back the free part of the propagator corresponding to the movement along the ${\cal OX}_3$ direction. Of course, for $\omega_0=0$ we reobtain the propagator found in our first example and for ${\bf B}={\bf 0}$ we reobtain the propagator for a bidimensional oscillator in the ${\cal OX}_1{\cal X}_2$ plane multiplied by a free propagator in the ${\cal OX}_3$ direction, as can be easily checked.

Regarding the gauge dependence of the propagator, the same comments done before are still valid here, namely, the above expression is written for a generic gauge, the transformation law for the propagator under a gauge transformation is the same as before, etc. We finish this section, extracting from the previous propagator, the corresponding energy spectrum. With this purpose, we first compute the trace of the propagator,

$$\begin{aligned}
\int_{-\infty}^\infty\!\!\! dx_1\!\!
 \int_{-\infty}^\infty\!\!\! dx_2 \,
 K_\perp^{\,\prime}(x_1,x_1,x_2,x_2;\tau) &=& {m\Omega\over 2\pi
i\hbar\,\sin(\Omega\tau) }
 \int_{-\infty}^\infty\!\!\! dx_1
 \!\!\int_{-\infty}^\infty\!\!\! dx_2
 \exp\biggl\{ {im\Omega\over
2\hbar\,\sin(\Omega\tau)}\left[
2\Bigl(\mbox{cos}(\Omega\tau)-\mbox{cos}(\omega\tau)\Bigr)(x_1^2+x_2^2)\right]
\biggr\} \nonumber\\
 &=& {1\over 2[\mbox{cos}(\Omega\tau)-\mbox{cos}(\omega\tau)]}\; ,\end{aligned}$$

where we used the well known result for the Fresnel integral. Using now the identity $$\cos(\Omega\tau)-\cos(\omega\tau) = -2\,\sin[(\Omega+\omega)\tau/2]
\,\sin[(\Omega-\omega)\tau/2)]\, ,$$ we get for the corresponding energy Green function $$\begin{aligned}
\label{FGreenG(E)}
\!\!&{\cal G}&\!\!\!(E) =\! -i\!\!\int_0^\infty\!\!\! d\tau\,
e^{{i\over\hbar}E\tau} \!\!\int_{-\infty}^\infty\, \!\!\!\!
dx_1\!\!\int_{-\infty}^\infty\!\!\!\! dx_2\,
K_\perp^{\,\prime}(x_1,x_1,x_2,x_2;\tau)\cr\cr
&=&{i\over4}\int_0^\infty d\tau {e^{{i\over\hbar}E\tau}\over
\,\mbox{sen}({\Omega\tau\over2}\tau)\,\mbox{sen}({\Omega-\omega\over2}\tau)}\cr\cr
