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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.

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.

Use position eigenstates normalized by $⟨q|q^{\prime }⟩=\delta (q-q^{\prime })$, and the free Hamiltonian operator $\hat{H}={\hat{p}}^2/(2m)$. The time-sliced path integral is the matrix element of the time-evolution operator:Insert momentum eigenstates, with $⟨q|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipq/\hslash }$. The free-particle propagator becomesComplete the square in the exponent:The Fresnel integral, defined by the damping prescription $T\to T-i0$, gives, for $T>0$,The square-root branch is $i^{-1/2}=e^{-i\pi /4}$; it is fixed by the regulated Gaussian integral and the short-time Dirac delta distribution limit, rather than an arbitrary phase.

The Euler-Lagrange equation for the free particle is $m\ddot{q}=0$. Its unique classical path satisfying the endpoint boundary conditions is $q_{\mathrm{cl}}(t)=q^{(i)}+\Delta qt/T$. Therefore its classical action isSubstituting into the operator result yields $\mathcal{I}=C(T)e^{i{S}_{\mathrm{cl}}/\hslash }$, with $C(T)=(m/(2\pi i\hslash T){)}^{1/2}$ independent of the endpoint positions.

The same independence is transparent directly in the path integral. Write $q=q_{\mathrm{cl}}+\eta$, with $\eta (0)=\eta (T)=0$. The cross term in the action is $m{\dot{q}}_{\mathrm{cl}}{\int }_0^T\dot{\eta }dt=0$, so $S[q]=S_{\mathrm{cl}}+(m/2){\int }_0^T{\dot{\eta }}^{2}dt$. The fluctuation integral depends only on $m,T,\hslash$. Because the action is quadratic, the factorization is exact; the semiclassical propagator already equals the full answer.

The Euler-Lagrange equation is $m\ddot{x}=0$. The unique path with the prescribed endpoints is thereforeIt is a minimum of the Euclidean action and a stationary point of the real-time action. Its classical action isThe principle of stationary action consequently fixes the position-dependent phase of the semiclassical propagator asBecause the action is quadratic, the stationary-phase evaluation of the path integral is exact. Composition of propagators, or the Van Vleck determinant, gives $C(T)=\sqrt{m/(2\pi i\hslash T)}$, reproducing part i.

With a nonconstant potential energy, momentum no longer diagonalizes the Hamiltonian operator. The operator derivation must use the energy eigenfunctions of the full Hamiltonian, a Dyson series, or a time-sliced Trotter product formula. In the classical derivation, the straight paths are replaced by every solution of the nonlinear Euler-Lagrange equation with the specified endpoints. The semiclassical propagator becomes a sumwhere the prefactor is the Van Vleck determinant and ${\mu }_{\mathrm{cl}}$ is a Maslov index. Unlike a quadratic theory, the classical-path sum is generally only an asymptotic approximation: the exact path integral includes fluctuations of every order. On the circle, the sum must still include all winding number sectors.

The Van Vleck determinant is the determinant of mixed endpoint derivatives of a classical action. It measures how nearby classical trajectories focus and supplies the fluctuation prefactor of the semiclassical propagator.