# A A Realization of a Basic Wigner's Friend Type Experiment

#### DarMM

Gold Member
Hm, that sounds also strange. Either the observer has obtained a fixed result or not. No superobserver can change it, because "fixed" means it's irreversible. I guess, it's hard to discuss without a concrete example.
The standard Wigner's friend scenario is an example, the canonical example.

Imagine a lab that consists of a microscopic spin-1/2 system, a device that can measure it and the rest of the lab environment.

Somebody inside this lab measures a particle in the state:
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)$$
they observe $\uparrow$ let's say and thus their device shows $D_{\uparrow}$ let's say, just symbolic for however the $\uparrow$ outcome is displayed on the set up.

Wigner outside the lab models the lab as:
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)\otimes|D_{0}\rangle\otimes |L_{0}\rangle$$
where $D_{0}$ represents the device in the "ready" state before any readings and $L_{0}$ is the initial state of the lab.

After under unitary evolution the entire lab evolves for Wigner as:
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle + |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right)$$
where $\{|D_{\downarrow}\rangle ,|D_{\uparrow}\rangle\}$ represents indicator states of the device and $\{|L_{\downarrow}\rangle ,|L_{\uparrow}\rangle\}$ are corresponding states of the lab.

Wigner can then measure the entire lab system in the basis:
$$\left\{\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle + |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right),\\ \frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle - |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right)\right\}$$
corresponding to some observable $\mathcal{X}$. Assigning $\{E_{+},E_{-}\}$ to indicate the two outcomes for brevity, he will have
$$P(E_{+}) = 1$$
in seeming contradiction with the device having recorded some result.

#### DarMM

Gold Member
One question: Typically a family of histories is a projective decomposition of the identity, such that the probabilities of the histories sum to one. For the families above, I compute a probability of 0.25 for each history, and two histories in each family mean their probabilities sum to 0.5? Although maybe there are more general accounts of families beyond PDIs I'm not aware of.
Sorry a total error on my part. We have five families.

The first:
$$\left\{\left[ 0\right]\otimes\left[W_{0}\right] , \left[ 1\right]\otimes\left[W_{0}\right], \left[ 0\right]\otimes\left[W_{1}\right], \left[ 1\right]\otimes\left[W_{1}\right] \right\}$$
Only those with matching values have non-zero probability.

The second:
$$\left\{\left[ 0\right], \left[ 1\right] \right\}$$
both with 0.5

The third:
$$\left\{\left[W_{0}\right], \left[W_{1}\right] \right\}$$
both with 0.5

Obviously the second and third can be seen as a coarse-graining of the first.

The fourth:
$$\left\{\left[ 0\right]\otimes\left[W_{+}\right] , \left[ 1\right]\otimes\left[W_{+}\right], \left[ 0\right]\otimes\left[W_{-}\right], \left[ 1\right]\otimes\left[W_{-}\right] \right\}$$
with outcomes all having a probability of 0.25

The fifth:
$$\left\{\left[W_{+}\right], \left[W_{-}\right] \right\}$$
with the first outcome having $P\left(\left[W_{+}\right]\right) = 1$

Now the second family can be seen as a coarse graining of the fourth one, but the fifth is incompatible with the fourth.

The first and third are incompatible with both the fourth and fifth.

Or more succintly:
$$\mathcal{F}_{1} \equiv \mathcal{F}_{2} \equiv \mathcal{F}_{3}\\ \mathcal{F}_{2} \equiv \mathcal{F}_{4}\\ \mathcal{F}_{5}$$

• Morbert

#### Morbert

$$\left\{\left[ 0\right]\otimes\left[W_{0}\right] , \left[ 1\right]\otimes\left[W_{0}\right], \left[ 0\right]\otimes\left[W_{1}\right], \left[ 1\right]\otimes\left[W_{1}\right] \right\}$$
Only those with matching values have non-zero probability.
Yeah this is what some consistent historians call a "measurement scenario", whereby the property/event that is measured correlates with the event that does the measuring, such that (as you mentioned) $P(\left[ 0\right] \mid \left[W_{0}\right] ) = P(\left[ 1\right] \mid\left[W_{1}\right] ) = 1$ and $P(\left[ 0\right] \mid \left[W_{1}\right] ) = P(\left[ 1\right] \mid\left[W_{0}\right] ) = 0$

$$\left\{\left[ 0\right]\otimes\left[W_{+}\right] , \left[ 1\right]\otimes\left[W_{+}\right], \left[ 0\right]\otimes\left[W_{-}\right], \left[ 1\right]\otimes\left[W_{-}\right] \right\}$$
What's interesting about this family is it also highlights the how the probabilities/consistency of families can depend on the preparation as well as just the dynamics. E.g. if the particle is prepared in the state
$$\frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle\right)$$
as is usually the case, then the probabilities will all be 0.25, but the decoherence functional will (I think) report interference. E.g. $\mathcal{D}( \left[ 0\right]\otimes\left[W_{+}\right] , \left[ 1\right]\otimes\left[W_{+}\right]) = 0.25$ If, however, the particle is prepared in a state $|0\rangle$, then the interference all goes away and the family becomes consistent (with a probability of 0.5 for two and 0 for the others)

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• DarMM

#### kimbyd

Gold Member
2018 Award
I think this is the critical premise of the argument, and in fact of the entire discussion about "superobservers" and such scenarios. Basically, a "superobserver" is an entity that can perform arbitrary unitary transformations on anything whatever, including "observers" like people. But this capability is equivalent to the capability to undo decoherence.
That's going pretty far beyond the requirements of the problem. It is not necessary to undo decoherence in general. It is only necessary for it to be possible for there to be an experimental apparatus that effectively does this in some situations. That's only non-viable if decoherence is both discontinuous and occurs identically for all possible observers.

Basically, it's a question of whether or not there is such a thing as objective wavefunction collapse. If there is, then the Wigner's Friend-type experiments cannot ever measure quantum effects after this collapse has occurred. If there is no such objective collapse, or if said collapse long after decoherence has made it impossible to measure, then we should see a regime where the quantum effects are visible, and the visibility of said effects should gradually disappear as the experimental apparatus is moved out of that regime.

#### PeterDonis

Mentor
It is not necessary to undo decoherence in general. It is only necessary for it to be possible for there to be an experimental apparatus that effectively does this in some situations.
If the "some situations" includes situations involving conscious observers, or macroscopic devices that record measurement results, then the issue I raised still applies. Our common sense understanding of what it means to "observe" something or to "record a measurement result" would no longer apply in these situations.

Also, I'm not sure I see a realistic limitation on the situations. It seems highly unlikely to me that there could exist a superobserver that can only do the particular unitary operations on my brain that correspond to "undoing" (erasing, destroying, whatever) my observation of the results of certain particular QM experiments, but not anything else.

• DarMM

#### vanhees71

Gold Member
If superobservers exist, the observer's observation of a result is not irreversible, because the superobserver has the ability to perform arbitrary unitary operations on the observer, including ones that reverse or destroy the observer's observation of the result. For example, if the observer observes, say, spin-z up, the superobserver could perform a unitary operation on the observer that took him from the "observed spin-z up" state to the state

$$\frac{1}{\sqrt{2}} \left( | \text{observed spin-z up} \rangle + | \text{observed spin-z down} \rangle \right)$$

which is a state in which the observer has not observed any definite result at all.
Sure, you can always do another SG experiment, but then you change inevitably the state. Superobservers shouldn't be allowed to be outside the rules of QT, if you want to make an argument within QT. That's what always comes out, when attempts are made to beat QT: At one point some "godlike" entity was envoked in the argument which can do measurements violating the principles of QT.

#### vanhees71

Gold Member
The standard Wigner's friend scenario is an example, the canonical example.

Imagine a lab that consists of a microscopic spin-1/2 system, a device that can measure it and the rest of the lab environment.

Somebody inside this lab measures a particle in the state:
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)$$
they observe $\uparrow$ let's say and thus their device shows $D_{\uparrow}$ let's say, just symbolic for however the $\uparrow$ outcome is displayed on the set up.

Wigner outside the lab models the lab as:
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)\otimes|D_{0}\rangle\otimes |L_{0}\rangle$$
where $D_{0}$ represents the device in the "ready" state before any readings and $L_{0}$ is the initial state of the lab.

After under unitary evolution the entire lab evolves for Wigner as:
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle + |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right)$$
where $\{|D_{\downarrow}\rangle ,|D_{\uparrow}\rangle\}$ represents indicator states of the device and $\{|L_{\downarrow}\rangle ,|L_{\uparrow}\rangle\}$ are corresponding states of the lab.

Wigner can then measure the entire lab system in the basis:
$$\left\{\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle + |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right),\\ \frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle - |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right)\right\}$$
corresponding to some observable $\mathcal{X}$. Assigning $\{E_{+},E_{-}\}$ to indicate the two outcomes for brevity, he will have
$$P(E_{+}) = 1$$
in seeming contradiction with the device having recorded some result.
But after the superobservers unitary transformation the state is given by that you wrote in the curly brackets, and that implies that the original spin is not in a state determined by the observer's measurement. In this case you applied the standard rules of QT, and that tells you that with the intervention of the friend the state changes from that prepared within the lab before.

#### DarMM

Gold Member
But after the superobservers unitary transformation the state is given by that you wrote in the curly brackets, and that implies that the original spin is not in a state determined by the observer's measurement. In this case you applied the standard rules of QT, and that tells you that with the intervention of the friend the state changes from that prepared within the lab before.
That unitary transformation is the time evolution representing the friend's measurement though, i.e. his measurement is complete at that stage. It's not a transformation the superobserver applies himself just his description of the measurement process. To the superobserver that is the state after the measurement has been performed. The superobserver has not seen any outcome at this point so he will not alter the state at that stage and he is still free to perform a measurement on the $\mathcal{X}$ observable.

#### vanhees71

Gold Member
But there's the measurement process within the lab. This should also be taken into account in the unitary time evolution by Wigner's friend, right?

#### DarMM

Gold Member
But there's the measurement process within the lab. This should also be taken into account in the unitary time evolution by Wigner's friend, right?
How would it be taken into account? In other words how is it not taken into account in the description above.

If the particle was originally in the state $|\uparrow\rangle$ then the state of the lab after the measurement is:
$$|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle$$
If it was originally $|\downarrow\rangle$ then the state after measurement is:
$$|\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle$$

How would Wigner, who sits outside the lab, model the lab from an initial state of $\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)$ for the system but by
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |D_{\uparrow}\rangle \otimes |L_{\uparrow}\rangle + |\downarrow\rangle \otimes |D_{\downarrow}\rangle \otimes |L_{\downarrow}\rangle\right)$$

What different state do you think he should be using?

#### DarMM

Gold Member
E.g. if the particle is prepared in the state
$$\frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle\right)$$
as is usually the case, then the probabilities will all be 0.25, but the decoherence functional will (I think) report interference. E.g. $\mathcal{D}( \left[ 0\right]\otimes\left[W_{+}\right] , \left[ 1\right]\otimes\left[W_{+}\right]) = 0.25$ If, however, the particle is prepared in a state $|0\rangle$, then the interference all goes away and the family becomes consistent (with a probability of 0.5 for two and 0 for the others)
To check my reasoning I got that they don't have interference because the chain kets are proportional to $|0,+\rangle$ and $|1,+\rangle$ respectively in the Wigner-Friend product space, which are orthogonal. Not confident in by reasoning though, these encapsulated histories don't have much literature behind them at the moment.

#### vanhees71

Gold Member
Then I don't understand, where the problem is. I thought Wigner's friend is about an apparent contradiction, which however is resolved when the proper notion of states:

https://en.wikipedia.org/wiki/Wigner's_friend

Of course, what's wrong is the collapse postulate!

#### Morbert

To check my reasoning I got that they don't have interference because the chain kets are proportional to $|0,+\rangle$ and $|1,+\rangle$ respectively in the Wigner-Friend product space, which are orthogonal. Not confident in by reasoning though, these encapsulated histories don't have much literature behind them at the moment.
Ah ok. Would you also need to include $|0,-\rangle$ and $|1,-\rangle$ for completeness? (-actually nvm that's not the problem, I have to think about it). I just used $|+\rangle$ and $|-\rangle$ which might not have been kosher. And yeah they're super weird to work with. A lot of intuition disappears.

#### DarMM

Gold Member
Ah ok. Would you also need to include $|0,-\rangle$ and $|1,-\rangle$ for completeness? (-actually nvm that's not the problem, I have to think about it). I just used $|+\rangle$ and $|-\rangle$ which might not have been kosher. And yeah they're super weird to work with. A lot of intuition disappears.
Yeah I missed them in my original post, but have them in #202 because I think they are also orthogonal. I'm not sure though.

#### DarMM

Gold Member
Then I don't understand, where the problem is. I thought Wigner's friend is about an apparent contradiction, which however is resolved when the proper notion of states:
What's the resolution though? What do you mean by the "proper notion of states" and how does it resolve the issue?

Wigner's friend has several resolutions but they are interpretational dependent. Can you state explicitly what your resolution is?

#### DarMM

Gold Member
Of course, what's wrong is the collapse postulate!
This is wrong in my opinion. Despite the problems with physical collapse in other cases, it offers a clear resolution to Wigner's friend. So it's not what's wrong in this case.

It's hard (but not impossible) to come up with a solution that avoids collapse without being Bohmian Mechanics or Many Worlds. That's why I'm interested to hear your resolution.

#### vanhees71

Gold Member
Of course I refer to the minimal interpretation. There's no contradiction between the probabilities you get from either using the pure state including Wigner's friend's pointer states and the reduced density matrices that describe the spin alone.

Of course, choosing subensembles according to Wigner's friend measurement outcomes changes the statistics due to entanglement. It's the same thing we discuss again and again concerning measurements on photon-Bell states, including "teleportation", "entanglement swapping", "delayed choice", and all that.

In short: Everywhere, where the collapse interpreters say "the state is collapsed" you have to concretely state what's measured: The "collapse" usually is just the selection of a subensemble depending on a measurement. In the minimal interpretation all these issues resolve in looking at the right ensembles and subensembles prepared and measured in the concrete experiments.

In the above example you say

Wigner has prepared his system in the state
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|0 \rangle +|1 \rangle).$$
At the same time you say, Wigner's friend has prepared the system in the state
$$|\Psi \rangle =\frac{1}{\sqrt{2}} (|0 \rangle \otimes |0 \rangle + |1 \rangle \otimes |1 \rangle).$$
That's a contradiction in adjecto, because if the latter preparation procedure is done, for Wigner's system the state is the reduce density matrix, tracing out the friend's pointer state, i.e.,
$$\hat{\rho}_W=\frac{1}{2} \hat{1}.$$
As long as W measured only the observable given by the corresponding basis, i.e., the observable described by $$\hat{O}=|1 \rangle \langle 1|,$$
of course both descriptions are compatible, because both predict for W's outcome of a measurement of $O$ that he gets with 50% probability 0 and with 50% probability 1.

However, both states are of course not the same, and W could figure out that his system was not prepared in the pure state but in the mixture, if F had in fact prepared the system in the state $|\Psi \rangle$.

The point is that F cannot prepare the system+pointer state without disturbing the system itself. According to QT she cannot be a godlike creature that can determine the state of the system + pointer without disturbing the system.

I'm not so sure, where the resolution of this Frauchiger et al example really is, because I find their paper very confusing, but I'm pretty sure that there cannot be insconsistencies of the QT formalism within the minimal interpretation. I'm pretty inclined to believe in the argument by Aaronson. At least this was my first suspicion when reading the Frauchiger paper, but not being able to point my finger on the specific point of the argument without taking a lot of efford to translate the writing into math ;-)):

But I reject an assumption that Frauchiger and Renner never formalize. That assumption is, basically: “it makes sense to chain together statements that involve superposed agents measuring each other’s brains in different incompatible bases, as if the statements still referred to a world where these measurements weren’t being done.” I say: in QM, even statements that look “certain” in isolation might really mean something like “if measurement X is performed, then Y will certainly be a property of the outcome.” The trouble arises when we have multiple such statements, involving different measurements X1, X2, …, and (let’s say) performing X1 destroys the original situation in which we were talking about performing X2.
https://www.scottaaronson.com/blog/?m=201809

He rightfully quotes Peres with his famous dictum: "Unperformed measurements have no results."

#### DarMM

Gold Member
Wigner has prepared his system in the state
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|0 \rangle +|1 \rangle).$$
At the same time you say, Wigner's friend has prepared the system in the state
$$|\Psi \rangle =\frac{1}{\sqrt{2}} (|0 \rangle \otimes |0 \rangle + |1 \rangle \otimes |1 \rangle).$$
That's a contradiction in adjecto, because if the latter preparation procedure is done, for Wigner's system the state is the reduce density matrix, tracing out the friend's pointer state, i.e.,
$$\hat{\rho}_W=\frac{1}{2} \hat{1}.$$
As long as W measured only the observable given by the corresponding basis, i.e., the observable described by $$\hat{O}=|1 \rangle \langle 1|,$$
of course both descriptions are compatible, because both predict for W's outcome of a measurement of $O$ that he gets with 50% probability 0 and with 50% probability 1.
To be clear on two things:
1. The friend is the person in the lab measuring the system. Wigner is the one outside.

2. I'm only talking about the basic Wigner's friend, not Frauchiger-Renner which is a more advanced version
What you discuss above is fine if Wigner measures the friends device or the system itself, but not if he measures the friend's entire lab. There's no contradiction in the case you've discussed for the reasons you've mentioned, but it doesn't resolve the case of measuring the entire lab state which is what the paradox is about.

The whole point is that Wigner has access to observables for the whole lab that are complimentary to the fact that the friend performed a measurement. That's the case you have to deal with.

#### DarMM

Gold Member
Wigner has prepared his system in the state
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|0 \rangle +|1 \rangle).$$
At the same time you say, Wigner's friend has prepared the system in the state
$$|\Psi \rangle =\frac{1}{\sqrt{2}} (|0 \rangle \otimes |0 \rangle + |1 \rangle \otimes |1 \rangle).$$
I only realised this now. This isn't a contradiction because you have the scenario confused.

The friend prepares the state in:
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|0 \rangle +|1 \rangle).$$

To Wigner outside this corresponds to the initial state:
$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle + |\downarrow\rangle\right)\otimes|D_{0}\rangle\otimes |L_{0}\rangle$$

Thus tracing gives you the superposed state. This is the preparation and there is no contradiction here.

The difficulty comes after the measurement as I discussed above in #201

#### vanhees71

Gold Member
To be clear on two things:
1. The friend is the person in the lab measuring the system. Wigner is the one outside.

2. I'm only talking about the basic Wigner's friend, not Frauchiger-Renner which is a more advanced version
What you discuss above is fine if Wigner measures the friends device or the system itself, but not if he measures the friend's entire lab. There's no contradiction in the case you've discussed for the reasons you've mentioned, but it doesn't resolve the case of measuring the entire lab state which is what the paradox is about.

The whole point is that Wigner has access to observables for the whole lab that are complimentary to the fact that the friend performed a measurement. That's the case you have to deal with.
What do you mean by "entire lab state". Then also the rules of QM apply. I don't see the paradox (at least not within the minimal statistical interpretation). As explained in the Wikipedia version, the two-spin state (let's call them Alice and Bob, i.e., A is measuring the spin in the lab and B entangles it with another spin ("pointer")) is all the fuss is about, and that I've discussed above.

It's simply wrong to say, when in this case you have
$$|\Psi \rangle =\frac{1}{\sqrt{2}}(\alpha |0 \rangle_A \otimes |0 \rangle_B + \beta |0 \rangle_A \otimes |0 \rangle_B),$$
then A's spin is described by the pure state
$$|\psi \rangle = \frac{1}{\sqrt{2}} (\alpha |0 \rangle_A + \beta | 0 \rangle_A),$$
but it's in a mixed state you get from taking the partial trace over B from $$\hat{\rho} = |\Psi \rangle \langle \Psi|$$.

#### DarMM

Gold Member
What you're describing there is not related to the paradox. I'm not saying that when the two particle system is in a pure state a one particle subsystem is in a pure state.

That is obviously wrong, but I'm not saying it nor does it really relate to the paradox.

Let me put it this way, most of the people working in Foundations are not complete idiots. The problem is not due to thinking a traced system gives a pure state as that point is obvious and clearly resolved.

#### vanhees71

Gold Member
Ok, then where can I find a clear definition of the paradox related to "Wigner's friend".

I'm also not claiming that people working in Foundations are complete idiots. Physicists usually are not... ;-))

#### DarMM

Gold Member
Ok, then where can I find a clear definition of the paradox related to "Wigner's friend".

I'm also not claiming that people working in Foundations are complete idiots. Physicists usually are not... ;-))
Don't worry I know you're not, it's just to summarise that the paradox is not so simple. There's a good description of the paradox in this paper section B:

#### vanhees71

Gold Member
Ehm, is this journal "entropy" ok? I was a bit wondering recently (not related with the article you quote here, which I don't know yet).

#### DarMM

Gold Member
Ehm, is this journal "entropy" ok? I was a bit wondering recently (not related with the article you quote here, which I don't know yet).
I'm not sure. I have been skeptical of it myself at times. Several good people publish in it and I've never seen "crank" material, but some papers are not far above lecture notes, i.e. they're not wrong but they seemed basically just a nice way to present something already known.

I just link that paper because Brukner is a good researcher and his explanation is correct and succinct.

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