Is the wave function objective or subjective?

Question

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.

Related: What are the differences between a $\psi$-epistemic ontological model and a $\psi$-ontic model of quantum mechanics, exactly? — Emilio Pisanty, Apr 01 '19 at 11:49

score 5 · Answer 1 · answered May 22 '11 at 19:18

5

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?

answered May 22 '11 at 19:18

anna v

233,453

Hear! Hear! For "navel gazing". – dmckee --- ex-moderator kitten May 22 '11 at 19:22
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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 $\theta$. – Zo the Relativist May 25 '13 at 00:21
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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

score 3 · Answer 2 · answered May 22 '11 at 17:57

3

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.

answered May 22 '11 at 17:57

dmckee --- ex-moderator kitten

86,062

A double slit picture is the wave function squared so it is observable. – Vladimir Kalitvianski May 22 '11 at 19:09
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No. $\Psi^2$ is an observable, but $\Psi$ is not. You can never learn the absolute phase. – dmckee --- ex-moderator kitten May 22 '11 at 19:20
Who cares about the total phase? – Vladimir Kalitvianski May 22 '11 at 19:57
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@Vlad: You don't, because it is not an observable. – dmckee --- ex-moderator kitten 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
$|\psi|^2$ is not observable. What is observed is the effect of the electrons momentum and charge. are. – my2cts Apr 01 '19 at 20:24

score 1 · Answer 3 · answered Feb 28 '15 at 15:09

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.

Vladimir Kalitvianski · Answer 4 · 2011-05-22T21:52:04.480

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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.

edited May 22 '11 at 21:52

answered May 22 '11 at 17:55

Vladimir Kalitvianski

13,885

(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 05 '15 at 20:12

score 0 · Answer 5 · answered Apr 01 '19 at 11:36

I think the question here is whether the amplitudes and probabilities in the wave function are proper to the observer or whether they are proper to the system. In other words, can different observers "perceive" a different wave function for the same system.

Well, that depends on your interpretation of quantum mechanics. In the classical, "Copenhagen" interpretation of quantum mechanics, the wave function actually collapses when observed by some kind of intelligent being. All other beings in the universe will then see that the state has collapsed to the new state. Although this would be a non-local effect, it cannot be used to transfer information, and is, for all practical purposes, a self-consistent definition of quantum-mechanics. The down side is that it is not very elegant, and that there is no clear-cut definition of what kind of "intelligent being" is required to collapse a wave function.

However, in "many worlds" quantum mechanics, the wave function becomes entangled with the observer and only tells you about correlations between what the observer sees and what "collapsed state" the system is in. Since the observer only feels like they are seeing one thing, the wave function in this case will be subjective in the sense that the observer is in a superposition of many different, approximately orthogonal states, each associated with a different, collapsed, "Copenhagen wave function". On the other, hand, in each of these "branches" of the total wave function, the wave function of the system will appear to be objective because every other observer in the universe will also be in the same branch of the wave function relative to the observer.

A confusing way to say this is that the system is in an objective state relative to the universe, but the entire universe will itself be in a subjective state relative to the observer. The downside to this is that we have to believe in zillions of "copies" of ourselves, with no objective reality, only a wave of many simultaneous realities. Also not particularly aesthetic.

Those, and their variations, are pretty much the only choices in the market. Maybe some day we'll have a better understanding of quantum mechanics that is both elegant and corresponds to experience.

score 0 · Answer 6 · answered Nov 22 '13 at 02:56

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.

Is the wave function objective or subjective?

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