CHSH

 It's back to Alice and Bob, this time with the CHSH game. The game is 

presented as a non-local game wherein two accomplices are trying for an advantage

without being able to communicate with each other. More fundamentally, this

game is not a game at all but a Bell State test, showing that quantum computing

has an advantage over classical computing.

The set-up: Alice and Bob can agree on a strategy beforehand, but cannot communicate

during the game. Each is handed a bit of information, 0 or 1 and outputs a 0 or 1

in return. The winning combinations for the pair are as follows, the XOR truth

table. A quick glance at the table shows they can win 75% of the time by agreeing

to always output 0.                                                         

 

By sharing an ebit, they can up their winning chances to 85.4%.

Checked in with Meta on this, to try to get an overview on how this might be done.

Quantum Computing, the CHSH game; what is the winning quantum strategy for Alice and Bob and why does it involve using different measurement basis?

Quantum Strategy for the CHSH Game

In the CHSH (Clauser-Horne-Shimony-Holt) game, Alice and Bob can use quantum entanglement to achieve a winning probability of cos^2(π/8) ≈ 0.854, which is higher than the classical limit of 3/4.

Quantum Strategy:

Alice and Bob share an entangled state: |Φ+ = (|00+ |11)/√2

Alice receives input x ∈ {0, 1} and Bob receives input y ∈ {0, 1}

Alice measures her qubit in the basis {∣∣+x, ∣∣−x}, where ∣∣+x = (|0+ e^(iπx/2)|1)/√2 and ∣∣−x = (|0- e^(iπx/2)|1)/√2

Bob measures his qubit in the basis {∣∣+y, ∣∣−y}, where ∣∣+y = (|0+ e^(iπy/2)|1)/√2 and ∣∣−y = (|0- e^(iπy/2)|1)/√2

Alice and Bob respond with their measurement outcomes a, b ∈ {0, 1}

Why different measurement bases?

Using different measurement bases allows Alice and Bob to exploit the entanglement and correlations in the shared state. By choosing bases that depend on their inputs, they can create a correlation between their outcomes that is stronger than what's possible classically.

Specifically, the quantum strategy works because:

When x = y, Alice and Bob's measurement bases are aligned, and their outcomes are correlated.

When x ≠ y, their measurement bases are not aligned, but the entanglement still provides some correlation.

This combination of alignment and entanglement allows Alice and Bob to achieve a higher winning probability than what's possible with classical strategies.

Note: This is a simplified explanation of the quantum strategy for the CHSH game. If you'd like more details or want to explore further, feel free to ask!

But all they are given is 0 or 1, how can they output in terms of+i and -i

Excellent follow-up question!

In the quantum strategy for the CHSH game, Alice and Bob don't directly output +i and -i. Instead, they output 0 or 1, which corresponds to the outcomes of their measurements.

The +i and -i come into play in the way they perform their measurements. Specifically:

When Alice measures her qubit in the basis {∣∣+x, ∣∣−x}, she's effectively asking the question: "Is my qubit in the state ∣∣+x or ∣∣−x?"

When Bob measures his qubit in the basis {∣∣+y, ∣∣−y}, he's asking: "Is my qubit in the state ∣∣+y or ∣∣−y?"

The states ∣∣+x and ∣∣+y are defined as:

**∣∣+x = ( 0+ e^(iπx/2) 1)/√2**

Similarly, ∣∣−x and ∣∣−y are defined as:

**∣∣−x = ( 0- e^(iπx/2) 1)/√2**

Notice the e^(iπx/2) and e^(iπy/2) terms? These are where the +i and -i come in. When Alice and Bob perform their measurements, they're effectively projecting their qubits onto these states, which involve complex phases.

The measurement outcomes 0 or 1 correspond to the probabilities of finding the qubit in the ∣∣+x or ∣∣−x state (for Alice) or ∣∣+y or ∣∣−y state (for Bob). These outcomes are then used to determine their responses to the game.

So, while Alice and Bob don't directly output +i and -i, the complex phases play a crucial role in their measurement bases and the correlations they exploit to win the game.

How are ∣+x or ∣∣−x state (for Alice) or ∣∣+y or ∣∣−y state (for Bob) different from each other?

The states ∣∣+x, ∣∣−x, ∣∣+y, and ∣∣−y are different in their relative phases, which are determined by the inputs x and y.

To see this, let's rewrite the states using Euler's formula:

**∣∣+x = ( 0+ e^(iπx/2) 1)/√2 = ( 0+ i^x 1)/√2**

**∣∣+y = ( 0+ e^(iπy/2) 1)/√2 = ( 0+ i^y 1)/√2**

Now, notice the following:

When x = 0, ∣∣+0 = (|0+ |1)/√2 and ∣∣−0 = (|0- |1)/√2

When x = 1, ∣∣+1 = (|0+ i|1)/√2 and ∣∣−1 = (|0- i|1)/√2

Similarly, for Bob's states:

When y = 0, ∣∣+0 = (|0+ |1)/√2 and ∣∣−0 = (|0- |1)/√2

When y = 1, ∣∣+1 = (|0+ i|1)/√2 and ∣∣−1 = (|0- i|1)/√2

The key difference lies in the relative phases:

When x = 0 or y = 0, the states ∣∣+ and ∣∣− have a relative phase of 0 or π (i.e., they are in-phase or out-of-phase).

When x = 1 or y = 1, the states ∣∣+ and ∣∣− have a relative phase of ±π/2 (i.e., they are in-quadrature).

This difference in relative phases allows Alice and Bob to exploit the entanglement and correlations in the shared state, leading to a higher winning probability in the CHSH game.