I Does MWI Agree on Measurement Apparatus Pointer?

[entropy1]

Does MWI just say that everyone agrees about what the pointer of the measurement apparatus shows?

 

[A. Neumaier]

Does MWI just say that everyone agrees about what the pointer of the measurement apparatus shows?

No, MWI leaves that an unanswered question.

 

[entropy1]

I mean, given a particular branch.

 

[A. Neumaier]

It doesn't even model multiple observers, as far as I know.

 

[Demystifier]

I mean, given a particular branch.

Yes, in a given branch all observers agree what the pointer shows.

 

[Demystifier]

It doesn't even model multiple observers, as far as I know.

It does.

 

[A. Neumaier]

It does.

How? Where is it modeled?

 

[Demystifier]

How?

Well, if you accept that MWI models one observer, then extension to many observers is trivial. For instance, if one observer called Alice is modeled as
$$|\Psi\rangle=\sum_k c_k|k\rangle|{\rm Alice}_k\rangle$$
(with self-explaining notation), then two observers, Alice and Bob, can be modeled as
$$|\Psi\rangle=\sum_k c_k|k\rangle|{\rm Alice}_k\rangle|{\rm Bob}_k\rangle$$
or
$$|\Psi\rangle=\sum_{k,l} c_{kl}|k\rangle|l\rangle|{\rm Alice}_k\rangle|{\rm Bob}_l\rangle$$
Is there something which you find clear in the first equation but unclear in the second or third equation?

 

[A. Neumaier]

Well, if you accept that MWI models one observer, then extension to many observers is trivial. For instance, if one observer called Alice is modeled as
$$|\Psi\rangle=\sum_k c_k|k\rangle|{\rm Alice}_k\rangle$$
(with self-explaining notation), then two observers, Alice and Bob, can be modeled as
$$|\Psi\rangle=\sum_k c_k|k\rangle|{\rm Alice}_k\rangle|{\rm Bob}_k\rangle$$
or
$$|\Psi\rangle=\sum_{k,l} c_{kl}|k\rangle|l\rangle|{\rm Alice}_k\rangle|{\rm Bob}_l\rangle$$
Is there something which you find clear in the first equation but unclear in the second or third equation?

Yes. Where does the second set of kets labeled l come from? Alice and Bob look at the same quantum system and the same pointer, but may get different readings because of subjective uncertainty.

 

[Demystifier]

Yes. Where does the second set of kets labeled l come from? Alice and Bob look at the same quantum system and the same pointer, but may get different readings because of subjective uncertainty.

The second equation (without l) corresponds to the case in which Alice and Bob measure the same observable, while the third equation (with l) corresponds to the case in which they measure different observables. So for your purpose you can ignore the third equation. Hence the state in one branch is
$$c_k|k\rangle|{\rm Alice}_k\rangle|{\rm Bob}_k\rangle$$
so there is no any uncertainty within one branch. Alice and Bob in the same branch get the same readings.

 

[A. Neumaier]

The second equation (without l) corresponds to the case in which Alice and Bob measure the same observable, while the third equation (with l) corresponds to the case in which they measure different observables. So for your purpose you can ignore the third equation. Hence the state in one branch is
$$c_k|k\rangle|{\rm Alice}_k\rangle|{\rm Bob}_k\rangle$$
so there is no any uncertainty within one branch. Alice and Bob in the same branch get the same readings.

But why does the state of the universe decompose in the way you claim? The general state in a tensor product of the state spaces of System, Alice and Bob is not what you write, but
$$\psi=\sum_{klm} c_{klm}|k\rangle|{\rm Alice}_l\rangle|{\rm Bob}_m\rangle!$$

 

[AlexCaledin]

The universe state decomposes into superposition of classical/macroscopic variants that are including A's and B's brains variants, right?

So it seems that some confusion of their readings is possible but improbable.

 

[Demystifier]

But why does the state of the universe decompose in the way you claim? The general state in a tensor product of the state spaces of System, Alice and Bob is not what you write, but
$$\psi=\sum_{klm} c_{klm}|k\rangle|{\rm Alice}_l\rangle|{\rm Bob}_m\rangle!$$

But we are not talking about the general state. We are talking a special state corresponding to a situation in which both Alice and Bob measure the same observable with eigenvalues $|k\rangle$. It is convenient to model such measurements with a state I have written. It is possible to have a more general model based on your state above, but even such a general model can be thought of as a model of two observers in MWI.

 

[A. Neumaier]

We are talking a special state corresponding to a situation in which both Alice and Bob measure the same observable with eigenvalues $|k\rangle$. It is convenient to model such measurements with a state I have written.

But there you already assume what is to be derived, namely that Alice and Bob always agree on the value $k$ measured.

 

[Demystifier]

But there you already assume what is to be derived, namely that Alice and Bob always agree on the value $k$ measured.

To understand what I assume, consider first a single observer. You would argue that in general one can have
$$|\Psi\rangle=\sum_{k,l}c_{kl}|k\rangle|{\rm Alice}_l\rangle$$
That is true, but such a general state does not correspond to the case in which Allice performs a reliable measurement of the observable with eigenstates $|k\rangle$. Instead, if Alice performs a reliable measurement of that observable, then $c_{kl}=c_k\delta_{kl}$. Do you agree so far?

Now by analogy, with two observers in general we have
$$|\Psi\rangle=\sum_{k,l,m}c_{klm}|k\rangle|{\rm Alice}_l\rangle|{\rm Bob}_m\rangle$$
The requirement that Alice performs a reliable measurement means
$$c_{klm}=a_{km}\delta_{kl}$$
while the requirement that Bob performs a reliable measurement means
$$c_{klm}=b_{kl}\delta_{km}$$
So the only way to satisfy both requirements at once is that
$$c_{klm}=c_{k}\delta_{kl}\delta_{km}$$
In other words, what I assume is that both observers perform reliable measurements, while the fact that their results match is derived from that assumption.

 

[A. Neumaier]

To understand what I assume, consider first a single observer. You would argue that in general one can have
$$|\Psi\rangle=\sum_{k,l}c_{kl}|k\rangle|{\rm Alice}_l\rangle$$
That is true, but such a general state does not correspond to the case in which Alice performs a reliable measurement of the observable with eigenstates $|k\rangle$. Instead, if Alice performs a reliable measurement of that observable, then $c_{kl}=c_k\delta_{kl}$. Do you agree so far?

Now by analogy, with two observers in general we have
$$|\Psi\rangle=\sum_{k,l,m}c_{klm}|k\rangle|{\rm Alice}_l\rangle|{\rm Bob}_m\rangle$$
The requirement that Alice performs a reliable measurement means
$$c_{klm}=a_{km}\delta_{kl}$$
while the requirement that Bob performs a reliable measurement means
$$c_{klm}=b_{kl}\delta_{km}$$
So the only way to satisfy both requirements at once is that
$$c_{klm}=c_{k}\delta_{kl}\delta_{km}$$
In other words, what I assume is that both observers perform reliable measurements, while the fact that their results match is derived from that assumption.

Ok, that's reasonable. But the argument is needed, the formula cannot simply be assumed!

 

[Demystifier]

Ok, that's reasonable. But the argument is needed, the formula cannot simply be assumed!

Do you need some additional argument, or do you just say that the original statement needed argument which is now provided?

 

[A. Neumaier]

Do you need some additional argument, or do you just say that the original statement needed argument which is now provided?

I referred to the argument you provided. It is clear and robust under approximations, hence satisfactory.

 

[AlexCaledin]

Wow, so it's a nice toy universe where nothing ever disturbs Alice and Bob's little game.