 |
| Figure 1: Alice and Bob (in boxes) with observers Ada and Bertrand |
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
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.
No comments:
Post a Comment