# Wave Function Collapse Versus Decoherence

I'm aware that wave function collapse is still a topic of debate-and that decoherence is a pretty good explanation for how things might approach wave function collapse, in some sense. But the way I've seen it be explained, it seems that the reason things collapse is that upon interacting with macroscopic things, the wave function decoheres until there is only one component left. But in this light, it's just that the remaining components are arbitrarily small, correct? The analogy I saw was that you wouldn't be able to reconstruct the waves in a far off country by looking at the ripples at your local beach- but theoretically you could, right?

• What do you mean by "component" of a wavefunction? A Fourier mode? – ACuriousMind Dec 2 '14 at 9:47
• I am answering you instead of Anthony: he means that one develops the wave-function into a coherent superposition of eigenstates of some observable, the observable one wishes to measure. – Sofia Dec 2 '14 at 15:30

But the way I've seen it be explained, it seems that the reason things collapse is that upon interacting with macroscopic things, the wave function decoheres until there is only one component left.

No. That's not the way decoherence works at all. A system $S$ interacts with the environment $E$ with a Hamiltonian that does the following: $$|k\rangle_S|0\rangle_E\to|k\rangle_S|k\rangle_E.$$ So a superposition evolves so that $$\sum_k\alpha_k|k\rangle_S|0\rangle_E\to\sum_k\alpha_k|k\rangle_S|k\rangle_E.$$

In the latter state you can't do interference between the $k$ values using the system $S$ alone whereas in the former state you could do that. So owing to the interaction between the environment and the system the system doesn't evolve coherently. This process does not change the amplitudes of the different possible outcomes.

But in this light, it's just that the remaining components are arbitrarily small, correct? The analogy I saw was that you wouldn't be able to reconstruct the waves in a far off country by looking at the ripples at your local beach- but theoretically you could, right?

In order to undo the decoherence you have to do two things. (1) You would need to know exactly what interaction took place between them. (2) You would need to control both the system and the environment well enough to reverse the interaction. In most situations neither of those requirements is met, so according to the theory you can't undo the decoherence and restore the original state.

The reason why you only see one outcome has nothing to do with the amplitude of the components. Rather, after the measurement there are multiple versions of you and of the measured system and the measuring apparatus. Those different versions can't interact with one another and so you can't see them. It is possible to tell that systems exist in multiple versions by doing specially prepared experiments like single particle interference and the EPR experiment.

• How/why does the superposition evolve like that? Just 'from interactions'? – user24082 Dec 2 '14 at 17:21
• The Schrodinger equation is linear. So if the evolution in the top expression occurs, then so does the one on the bottom. – alanf Dec 2 '14 at 17:32

Decoherence is not a sufficient explanation for the collapse. Decoherence means that the phases between the components of the wave-function, (when the latter is expanded as a quantum superposition), are destroyed. But WHY in a measurement we get a certain ONE from these components, the decoherence doesn't explain. And what happens with the other components after the measurement, we don't know. We remain only aware of the component that we detected.