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Here is a question I am curious about.

Is the wave function objective or subjective, or is such a question meaningless?

Conventionally, subjectivity is as follows: if a quantity is subjective then it is possible for two different people to legitimately give it different values. For example, in Bayesian probability theory, probabilities are considered subjective, because two agents with access to different data will have different posteriors.

So suppose two scientists, A and B, have access to different information about the same quantum system. If A believes it has one wavefunction and B believes it has another, is one of them necessarily "right" and the other "wrong"? If so then the wavefunction is objective, but otherwise it must contain some subjective element.

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You're allowed to ask the question if you can clearly define what it means for a wavefunction to be objective or subjective - for instance, propose an experiment that would determine it, and ask what the result would be. You haven't done that here, so this question isn't well-formed enough for this site. – David Z May 22 '11 at 19:32
@DavidZaslavsky subjectivity has a fairly well-recognised and standard definition, so I've edited it into the question. – Nathaniel May 19 '13 at 1:09
@Weissman I hope I haven't changed the meaning of your question too much by editing it. Please feel free to roll back the edit if I have. – Nathaniel May 19 '13 at 1:10
@EmilioPisanty I know. But this other question seemed in danger of being marked a duplicate of this one. Since this question is closed, it means the question cannot be asked. But it's a reasonable question that probably has a definite answer, so that would be a bit of a shame. – Nathaniel May 19 '13 at 3:31
@EmilioPisanty: Reopening questions aren't meant only for the question authors. It's just that the question seems to be reasonable and it can be improved (either the question itself) or with other answers ;-) – Waffle's Crazy Peanut May 19 '13 at 5:28

The wave function is a solution of an equation.

It is as subjective and as objective as all mathematical solutions to equations describing physical fields, whether classical or quantum mechanical.

Certainly as a solution it is objective, a formula written on paper.

Subjectivity enters in the choice of the equation to be solved and thus the choice of the specific solution.

Objectivity enters again because the particular equation was chosen due to its being appropriate to the problem, having fitted previous experimental observations.

I consider all this as navel gazing.

Are the solutions of Maxwell's equations objective or subjective? What about the gravitational orbit solutions?

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Hear! Hear! For "navel gazing". – dmckee May 22 '11 at 19:22
Well, the wave fucntion is only unique up to a ray in complex space. If $\psi$ is a solution to Schrodinger's equation, then $e^{i\theta}\psi$ is a solution to Schrodinger's equation for any constant $\theta$. – Jerry Schirmer May 25 '13 at 0:21
I mostly agree with the sentiment here, but I disagree that the state vector is like all the other mathematical constructs through which we discuss physics. There is a useful sense in which the state vector changes if I assume I don't have access to a subset of the total system's degrees of freedom; one traces the full density matrix over the inaccessible degrees of freedom. To use Vladimi's language, you have to account for any physically existing but not-available-to-you information. – DanielSank Feb 28 '15 at 17:09

The wave function is not an observable, so you can characterize it however you want.

However, any alternative explanation you want to put forth must agree with experimental reality, which in the end will mean it must be mathematically equivalent to the usual approach and will accordingly have some object that is isomorphic to the wave function.

That makes it real enough for my purposes.

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A double slit picture is the wave function squared so it is observable. – Vladimir Kalitvianski May 22 '11 at 19:09
No. $\Psi^2$ is an observable, but $\Psi$ is not. You can never learn the absolute phase. – dmckee May 22 '11 at 19:20
Who cares about the total phase? – Vladimir Kalitvianski May 22 '11 at 19:57
@Vlad: You don't, because it is not an observable. – dmckee May 22 '11 at 20:09
OK, $\Psi(\vec{r})$ is also not interesting at a given $\vec{r}$. What is interesting is the whole field $\Psi(\vec{r})$ space variations. – Vladimir Kalitvianski May 22 '11 at 20:23

According to the Pusey-Barrett-Rudolph theorem, if scientists Alice and Bob disagree about their beliefs of the wave function, at least one of them has to be wrong. By your definition of objectivity, this makes the wave function "objective".

Note the PBR theorem doesn't apply to mixed density states, as in Wigner's friend scenarios. This kind of implies there is a conceptual difference between wave functions and density states.

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A wave function is as objective as a photo picture of somebody. It encodes an objective information about something. It consist, as any information, of many bits of information.

At the same time is is subjective because it depends on our perception. For one some information tells nothing, for somebody else it may be very speaking.

Concerning our minds, yes, everybody believes in the wrong wave function due to lack of knowledge.

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(This comment was previously erroneously posted under a different answer) "A wave function is as objective as a photo picture of somebody. It encodes an objective information about something." Agreed, and since the amount of information available to different perspectives, the state vector is subjective by OP's definition. – DanielSank Sep 5 '15 at 20:12

Consider an EPR experiment where an entangled pair of electrons is created. One of them hits a detector which finds its spin to be up. The other hits a detector at some distance. The first detector sends a light signal to the vicinity of the distant detector. That signal arrives before the other electron. Near that second detector sit two observers. The first is able to see the light signal and the second detector, the second observer is only able to see the detector but is shielded from the light signal. In the split second after the light signal arrives but before the second electron arrives, the wave function collapses for the first observer but not for the second. When the electron arrives, it collapses for the second observer too. But during that short interval the wave function is subjective in the sense that it is different for the two observers. Empirically both observers' observations agree with their version of the wave function: For the first observer the spin is always up or down as predicted by the light signal, for the second it is 50% of the time up and 50% down, agreeing with the uncollapsed version of the wave function.

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