillator, expand the action about a classical path $x$, writing $q=x+\xi$ with homogeneous Dirichlet boundary conditions on $\xi$. The only mixed term isThe boundary term vanishes and the Euler-Lagrange equation sets the remaining integrand to zero. Thus $S[x+\xi ]=S[x]+S[\xi ]$ exactly, rather than only to quadratic approximation. Translation of each integrated coordinate has unit Jacobian determinant; in the continuum measure convention of the question this is $\mathcal{D}q=\mathcal{D}\xi$. The classical-path factorization of a quadratic path integral follows:The fluctuation action, measure and boundary values depend on $T,m,\omega$ but not on $\alpha ,\beta$. For real $T$ with $\sin \omega T\ne 0$ the boundary-value classical path exists uniquely. At the conjugate times $\omega T=k\pi$, generic endpoints admit no such path and the kernel is a delta distribution; the formula is interpreted by the same regulated boundary limit, not as an ordinary finite prefactor times a nonexistent classical solution.

Work in units $\hslash =1$ and with the particle mass absorbed into the stated kinetic term. Divide the interval into $N$ pieces of length $\Delta =T/N$. For a suitable self-adjoint Hamiltonian operator, the Trotter product formula gives the regulated limitInsert $1=\int d{q}_j|q_j⟩⟨q_j|$ between neighboring factors. Because $V(\hat{q})$ is diagonal in the position operator basis, the given free-particle propagator yieldsThe exponent is the time-sliced action. Its continuum notation isThe measure is defined by the displayed finite-dimensional limits and their normalization, not by a translation-invariant flat measure on an infinite-dimensional space. The oscillatory limit needs a damping/analytic-continuation prescription. This is the time-sliced configuration-space path integral derived from the operator kernel.

For a quadratic potential energy, write $q=q_c+\eta$, where the classical path satisfies ${\ddot{q}}_c+V^{\prime }(q_c)=0$ and the endpoint conditions, and $\eta (0)=\eta (T)=0$. Expansion is exact:Integration by parts makes the linear term equal to $-[{\int }_0^T({\ddot{q}}_c+V^{\prime }(q_c))\eta dt]$, since the endpoint term vanishes. It is zero by the Euler-Lagrange equation. Also $V^{\prime \prime }$ is constant, independent of the endpoints. Thus classical-path factorization of a quadratic path integral is exact:Only the classical action carries the endpoint dependence.

Let $D={\partial }_t^2+V^{\prime \prime }$ on homogeneous Dirichlet boundary conditions. Expand $\eta$ in its normalized eigenfunctions, with eigenvalues ${\lambda }_n$. The fluctuation integral becomes a product of regulated integrals $\int d{a}_n\exp (-i{\lambda }_n{a}_n^2/2)$. Each is proportional to ${\lambda }_n^{-1/2}$, with a prescribed complex phase. Relative to $D_0={\partial }_t^2$, the normalization is thereforeThe subscript denotes the Dirichlet operator domain, and the square-root branch is continued from the free kernel with its convergence prescription. Equivalently one can use determinants of $-D$ and $-D_0$. For $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isThis also checks the free limit. At conjugate times, the fluctuation operator has a zero eigenvalue and the ordinary finite-prefactor expression fails; the kernel is obtained by distributional continuation. Away from these times the classical boundary problem is unique and the stated factorization applies directly.

For the infinite-time oscillator with an external source, use vacuum boundary conditions and the Feynman i-epsilon prescription. For a smooth source of compact support, defineThe imaginary term damps the real-field integral. Integration by parts gives $S_{ϵ}[q]=\frac{1}{2}q{A}_{ϵ}q+Jq$, with the pairings understood as time integrals. Completing the square givesTranslate the integration variable. The source-independent Gaussian functional integral is $N$, soOn Fourier modes $e^{i\omega t}$, $A_{ϵ}$ has eigenvalue ${\omega }^2-m^2+iϵ$, henceFor $m>0$, closing the contour above for $t>0$ and below for $t<0$ gives $G(t)=-i{e}^{-im|t|}/(2m)$. The physical time-ordered oscillator quantum field theory propagator is $iG$, not $G$ itself. The shifted stationary path is $q_c=-GJ$ and obeys ${\ddot{q}}_c+m^2{q}_c=J$ with the same vacuum prescription. A retarded inverse would define a different boundary problem. Dividing by the zero-source integral removes $N$ and gives the vacuum source functional of a quantum oscillator. At zero mass an additional infrared prescription is needed for the inverse on constant modes.

Use units $\hslash =1$. The fluctuations about a classical path satisfy $f(0)=f(T)=0$. Expanding the action functional and integrating the term $m{\dot{q}}_c\dot{f}$ by parts givesThe linear term vanishes by the Euler-Lagrange equation and the endpoint conditions. Integrating the quadratic kinetic term by parts yields the Dirichlet fluctuation determinant of a semiclassical propagator:A finite time-slicing first turns this into a product of ordinary Gaussian integrals, whose limit is proportional to $(\det {\mathcal{O}}_c{)}^{-1/2}$. A useful normalization relative to the free operator ${\mathcal{O}}_0=-m{d}^2/d{t}^2$ isThe oscillatory square root is fixed by the short-time limit and the Feynman i-epsilon prescription. Away from zero modes this gives the semiclassical propagator; if several classical paths contribute, their saddle contributions are summed with their corresponding determinant phases.

For a general potential the prefactor depends on the endpoints through $q_c$; writing only $D(T)$ suppresses that dependence. If $V(q)=a{q}^2/2+bq+c$, the Taylor expansion has no terms beyond quadratic and $V^{\prime \prime }=a$ is independent of $q_c$. The Gaussian functional integral is then exact, and its prefactor genuinely depends only on $T$ and the potential parameters. Singular focusing times are understood by continuation of the kernel, rather than by assigning a finite value to a zero-mode determinant.

For the free-particle propagator, $q_c(t)=q_0+(q_1-q_0)t/T$ and $S_c=m(q_1-q_0{)}^2/(2T)$. The free determinant supplies the factor $\sqrt{m/(2\pi iT)}$. Its normalization can also be checked by the momentum-space representationwhere completing the square evaluates the Gaussian integral and the short-time limit gives the position-space delta function.

For the gravitational potential, the original PDF contains $-mgq$; the TeX aid drops the final $q$. The Euler-Lagrange equation is ${\ddot{q}}_c=g$, and its endpoint solution isSubstitution into the kinetic and potential terms givesSince $V^{\prime \prime }=0$, the fluctuation operator is the free one. Thus the quantum-mechanical propagator in a constant force is exactly

The quantum-mechanical propagator is $⟨q_1|e^{-iHT}|q_0⟩$, with Hamiltonian operator $H=-{\partial }_q^2/(2m)+V(q)$. For imaginary time $-i\tau$, insert an orthonormal complete set of energy eigenstates to obtainHere ${\psi }_n(q)=⟨q|n⟩$. This discrete sum assumes a discrete complete spectrum; continuous spectral components require the corresponding spectral integral. The heat-semigroup expression is well defined for an appropriate self-adjoint Hamiltonian bounded below. It is this spectral condition, not an unrestricted formal rotation for every conceivable potential, that justifies the imaginary-time kernel.

A path on the circle lifts to a path on the real line ending at $q_1+2\pi w$, where the integer $w$ is its winding number. The periodic quantum theory sums all these winding sectors with equal weight, giving the real-time winding representation of a quantum rotor kernel. For imaginary time, put $x=q_1-q_0$ and $y=\tau /m$ in the supplied Gaussian summation identity:Comparison with the spectral expansion determinesThe normalization uses $0\le q<2\pi$. The ground state is nondegenerate; levels with $n\ne 0$ have the degeneracy $n\leftrightarrow -n$. Returning to real time replaces $e^{-\tau E_n}$ by $e^{-iT{E}_n}$, with the usual limiting prescription. No twisted boundary condition or flux phase is present in the stated image sum.