22.1 Two entangled laboratories and superobserver basis choices
22.1.1 Comparing the Frauchiger–Renner protocol with local-friendliness theorems
A single Wigner’s friend laboratory (Chapter 21) does not by itself force a Bell-type contradiction: one can always assign a joint probability distribution to the friend’s single outcome and Wigner’s single outcome because the two measurements are never performed together. By coupling two isolated Wigner’s-friend laboratories—\(\mathrm{Lab}_1\) containing friend \(F_1\) and \(\mathrm{Lab}_2\) containing friend \(F_2\)—and allowing external superobservers \(W_1\) and \(W_2\) to measure those laboratories in complementary bases, Časlav Brukner (2018), Daniela Frauchiger and Renato Renner (2018), and Bong et al. (2020) proved extended Wigner’s-friend no-go theorems. These theorems show that one cannot simultaneously maintain universal unitary quantum mechanics, locality/no-superdeterminism, and the assumption that every internal measurement produces a single context-independent “absolute” outcome that can be combined into a joint four-party probability table \(P(f_1, f_2, w_1, w_2)\).
22.1.2 Internal record basis states across two coupled laboratories
Let \(|0\rangle_{L_k}\) and \(|1\rangle_{L_k}\) denote the orthonormal composite states of laboratory \(k \in \{1, 2\}\) in which the internal system and the friend’s memory register record outcome \(0\) (e.g., “heads” or \(-1\)) or outcome \(1\) (e.g., “tails” or \(+1\)):
Each superobserver \(W_k\) can either open the door to read the friend’s record in the computational basis \(\{|0\rangle_{L_k}, |1\rangle_{L_k}\}\) or perform a coherent reversal measurement in the Hadamard/superobserver basis
22.1.3 Tracking amplitude interference separately from modal agent reasoning
Extended-friend papers often interleave two very different kinds of derivation: (1) standard Hilbert-space inner products on the two-laboratory state \(|\Psi_{L_1 L_2}\rangle\), which are completely unambiguous, and (2) modal logic rules describing how one agent reasons about another agent’s certainty across different measurement contexts. Keeping the exact Hilbert-space amplitudes separate from the modal inference rules makes it immediately transparent why the contradiction arises.
Visual animation · Section 22.1
Two entangled laboratories and superobserver basis choices
Interactive visual model for 22.1 Two entangled laboratories and superobserver basis choices (Comparing the Frauchiger–Renner protocol with local-friendliness theorems · Internal record basis states across two coupled laboratories · Tracking amplitude interference separately from modal agent reasoning). Use Animate or Step to follow the physical state evolution and quantitative predictions.
22.2 Three-term amplitude cancellation and the one-twelfth probability
22.2.1 Evaluating the non-maximally entangled two-lab state in the outside basis
In the Frauchiger–Renner state preparation, friend \(F_1\) measures a biased coin with \(P(0) = 1/3\) and \(P(1) = 2/3\); conditioned on \(0\), \(F_1\) sends \(|0\rangle_{S_2}\) to \(F_2\), and conditioned on \(1\), \(F_1\) sends \((|0\rangle_{S_2} + |1\rangle_{S_2})/\sqrt{2}\) to \(F_2\). Under universal unitary evolution of both isolated laboratories, the joint state of \(\mathrm{Lab}_1 \otimes \mathrm{Lab}_2\) is the Hardy-type non-maximally entangled state
where the fourth computational basis term \(|0,1\rangle\) has amplitude strictly zero: \(\langle 0,1|\Psi_{L_1 L_2}\rangle = 0\).
Now evaluate the amplitude for both superobservers \(W_1\) and \(W_2\) to obtain \(|\mathrm{ok}\rangle = (|0\rangle - |1\rangle)/\sqrt{2}\) simultaneously. Since
22.2.2 Full joint probability table for superobserver outcomes
Projecting \(|\Psi_{L_1 L_2}\rangle\) onto all four basis states \(\{|\mathrm{ok}\rangle, |\mathrm{fail}\rangle\}^{\otimes 2}\) yields the normalized superobserver distribution:
Notice that the \(|0,0\rangle\) and \(|1,0\rangle\) amplitudes cancel destructively in \(\langle\mathrm{ok},\mathrm{ok}|\Psi_{L_1 L_2}\rangle\) (\(+1 - 1 = 0\)), leaving only the \(|1,1\rangle\) contribution \(1/(2\sqrt{3})\).
22.2.3 Where chained certainty inferences cross incompatible contexts
Why do the chained agent inferences predict \(P(\mathrm{ok}_1, \mathrm{ok}_2) = 0\) when the quantum calculation gives \(1/12\)? Consider the three certainty conditions encoded in \(|\Psi_{L_1 L_2}\rangle\):
Because \(\langle 0,1|\Psi_{L_1 L_2}\rangle = 0\), if \(F_1\) records \(1_{L_1}\), then \(\mathrm{Lab}_1\) is definitely in \(|1\rangle_{L_1}\)? No: if \(F_2\) records \(1_{L_2}\), the only surviving computational branch is \(|1,1\rangle\), so \(L_1\) is in \(|1\rangle_{L_1}\).
Writing \(|\Psi_{L_1 L_2}\rangle = \sqrt{\frac{2}{3}}\,|+\rangle_{L_1}|0\rangle_{L_2} + \frac{1}{\sqrt{3}}\,|1\rangle_{L_1}|1\rangle_{L_2}\) shows that \(\langle\mathrm{ok}_1, 0_{L_2}|\Psi_{L_1 L_2}\rangle = 0\): conditioned on \(W_1\) obtaining \(\mathrm{ok}_1\), \(\mathrm{Lab}_2\) would yield \(1_{L_2}\) with certainty if measured in the \(Z_{L_2}\) basis.
Writing \(|\Psi_{L_1 L_2}\rangle = \frac{1}{\sqrt{3}}\,|0\rangle_{L_1}|0\rangle_{L_2} + \sqrt{\frac{2}{3}}\,|1\rangle_{L_1}|+\rangle_{L_2}\) shows that \(\langle 1_{L_1}, \mathrm{ok}_2|\Psi_{L_1 L_2}\rangle = 0\): conditioned on \(1_{L_1}\), \(W_2\) would obtain \(\mathrm{fail}_2\) with certainty.
Chaining (2) \(\mathrm{ok}_1 \Rightarrow 1_{L_2}\), (1) \(1_{L_2} \Rightarrow 1_{L_1}\), and (3) \(1_{L_1} \Rightarrow \mathrm{fail}_2\) treats the unperformed intermediate measurements \(Z_{L_2}\) and \(Z_{L_1}\) as possessing definite values inside the run where \(W_1\) and \(W_2\) actually measure \(X_{L_1}\) and \(X_{L_2}\). Just as in Hardy’s paradox and Chapter 18, combining counterfactual certainties from noncommuting contexts eliminates the destructive interference between \(|0,0\rangle\) and \(|1,0\rangle\).
Visual animation · Section 22.2
Three-term amplitude cancellation and the one-twelfth probability
Interactive visual model for 22.2 Three-term amplitude cancellation and the one-twelfth probability (Evaluating the non-maximally entangled two-lab state in the outside basis · Full joint probability table for superobserver outcomes · Where chained certainty inferences cross incompatible contexts). Use Animate or Step to follow the physical state evolution and quantitative predictions.
22.3 Dissecting no-go assumption packages
22.3.1 Universal unitary validity, single outcomes, and cross-agent consistency
The Frauchiger–Renner theorem proves that at least one of three assumptions must be dropped: (Q) universal validity of unitary quantum mechanics at the superobserver scale, (S) each measurement has a single outcome, and (C) agents can paste together certainty claims derived in incompatible measurement contexts as if a single global history of internal and external outcomes existed.
22.3.2 Absoluteness of observed events in local-friendliness inequalities
The local-friendliness theorem (Bong et al., 2020) replaces modal agent logic with a strictly operational Bell-like test. In each run, superobserver \(W_k\) chooses either \(x_k = 0\) (open the door and copy friend \(F_k\)’s record \(w_k = f_k\)) or \(x_k = 1\) (measure \(\mathrm{Lab}_k\) in a rotated basis). The assumption of Absoluteness of Observed Events (AOE) states that in every run there exists a joint probability distribution \(P(w_1, w_2, f_1, f_2 \mid x_1, x_2)\) whose marginals recover the observed outcome when \(x_k = 0\). Together with Locality and No-Superdeterminism, AOE implies that the observed correlations \(E(x_1, x_2)\) must satisfy the standard CHSH and local-friendliness polytope bounds (such as \(|S| \le 2\)), whereas unitary quantum mechanics on two entangled laboratories violates those bounds up to \(2\sqrt{2}\).
22.3.3 Why destroying internal records during superobserver readout matters
Measuring \(\mathrm{Lab}_k\) in the \(|\mathrm{ok}\rangle/|\mathrm{fail}\rangle\) basis requires applying the inverse of the friend’s measurement unitary and erasing the internal record \(f_k\). Consequently, in any run where \(W_1\) and \(W_2\) actually measure \(\mathrm{ok}/\mathrm{fail}\), no durable physical record of \((f_1, f_2)\) survives at the end of the experiment.
Visual animation · Section 22.3
Dissecting no-go assumption packages
Interactive visual model for 22.3 Dissecting no-go assumption packages (Universal unitary validity, single outcomes, and cross-agent consistency · Absoluteness of observed events in local-friendliness inequalities · Why destroying internal records during superobserver readout matters). Use Animate or Step to follow the physical state evolution and quantitative predictions.
Explore the Apparatus
Inspect the four computational branch amplitudes of the two-laboratory state \(|\Psi_{L_1 L_2}\rangle = (|0,0\rangle + |1,0\rangle + |1,1\rangle)/\sqrt{3}\) and adjust the coherence factor \(W \in [0, 1]\) of the superobserver reversal measurements. Compare the unitary quantum probability \(P(\mathrm{ok}_1, \mathrm{ok}_2) = 1/12\) against the decohered mixture (\(W = 0\), where \(P(\mathrm{ok}_1, \mathrm{ok}_2) = 1/4\)) and trace how the \(|0,0\rangle\) and \(|1,0\rangle\) amplitudes cancel destructively.
Interactive model · Idealized predictions
Add signed amplitudes
On a small screen, swipe the diagram to inspect its components. Moving marks illustrate the model; they are not measured trajectories.
Read the values as a table
Which assumptions make different observers’ certainty claims conflict? The text and worked examples remain available without JavaScript. Animation starts only when requested; Step gives a motion-free way to follow the model.
Worked Check
Problem: Suppose the two laboratories are prepared in \(|\Psi_{L_1 L_2}\rangle = (|0,0\rangle + |1,0\rangle + |1,1\rangle)/\sqrt{3}\), and each laboratory undergoes partial dephasing in the friend’s record basis so that off-diagonal elements between \(|0\rangle_{L_k}\) and \(|1\rangle_{L_k}\) are damped by \(W \in [0,1]\) independently in each lab. Find \(P(\mathrm{ok}_1, \mathrm{ok}_2)\) as a function of \(W\), and evaluate it at \(W = 1\) (ideal unitary labs) and \(W = 0\) (collapsed labs).
Solution:
1. Since \(|\mathrm{ok}\rangle\langle\mathrm{ok}| = \frac{1}{2}(I - X)\), independent dephasing \(X_{L_k} \mapsto W X_{L_k}\) replaces the projector expectation by
\[P(\mathrm{ok}_1, \mathrm{ok}_2) = \frac{1}{4}\Big\langle (I - W X_{L_1})\otimes (I - W X_{L_2})\Big\rangle_{\Psi}.\]
2. On \(|\Psi_{L_1 L_2}\rangle = (|0,0\rangle + |1,0\rangle + |1,1\rangle)/\sqrt{3}\), direct evaluation gives
Brukner’s and Frauchiger–Renner’s extended Wigner’s friend scenarios couple two entangled laboratories so four observables \(A_1, A_2, B_1, B_2\) cannot share a single joint probability distribution \(P(A_1,A_2,B_1,B_2)\).
No-go theorems prove that four assumptions—universal unitary validity (Q), observer-independent single outcomes (A), local setting independence (L), and consistency across agents (C)—cannot all hold simultaneously.
Laboratory demonstrations using single-photon or qubit ‘friends’ test two-qubit unitary interference and Bell violations, not conscious macroscopic observers.
Practice Problems
Problem 22.1: Verifying the zero-amplitude Hardy condition
Starting from the state \(|\Psi_{L_1 L_2}\rangle = (|0,0\rangle + |1,0\rangle + |1,1\rangle)/\sqrt{3}\) and \(|\mathrm{ok}\rangle = (|0\rangle - |1\rangle)/\sqrt{2}\), verify explicitly that \(\langle\mathrm{ok}_1, 0_{L_2}|\Psi_{L_1 L_2}\rangle = 0\) and \(\langle 1_{L_1}, \mathrm{ok}_2|\Psi_{L_1 L_2}\rangle = 0\).
Hint
Expand \(\langle\mathrm{ok}_1, 0_{L_2}| = (\langle 0,0| - \langle 1,0|)/\sqrt{2}\) and take the inner product with \(|\Psi_{L_1 L_2}\rangle\).
Problem 22.2: Superobserver marginal probability \(P(\mathrm{ok}_1)\)
Using \(|\Psi_{L_1 L_2}\rangle = \sqrt{\frac{2}{3}}\,|\mathrm{fail}\rangle_{L_1}|0\rangle_{L_2} + \frac{1}{\sqrt{3}}\,|1\rangle_{L_1}|1\rangle_{L_2}\), find the marginal probability \(P(\mathrm{ok}_1)\) that superobserver \(W_1\) obtains outcome \(\mathrm{ok}_1\) regardless of what \(W_2\) measures.
Hint
Use \(\langle\mathrm{ok}|\mathrm{fail}\rangle = 0\) and \(|\langle\mathrm{ok}|1\rangle|^2 = 1/2\).
Solution
Projecting onto \(|\mathrm{ok}\rangle_{L_1}\) annihilates the first term and leaves \(\frac{1}{\sqrt{3}}\langle\mathrm{ok}|1\rangle_{L_1}|1\rangle_{L_2} = -\frac{1}{\sqrt{6}}|1\rangle_{L_2}\), whose norm squared is \(P(\mathrm{ok}_1) = 1/6 = 2/12\) (which equals \(P(\mathrm{ok}_1, \mathrm{ok}_2) + P(\mathrm{ok}_1, \mathrm{fail}_2) = 1/12 + 1/12\)).
Problem 22.3: Why \(P(\mathrm{ok}_1, \mathrm{ok}_2) = 1/4\) when both friends’ records collapse
Suppose both friends’ measurements irreversibly collapse the state to the diagonal mixture \(\rho_{\mathrm{mix}} = \frac{1}{3}(|0,0\rangle\langle 0,0| + |1,0\rangle\langle 1,0| + |1,1\rangle\langle 1,1|)\). Show directly that \(\langle\mathrm{ok}, \mathrm{ok}|\rho_{\mathrm{mix}}|\mathrm{ok}, \mathrm{ok}\rangle = 1/4\).
Hint
For every computational basis state \(|a,b\rangle\), \(|\langle\mathrm{ok},\mathrm{ok}|a,b\rangle|^2 = (1/2)(1/2) = 1/4\).
Solution
Because \(|\langle\mathrm{ok},\mathrm{ok}|a,b\rangle|^2 = 1/4\) for every \((a,b) \in \{0,1\}^2\), any diagonal mixture \(\sum_{a,b} p_{ab}|a,b\rangle\langle a,b|\) yields \(P(\mathrm{ok}_1, \mathrm{ok}_2) = \frac{1}{4}\sum_{a,b} p_{ab} = 1/4\). The reduction from \(1/4\) down to \(1/12\) in the unitary case is a direct signature of destructive quantum interference between \(|0,0\rangle\) and \(|1,0\rangle\).
Key Takeaways
For the two-laboratory state \(|\Psi_{L_1 L_2}\rangle = (|0,0\rangle + |1,0\rangle + |1,1\rangle)/\sqrt{3}\), unitary quantum mechanics predicts \(P(\mathrm{ok}_1, \mathrm{ok}_2) = |\frac{1}{2\sqrt{3}}(1 - 0 - 1 + 1)|^2 = 1/12\).
The apparent paradox (\(P = 0\) vs.\ \(P = 1/12\)) arises from chaining counterfactual certainty statements across noncommuting measurement contexts (\(Z_{L_k}\) vs.\ \(X_{L_k}\)), which ignores the destructive interference between the \(|0,0\rangle\) and \(|1,0\rangle\) amplitudes.
Local-friendliness theorems show that universal unitarity, locality, and the existence of a single joint probability distribution over both internal and external outcomes cannot all hold simultaneously.