Can quantum theory consistently describe the use of itself?

Heads, tails, or something else?

In my previous post I described the Wigner-Deutsch thought experiment. The thought experiment showed that the friend was able to send a message to Wigner indicating that she had performed a measurement and observed a definite result.

Now once the experiment has been completed, including the reverse unitary transformation there described, what result will the friend predict that Wigner will see?

On a non-unitary interpretation (which involves an objective collapse and a single definite result), she will predict a 50% chance of seeing heads (since the definite result obtained in the experiment is now in superposition again). Whereas on a unitary interpretation, she will predict a 100% chance of seeing heads (since her lab which was in superposition during the experiment has now been restored to its original state).

A similar reasoning step occurs in the Frauchiger-Renner thought experiment where Alice (one of the friends), on account of seeing tails, concludes that Bertrand (one of the Wigners) will see \(\boldsymbol{|+\rangle}\).

But that is only true if her measurement is non-unitary. However if her lab is in superposition with respect to Bertrand (and Alice) then her lab will subsequently be reversed to its original state. As such, she can't conclude what Bertrand will see from her own measurement. Similarly, Ada can't conclude what Bob and Alice saw since, at the time of Ada's measurement, the friends' measurements will have been reversed.[1] Thus if Ada measures \(\boldsymbol{|-\rangle}\) she can only predict, per quantum theory, that Bertrand could also measure \(\boldsymbol{|-\rangle}\) with probability \(\frac{1}{12}\).

So, on the assumption of unitarity, there is no inconsistency.

How about on the assumption of non-unitarity? In this case, Ada can conclude that Bertrand will see \(\boldsymbol{|+\rangle}\). Again there is no inconsistency since quantum theory doesn't apply to this case - the state would have collapsed at an earlier time.

Inconsistency only arises when combining Ada's inferential reasoning (which assumes non-unitarity) and the prediction from quantum theory (which assumes unitarity). So the solution to the paradox is to recognize that measurements will be either unitary or non-unitary, but not both at the same time (i.e., depending on whether one is inside or outside the lab).

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[1] As an analogy, consider a qubit that has been been measured as \(|0\rangle\) in the \(\{|0\rangle,|1\rangle\}\) basis. Suppose the qubit is then subsequently measured as \(|+\rangle\) in the \(\{|+\rangle,|-\rangle\}\) basis. If the qubit is again measured in the \(\{|0\rangle,|1\rangle\}\) basis, there is a 50% chance that the result will be \(|1\rangle\). That is, the original measurement result has been lost and can no longer be used in our reasoning.

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

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[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)

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[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.

The GHZ Experiment

A GHZ gate

Consider a coin that can represent the behaviour of a quantum particle. This quantum coin [1] can be placed into an equal superposition of heads and tails, which is represented in Dirac (Ket) notation as follows:

\[(1) \hspace{5 mm}|\psi\rangle = \frac{1}{\sqrt2}(|H\rangle + |T\rangle)\]

\(|\psi\rangle\) (read: psi) represents the state of the quantum coin. \(|H\rangle\) represents the head component and \(|T\rangle\) represents the tails component. 1/√2 is the probability amplitude for each component of the superposition. Squaring this number gives the probability that a particular component will be the result when measured. That is, there is a (1/√2)² = 0.5 (or 50%) probability that heads will be measured. Similarly, in this case, for tails.

Quantum coins can also be entangled. The following state, which represents two entangled quantum coins, is called a Bell state (see also the quantum circuit):

\[(2) \hspace{5 mm}|\psi\rangle = \frac{1}{\sqrt2}(|HT\rangle - |TH\rangle)\]

When the coins are entangled, they are not in independent superpositions, but in a collective superposition. Which is to say, if one of the coins is measured as heads then the other coin, when measured, will be tails.

Quantum coins can be measured from any angle (and are still measured as only heads or tails - there are no 'side of coin' measurements, so-to-speak). To rotate the state in equation (2) from the default z-axis onto the x-axis, a Hadamard gate is applied to each coin (note: a Hadamard gate is conventionally symbolized as \(\boldsymbol H\), not to be confused with the ket \(|H\rangle\) representing heads in this post) . The Hadamard gate rotates the \(|H\rangle\) component to \(\frac{1}{\sqrt2}(|H\rangle + |T\rangle)\) and the \(|T\rangle\) component to \(\frac{1}{\sqrt2}(|H\rangle - |T\rangle)\).[2] This results in the following state (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H)}(|HT\rangle - |TH\rangle) \\ &= (|H\rangle + |T\rangle)(|H\rangle - |T\rangle) - ((|H\rangle - |T\rangle)(|H\rangle + |T\rangle) \\ &= |HH\rangle - |HT\rangle + |TH\rangle - |TT\rangle - (|HH\rangle + |HT\rangle - |TH\rangle - |TT\rangle) \\ &= |HH\rangle - |HT\rangle + |TH\rangle - |TT\rangle - |HH\rangle - |HT\rangle + |TH\rangle + |TT\rangle \\ &= -(|HT\rangle - |TH\rangle)\end{aligned}\]

This is the same Bell state as in equation (2) except that a global phase (the minus sign, which is equivalent to π radians, or a 180° rotation around the axis origin) has been added. This global phase makes no difference to the calculated probabilities. A similar result occurs if the state is rotated onto the y-axis, or onto any arbitrary axis. In other words, that Bell state (called the singlet state) is rotationally invariant, ignoring the global phase which has no physical meaning. The rotation onto the y-axis uses the Phase gate (conventionally symbolized as \(\boldsymbol S\)) and Hadamard gate (conventionally symbolized as \(\boldsymbol H\), as noted above) to rotate the \(|H\rangle\) component to \(\frac{1}{\sqrt2}(|H\rangle + |T\rangle)\) and the \(|T\rangle\) component to \(\frac{i}{\sqrt2}(|H\rangle - |T\rangle)\).[3] This results in the following state (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H)(S \otimes S)}(|HT\rangle-|TH\rangle) \\ &= (|H\rangle + |T\rangle)(i|H\rangle - i|T\rangle) - ((i|H\rangle - i|T\rangle)(|H\rangle + |T\rangle) \\ &= i|HH\rangle - i|HT\rangle + i|TH\rangle + -i|TT\rangle - (i|HH\rangle + i|HT\rangle + -i|TH\rangle + -i|TT\rangle) \\ &= i|HH\rangle - i|HT\rangle + i|TH\rangle + -i|TT\rangle - i|HH\rangle - i|HT\rangle + i|TH\rangle + i|TT\rangle \\ &= -i(|HT\rangle - |TH\rangle)\end{aligned}\]

Again, this is the same Bell state as in equation (2) except that a global phase (the imaginary −i, which is equivalent to 3*π/2 radians, or a 270° rotation around the axis origin) has been added.

Now how does the entanglement work? For example, if Alice and Bob take one of two quantum coins entangled as per equation (2) to distant locations, and measure their coin on a predetermined axis (say, the y-axis), how do they end up being measured in opposite orientations? This is the question that the EPR paradox raises. Einstein's answer was that there must be hidden "elements of reality" corresponding to the coin's orientation on any given axis. I discuss Bell's Theorem which demonstrates the constraints on that position - namely, that hidden variables can't be used to explain the measurements without also violating locality (i.e., cause/effect not greater than the speed of light). In this post, I want to outline an intuitive demonstration of those constraints via the GHZ experiment.

To begin, a three qubit entangled state is created, termed the GHZ state, as follows:

Figure 1: The GHZ state

In terms of our quantum coins, this can be represented as:

\[(3) \hspace{5 mm}|\psi\rangle = \frac{1}{\sqrt2}(|HHH\rangle + |TTT\rangle)\]

If Alice, Bob and Charlie each take one of the three quantum coins, measure it, and then compare results, they will find that they have all measured heads, or have all have measured tails.

Figure 2: xxx basis

Now suppose that Alice, Bob and Charlie instead rotate their respective coins onto the x-axis.[4] The resulting state is:

\(\begin{aligned}(4) \hspace{5 mm}|\psi\rangle = \frac{1}{2}(&|HHH\rangle + |HTT\rangle + \\ & |THT\rangle + |TTH\rangle)\end{aligned}\)

If we assign heads the value of 1 and tails the value of -1 then, for each component, the product of the three coin measurements will be 1. For example, \(|HTT\rangle\) gives 1 * -1 * -1 = 1. Similarly for the other components. This is a mathematical representation of the fact that for each of the four measurement possibilities, there will be an odd number of heads measured (1 or 3) and an even number of tails measured (0 or 2).

Figure 3: yyx basis

Now suppose that only Alice rotates her coin onto the x-axis, and instead Bob and Charlie rotate their coins onto the y-axis.[5] The resulting state is:

\(\begin{aligned}(5) \hspace{5 mm}|\psi\rangle = \frac{1}{2}(&|HHT\rangle + |HTH\rangle + \\ & |THH\rangle + |TTT\rangle)\end{aligned}\)

In this case, if we assign heads the value of 1 and tails the value of -1 then, for each component, the product of the three coin measurements will be -1. For example, \(|HHT\rangle\) gives 1 * 1 * -1 = -1. Similarly for the other components. This is a mathematical representation of the fact that for each of the four measurement possibilities, there will be an even number of heads measured (0 or 2) and an odd number of tails measured (1 or 3). Note that this is the opposite of the possible outcomes for state (4).

Symmetrically, if it is only Bob or Charlie, rather than Alice that rotates their coin onto the x-axis, while the others instead rotate their coins onto the y-axis, the resulting state is the same as equation (5). We can now list the product equations for each of the quantum states:

\((6) \hspace{5 mm}Alice_x * Bob_x * Charlie_x = 1 \hspace{5 mm}[from (4)]\)

\((7) \hspace{5 mm}Alice_x * Bob_y * Charlie_y = -1 \hspace{5 mm}[from (5)]\)

\((8) \hspace{5 mm}Alice_y * Bob_x * Charlie_y = -1 \hspace{5 mm}[from (5)]\)

\((9) \hspace{5 mm}Alice_y * Bob_y * Charlie_x = -1 \hspace{5 mm}[from (5)]\)

Now the product of the left-sides of the above equations is:

\(\begin{aligned}LHS = & Alice_x * Bob_x * Charlie_x * \\ & Alice_x * Bob_y * Charlie_y * \\ & Alice_y * Bob_x * Charlie_y * \\ & Alice_y * Bob_y * Charlie_x\end{aligned}\)

As each element appears twice, the equation can be rearranged as:

\(LHS = {Alice_x}^2 * {Bob_x}^2 * {Charlie_x}^2 * {Alice_y}^2 * {Bob_y}^2 * {Charlie_y}^2\)

Since LHS is the product of squares, LHS = 1. But the product of the RHS is:

\(RHS = 1 * -1 * -1 * -1 = -1\)

Thus the equations can't be jointly satisfied. This proves that there can't be hidden elements of reality (i.e., unknown predefined values) for each possible measurement combination since that would lead to contradiction. That is, quantum mechanics experimentally rules out local hidden variable theories.[6]

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[1] Our quantum coin is a qubit, where \(|0\rangle\) and \(|1\rangle\) have been replaced by \(|H\rangle\) and \(|T\rangle\).

[2] See the Bloch Sphere in Figure 5 of this earlier blog post to visualize the equivalent qubit rotation.

[3] Derivation for the matrix product of \(\boldsymbol H\) and \(\boldsymbol S\):

\[\hspace{10 mm}\boldsymbol{HS} = \begin{bmatrix} 1 & 1 \\ 1 & −1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & i \end{bmatrix} = \begin{bmatrix} 1*1 + 0*1 & 0*1 + i*1 \\ 1*1 + 0*−1 & 0*1 + i*−1 \end{bmatrix} = \begin{bmatrix} 1 & i \\ 1 & −i \end{bmatrix}\]

[4] Derivation for Alice, Bob, and Charles rotating their coins from the z-axis onto the x-axis (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H \otimes H)}(|HHH\rangle + |TTT\rangle) \\ & = (|H\rangle + |T\rangle)(|H\rangle + |T\rangle))(|H\rangle + |T\rangle) + ((|H\rangle - |T\rangle)(|H\rangle - |T\rangle)(|H\rangle - |T\rangle)) \\ & = |HHH\rangle + |HHT\rangle + |HTH\rangle + |HTT\rangle + |THH\rangle + |THT\rangle + |TTH\rangle + |TTT\rangle + \\ & \hspace{5 mm} |HHH\rangle - |HHT\rangle - |HTH\rangle + |HTT\rangle - |THH\rangle + |THT\rangle + |TTH\rangle - |TTT\rangle \\ & = |HHH\rangle + |HTT\rangle + |THT\rangle + |TTH\rangle\end{aligned}\]

[5] Derivation for Alice, Bob, and Charles rotating their coins from the z-axis onto the y-axis, y-axis, and x-axis respectively (excluding the \(\frac{1}{\sqrt2}\)):

\[\begin{aligned}\boldsymbol{(H }&\boldsymbol{\otimes H \otimes H)(S \otimes S\otimes I)}(|HHH\rangle + |TTT\rangle) \\ & = (|H\rangle + |T\rangle)(|H\rangle + |T\rangle))(|H\rangle + |T\rangle) + \\ & \hspace{5 mm} ((i|H\rangle + -i|T\rangle)(i|H\rangle + -i|T\rangle)(|H\rangle - |T\rangle)) \\ & = |HHH\rangle + |HHT\rangle + |HTH\rangle + |HTT\rangle + \\ & \hspace{5 mm}|THH\rangle + |THT\rangle + |TTH\rangle + |TTT\rangle + \\ & \hspace{5 mm} ii|HHH\rangle + -ii|HHT\rangle + -ii|HTH\rangle + --ii|HTT\rangle + \\ & \hspace{5 mm} -ii|THH\rangle + --ii|THT\rangle + --ii|TTH\rangle + ---ii|TTT\rangle \\ & = |HHH\rangle + |HHT\rangle + |HTH\rangle + |HTT\rangle + \\ & \hspace{5 mm} |THH\rangle + |THT\rangle + |TTH\rangle + |TTT\rangle + \\ & \hspace{5 mm} -|HHH\rangle + |HHT\rangle + |HTH\rangle + -|HTT\rangle + \\ & \hspace{5 mm} |THH\rangle + -|THT\rangle + -|TTH\rangle + |TTT\rangle \\ & = |HHT\rangle + |HTH\rangle + |THH\rangle + |TTT\rangle\end{aligned}\]

[6] Note that quantum mechanics rules out predefined measurement values for non-local hidden variable theories as well. However, if a theory is non-local then, once a measurement has been made for one coin, non-local variables can potentially be updated instantaneously thus potentially affecting subsequent measurements of the other coins.

The Delayed-Choice Quantum Eraser

Figure 1: The delayed-choice quantum eraser experiment

Suppose a series of photons are sent through a double-slit apparatus. As you may know, an interference pattern is formed on the back screen. Each photon's trajectory can be represented as a wave that passes through both slits and is finally absorbed at a particular position on the back screen (with a specific probability).

Adding a splitting crystal (BBO in Figure 1) immediately after the slits converts the photon into two entangled photons, each with half the energy of the original photon. One photon (called the signal photon) goes to the back screen (D0) which, after a series of such experiments, builds up a blob pattern that does not exhibit interference. Why not? Because the wave function depends on the location where the photon pairs were created (as represented by the green line apex at the upper slit or the red line apex at the lower slit) and thus on which slit the original photon went through. This constitutes "which-way" information which destroys interference.

So far so good. Now suppose the second photon (called the idler photon) is sent on a long journey - a much longer journey than the signal photon took to get to the back screen - and then, optionally, a beam splitter is placed in its path. This is the delayed-choice aspect of the experiment.

If the beam splitter is present (as in Figure 1), then the idler photon hits one side of the beam splitter according to which slit the original photon went through (i.e., the red path comes from the lower slit and hits the lower side of the beam splitter, the green path comes from the upper slit and hits the upper side of the beam splitter). Analogous to the double-slit interaction, this beam splitter interaction can be represented as a wave that is reflected by and also passes through the beam splitter. Subsequently detecting the photon on the lower side of the beam splitter (D1) or the upper side (D2) will tell you nothing about which side of the beam splitter the idler photon came from, and thus nothing about which slit the original photon went through. That is, the "which-way" information has been lost. This is the eraser aspect of the experiment.

Given that the "which-way" information is erased when the beam splitter is present, what kind of pattern do you predict will be seen on the back screen (i.e., by the signal photons at D0)?

To see the answer, go back to the second paragraph of this post. What pattern was exhibited on the back screen then? A blob pattern that did not exhibit interference, and the same answer remains true here. The rest of the setup with the idler photon makes no difference at all. There is no mysterious "backwards-in-time" effect that changes the pattern (as is sometimes suggested), since the pattern is the same regardless of what subsequently occurred elsewhere.

Figure 2: D0 = D1 + D2 (with the beam splitter)

Now that is not quite the end of the story. It is possible to match up the idler photon detected at one of the final detectors (D1 or D2) with its entangled signal photon partner on the back screen (using a mechanism called coincidence counting). If the beam splitter is in place and the signal photons are later highlighted according to which detector the idler photon partner was detected at, then an interference pattern is revealed for each highlighted group (see Figure 2).

So does this imply a "backwards in time" effect, albeit hidden? No, the two interference patterns are always encoded in the total signal photon pattern and can even be used to predict the probability of measuring the idler photon at D1 or D2. For example, suppose a signal photon was detected at D0 at the location indicated by the green bar in Figure 2. Note that there is a peak for D1 and a trough for D2. This predicts that the idler photon will, with near

Figure 3: D0 = D1 + D2 (without the beam splitter)

certainty, be detected at D1. The correlation is due simply to the entanglement between the signal and idler photons which the measurement after the beam splitter verifies.

If the beam splitter were not present, then D1 and D2 would instead detect which slit the original photon passed through (i.e., the "which-way" information). So if the signal photons were highlighted according to which detector the idler photon partner was detected at, then a non-interference pattern would be revealed for each highlighted group (see Figure 3).

References:

A Delayed Choice Quantum Eraser - experiment by Kim et al., in 1999.

The Notorious Delayed-Choice Quantum Eraser - blog post by Sean Carroll

The delayed choice quantum eraser, debunked - blog post by Sabine Hossenfelder

Delayed Choice Quantum Eraser - slides by Anita Kulkarni

Delayed-choice quantum eraser - Wikipedia