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In the "Superposition" Microsoft quantum katas (https://github.com/microsoft/QuantumKatas/blob/master/Superposition/ReferenceImplementation.qs) the solution for Task 7 looks like this:

// ------------------------------------------------------
// Task 7. All Bell states
// Inputs:
//      1) two qubits in |00⟩ state (stored in an array of length 2)
//      2) an integer index
// Goal: create one of the Bell states based on the value of index:
//       0: |Φ⁺⟩ = (|00⟩ + |11⟩) / sqrt(2)
//       1: |Φ⁻⟩ = (|00⟩ - |11⟩) / sqrt(2)
//       2: |Ψ⁺⟩ = (|01⟩ + |10⟩) / sqrt(2)
//       3: |Ψ⁻⟩ = (|01⟩ - |10⟩) / sqrt(2)

operation AllBellStates_Reference (qs : Qubit[], index : Int) : Unit is Adj {
H(qs[0]);
CNOT(qs[0], qs[1]);

// now we have |00⟩ + |11⟩ - modify it based on index arg
if (index % 2 == 1) {
    // negative phase
    Z(qs[1]);
}
if (index / 2 == 1) {
    X(qs[1]);
}

}


And my alternative solution looks like this:

// Task 7. All Bell states 
// Inputs: 
//      1) two qubits in |00⟩ state (stored in an array of length 2) 
//      2) an integer index 
// Goal: create one of the Bell states based on the value of index: 
//    0: |Φ⁺⟩ = (|00⟩ + |11⟩) / sqrt(2) 
//    1: |Φ⁻⟩ = (|00⟩ - |11⟩) / sqrt(2) 
//    2: |Ψ⁺⟩ = (|01⟩ + |10⟩) / sqrt(2) 
//    3: |Ψ⁻⟩ = (|01⟩ - |10⟩) / sqrt(2)

operation AllBellStates (qs : Qubit[], index : Int) : Unit 
{
H(qs[0]);
CNOT(qs[0], qs[1]);

if (index == 0){ 
} elif(index == 1) {
    Z(qs[0]);
} elif(index == 2){
    X(qs[0]);
} elif(index == 3){
    X(qs[0]);
    Z(qs[1]);
}

}


Yet BOTH solutions pass the unit tests. How is that possible, since the X-gate and Z-gate operators are being applied to the other qubit in the pair? Could the unit test logic be incorrect?

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The unit test logic is correct, no worries there! One of the really neat things about the four Bell states is that you can transform between each using single-qubit operations. For example, consider the Bell state $\left|\Phi^{+}\right\rangle = \frac{1}{\sqrt{2}} \left(\left|00\right\rangle + \left|11\right\rangle\right)$. Flipping the first qubit with an X instruction transforms the state of your qubits to $\frac{1}{\sqrt{2}} \left(\left|10\right\rangle + \left|01\right\rangle\right)$, while flipping the second qubit with X(qs[1]) transforms your qubits into the state $\frac{1}{\sqrt{2}} \left(\left|01\right\rangle + \left|10\right\rangle\right)$. Since addition is commutative (that is, $a + b = b + a$), these two states are equal.

Thus, even though X(qs[0]) and X(qs[1]) are very different instructions, they do the same thing in the special case of the Bell state $\left|\Phi^{+}\right\rangle$. In the same way, Z(qs[0]) and Z(qs[1]) do the same thing to $\left|\Phi^{+}\right\rangle$. Another way of thinking of why this is is that the quantum programs ApplyToEach(X, qs) and ApplyToEach(Z, qs) do nothing to qubits in the state $\left|\Phi^{+}\right\rangle$, similarly to how Z(q) does nothing when q is in the state |0⟩. If you're interested, there's a beautiful mathematical framework known as the stabilizer formalism for helping to track these kinds of symmetries of quantum states.

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  • 1
    $\begingroup$ Thnx for the explanation. $\endgroup$ – Oke Uwechue Sep 9 at 0:31

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