The Wigner-Deutsch thought experiment

Figure 1: Wigner's lab in superposition

In my previous post I introduced the Frauchiger-Renner thought experiment. Before discussing the mistake in Ada's reasoning, I'd like to introduce three earlier thought experiments that provide some necessary background for the discussion.

The first is the Schrödinger's Cat thought experiment posed by Erwin Schrödinger in 1935. It raises the question, "When does a quantum system stop existing as a superposition of states and become one or the other?" According to quantum mechanics, large-scale superpositions can potentially exist such as the superposition of a live cat and a dead cat. That state can be described as follows:

\[|cat\rangle = \frac{|alive\rangle + |dead\rangle}{\sqrt2}\]

In the thought experiment, the cat is in a box that is isolated from the rest of the world. When an observer opens the box, the cat's state reduces (collapses) to one of the component states. That is, the cat is either observed to be alive or observed to be dead, but not both at the same time.

The Wigner's friend thought experiment, posed by physicist Eugene Wigner in 1961, extends this idea by considering the possibility of the observer also being in superposition. In this case, consider a quantum coin that is initially heads and then flipped to be in a superposition of heads and tails. Wigner's friend is a scientist in an isolated laboratory who is tasked with observing the quantum coin. Prior to her measurement, the quantum state is:

\[\frac{|heads\rangle + |tails\rangle}{\sqrt2}|friend\rangle\]

When the friend observes the coin, the state becomes:

\[\frac{|heads\rangle|friend_{heads}\rangle + |tails\rangle|friend_{tails}\rangle}{\sqrt2}\]

From the friend's point-of-view, the coin's state has collapsed to either heads or tails (since she observed either heads or tails). However from Wigner's point-of-view, the friend is in a superposition of having observed heads and having observed tails! [1] When Wigner enters the lab ten minutes later, the coin/friend superposition will collapse into one of the component states. Wigner's friend will report to Wigner that the coin is either heads or tails and, furthermore, that it collapsed to that state ten minutes prior when she observed it. This creates a paradox regarding when the state actually collapsed. Wigner's position was that the friend collapsed the wavefunction, and thus the postulates of quantum mechanics needed to be revised. An alternative position is that collapse never occurs and thus even Wigner's observation is part of a wider superposition.

This brings us to the Wigner-Deutsch thought experiment, posed by David Deutsch in 1985, providing a potential test for the above theories. In Deutsch's extended thought experiment, the friend writes a note stating that she has observed the coin's orientation, but without stating what that orientation is. Prior to the friend's measurement, the quantum state is:

\[\frac{|heads\rangle + |tails\rangle}{\sqrt2}|friend\rangle|not\ noted\rangle\]

When the friend observes the coin, the state becomes:

\[\frac{|heads\rangle|friend_{heads}\rangle + |tails\rangle|friend_{tails}\rangle}{\sqrt2}|not\ noted\rangle\]

The friend then notes that she has observed the coin's orientation (but doesn't specify what it is):

\[\frac{|heads\rangle|friend_{heads}\rangle|noted\rangle + |tails\rangle|friend_{tails}\rangle|noted\rangle}{\sqrt2}\]

\[= \frac{|heads\rangle|friend_{heads}\rangle + |tails\rangle|friend_{tails}\rangle}{\sqrt2}|noted\rangle\]

Now Wigner applies a reverse unitary transformation that undoes the friend's measurement and restores the initial state of the coin, resulting in:

\[|heads\rangle|friend\rangle|noted\rangle\]

If quantum mechanics is universally applicable (i.e., unitary without objective collapse), then Wigner and his friend will observe the coin to be heads (the initial state prior to observation) with certainty. While the friend will have no memory of what her earlier coin observation was, she will have the record of the note indicating that she did in fact observe the coin's orientation.

On the other hand, if the friend's earlier observation of the coin irreversibly collapsed the coin's state, then the reverse unitary transformation will place the coin into superposition again. When Wigner and the friend observe the coin together, there will be a 50% chance of observing tails.

Thus there is an (in principle) observable distinction between objective collapse theories and unitary quantum mechanics.

This finally brings us back to the Frauchiger-Renner thought experiment. It is similar to the Wigner-Deutsch thought experiment in that Alice sends a message outside her isolated laboratory, in this case to Bob. However, in this case, it encodes partial information about the results of her experiment.

I'll leave that discussion for another post.

References:

Quantum theory as a universal physical theory - David Deutsch, p32

On the quantum measurement problem - Caslav Brukner, p15

A no-go theorem for observer-independent facts - Caslav Brukner, p2

Experimental test of local observer-independence - Proietti, et al.

Testing quantum theory with thought experiments - Nurgalieva, Renner, p17

--

[1] The friend's observation can be illustrated using qubits. The scenario prior to the friend's observation, where \(\boldsymbol{|0\rangle\ = |heads\rangle}\) and \(\boldsymbol{|1\rangle\ = |tails\rangle}\) for the first qubit (representing the coin) and \(\boldsymbol{|0\rangle\ = |friend_{heads}\rangle}\) and \(\boldsymbol{|1\rangle\ = |friend_{tails}\rangle}\) for the second qubit (representing the friend), is:

\[\frac{|0\rangle + |1\rangle}{\sqrt2}|0\rangle \]

Next, the friend observes the coin. The observation process is represented by a CNOT gate. In effect, it says that if the coin is heads, then leave the friend in the initial (observing heads) state. Otherwise, if the coin is tails, flip the friend's state (from observing heads to observing tails).

\[\begin{aligned}CNOT \frac{|0\rangle + |1\rangle}{\sqrt2}|0\rangle &= CNOT \frac{|00\rangle + |10\rangle}{\sqrt2} \\ &= \frac{|00\rangle + |11\rangle}{\sqrt2}\end{aligned} \]

This final state (called a Bell state) represents the entangled friend and coin.

[2] A circuit diagram for the Wigner-Deutsch thought experiment. The coin and friend qubit states are described in [1] above. \(\boldsymbol{|0\rangle\ = |not\ noted\rangle}\) and \(\boldsymbol{|1\rangle\ = |noted\rangle}\) for the third qubit (representing the note indicating whether the coin has been observed by the friend or not).

Figure 2: Wigner-Deutsch thought experiment

Section 1: The Hadamard gate represents the coin flip, placing the coin into superposition. The CNOT gate represents the friend's observation.

Section 2: The first three gates (NOT, CNOT, NOT, effectively a CNOT gate that is triggered on \(|0\rangle\) instead of \(|1\rangle\)) represents the condition that if the friend observed heads, i.e., \(|0\rangle\), then the note state is flipped (from not noted to noted, i.e., \(|0\rangle\) to \(|1\rangle\)). The fourth gate (CNOT) represents the condition that if the friend observed tails, i.e., \(|1\rangle\), then the note state is flipped (from not noted to noted, i.e., \(|0\rangle\) to \(|1\rangle\)).

Section 3: Both the Hadamard and CNOT gates are their own inverses. So applying them again reverses the operations in section 1.

The circles with radii represent the states of the qubits at each stage. Initially the state is \(|000\rangle\), then \(\frac{|00\rangle + |11\rangle}{\sqrt2}|0\rangle\) after section 1, then \(\frac{|00\rangle + |11\rangle}{\sqrt2}|1\rangle\) after section 2, then \(|001\rangle\) after section 3. (Be aware that the circuit diagram composer uses little endian notation, so describes the final state as \(|100\rangle\)).

Consider that the note is separable from the entangled coin/friend pair at the end of section 2. If only one of the CNOT gates had been applied in section 2, then the note would also be entangled. Ultimately the four section 2 gates are equivalent to a single NOT gate on the note qubit, but it's useful to visualize the independent contributions from the friend paths.

The Frauchiger-Renner thought experiment

Figure 1: Alice and Bob (in boxes) with observers Ada and Bertrand

The Frauchiger-Renner thought experiment combines aspects of the Schrödinger's Cat thought experiment, the Wigner's Friend thought experiment, Hardy's Paradox and Bell's Theorem. The reflective reasoning of agents in the experiment, using the rules of quantum mechanics, seem to lead to contradiction. Hence the title of the FR paper, "Quantum theory cannot consistently describe the use of itself" (Nature Communications, 2018).

Let's go through the thought experiment and see how the argument goes.

As illustrated in Figure 1, Alice and Bob each have the role of a Wigner's friend conducting an experiment in an isolated lab (visually indicated by the boxes). The two observer's have the role of Wigner for each of the friends, i.e., as Alice's observer (named Ada) and Bob's observer (named Bertrand) respectively.

To begin, Alice tosses a quantum coin that is twice as likely to land tails as heads. The coin is described by the following quantum state:

\[(1) \hspace{5 mm}|coin\rangle = \frac{1}{\sqrt3}|heads\rangle + \frac{2}{\sqrt3}|tails\rangle\]

Per the Born rule, squaring the coefficients gives the probabilities for each outcome, in this case 1/3 heads and 2/3 tails, respectively.

If Alice observes heads, she prepares the spin state of a particle as spin-down. Otherwise, if tails, she prepares the spin state of the particle as spin-right (i.e., a superposition of spin-down and spin-up). This preparation \((\boldsymbol{U})\) is a unitary process, described as follows: [1]

\[\begin{aligned}(2) \hspace{5 mm}|coin \& particle\rangle & = \boldsymbol {U}|coin\rangle|down\rangle \\ &= \frac{1}{\sqrt3}|heads\rangle|down\rangle + \frac{2}{\sqrt3}|tails\rangle(\frac{1}{\sqrt2}(|down\rangle + |up\rangle)) \\ &= \frac{1}{\sqrt3}|heads\rangle|down\rangle + \frac{1}{\sqrt3}|tails\rangle|down\rangle + \frac{1}{\sqrt3}|tails\rangle|up\rangle \\ &= \frac{|tails\rangle|down\rangle + |tails\rangle|up\rangle + |heads\rangle|down\rangle}{\sqrt3} \end{aligned}\]

The resulting entangled state is called a Hardy state. Its qubit representation, where \(\boldsymbol{|0\rangle\ = |tails\rangle}\) and \(\boldsymbol{|1\rangle\ = |heads\rangle}\) for the first qubit (representing the coin) and \(\boldsymbol{|0\rangle\ = |down\rangle}\) and \(\boldsymbol{|1\rangle\ = |up\rangle}\) for the second qubit (representing the particle), is:

\[(CC) \hspace{5 mm}|coin \& particle\rangle = \frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt3} \]

The qubits above are expressed in the computational \(\{|0\rangle,|1\rangle\}\) basis. The qubits can also be expressed in alternative bases where C and S indicate the computational \(\{|0\rangle,|1\rangle\}\) and superposition \(\{|+\rangle,|-\rangle\}\) bases respectively. [2]

\[(SC) \hspace{5 mm}|coin\pm \& particle\rangle = \frac{2|{+0}\rangle + |{+1}\rangle + |{-1}\rangle}{\sqrt6} \]

\[(CS) \hspace{5 mm}|coin \& particle\pm\rangle = \frac{2|{0+}\rangle + |{1+}\rangle + |{1-}\rangle}{\sqrt6} \]

\[(SS) \hspace{5 mm}|coin\pm \& particle\pm\rangle = \frac{3|{++}\rangle + |{+-}\rangle + |{-+}\rangle - |{--}\rangle}{\sqrt{12}} \]

Alice now sends the particle (the second qubit) to Bob. The particle is, in effect, a message entangled with and containing partial information about Alice's coin toss (the first qubit).

Bob measures the particle, per (CC). Ada (Alice's observer) then measures Alice's system (with the coin) in the superposition basis, per (SC). Bertrand (Bob's observer) then measures Bob's system (with the particle) in the superposition basis, per (CS).

The experiment has been completed and it's now time to analyse the deductions that each person can make, given that they know both quantum theory and how the experiment was conducted (though not necessarily the measured results).

Suppose Alice saw tails, i.e., \(|0\rangle\), per (CC). That reduces the state to \(\frac{|00\rangle + |01\rangle}{\sqrt2}\), so she concludes that Bertrand will see \(|+\rangle\), since \(\frac{|0\rangle + |1\rangle}{\sqrt2} = |+\rangle\). [3]

Suppose Bob saw spin-up, i.e., \(|1\rangle\), per (CC). That reduces the state to \(|01\rangle\), so he concludes that Alice saw tails, i.e., \(|0\rangle\). As with Alice, he will also conclude that Bertrand will see \(|+\rangle\).

Suppose Ada sees \(|-\rangle\), per (SC). That reduces the state to \(|-1\rangle\), so she concludes that Bob saw spin-up, i.e., \(|1\rangle\). As with Bob, she will conclude that Alice saw tails, i.e., \(|0\rangle\). As with Bob and Alice, she will also conclude that Bertrand will see \(|+\rangle\).

Ada now tells Bertrand that since she saw \(|-\rangle\) and reasoned as above, that he will therefore see \(|+\rangle\).

However Ada and Bertrand measure Alice and Bob's labs per (SS). There is a one in twelve chance that they both will measure \(|-\rangle\), contrary to Ada's reasoning.

So that's the paradoxical conclusion - Ada has apparently used quantum theory to derive a contradiction. What is the mistake in Ada's reasoning? I'll leave a discussion of that question for another post.

References:

Quantum theory cannot consistently describe the use of itself - Frauchiger and Renner

New Quantum Paradox Clarifies Where Our Views of Reality Go Wrong - Anil Ananthaswamy

Beyond Schrödinger's Cat - Renato Renner

It’s hard to think when someone Hadamards your brain - Scott Aaronson

Watching the Watchmen: Demystifying the Frauchiger-Renner Experiment - George Musser

If your interpretation of quantum mechanics has a single world but no collapse, you have a problem - Mateus Araújo (and further comment)

--

[1] Specifically, an anti-controlled Hadamard two-qubit gate. It performs a Hadamard transformation on the second qubit if the first qubit is zero (as opposed to one for a controlled Hadamard gate), otherwise it leaves the second qubit unchanged.

[2] This footnote shows the working to express the Hardy state in superposition (Hadamard) bases. This is achieved by substituting \(\frac{|+\rangle + |-\rangle}{\sqrt2}\) for \(|0\rangle\) and \(\frac{|+\rangle - |-\rangle}{\sqrt2}\) for \(|1\rangle\) for the first qubit only (SC), the second qubit only (CS), and both qubits (SS), and then simplifying. (For reference, in the FR paper, \(|fail\rangle\ = |+\rangle\) and \(|ok\rangle\ = |-\rangle\).)

\[\small{(CC) \hspace{5 mm}|coin\&particle\rangle = \frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt3}}\]

\[\small{\begin{aligned}(SC) \hspace{5 mm}|coin{\pm}\&particle\rangle &= \frac{(|+\rangle + |-\rangle)|0\rangle + (|+\rangle + |-\rangle)|1\rangle + (|+\rangle - |-\rangle)|0\rangle}{\sqrt3.\sqrt2} \\ &= \frac{|{+0}\rangle + |{-0}\rangle + |{+1}\rangle + |{-1}\rangle + |{+0}\rangle - |{-0}\rangle}{\sqrt6} \\ &= \frac{2|{+0}\rangle + |{+1}\rangle + |{-1}\rangle}{\sqrt6} \end{aligned}}\]

\[\small{\begin{aligned}(CS) \hspace{5 mm}|coin\&particle{\pm}\rangle &= \frac{|0\rangle(|+\rangle + |-\rangle) + |0\rangle(|+\rangle - |-\rangle) + |1\rangle(|+\rangle + |-\rangle)}{\sqrt3.\sqrt2} \\ &= \frac{|{0+}\rangle + |{0-}\rangle + |{0+}\rangle - |{0-}\rangle + |{1+}\rangle + |{1-}\rangle}{\sqrt6} \\ &= \frac{2|{0+}\rangle + |{1+}\rangle + |{1-}\rangle}{\sqrt6} \end{aligned}}\]

\[\small{\begin{aligned}&(SS) \hspace{5 mm}|coin{\pm}\&particle{\pm}\rangle \\ &= \frac{(|+\rangle + |-\rangle)(|+\rangle + |-\rangle) + (|+\rangle + |-\rangle)(|+\rangle - |-\rangle) + (|+\rangle - |-\rangle)(|+\rangle + |-\rangle)}{\sqrt3.\sqrt2.\sqrt2} \\ &= \small{\frac{|{++}\rangle + |{+-}\rangle + |{-+}\rangle + |{--}\rangle + |{++}\rangle - |{+-}\rangle + |{-+}\rangle - |{--}\rangle + |{++}\rangle + |{+-}\rangle - |{-+}\rangle - |{--}\rangle}{\sqrt{12}}} \\ &= \frac{3|{++}\rangle + |{+-}\rangle + |{-+}\rangle - |{--}\rangle}{\sqrt{12}} \end{aligned}}\]

[3] See Hadamard operations.