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After the shift $x(\tau )=x_0(\tau )+y(\tau )$, the endpoint values of $y$ are both zero. Expanding the Euclidean action givesBy integration by parts, the cross term isThe first term vanishes by the homogeneous Dirichlet boundary conditions, and the second by the Euler-Lagrange equation. The translation has unit Jacobian determinant in the time-sliced configuration-space path integral, so $\mathcal{D}x=\mathcal{D}y$. Consequently classical-path factorization of a quadratic path integral is exact here:All dependence on the common endpoint $x$ lies inThe endpoint domain and the integrand defining $Z$ are the same for every $x$. This $Z$ is a zero-endpoint Gaussian path integral; the canonical partition function instead integrates the entire diagonal kernel over $x$. The final sentence of the printed part omits $/\hslash$ from its exponential: the physical Boltzmann factor has the dimensionless exponent $-S/\hslash$.

An optional normalization check uses the Dirichlet oscillator fluctuation determinant. Its eigenvalues are $m[(n\pi /\mathcal{T}{)}^2+{\omega }^2]$, $n\ge 1$. Relative to the free-particle Gaussian path integral, the hyperbolic-sine infinite product givesThus $Z=\sqrt{m\omega /[2\pi \hslash \sinh (\beta \hslash \omega )]}$. Integrating the resulting diagonal thermal kernel of a quantum harmonic oscillator by a Gaussian integral yields $\mathrm{Tr}{e}^{-\beta H}=[2\sinh (\beta \hslash \omega /2){]}^{-1}$, including the zero-point energy.

Write $x=x_0+y$, so the fluctuation has homogeneous Dirichlet boundary conditions $y(0)=y(T_E)=0$. Because ${\dot{x}}_0$ is constant, the cross term in the Euclidean action vanishes exactly:ConsequentlyTranslation of each intermediate integration variable has Jacobian determinant one in the time-sliced configuration-space path integral. This is an exact classical-path factorization of a quadratic path integral, givingThe fluctuation Euclidean action, the homogeneous boundary conditions, and the measure normalization contain neither $x_i$ nor $x_f$. Hence $\mathcal{N}$ is independent of both endpoint positions. It depends on $\beta ,m,\hslash$.

With the standard position-state normalization, the Euclidean free-particle bridge amplitude isFor example, a momentum resolution gives a one-dimensional Gaussian integral,whose evaluation gives the same exponent and prefactor. In particular, the result integrates to one over $x_f$ at fixed $x_i$, as required for its heat kernel interpretation.

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.

Measure intermediate time $u$ from $t^{\prime }$. The classical path with the fixed endpoints isIt satisfies $m{\ddot{x}}_c=-\alpha$. Direct integration of $m{\dot{x}}_c^2/2-\alpha x_c$ givesFor an arbitrary path write $x=x_c+\eta$ with $\eta (0)=\eta (T)=0$. Integrating the cross term by parts gives $-\int (m{\ddot{x}}_c+\alpha )\eta du=0$. Thus $S[x]=S_c+\int m{\dot{\eta }}^{2}du/2$ exactly. The fluctuation path integral is the free endpoint-fixed one, so its determinant supplies $\sqrt{m/(2\pi i\hslash T)}$. Multiplication by $\theta (T)e^{i{S}_c/\hslash }$ reproduces the previous kernel. This exact classical-path factorization of a quadratic path integral is possible because a linear potential introduces no additional quadratic fluctuation term.

In units $\hslash =1$, a particle of mass $m$ with potential $V(q)=-Fq$ has classical path $q_c(t)=q_0+[(q_1-q_0)/T-FT/(2m)]t+F{t}^2/(2m)$. Its action and exact quantum-mechanical propagator areThe fluctuation operator is the free one because the potential is linear. This is a special case of classical-path factorization of a quadratic path integral.