Sure, I'd be happy to help! Let's take a look at your work and see where the mistake might be.

The expression you provided involves the CNOT gate operation acting on a tensor product of four qubits. The CNOT gate is a two-qubit gate that flips the second qubit if the first qubit is in the state |1⟩. In this case, the gate is applied to qubits 2 and 4, with qubit 3 being the control qubit and qubit 4 being the target qubit.

Your first step is correct, where you apply the CNOT gate to qubits 3 and 4, resulting in the following expression:

|Ο•β‚βŸ© βŠ— |Οˆβ‚‚βŸ© βŠ— (|ΞΎβ‚ƒβŸ© βŠ— |Ξ·β‚„βŸ© if |c⟩ = 0, X|ΞΎβ‚ƒβŸ© βŠ— X|Ξ·β‚„βŸ© if |c⟩ = 1)

However, your subsequent steps are incorrect. The expression you provided for the normalized state is missing the normalization factor, which is the square root of the sum of the probabilities of the two possible outcomes. The probabilities can be calculated as follows:

P(0) = |α₁β₂γ₃δ₄|^2 (assuming |c⟩ = 0)

P(1) = |α₁β₂X(γ₃)X(Ξ΄β‚„)|^2 (assuming |c⟩ = 1)

The normalization factor is then given by:

N = √(P(0) + P(1))

The normalized state is then:

(1/N) * (|Ο•β‚βŸ© βŠ— |Οˆβ‚‚βŸ© βŠ— (|ΞΎβ‚ƒβŸ© βŠ— |Ξ·β‚„βŸ© if |c⟩ = 0, X|ΞΎβ‚ƒβŸ© βŠ— X|Ξ·β‚„βŸ© if |c⟩ = 1)

I hope this helps! Let me know if you have any further questions.

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That sounds a little better

Sure, I'd be happy to help! From what I can see, the math looks correct. The CNOT gate operation is applied to qubits 2 and 4, with qubit 3 as the control and qubit 4 as the target. The resulting state is a combination of the original state and the state resulting from the CNOT gate operation.

The normalization factor is calculated as the square root of the sum of the probabilities of measuring the state in either the 0 or 1 state. The probability amplitudes are then divided by the normalization factor to obtain the normalized state.

Let me know if you have any further questions or if you need help with anything else!