I don't think it is actually possible to have complete destructive interference everywhere in quantum mechanics (unless the state you started with has zero amplitude). The wavefunction of a particle contains all the information about that particle, including everything needed to calculate what it is going to do in the future. This means that a right propagating wave has a different wave function to a left propagating wave, and so they cannot totally destructively interfere.
This is possible because the wave function is a complex valued function. We can write this as \begin{equation} \psi(x,t) = R(x,t)e^{\imath\,\theta(x,t)}\end{equation} Where $R$ and $\theta$ are real valued functions. The magnitude of the wavefunction, $R$ tells us the probability of finding the particle in a small region \begin{equation}P(x_0<x<x_0+\mathrm{d}x) = R(x_0,t)^2\mathrm{d}x\end{equation} The phase $\theta$ does not tell us anything directly measurable, but becomes important when we are calculating how the wavefunction changes in time.
For example say we have two plane waves propagating in different directions \begin{equation}\psi_r = e^{-\imath(\omega t - kx)} \end{equation} propagating to the right and \begin{equation}\psi_l = e^{-\imath(\omega t + kx)} \end{equation} propagating to the left. We can right a superposition of these states as \begin{equation}\Psi = \alpha\psi_r + \beta\psi_l\end{equation} If we choose say $\alpha = \frac{1}{2}$, $\beta = -\frac{1}{2}$ we find \begin{align} \Psi &= \frac{1}{2} \left( e^{-\imath(\omega t - kx)} - e^{-\imath(\omega t + kx)} \right)\\ &= \imath \sin(kx)e^{-\imath \omega t} \ne 0 \end{align}
In general if the two wavefunction are going to evolve differently in the future they must have different complex phases, and so they cannot destructively interfere everywhere. If they did they would be the same wave function, and so would remain the same forever, and you would have a wave of 0 amplitude.