I know this may be beating a dead-horse, but I'm still puzzled by this topic even after reading so many other related SE questions/answers and online articles. I will be using the Copenhagen interpretation, mostly. Also, I have included a lot of information so that a wise responder may beat out every misconception I have on this topic.

First, preliminary information. The wavefunction is a calculational tool in quantum mechanics. It represents the information about a system known by an observer, and the (unitary) evolution of this information obeys Schrodinger's equation. When an observer makes a measurement on the system, what was previously a probability distribution now becomes a definite quantity, and thus the wavefunction "collapses" into the eigenspace of a particular observable. Because the observer can now say that the system has that certain quantity, they then change their wavefunction in order to reflect that appropriately, but nothing forces them to do so. The wavefunction, collapsed or uncollapsed, will still evolve according to Schrodingers equation and still give probability amplitudes for subsequent measurements.

But is the wavefunction then unique to the observer, or must it be common to every potential observer? I know this sounds like a bunch of unimportant hocus-pocus, but because I don't have a good answer to it I don't feel so confident with quantum mechanics as a whole. Please do not misconstrue my words though - I am not saying I do not think nature is quantum. Of course it is, and experiment proves it.

As a tangible example, consider the standard example of two electrons infinitely separated from each other which are together in the singlet state. Using the standard basis defined by an arbitrary "z" axis,

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|\uparrow \downarrow\rangle + |\downarrow\uparrow\rangle\right)$$

Now suppose you make a measurement. One of the states is realized.

$$|\psi\rangle = |\uparrow\downarrow\rangle$$

But another experimenter/observer that has not made such a measurement will still describe the system with the first wavefunction, whereas you will describe the particle using the second wavefunction. What's up with this? This seems to imply that, for the other observer who has not yet measured the electron spins, there is such a thing as counterfactual definiteness (this means the state of a system was well-defined prior to measurement). If they measures the electron spins, the first observer can say "I already knew that the electrons were in that state! You could have just asked me!". However, Bell's theorem says that quantum mechanics cannot have counterfactual definiteness (since it is local)!

I feel like somebody may tell me "It doesn't matter which wavefunction you use. They both contain essentially the same information, except that with the second one you will already know the spin-states of the electrons." But I believe this problem goes deeper. The crux of Bell's original paper was that if we assume nature is local, then quantum mechanics can not counterfactually-definite - i.e. the results of experiments can not be defined before the measurements themselves have been made. If the measurements were counterfactually-definite, then they would give different statistics which would satisfy Bell's inequality.

But as for my two aforementioned observers, the electron spins are counterfactually-definite for the first and are not for the second. Why should nature care who measures first?


3 Answers 3


You should distinguish between wavefunctions, which represent pure states, and density matrices, which represent mixed states. If a system is in a pure state for two observers, then both observers have to agree which pure state it's in, but two observers need not always agree on mixed states, or even on whether a system is in a mixed state or a pure one. The reason for this is that mixed states include aspects of the observer's knowledge about the state.

I want to modify your example slightly, to make it simpler. Suppose we have just one electron, and that we prepare it in the state $\left|\uparrow\right\rangle$. Suppose we both saw it being prepared and both agree that it's in state $\left|\uparrow\right\rangle$. Now, if we measure it in the $y$ direction we will always measure it as having a positive spin, 100% of the time. In the density matrix formalism this state is represented by $\left|\uparrow\right\rangle\left\langle\uparrow\right|$.

However, suppose I now measure the electron in the $x$ direction and don't tell you the result. (But you did see me make the measurement, so you know I measured in the $x$ direction.) There's a 50% chance that I measured a positive spin, leaving it in state $\left|\rightarrow\right\rangle\left\langle\rightarrow\right|$, and a 50% chance that I measured a negative spin, leaving it in state $\left|\leftarrow\right\rangle\left\langle\leftarrow\right|$. For me it's either in one of those states or the other, but since you don't know which you have to represent it by a mixed state given by $\frac12\left|\rightarrow\right\rangle\left\langle\rightarrow\right|+\frac12\left|\leftarrow\right\rangle\left\langle\leftarrow\right|$. This is not the same as a quantum superposition, and there's no way to represent this as a pure state. You are now unable to predict the result of any spin measurement on the electron, whereas I can predict the result of measurements in the $x$ direction, because I know that if I measure it again I'll get the same result I got just now.

If this seems mysterious, a classical analogy can make it less so. Let's imagine that instead of an electron we have a classical object with a definite orientation, e.g. a ping-pong ball with a dot painted on it. If each of us believes that we know, with 100% certainty, the orientation of the ball then we have to agree. If we disagree then one of us must just be wrong, since we can't both be right. But it's possible for one of us not to know the orientation of the ball while the other one does, and in that case one of us can make predictions about measurements while the other one can't. This is all that the density matrix formalism is doing, it's giving us a way to represent this kind of lack of knowledge about quantum systems.

In short, the wavefunction is usually thought of as the physical state of the system, and the uncertainty it embodies is of a "fundamental" kind that's shared by all observers. (Different interpretations differ in the meaning of this "fundamental" uncertainty, but its observer-indepndence is a property of the mathematics.) However, it's also possible to have the more classical kind of uncertainty that comes from an observer's lack of knowledge, and if you need to represent both types of uncertainty at the same time you have to use the density matrix formalism.

  • $\begingroup$ Thanks, this makes sense. Could you give a response to my last paragraph, on the possible implications this would have for a Bell-test? I haven't precisely sorted out how everything you said would give a sound answer for that. $\endgroup$ Dec 6, 2016 at 4:21
  • 1
    $\begingroup$ @ArturodonJuan I intend to update the answer as requested when I have time - please ping me if I don't get around to it in a week or so. (It's a little involved.) $\endgroup$
    – N. Virgo
    Dec 7, 2016 at 8:29
  • 1
    $\begingroup$ It's because I feel like the lack-of-knowledge type of uncertainty, which as you say is characteristic of density matrices (mixed states), is the type of uncertainty that could be encompassed by a classical correlation (i.e. one that could fit within the parameters of a hidden variable theory). Clearly that isn't the case though, as entangled/mixed states violate the Bell-CHSH inequality. This seeming misalignment in my brain bothers me. :( $\endgroup$ Dec 8, 2016 at 9:38

An experimentalist's view, not a philosopher's nor a theorist's:

The wavefunction is defined by the boundary conditions, and the potentials entering the problem, it is a solution of a differential equation after all.

But another experimenter/observer that has not made such a measurement will still describe the system with the first wavefunction, whereas you will describe the particle using the second wavefunction. What's up with this?

It is a fact that an observer/measuring_instrument changes the boundary conditions. Change of boundary conditions changes the wavefunction and the probabilities of outcome. If you have two observers, you should introduce them to the boundary conditions plus any changes these imply in the potentials used , there will be one wavefunction giving probabilities for all outcomes with two observers in the problem, and if they persist in repeating the measurements the probabilities will match the square of the common wavefunction. Maybe it would help you if you read this experiment elucidating a different conundrum .

That is all.


The Copenhagen interpretation is wrong and confusing. The wave function is not just a calculational tool, it is a partial representation of what is happening in reality. The whole of reality is represented by the Heisenberg picture descriptors of a system, as explained here:


The whole of physical reality includes a load of things each of which acts approximately like the universe as described by classical physics. In single particle interference and EPR type experiments, this approximation breaks down. the wavefunction represents the part of that reality that you can directly interact with, i.e. - in the single universe approximation it just represents what universe you are in.

Your post does not fully analyse the situation. In particular, it omits any description of the interactions between the observer and the system. so the system starts out in the state $$\tfrac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle_s+|\downarrow\uparrow\rangle_s).$$

The observers start in the blank state $|0\rangle_1,|0\rangle_2$. When observer 1 measures the state the state becomes $$\tfrac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle_s|\uparrow\downarrow\rangle_1+|\downarrow\uparrow\rangle_s|\downarrow\uparrow\rangle_1).$$ The observer is present in two versions who can't undergo interference and are somewhat independent. So it is often useful to say each version is in a relative state in which he saw one outcome or the other. You can do a similar analysis for observer 2. You can include the environment with which the system and the observer interact and so on, see


and references therein.

If you want to avoid confusion about quantum mechanics, apply it consistently.

  • $\begingroup$ saying that one interpretation of qm is "right" and another is "wrong" is incorrect and misleading. They are deemed "interpretations" for a reason. $\endgroup$ Dec 6, 2016 at 19:41
  • $\begingroup$ They are deemed interpretations because of bad philosophy that treats explanations and formalism as separate. This is wrong because a bad explanation of the formalism can be criticised for being a bad explanation. If I proposed that GR is actually about a flat spacetime where invisible pixies push stuff around as if space-time was curved I would be criticized. The Copenhagen interpretation amounts to claiming we should use quantum mechanics to do calculations while denying its implications, i.e. - that invisible pixies push stuff around as if the multiverse exists. $\endgroup$
    – alanf
    Dec 7, 2016 at 8:39

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