Propagator for harmonic potential

For the harmonic oscillator, the action looks like:
\[S = \int_{t_{I}}^{t_{F}} dt \, \left[-\dot{q} \cdot p + \frac{1}{2m}p^{2} + \frac{1}{2} m \omega^{2} q^{2}\right]\]
The equations of motion have trigonometric solutions:
\[\ddot{q}_{cl} + \omega^{2} q_{cl} = 0 \Longrightarrow q_{cl}(t) = A \cos{(\omega t)} + B \sin{(\omega t)}\]
We fix the coefficients by specifying initial and final data:
\[q_{cl} (t_{I}) = q_{I} \qquad q_{cl}(t_{F}) = q_{F}\]
\[A = \frac{q_{I} \sin{(\omega t_{F})} - q_{F} \sin{(\omega t_{I})}}{\sin{(\omega \Delta t)}} \qquad B = \frac{q_{F} \cos{(\omega t_{I})} - q_{I} \cos{(\omega t_{F})}}{\sin{(\omega \Delta t)}}\]
The classical action and Van Vleck ``matrix" give
\[S_{cl} = -\frac{m u}{2\Delta t} \left(\frac{(q_{F}^{2} + q_{I}^{2}) \cos{u} - 2 q_{I} q_{F}}{\sin{u}}\right) \qquad \mathcal{V} \equiv -i\frac{\partial^{2} S_{cl}}{\partial q_{F} \partial q_{I}} = -i\frac{m u}{ \Delta t \sin{u}} \qquad u \equiv \omega \Delta t\]
which lead to the transition amplitude
\[\left\langle q_{F}, \, t_{F} | q_{I}, \, t_{I} \right\rangle = \sqrt{-i\frac{m u}{ \Delta t\sin{u}}} \, \exp{\left(i\frac{m u}{2 \Delta t} \left(\frac{(q_{F}^{2} + q_{I}^{2}) \cos{u} - 2 q_{I} q_{F}}{\sin{u}}\right)\right)}\]
Note that since the Lagrangian (and Hamiltonian) explicitly depend on the coordinates, we have no momentum conservation. In order to obtain the free point-particle we have written everything in terms of $\Delta t$ and the dimensionless variable $u$. For the short time limit we should write everything in terms of $\omega$ and $u$. The free point-particle is obtained in the limit when $u \rightarrow 0$ since
\[\lim_{u \longrightarrow 0} \frac{u}{\Delta t \sin{u}} = \frac{1}{\Delta t}\]

From the classical action we can obtain the momenta:
\[p_{F} = \frac{m u}{\Delta t \sin{u}} (q_{F} \cos{u} - q_{I}) \qquad p_{I} = -\frac{m u}{\Delta t \sin{u}} (q_{I} \cos{u} - q_{F})\]
\[p_{F} - p_{I} = \frac{m u }{\Delta t \sin{u}} (\cos{u} - 1) (q_{F} + q_{I})\]
These relations can be solved for the coordinates:
\[q_{F} = \frac{\Delta t}{m u \sin{u}} \left(p_{I} - p_{F} \cos{u}\right) \qquad q_{I} = -\frac{\Delta t}{m u \sin{u}} \left(p_{F} - p_{I} \cos{u}\right)\]
Now we can write down $\tilde{S}_{cl}$ and $\mathcal{J}$:
\[\tilde{S}_{cl} = -\frac{\Delta t}{2m u} \left(\frac{(p_{F}^{2} + p_{I}^{2}) \cos{u} - 2 p_{I} p_{F}}{\sin{u}}\right)\]
\[\mathcal{J} = \left( \begin{array}{c c}
\displaystyle i \frac{m u}{\Delta t \tan{(u)}} & \displaystyle -i \frac{m u}{\Delta t \sin{(u)}} \\
& \\
\displaystyle -i \frac{m u}{\Delta t \sin{(u)}} & \displaystyle i \frac{m u}{\Delta t \tan{(u)}}
\end{array} \right) \qquad \det{\mathcal{J}} = \frac{m^{2}u^{2}}{\Delta t^{2}}\]
Then
\[\sqrt{\det{\tilde{\mathcal{V}}}} \equiv \frac{\sqrt{\det{\mathcal{V}}}}{\sqrt{\det{\mathcal{J}}}} = \sqrt{-i \frac{\Delta t}{m u \sin{u}}}\]

The semiclassical transition amplitude in the momentum basis is:
\[\left\langle p_{F}, \, t_{F} | p_{I}, \, t_{I} \right\rangle = \sqrt{-i\frac{\Delta t}{m u \sin{u}}} \exp{\left[i\frac{\Delta t}{2 m u} \left(\frac{(p_{F}^{2} + p_{I}^{2}) \cos{u} - 2 p_{I} p_{F}}{\sin{u}}\right)\right]}\]
When $u$ is small we have
\[\cos{u} \approx 1 - \frac{1}{2} u^{2} + \mathcal{O}(u^{4}) \qquad \sin{u} \approx u + \mathcal{O}(u^{3})\]
The action takes the form
\[\tilde{S}_{cl} \approx -\frac{\Delta t}{2m u^{2}} (p_{F} - p_{I})^{2} + \frac{1}{6m} (p_{F}^{2} + p_{F} p_{I} + p_{I}^{2})\Delta t\]
and the transition amplitude becomes
\[\left\langle p_{F}, \, t_{F} | p_{I}, \, t_{I} \right\rangle \approx \sqrt{-i\frac{\Delta t}{m u^{2}}} \exp{\left[ -i\frac{\Delta t}{6m}(p_{F}^{2} + p_{F} p_{I} + p_{I}^{2}) + i\frac{\Delta t}{2 m u^{2}} (p_{F} - p_{I})^{2}\right]}\]
In the limit when $u \rightarrow 0$ we obtain the expected momentum-conserving Dirac delta function from the second term in the exponential. The prefactor is important for this limit:
\[\left\langle p_{F}, \, t_{F} | p_{I}, \, t_{I} \right\rangle = \delta^{D} (p_{F} - p_{I}) \, \exp{\left( -\frac{i}{\hbar} \frac{\Delta t}{2m} p^{2}_{F} \right)}\]