By establishing a dictionary between the QM harmonic oscillator and the Collatz process, it reveals very important clues as to why the Collatz conjecture most likely is true. The dictionary requires expanding any integer $ n $ into a binary basis (bits) $ n = \sum a_{nl} 2^l $ ($l$ ranges from $ 0 $ to $ N - 1$) that allows to find the correspondence between every integer $ n $ and the state $ | \Psi_n \rangle $, obtained by a superposition of bit states $ | l \rangle $, and which are related to the energy eigenstates of the QM harmonic oscillator. In doing so, one can then construct the one-to-one correspondence between the Collatz iterations of numbers $ n \rightarrow { n \over 2 }$ ($n$ even); $ n \rightarrow 3 n + 1$ ($n$ odd) and the operators $ {\bf L}_{ { n \over 2} }; { \bf L}_{ 3 n + 1 } $, which map $ \Psi_n $ to $ \Psi_{ { n \over 2 } }$, or to $ \Psi_{ 3 n + 1 } $, respectively, and which are constructed explicitly in terms of the creation $ {\bf a}^\dagger$, annihilation $ {\bf a }$, and unit operator $ { \bf 1 } $ of the QM harmonic oscillator. A rigorous analysis reveals that the Collatz conjecture is most likely true, if the composition of a chain of $ {\bf L}_{ { n \over 2} }; { \bf L}_{ 3 n + 1 } $ operators (written as $ L_*$ in condensed notation) leads to the null-eigenfunction conditions $ ( {\bf L_* L_* \ldots L_* } - {\cal P } ) \Psi_n = 0 $, where $ {\cal P} $ is the operator that $projects$ any state $ \Psi_n $ into the ground state $ \Psi_1 \equiv | 0 \rangle $ representing the zero bit state $ | 0 \rangle$ (since $2^0 = 1$). In essence, one has a realization of the integer/state correspondence typical of QM such that the Collatz paths from $ n $ to $ 1$ are encoded in terms of quantum transitions among the states $ \Psi_n$, and leading effectively to an overall downward cascade to $ \Psi_1$. The QM oscillator approach explains naturally why the Collatz conjecture fails for negative integers because there are no states below the ground state.