Quantum Spin: Is it Random? Alice & Bob's Test

[entropy1]

I got confronted with this issue:

Suppose we have Alice and Bob, each of them measuring the quantum spin on one of a pair of electrons along parallel axes, thus yielding an identical spin for both with each measurement.

Now Alice's measurement is done earlier than Bob's.

Can we now predict Bob's measurement outcome given Alice's outcome?

And if we do, how is this random?

I must be missing an elephant here...

 

[entropy1]

I guess there is always a tiny chance that Alice and Bob's outcomes differ...

 

[Nugatory]

Can we now predict Bob's measurement outcome given Alice's outcome?

We have a single quantum system with two possible measurement outcomes: “Alice up, Bob down” and “Alice down, Bob up”. When we measure, we’ll find one of those two outcomes; which one is completely random.

You have allowed the English-language words “pair of electrons” to mislead you into thinking that Alice is measuring her electron instead of the state of the two-electron system.

An analogy: you toss a coin, it lands heads-up. You know without looking that the facedown side is tails... but the outcome of the coin toss is still random.

 

[entropy1]

We have a single quantum system with two possible measurement outcomes: “Alice up, Bob down” and “Alice down, Bob up”. When we measure, we’ll find one of those two outcomes; which one is completely random.

You have allowed the English-language words “pair of electrons” to mislead you into thinking that Alice is measuring her electron instead of the state of the two-electron system.

An analogy: you toss a coin, it lands heads-up. You know without looking that the facedown side is tails... but the outcome of the coin toss is still random.

I am reasonably acquainted with the concept of entanglement.

So, do you mean that if Alice's measurement outcome is random (between two possibilities), Bob's is too?

What I mean is that given Alice's outcome, Bob's outcome seems fixed. The question about randomness concerns Bob's measurement in this case.

 

[Nugatory]

What I mean is that given Alice's outcome, Bob's outcome seems fixed.

Sure, because “Alice measured up” and “Bob measured down” are just different words describing the same thing (in the context of the experimental setup you have chosen).

 

[phyzguy]

What I mean is that given Alice's outcome, Bob's outcome seems fixed. The question about randomness concerns Bob's measurement in this case.

You might want to study the case where Bob abd Alice measure along different axes. Then it is not true to say that "given Alice's outcome, Bob's outcome is fixed". There is some randomness to Bob's result, and yet, depending on the angle between Bob's measurement and Alice's measurement, they are not independent.

 

[nrqed]

I am reasonably acquainted with the concept of entanglement.

So, do you mean that if Alice's measurement outcome is random (between two possibilities), Bob's is too?

What I mean is that given Alice's outcome, Bob's outcome seems fixed. The question about randomness concerns Bob's measurement in this case.

You say that "the question about randomness concerns Bob's measurement in this case."

This is not a good way to think about it, because one can set up the experiment such that the order in which the two measurements are made is not the same in all frames, so one cannot even say in a meaningful way if is is Alice or Bob who made the first measurement. One must consider the two measurements: if we know one of the two, we know for sure the second one. But whether it is Alice = up and Bob =down or the other case is random.

 

[vanhees71]

I got confronted with this issue:

Suppose we have Alice and Bob, each of them measuring the quantum spin on one of a pair of electrons along parallel axes, thus yielding an identical spin for both with each measurement.

Now Alice's measurement is done earlier than Bob's.

Can we now predict Bob's measurement outcome given Alice's outcome?

And if we do, how is this random?

I must be missing an elephant here...

This question cannot be answered the information given. You have to specify a state your electron pair is prepared in when measured. Otherwise we cannot calculate any probabilities for the outcome of any measurements!

 

[entropy1]

@vanhees71 - I think the maximally entangled state if I remember correctly.

I thought of this: often with these kinds of experiments, some speak of "the entanglement being broken after measurement". So in this case, when Alice measures her particle of the pair, the entanglement should be broken. This means that if Alice measures up, the other particle IS down, and similarly if she measures down.

No, we don't know if it is going to be up or down, and we can't force the spin into a specific state, so we can't send information.

But I am wondering when that entanglement would actually be broken, since Alice and Bob may be lightyears apart in theory, and then what is (the entanglement being) "simultaneously" (broken), following @nrqed .

Considering this, I think one can't speak of "entanglement being broken by measurement". But I think in MWI this might be the correct conclusion.

 

[vanhees71]

Which one? There are two spin-entangled states $|1/2,-1/2 \rangle + |-1/2,1/2 \rangle$ and $|1/2,-1/2 \rangle -|-1/2,1/2 \rangle$ (modulo normalization).

Entanglement is usually "broken" by random interactions of the electrons with "the environment" (including the interaction with a macroscopic measurement device). Unfortunately this "decoherence" is a very efficient mechanism, particularly for charged particles like electrons.

 

[entropy1]

Which one? There are two spin-entangled states $|1/2,-1/2 \rangle + |-1/2,1/2 \rangle$ and $|1/2,-1/2 \rangle -|-1/2,1/2 \rangle$ (modulo normalization).

I thought the latter. I can't recall the name, it has a specific name but I can't remember.

Entanglement is usually "broken" by random interactions of the electrons with "the environment" (including the interaction with a macroscopic measurement device). Unfortunately this "decoherence" is a very efficient mechanism, particularly for charged particles like electrons.

Yes, but isn't it the case that in MWI, decoherence does not exclude outcomes, but includes ALL outcomes? (that are possible given the operator used)

 

[vanhees71]

I don't follow the MWI since I've no clue what it means. Please, don't let us get into interpretation discussions. Then this thread has to be moved to the interpretational section of this forum. I'd then not participate in it anymore.

The latter state is called the "spin singlet state". It's a eigenvector of the total spin with $S=0$.

 

[entropy1]

I don't follow the MWI since I've no clue what it means. Please, don't let us get into interpretation discussions. Then this thread has to be moved to the interpretational section of this forum. I'd then not participate in it anymore.

Agreed. But... I thought decoherence is a feature of MWI. They move around stuff very fast here if you are not careful though. I am not insisting on discussing interpretational issues in this thread.

The latter state is called the "spin singlet state". It's a eigenvector of the total spin with S=0S=0.

Yes, that's the one.

 

[vanhees71]

Then we can answer the question in #1. If A measures the spin of her electron (say in $z$ direction) and her outcome is $+1/2$, then she knows that Bob with certainty measures $-1/2$ if he also measures the $z$-component of his electron's spin. This happens with a probability of 50% (just calculate the probability using Born's rule!). The same holds if she measures $-1/2$: then Bob for sure finds $-1/2$, and also this happens with 50% probability.

A and B both just have completely unpolarized electrons. They can meausure their electron's spin in any temporal order. This doesn't change the outcomes including this 100% correlation. There's no necessity for causal influences of one of the measurements on the other. The 100% correlation is due to the preparation of the electron pair in this spin-entangled state.

 

[entropy1]

A and B both just have completely unpolarized electrons. They can meausure their electron's spin in any temporal order. This doesn't change the outcomes including this 100% correlation. There's no necessity for causal influences of one of the measurements on the other. The 100% correlation is due to the preparation of the electron pair in this spin-entangled state.

@vanhees71 - I think the maximally entangled state if I remember correctly.

I thought of this: often with these kinds of experiments, some speak of "the entanglement being broken after measurement". So in this case, when Alice measures her particle of the pair, the entanglement should be broken. This means that if Alice measures up, the other particle IS down, and similarly if she measures down.

No, we don't know if it is going to be up or down, and we can't force the spin into a specific state, so we can't send information.

But I am wondering when that entanglement would actually be broken, since Alice and Bob may be lightyears apart in theory, and then what is (the entanglement being) "simultaneously" (broken), following @nrqed .

Considering this, I think one can't speak of "entanglement being broken by measurement". But I think in MWI this might be the correct conclusion.

I think the issue here is the "entanglement being broken by measurement". If that would happen, one might assert that at a certain point, the entanglement gets broken and the spin value of the remaining particle becomes fixed. I'm afraid that we have to invoke MWI here to stay out of the contradictions, with respect to that Bob's particle spin does not become fixed, and if Alice has measured her spin and got a value, it must be ALL values, as in MWI, because Bob's (still traveling) particle keeps its undefined spin value.

 

[vanhees71]

As I said, I've no clue what MWI is good for. For me there's a physical theory, predicting the probabilities for the outcome of measurements for a given situation. That's all there is from a physics view point to quantum theory. That's called the minimal statistical interpretation. Everything in addition is simply outside of science entering the realm of personal believe (or religion if you wish). I think that's entirely off-topic and irrelvant in a science forum, and that's why I want to stay out of such discussions, because they don't lead to anything concerning the understanding of physics, which imho is the goal of this forum.

 

[Nugatory]

But... I thought decoherence is a feature of MWI.

No. Decoherence is part of quantum mechanics and there no matter what interpretation you use.

 

[entropy1]

As I said, I've no clue what MWI is good for. For me there's a physical theory, predicting the probabilities for the outcome of measurements for a given situation. That's all there is from a physics view point to quantum theory. That's called the minimal statistical interpretation. Everything in addition is simply outside of science entering the realm of personal believe (or religion if you wish). I think that's entirely off-topic and irrelvant in a science forum, and that's why I want to stay out of such discussions, because they don't lead to anything concerning the understanding of physics, which imho is the goal of this forum.

Well, about the math I don't have much questions, for I assume it passed the test and does fine. But I feel I should not need to be a professional physisist, knowing all the math, for I would have no questions left to ask, considering the Shup Up (and Calculate) principle. For me it seems to make more sense asking about other things. The math is immaculate, it is just about discovering how to interpret it. Well, more like I don't know how to put it. I can imagine an underlying structure in the math also.

Perhaps it is like a discrete signal consisting of samples. One could also take a different viewpoint, for instance a description in the frequency spectrum. This is a different way of putting a thing in math. Perhaps something could be discovered in QM math that is a whole different approach. If is does, I won't be discovering it. I just want a few things explored to enlighten my understanding instead of studying for becoming a physisist. There already are enough physisists that can help me.

 

[Nugatory]

If that would happen, one might assert that at a certain point, the entanglement gets broken...

Yes, but do remember that “the entanglement gets broken” is not what the math says, it’s just the best we can do trying to find English-language words that come close to conveying what the math says.

...and the spin value of the remaining particle becomes fixed.

No, that is not what the happens. The quantum state of the two-particle system becomes such that we can predict the result of a measurement on one particular spin axis with certainty, but that is not the same thing as “the spin value of the remaining particle becomes fixed”.

I'm afraid that we have to invoke MWI here to stay out of the contradictions, with respect to that Bob's particle spin does not become fixed, and if Alice has measured her spin and got a value, it must be ALL values, as in MWI, because Bob's (still traveling) particle keeps its undefined spin value.

You could save yourself just a ton of unnecessary confusion if you would remember that there is no such thing as “the spin value of Bob’s particle”. Every time you find yourself saying that... STOP, you’re about to make a mistake. Try again using the words “the measurement result recorded at Bob’s spin detector” and see what happens to the contradiction you’re thinking of.

 

[entropy1]

You could save yourself just a ton of unnecessary confusion if you would remember that there is no such thing as “the spin value of Bob’s particle”. Every time you find yourself saying that... STOP, you’re about to make a mistake. Try again using the words “the measurement result recorded at Bob’s spin detector” and see what happens to the contradiction you’re thinking of.

Ok, so I think you mean that we should only speak of pointer states concerning entanglement? That could solve the paradox indeed I suspect.

 

[Nugatory]

Ok, so I think you mean that we should only speak of pointer states concerning entanglement?

No, we do not have to limit ourselves to pointer states... and a good thing too, because the two-electron state that you mentioned in the first post of this thread is not one.

Quantum mechanics is a theory that predicts measurement results. Use it that way.

 

[vanhees71]

Well, the math also tells you how entanglement gets "lost" namely by interactions of the entangled system with something else. There's nothing mysterious in this and there's no need for esoterical non-physical additions, which are maybe nice entertainment as some science fiction stories are, but it's not part of science.

 

[PeterDonis]

The math is immaculate, it is just about discovering how to interpret it.

For posts in this forum (the regular quantum physics forum), there is only one interpretation you can use, the "minimal" one described in this Insights article:

https://www.physicsforums.com/threads/the-7-basic-rules-of-quantum-mechanics.971724/

This is sufficient to calculate all predictions for experimental results.

If you want to discuss anything beyond that as regards interpretations (such as how the MWI, for example, interprets a particular scenario), you need to start a separate thread in the quantum interpretations forum.

 

[Demystifier]

And if we do, how is this random?

Do you know what is conditional probability?

 

[entropy1]

Do you know what is conditional probability?

Yes. P(Bob-down|Alice-up)=1.

 

[Demystifier]

Yes. P(Bob-down|Alice-up)=1.

Yes. And at the same time, a priori joint probability obeys P(Bob-down,Alice-up)<1. Usually, when probability obeys 0<p<1 we call it random, but when p=1 we call it non-random. So how can an event be both random and non-random at the same time? The answer is that the notion of "random" is ambiguous. Sometimes it means objectively random, and sometimes it means subjectively random. In a sense, a priori probability can often (but not always) describe objective randomness, while conditional probability expresses our (lack of) knowledge. The simplest way to think of it is to imagine that all probabilities are fundamentally subjective (in QM it means the existence of deterministic hidden variables), in which case there is nothing strange with the idea that a change of our knowledge (e.g. that Alice is up) may turn something random into non-random. Otherwise, if you want to retain some notion of objective randomness in QM, you must deal with hard quantum interpretation problems. You must separate objective randomness from subjective randomness, which is hard.

 

[DrChinese]

Yes, that's the one [referring to the singlet state].

The possible random outcomes of a measurement on a singlet state (following the idea of post #1) are:

a: [−1/2 , +1/2]
-or-
b: [+1/2 , -1/2]

So when you talk about Alice's measurement giving a random result when that measurement occurs first, and then Bob's can't be random: you are talking AS IF the measurement outcomes were from completely separate and independent particles. You are in fact measuring a component/components of a combined system which is entangled.

The outcomes a and b occur randomly. You can deduce [or predict] some information about one component of the system given some knowledge of the other component of an entangled system of electrons. Note that outcomes a and b do NOT contain any dependency on the ordering of the measurements on components of an entangled system.

 

[entropy1]

So when you talk about Alice's measurement giving a random result when that measurement occurs first, and then Bob's can't be random: you are talking AS IF the measurement outcomes were from completely separate and independent particles. You are in fact measuring a component/components of a combined system which is entangled.

The outcomes a and b occur randomly. You can deduce [or predict] some information about one component of the system given some knowledge of the other component of an entangled system of electrons. Note that outcomes a and b do NOT contain any dependency on the ordering of the measurements on components of an entangled system.

Yes. The particles make up a single wavefunction, AND we can only relate the measurements and NOT spin values that would exist. I think it is all about measurement outcomes right? That is how I understand it.

But if we say that measurement collapses the wavefunction, then what?

 

[PeterDonis]

if we say that measurement collapses the wavefunction

Then, as has already been pointed out, you are talking about an interpretation beyond the "minimal" interpretation that is the basis for discussion in this forum, and you need to start a separate thread in the interpretations forum to discuss something like this.

 

[Nugatory]

But if we say that measurement collapses the wavefunction, then what?

Then we are choosing to use a collapse interpretation in a situation where it is known to work badly (because of the conflict between relativity and instantaneous collapse everywhere) and making the problem unnecessarily confusing.

We choose to think in terms of a particular interpretation because it helps us with the problem at hand. If it doesn't help... don't choose it, for about the same reason that you generally choose not to poke yourself in the eye with a sharp stick.