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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.