&=&-i\!\!\int_0^\infty\!\!\!\! d\tau\,
e^{{i\over\hbar}E\tau}\left(\sum_{l=0}^\infty
e^{-(l+{1\over2})(\Omega+\omega)\tau}\!\!\right)\!\!
\left(\sum_{n=0}^\infty
e^{-i(n+{1\over2}(\Omega-\omega)\tau}\!\!\right) ,\nonumber\end{aligned}$$ where is tacitly assumed that $E\rightarrow E-i\varepsilon$ and we also used that (with the assumption $\nu\rightarrow \nu-i\epsilon$) $${1\over\,\mbox{sen}({\nu\over2}\tau)}=2i\sum_{n-0}^\infty
e^{-i(n+{1\over2}) \nu\tau}\; .$$ Changing the order of integration and summations, and integrating in $\tau$, we finally obtain $${\cal G}(E)=\sum_{l,n=0}^\infty{1\over E-E_{nl}}\; ,$$ where the poles of ${\cal G}(E)$, which give the desired energy levels, are identified as $$E_{nl}=(l+n+1)\hbar\Omega+(l-n)\hbar\omega\, ,\;\;\; (l,n = 0,1,...)\; .$$ The Landau levels can be reobtained from the previous result by simply taking the limit $\omega_0\rightarrow 0$: $$E_{nl}\longrightarrow(2l+1)\hbar\omega
 = (l+{1\over2})\hbar\omega_c\; ,$$ with $l=0,1,...$ and $\omega_c = eB/m$, in agreement to the result we had already obtained before.

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Final Remarks {#SectionFinalRemarks}
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In this paper we reconsidered, in the context of Schwinger’s method, the Feynman propagators of two well known problems, namely, a charged particle under the influence of a constant and uniform magnetic field (Landau problem) and the same problem in which we added a bidimensional harmonic oscillator potential. Although these problems had already been treated from the point of view of Schwinger’s action principle, the novelty of our work relies on the fact that we solved the Heisenberg equations for gauge invariant operators. This procedure has some nice properties, as for 