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When a spin 1/2-particle is in a superposition of spin-up and spin-down states:

Should one consider the particle spin undefined in the superposition state or should one consider the particle spin having both values at the same time?

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    $\begingroup$ It's a superposition state. What difference does it make if you say "it has both spins" or "it has neither spin"? $\endgroup$
    – ACuriousMind
    Commented Aug 20, 2015 at 16:24
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    $\begingroup$ @ACuriousMind I consider it a big difference whether a physical quantity is undefined or whether it has two different values at the same time: The latter viewpoint has to deal with a contradiction, the former with the limited validity of the physical quantity. $\endgroup$
    – Jo Wehler
    Commented Aug 20, 2015 at 16:28
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    $\begingroup$ It doesn't make a single difference to the physics what ever you decide to say (as far as I can see) - this appears to be pure philosophy. $\endgroup$
    – ACuriousMind
    Commented Aug 20, 2015 at 16:32
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    $\begingroup$ @jowehler ACuriousMind is right here. As you know QM has many interpretations, so to answer the question of "how to interpret superposition" does not have a unique answer, meaning physicists hold different positions on this. ACuriousMind's point is that, there's no operational meaning to such statements as mathematically they correspond to the same thing. A simple view would be: only when there's no way of knowing, not even in principle, which state the particle is occupying, do we say it is in superposition. $\endgroup$
    – Ellie
    Commented Aug 20, 2015 at 16:51
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    $\begingroup$ @Timaeus Title says "physical quantities", so they asked for operators :-D My point is, I would consider the "particle spin" as undefined only if the "particle state" is not in the domain of the "particle spin" operator. $\endgroup$
    – arivero
    Commented Aug 20, 2015 at 18:51

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How to interpret physical quantities in a superposition state?

Every superposition of spin up and spin down for the z component is an actual eigenstate of spin for some other direction.

So just think of it as spin up for some direction. Even spin down is just spin up for the opposite direction.

This is one case were superposition can't even pretend to be mysterious.

When something is spin up in one direction then the results you get for spin so called "measurement" will not be determined solely by the spin state. And they aren't measurements, they change the system.

Should one consider the particle spin undefined in the superposition state or should one consider the particle spin having both values at the same time?

You should consider the spin to be the spin state, that's why it was invented. It is possible that the spin state is entangled with other things. Regardless you can make it become entangled with new things, and those entanglements can have the separable branches become so independent dynamically that they each like a world to itself. You can call that a measurement, but it is a dynamical interaction, an evolution over time. Whether you get two branches or one and the relative size of the branch's will depend on the spin state and what you interact with. This will determined the frequencies of repeated measurements.

But there are different interactions that entangle a particles spin differently with its spatial state. These are still called spin measurements because they still have no branching if done again.

The spin state tells you how things interact with bother things. That is what you expect from a property what else do you want.

And measurements don't reveal preexisting things because they change things into things that act differently. For instance an eigenvector of $\sigma_z$ doesn't split into two branches for inhomogeneous magnetic fields in the z direction, that's what it means to be an eigenvector of $\sigma_z$ but if you pass it through an inhomogeneous field in the y direction it splits into two branches. Do before the split the particle had the property that It doesn't split when it encounters an inhomogeneous magnetic field in the z direction.

But when you split it with an inhomogenous field in the y direction the spin state for each of those branches is now an eigenvalue of $\sigma_y$ and this is a different spin state. You can tell becsuse it reacts differently. If you take the results of each of those branches then they each have a spin state that an eigenvalue of $\sigma_y$ and they now will split if they encounter an inhomogeneous magnetic field in the z direction.

So before they had the property of not splitting for magnetic fields in the z direction and after they were "measured" with an inhomogeneous magnetic field in the y direction they no longer had that property.

You have some properties, specifically to split. Doing a complementary interaction will change you into something that won't split under that interaction but noe you will split under the first type of interaction. So you have been changed.

Just accept that the spin state is the property you are looking for and that it tells you want to know.

If you want know which branch you see when you split. That is a question you don't even have to ask, but you can resolve it in the non relativistic case by saying it is determined by the initial configuration of every particle in the universe, the wavefunction and the choice of which interaction you do (since different devices that give the same correct ratios will entangled the particle differently with its own other properties such as position). But that just makes it determined by things you don't know and won't know so it doesn't accomplish much for scientists.

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  • $\begingroup$ I like where you say this is one case where superposition "can't even pretend to be mysterious." $\endgroup$ Commented Aug 20, 2015 at 17:23
  • $\begingroup$ "Every superposition of spin up and spin down for the z component is an actual eigenstate of spin for some other direction." This is the important point. $\endgroup$ Commented Aug 20, 2015 at 21:46
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I would reserve the word "undefined" for the case where the wavefunction is not in the domain of the observable.

Once the wavefunction is in the domain of the observable, you consider the spectrum of the operator; for each eigenvalue you have a probability. You could call any number $\epsilon$ between 0 and 1 and say something as that the particle "has with probability greater than $\epsilon$" any of a set of eigenvalues.

It seems reasonable to consider a word or a verb that discards precisely the eigenvalues with probability zero. This is linguistics, not philosophy. Some languages can have such word, some others not.

EDIT: while this answer is general for the question title, the OP refers to a simple finite Hilbert space, and in this case the ambiguity of the meaning of "undefined" does not appear. For a more complicated system where the problem arises, consider solutions of the Schrödinger equation over an interval $(0,L)$.

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  • $\begingroup$ Remind that an observable $O$ is a always a selfadjoint operator, which implies that its domain is the same that the domain of its adjoint; in this way we are granting that when $O \Psi$ exists we can project it against $\Psi^*$. $\endgroup$
    – arivero
    Commented Aug 20, 2015 at 17:40

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