In Wikipedia's article on quantum decoherence, it states that despite decoherence creating the appearance of wavefunction collapse,

A total superposition of the global or universal wavefunction still exists (and remains coherent at the global level), but its ultimate fate remains an interpretational issue.

Most of this makes sense to me, but what I'm struggling with is the claim made in the parentheses. Is the universal wavefunction globally coherent?

At first glance, it makes sense for it to be. Since the universal wavefunction describes everything, there's no external environment for it to interact with to cause decoherence. On the other hand, the fact that it's globally coherent would lead me to believe that the different global quantum states of the universe (describing parallel universes) can interfere with each other, which I highly doubt is the case.

I asked a similar question in the context of the Schrödinger's Cat thought experiment and the responses which I got there seemed to suggest that a quantum system can lose its global coherence just by interacting with itself, which I also highly doubt is the case.

What am I missing? Perhaps the relationship between the coherence of quantum states and their ability to interfere with each other is more complicated than I thought. How does this work?

Edit: I am aware of the fact that wavefunction collapse does not occur under the Many-Worlds Interpretation.

  • 1
    $\begingroup$ Why do you "highly doubt" that the different worlds in MWI can interfere? $\endgroup$ Aug 17, 2020 at 2:55
  • 1
    $\begingroup$ We see these effects all the time, for instance in the Hong-Ou-Mandel effect. $\endgroup$ Aug 17, 2020 at 6:48
  • 1
    $\begingroup$ My understanding of MWI (which may be incorrect) is that there is no quantum collapse. So I don't understand this claim.The alternative versions of you or me is perhaps a bit of a stretch. What we perceive as you and me may involve all the worlds together with these interference. It may just be the we cannot see these interference effects because they are too subtle. Therefore, we cannot proof or disprove any of the interpretations. $\endgroup$ Aug 18, 2020 at 5:41
  • 1
    $\begingroup$ @MarkMoralesII Suppose that the initial state is pure (not mixed), and suppose that the system is closed (which seems to be implied by the phrase "global or universal wavefunction"). Time-evolution is unitary in a closed system, so the state remains pure forever. If "coherent" means "pure," then the answer would be that simple. But is that what the author meant by "coherent"? Did the author give a definition? $\endgroup$ Aug 21, 2020 at 21:22
  • 1
    $\begingroup$ @ChiralAnomaly Hmmm, in this article, physicist Seth Lloyd is quoted as saying, "The universe as a whole is in a pure state. But individual pieces of it, because they are entangled with the rest of the universe, are in mixtures." $\endgroup$ Aug 22, 2020 at 4:04

2 Answers 2


In the MWI, the total quantum state never collapses. See this: https://thereader.mitpress.mit.edu/the-many-worlds-theory/.

The different "branches" of the world can and do interfere with each other. The double slit interferometer is a clear example: each path the particle takes represents a different world. In fact, I think it is correct to say that all quantum interference constitutes the interference between alternative "worlds'.

  • $\begingroup$ I am aware of the fact that wavefunction collapse does not occur under the Many-Worlds Interpretation and agree with your statement regarding interference. However, my question pertains to the possibility of interference between branches at the "global," universal level. $\endgroup$ Aug 18, 2020 at 23:16
  • $\begingroup$ Not sure what that would mean. At the "Global" level is where all the interference occurs. Can you describe a scenario or consequence that would tell you that the interference does occur? $\endgroup$
    – S. McGrew
    Aug 19, 2020 at 1:15

Considering only the many-worlds interpretation of quantum theory.

You can think of the universal wavefunction as a pure state (and if it is somehow not, just add qubits until it is one) and always stays that way. So if you have a wavefunction of the form $$|\Psi\rangle = \frac{1}{\sqrt{2}} \left(|\phi_{1}\rangle + |\phi_{2}\rangle \right)$$ then you can find $|\phi_{1}\rangle$ and $|\phi_{2}\rangle$ can interfere with each other just like normal.

When you start to think about observers it gets a bit more confusing but writing the universal wavefunction as: $$|\Psi(t)\rangle = \frac{1}{\sqrt{2}} \left(|o_{1}(t)\rangle\otimes |s_{1}(t)\rangle + |o_{2}(t)\rangle\otimes |s_{2}(t)\rangle \right).$$ Then the question becomes, can the systems $s_{j}$ interfere with each other, and the answer is yes, but only if/when the two observers match up with each other $$|o_{1}(t^*)\rangle = |o_{2}(t^*)\rangle.$$

If this happened then regardless of which path you took you'd have exactly the same thoughts at this time. It would also appear that this should only ever happen instantaneously, however when we are at times near $t^*$ we can always express $|o_{j}\rangle$ as some sum of the observer's state at the critical time $|0\rangle$ plus some small perturbation by state $|j\rangle$ that goes to zero as $t\rightarrow t^*$.

This argument is fairly simplified as the observer is made of well beyond trillions of qubits and so you probably don't have to worry about this looping procedure occurring and instead will only ever see interference if you can keep the coupling between the observer and the system sufficiently small (and so don't see interference arising due to interfering branches).

  • $\begingroup$ Thank you, this is very helpful. You win the bounty! :) $\endgroup$ Aug 27, 2020 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.