The answer to this question suggest that one can solve the measurement problem by decoherence. If I understand it correctly the decoherence appears when the quantum state interacts with the measurement device (and possibly the environment).

However I was wondering about locality. Bell tests tell us that measurements can change the quantum state globally (Although one cannot transfer information, one can still measure correlations).

I was wondering how decoherence can behave non local? So far all matter interacts only locally (In QFT one assumes this explicitly). Why can the measurement device then affect the wavefunction outside of the lightcone?

EDIT: Thank you all for your answers. Maybe I should rephrase the question (Maybe also the title is not suitable, but I didn't have any better idea). This question is not about the Bell test itself. I completely understood that a measurement changes the wave function globally, but this cannot be used to transfer information (because the reduced density matrix of system B is not affected by a measurement on system A).

This question is more about the measurement procedure itself. Decoherence tries to explain the measurement process in way like: From the microscopic point of view the wave function completely behaves according to the Schrödinger equation, however from the macroscopic point of view it looks like (what we call) "a measurement" happened.

I was wondering about the following: If I can explain the measurement procedure completely using ordinary quantum mechanics, how can a measurement procedure change the wave function outside of the lightcone?

Unfortunately I cannot explain this using the standard Schrödinger equation $i\partial_t |\psi\rangle = (-\frac{1}{2m}\nabla^2 + V(x)) |\psi\rangle$ since it is non relativistic.

So as a toy model let's use the Klein-Gordon equation for our particle wave function instead: $(\partial_\mu\partial^\mu + m^2) |\psi\rangle = 0$. To do a measurement we need to couple it to a measurement system. This will introduce a "source term" on the right side, something like this: $$(\partial_\mu\partial^\mu + m^2) |\psi\rangle = \hat{A}|\text{detector}\rangle $$ where $\hat{A}$ is some coupling operator and $|\text{detector}\rangle$ the detector state. One can solve this equation using the retarded Greens function. This tells me that the interaction with the detector will only affect the wave function $|\psi\rangle$ inside the lightcone. So in this toy model the decoherence process cannot introduce any correlations between spacelike separated regions. But these correlations exist (they are measured in the bell experiments).

Of course this was only a toy model. However all other matter fields I encountered so far (like the Dirac field or the photon field) also satisfy the Klein-Gordon equation (I have no idea about strong or weak interaction, but I guess it's the same there). This means whatever you do with the field at one point in spacetime - it will only effect the wavefunction inside the lightcone. Therefore (from a microscopic view) one cannot create any correlations between spacelike separated regions. How is it therefore possible to create these correlations at the macroscopic level?

  • 1
    $\begingroup$ Where in those answers are you getting the idea that a measurement device can affect the wavefunction outside of the lightcone? $\endgroup$ Commented Oct 9, 2018 at 13:24
  • $\begingroup$ The wavefunction is a non-local object and quantum mechanics is non-local, this should not be too shocking for you. Apart from that, decoherence does not solve the measurement problem, as explained here: physics.stackexchange.com/questions/373905/… $\endgroup$
    – Luke
    Commented Oct 9, 2018 at 13:26
  • $\begingroup$ Of course the wavefunction is a non-local object. But only because it is non local this does not mean that locality is violated. For example the electromagnetic field is defined everywhere in spacetime (therefore is a nonlocal object). However electromagnetism is a local theory because if I perturbe the field at a specific point (or region) in spacetime it will only cause effects inside the lightcone. $\endgroup$
    – toaster
    Commented Oct 9, 2018 at 16:58

2 Answers 2


For the record I do not necessarily want to claim that decoherence solves all the subtle interpretational issues that go under the "measurement problem", at least not without some extra ingredient(s) and/or interpretational disambiguation(s). It is however a fairly convincing model to describe what concretely happens to a quantum system when it undergoes a measurement, unpacking the instantaneous, idealized "quantum measurement" of the measurement postulate. As such, it can be used to answer various questions regarding measurement in quantum mechanics, namely those questions that do not, in fact, crucially depend on the above-mentioned interpretational difficulties.

So with that disclaimer, how do we apply this decoherence-inspired measurement model to a Bell experiment?

Let us produce an entangled state: $$|\Psi\rangle = \frac{1}{\sqrt{2}} \left( |0\rangle_{A} \otimes |0\rangle_{B} + |1\rangle_{A} \otimes |1\rangle_{B} \right)$$ and send the $_A$ labeled particle to Alice and the $_B$ labeled one to Bob, making sure that the region in which Alice will latter interact with the particle she received is cleanly spacelike separated from the region in which Bob will do the same with his.

Now Alice applies a detector1 to her particle, let's say along the $|0\rangle_{A}, |1\rangle_{A}$ basis, with the joint $(\text{detector}_{A} \otimes \text{particle}_{A})$ system undergoing the unitary transformation: $$M_{A}: \begin{array}{lll} |\text{init}\rangle_{A} \otimes |0\rangle_{A} & \mapsto & |0 \text{ detected}\rangle_{A} \otimes |0\rangle_{A}\\ |\text{init}\rangle_{A} \otimes |1\rangle_{A} & \mapsto & |1 \text{ detected}\rangle_{A} \otimes |1\rangle_{A} \end{array} $$ On the overall $(\text{detector}_{A} \otimes \text{particle}_{A} \otimes \text{particle}_{B})$ system, the operation has the form $M_{A} \otimes \mathbf{1}_{B}$, as appropriate for a local interaction happening entirely in region A. To double-check that nothing untoward is going on here, we can verify that the reduced density matrix of Bob's particle has been completely unaffected by Alice's operations: the complete state went from $|\text{init}\rangle_{A} \otimes |\Psi\rangle$ to: $$\frac{1}{\sqrt{2}} \left( |0 \text{ detected}\rangle_{A} \otimes |0\rangle_{A} \otimes |0\rangle_{B} + |1 \text{ detected}\rangle_{A} \otimes |1\rangle_{A} \otimes |1\rangle_{B} \right)$$ with $\text{particle}_B$ reduced density matrix remaining $\frac{1}{2}\left(|0\rangle\langle 0|_{B} + |1\rangle\langle 1|_{B}\right)$.

On the other hand, the reduced density matrix of $(\text{particle}_{A} \otimes \text{particle}_{B})$ has gone from $|\Psi\rangle\langle\Psi|$ to: $$\frac{1}{2} \left( |0\rangle\langle 0|_{A} \otimes |0\rangle\langle 0|_{B} + |1\rangle\langle 1|_{A} \otimes |1\rangle\langle 1|_{B} \right) \tag{*}\label{post-measure-pre-reading}$$ so decoherence has occurred and our initial quantum entangled state has been downgraded to a simple classical superposition. Again, this does not require any violation of locality or causality: the operation in region A has simply affected the correlations between regions A and B (turning quantum correlations into classical ones2), which is fine because those correlations are not located entirely within region B, they are a global property of the density matrix over the union of regions A and B.

But now Alice reads out her result and, with probability $\frac{1}{2}$, she may find it's $0$. If we think of the density matrix of $(\text{particle}_{A} \otimes \text{particle}_{B})$ as recording the accumulated knowledge we have of the system, it is appropriate to subsequently update it to: $$|0\rangle\langle 0|_{A} \otimes |0\rangle\langle 0|_{B}$$ so suddenly the reduced density matrix for $\text{particle}_B$ has jumped from $\frac{1}{2}\left(|0\rangle\langle 0|_{B} + |1\rangle\langle 1|_{B}\right)$ to just $|0\rangle\langle 0|_{B}$. Does it mean that the reading-out of the result is itself a non-local operation? I would think of it rather as revealing the pre-existing non-locality contained in the state of $(\text{particle}_{A} \otimes \text{particle}_{B})$: it is because of the non-local correlations encoded by (\ref{post-measure-pre-reading}) that determining the state of $\text{particle}_{A}$ gives us new information on the state of the far-away $\text{particle}_{B}$.

What exactly is entailed by the "reading out" step will depend on your favorite interpretation of quantum mechanics (decoherence is not per se an interpretation, and, being derived from vanilla quantum evolution, should be compatible with any reasonable interpretational narrative). But the important point is that this step is fully "transparent". By transparent, I mean that while it is convenient to perform it in preparation of future operations (specifically to compute the conditional probabilities of subsequent measurement results given $\text{particle}_{A}$ has been measured in state $0$), someone who doesn't know (yet) the intermediary result (e.g. Bob) would get fully consistent overall probabilities, because statistical superpositions crucially obey the formula of total probability: $$P(\text{result}_{B} = b) = \sum_a P(\text{result}_{B} = b | \text{result}_{A} = a) \, P(\text{result}_{A} = a)$$ (in contrast to quantum superpositions).

TL;DR: For Bob, who doesn't know the measurement result, his version of his partial density matrix is completely unaffected by the measurement going on in the spacelike separated region A: it will remain $\frac{1}{2}\left(|0\rangle\langle 0|_{B} + |1\rangle\langle 1|_{B}\right)$ until he perform a measurement of his own. As a side note, this provides a quick proof that quantum measurements cannot in fact be used to exchange of information faster than light.

1 I use the term "detector" or "measurement device" in a broad sense: it may be appropriate to include in it the lab environment, Alice's brain,...

2 Creating non-local correlations where there weren't any before would require a non-local operation but that's not what we have done here, we have only leveraged the existing quantum entanglement between $\text{particle}_{A}$ and $\text{particle}_{B}$.

  • $\begingroup$ Good. The decoherence approach though seems to force us to give a physical reality to the wavefunction beyond just a probability calculating aid of the Copenhagen interpretation. This then makes it difficult to justify tossing away the unreal parts of the wavefunction when they are no longer needed to describe "our universe". You can see then how some are led to MWI as the natural conclusion of the narrative you present. Not expecting you to argue this here - but are you sympathetic to that reasoning or could you expand more in your answer how the "dropping" is conceptually justified? $\endgroup$
    – isometry
    Commented Oct 10, 2018 at 19:45
  • $\begingroup$ @BruceGreetham "The decoherence approach seems to force us to give a physical reality to the wavefunction beyond just a probability calculating aid" --> I am not totally sure what makes you think so to be honest. As far as I am concerned, I very much think of the wavefunction (actually the density matrix) as a probability-computing/record-keeping device, and my answer above certainly reflects this perspective. Also, because decoherence is dependent on the subsystem partition, it suggests measurement is somewhat relative, which to me doesn't seem to sit well with a many-world postulate. $\endgroup$
    – Luzanne
    Commented Oct 10, 2018 at 20:04
  • $\begingroup$ Thanks, useful to hear your perspective on this - that's all I was asking. I will keep taking the therapy for my condition! $\endgroup$
    – isometry
    Commented Oct 10, 2018 at 20:22
  • $\begingroup$ It seems to me that the "Alice reads out her result" operation is a measurement, precisely what decoherence theory is supposed to elucidate. But here it is handled in the standard QM way, as triggering a redefinition (collapse) of the superposed state. If we want to stay consistent with respect to decoherence, we should consider Alice herself as entangled with the system along with her measurement apparatus, and since Bob is also entangled on his side, the usual questions remain: how do they eventually reach (each of them) a classical state, and why are these two states correlated? $\endgroup$ Commented Oct 10, 2018 at 21:23
  • $\begingroup$ @StéphaneRollandin At this stage, we have a statistical superposition (that's what decoherence does for us) so it is no different from updating a classical statistical probability distribution with new information. As mentioned in a comment above, this is from the perspective of the density matrix as a record-keeping device. Of course, this works only as long as we refrain from any further interaction between the particles and the apparatus keeping a record of the measurement: otherwise we could go and make interferences to reveal the quantum superposition of results in the global state. $\endgroup$
    – Luzanne
    Commented Oct 10, 2018 at 21:54

I was wondering how decoherence can behave non local?

In short: no. Decoherence, like all actions you could possibly take on side B of a bipartite entangled state $\Psi_{AB}$, is subject to the No-Communication Theorem, which tells you that no effect can be observed at side B until you allow for sufficient time for light to propagate.

Basically, if you have one half of an entangled state and you don't know what's happened to the other half (say, whether the particle's been kept isolated in a box or whether it's been allowed to interact with an environment that would induce decoherence), then your local description of your system is given by the reduced density matrix that you obtain when you trace out the other half, and this reduced density matrix is the maximally-mixed state, from which no information can be extracted.

Thus, as usual,

Why can the measurement device then affect the wavefunction outside of the lightcone?

it is possible to see measurements as affecting the wavefunction outside of the light-cone, but that's only if you know the outcome of the measurement. Since you can't know the outcome of the measurement outside of the light-cone, local experimental results are not affected.


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