When a qubit is measured, there is a ‘collapse of the wave-function’ as a result is randomly chosen.

If the qubit is entangled with others, this collapse will also effect them. And the way it affects them depends on the way we chose to measure our qubits.

From this it seems as though things we do on one qubit have instantaneous effects on another. Is this the case, or is the apparent effect more like a Bayesian update of our knowledge about the qubits?


2 Answers 2


If Alice and Bob have an entangled pair of qubits and Alice locally measures her qubit, it does not affect local state of the Bob's qubit in any way. Mathematically, if Alice measures but does not look at the measurement outcome, density matrix of the Bob's qubit does not change. The sole fact of Alice's measurement does not affect the Bob's qubit in any way. If Alice measures and knows the measurement outcome, then Alice has more information about the Bob's qubit than Bob, but this is pure classical situation described by conditional probabilities.

So the Alice's measurement can only instantly affect Alice's information about the Bob's qubit, and no more.

The above said does not explain the "spooky action at distance", we know the satisfying explanation does not exist. Still we can argue about entanglement and measurements avoiding paradoxes and contradictions, and so the answer to the question in the title:

No, it is not true.

  • 1
    $\begingroup$ Two points: 1) Alice looking at the measurement or knowing its result does not change a thing, I think we should avoid perpetuating the myth that consciousness is at all relevant to the discussion of quantum mechanics. 2) Alice's measurement does more than just affect her own information about Bob's qubit, it truly does collapse its state and thus affect the probability for Bob to measure a certain outcome. (But of course Bob doesn't know how this probability has changed.) So yes, the other qubits in the entangled state are affected instantaneously, but not in a way that transmits information. $\endgroup$
    – Betohaku
    May 12, 2018 at 13:54
  • $\begingroup$ @Betohaku 1) Alice looking at the measurement and ignoring its result is quite usual way of saying things in quantum information science problems; it has no relation to consciousness, and there is no need to avoid it. 2) The assertion that "the other qubits in the entangled state are affected instantaneously" contradicts special relativity and should be avoided. Personally I adopted subjective Bayesian view on probabilities; so if you are saying "probability has changed" I would ask "for whom?". Sometimes it is possible to think that probability is objective, but generally not. $\endgroup$
    – kludg
    May 12, 2018 at 14:29
  • 1
    $\begingroup$ You say that Alice's measurement does not affect Bob's qubit in any way, but I would argue that it actually does. While it is true that Bob's reduced state is the same before and after the measurement, before the measurement Bob's state is to be described as a mixture because of the arguably unpredictable outcome of Alice's measurement, while after A's measurement B's state is a mixture which represents what is now "purely classical" ignorance about the state. $\endgroup$
    – glS
    May 17, 2018 at 13:56
  • 1
    $\begingroup$ Stated in another way, if you say that A's measurement does not affect B's qubit then you also have to say that there is no correlation between A's and B's outcomes, which is not true. What is true is that, without the aid of an additional channel (e.g. Alice communicating the measurement outcome to Bob), Bob does not have any information about A's measurement outcome $\endgroup$
    – glS
    May 17, 2018 at 13:58
  • $\begingroup$ @glS correlation does not mean causation. Saying that A's measurement does not affect B's qubit does not mean that there is no correlation between A's and B's measurements. I understand very well your point of view, but as I said, I am not explaining the "spooky action at distance", I know it is not possible; and the explanation "A's measurement affects B's qubit" is wrong, because it contradicts special relativity. It is still possible to argue about entanglement without contradictions, though it involves inconspicuous change of the meaning of the words. $\endgroup$
    – kludg
    May 17, 2018 at 15:19

It is certainly true that, within the mathematical description of qubits, operations on one qubit can require the whole description to be updated. This therefore affects the description of every qubit.

Those who take a 'epistemic' view of this mathematical description might say that we are just updating our knowledge about the other qubits, and that it doesn't affect the qubits themselves. Those who take an 'ontic' view, however, regard the wave function described by the mathematics of quantum mechanics as being a physical property of the qubits. So they would certainly conclude that the the operation on one qubit instantly affected the others.

I think the ontic view is more prevalent these days, among those who have opinions on these things. Though most take the 'Shut up and calculate' option and don't think about them too much.

Another interesting issue is the fact that instantaneous effects cause problems for relativity. Different observers in different reference frames can disagree on the time ordering of events. So one observer might see one qubit being used to affect a second, whereas another observer might see the same events and conclude that the second qubit is affecting the first. Entanglement avoids direct confrontation with relativity by making sure that the affect cannot be used to send any information instantaneously. But nevertheless, they don't play very well together. So that's why we can be hesitant to state that very strongly that entanglement allows instantaneous effects.

The process of teleportation is, I think, a good one to argue that entanglement does indeed allow qubits to instantly affect each other, as well as showing how it compromises with relativity. It is a process by which the state of a qubit is instantaneously sent from one qubit to another, using entanglement. But the state being sent also gets 'scrambled' during the process. This means that it is impossible for the receiving end to even confirm that the qubit has been sent, never mind see what its state is. However, the transmitter can send a message to the receiver with instructions on how to unscramble the qubit. Once this is done, the receivers can confirm that the teleportation did indeed send the state of the qubit. So there was indeed an instantaneous effect, but non-instantaneous effects (sending the unscrambling information) were also needed to reveal the effect and make it useful.

  • $\begingroup$ "So one observer might see one qubit being used to affect a second, whereas another observer might see the same events and conclude that the second qubit is affecting the first. (...) So that's why we can be hesitant to state very strongly that entanglement allows instantaneous effects." Do you think we could say that what makes an interaction instantaneous is precisely this, that it is entirely frame dependent which seems to affect which? If A affects B with a speed-of-light delay, then there exists no frame within which B affects A. Only with zero delay does this become frame dependent. $\endgroup$
    – Betohaku
    May 14, 2018 at 10:37
  • $\begingroup$ A similar problem about the mismatch between finite dimensional quantum mechanics intuition and relativity is given here The role of type III factors in QFT. It requires understanding how states are different for type I and III factors. $\endgroup$
    – AHusain
    May 18, 2018 at 3:46

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