Superimposed state vs. zero amplitude state

Two equal amplitude wave pulses approaching each other through some medium such as a string may form a region of zero amplitude when they overlap completely. At this point, the location of overlap is (apparently) indistinguishable from any other region in the medium with zero amplitude. However, the two pulses will emerge from the blank region and continue to travel through the medium.

How is the region at which total destructive interference occurs different from any other region of zero amplitude in the medium? Where is the energy and information present within each wave pulse stored during superposition? I assume that the molecules in a string gain potential energy during superposition, but where are wave energy and information stored in superimposed states at the molecular and quantum levels?

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.

First to start, and to be clear, let's talk about a macroscopic string. Then we can talk about a quantum mechanical system.

For the macroscopic string, the state is more than just the position of each infinitesimal piece of the string - it is also the momentum of each infinitesimal unit of the string. So when you are looking at the spot where there has been destructive interference and are commenting that it is no different than a region where there is no motion, you are neglecting the momentum. Immediately adjacent to the spot of destructive interference the string is moving (in the simplest case) in equal and opposite directions. This is not the case in the region where there is no motion (by definition).

The same thing applies for quantum mechanical systems. We generally talk about psi(x) - which is just a function of position - but we know from the Schrodinger equation and Fourrier transform that we can convert this into as psi(p), the wavefunction for the momentum. And the same thing applies - if you imagine a "wide" region of zero amplitude wavefunction, and compare it to a "node" - a point of zero amplitude - it is analogous to the string description above.