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Use the prepoint prescription for the separable Hamiltonian. Its regulated exponent isEach momentum integration is an ordinary Gaussian integral:Therefore the measure written in the question gives exactlyThe resulting configuration integral is $\int \mathcal{D}{q}_{\mathrm{raw},N}{e}^{-S_{E,N}}$. This is Gaussian momentum integration in a phase-space path integral. Formally its exponent tends to the usual Euclidean kinetic-plus-potential action, but the slice-dependent factors in the measure must be retained.

For a normalized quantum-mechanical propagator, Fourier completeness uses $d{p}_i/(2\pi )$ rather than $d{p}_i$. With that normalization the configuration measure isThis differs from the raw measure by $(2\pi {)}^{-N}$. It makes the free single-step kernel integrate to one and tend to a delta distribution as the interval tends to zero. In these formulas $q_{N+1}$ is fixed and $q_1,\dots ,q_N$ are integrated for propagation from an initial wavefunction. If both endpoints are fixed, omit $d{q}_1$ while retaining its slice normalization factor. The continuum expression means the limit of these measures and exponents, not a flat product of $dq(t)$ with no time-step weights.

We now address the unheaded real-time continuation. Set $m=\hslash =1$. The normalized short-time Schrödinger kernel gives the final-slice recurrenceThe square-root branch is fixed by the usual damped Fresnel integral, or continuation from the Euclidean kernel. The hint's $d{q}^{\prime }/\sqrt{\Delta t}$ contains the essential time-step dependence; the constant $(2\pi i{)}^{-1/2}$ fixes the identity limit and is included at every step.

Put $\eta =q^{\prime }-q$. The normalized oscillatory Gaussian has momentsTaylor-expand the smooth wavefunction and the potential over one short step. Odd moments vanish, and potential-derivative corrections first contribute at order $(\Delta t{)}^2$. ThusSubtract the initial value, divide by $\Delta t$ and take the limit:This proves the Time-dependent Schrödinger equation by the Schrödinger equation from a short-time path integral argument. Nonuniform partitions give the same limit when their largest time step tends to zero.

For the quantum harmonic oscillator, set $\mu =0$ and $\Omega =\hslash \omega$. The normal symbol is $H_N(\bar{\psi },\psi )=\Omega (\bar{\psi }\psi +1/2)$, so the thermal coherent-state time slicing givesThe connection with a phase-space path integral uses the canonical real coordinatesTheir derivative term satisfies $\bar{\psi }\dot{\psi }=-ip\dot{q}/\hslash$ up to total derivatives that vanish for periodic paths. The normal and Weyl symbols of a harmonic oscillator must be distinguished: in midpoint phase-space time slicing, the Weyl ordering symbol of $a^{†}a$ is $\bar{\psi }\psi -1/2$. Thus the appropriate midpoint Hamiltonian is $H_W=p^2/(2m)+m{\omega }^2{q}^2/2$, with no additional constant. The change from the adjacent-label normal prescription to the midpoint prescription includes this ordering correction.

In physical imaginary time $u=\hslash \tau$, the resulting phase-space path integral isGaussian momentum integration in a phase-space path integral produces the usual oscillator configuration-space path integral. An exact check follows directly from the coherent kernel $⟨{\psi }^{\prime }|e^{-\beta \Omega a^{†}a}|\psi ⟩=\exp (e^{-\beta \Omega }{\bar{\psi }}^{\prime }\psi )$:This is the thermal partition function of a quantum harmonic oscillator. Keeping the $\Omega /2$ normal-ordering constant a second time after switching to the Weyl symbol would double count the zero-point energy.

For $H={\hat{p}}^2/(2m)+V(\hat{x})$, divide elapsed time $t$ into $J$ slices $\Delta t=t/J$ and apply the Trotter product formula. Inserting complete position and momentum bases gives the regulated phase-space path integralThe endpoint values are $x_0=x_i$, $x_J=x_f$. Each Gaussian momentum integration in a phase-space path integral can be performed explicitly, leaving the configuration-space path integralIts measure is the time-sliced limit of $(m/(2\pi i\hslash \Delta t){)}^{J/2}{\prod }_{j=1}^{J-1}d{x}_j$, with the usual infinitesimal convergence prescription. The integral sums amplitudes, not positive classical probabilities. Operator ordering remains part of the time-slice definition. Expanding a short-time kernel recovers the Schrödinger equation; composing kernels reproduces the time-evolution operator.

In the classical limit, rapid phase cancellation suppresses paths away from stationary action. Varying $S$ with fixed endpoints gives $m\ddot{x}=-V^{\prime }(x)$. The semiclassical result sums the classical-path phases multiplied by fluctuation functional determinants; for a quadratic Hamiltonian the fluctuation integral is Gaussian and this expansion is exact.

For quantum statistical mechanics, continue $t=-i\tau$ by a Wick rotation. The Euclidean action and thermal trace giveThe trace closes position-space paths. In a many-particle trace one also sums over endpoint permutations with the appropriate exchange signs. In a coherent state field formulation, bosonic fields are periodic and fermionic coherent states instead give antiperiodic Grassmann fields, hence bosonic and fermionic Matsubara frequencies $2\pi n/(\beta \hslash )$ and $(2n+1)\pi /(\beta \hslash )$. ^[ref]

The quantum-to-classical imaginary-time mapping views a local $d$-dimensional quantum problem as a classical-looking field theory in $d+1$ dimensions, with imaginary-time extent $\beta \hslash$. A genuine classical probability interpretation additionally requires a nonnegative weight; fermion signs and Berry terms can obstruct it.

For a smooth potential in the high-temperature or small-$\hslash$ limit, imaginary-time paths fluctuate only over the thermal de Broglie wavelength ${\lambda }_T=\sqrt{2\pi \beta {\hslash }^2/m}$. If this is short compared with the potential's variation scale, the diagonal free kernel givesThis classical phase-space limit of the thermal kernel is the classical phase space partition function for one degree of freedom. A dilute identical-particle gas also needs its $1/N!$ factor; when exchange is appreciable, classical statistics fails. As a check, a single quantum harmonic oscillator gives $Z=[2\sinh (\beta \hslash \omega /2){]}^{-1}\to 1/(\beta \hslash \omega )$, precisely the classical phase-space integral.