# The probability of measuring the control qubit in zero in a quantum circuit

I’m working on an assignment where I have to solve some questions about a quantum circuit. In particular, I have a quantum circuit with three qubits: $|0\rangle$(referenced to as the control qubit), $|\Psi_1\rangle$ and $|\Psi_2\rangle$. First, A Hadamard gate is applied to the control qubit. Next, a CSWAP operation is applied to the three qubits. The CSWAP operation swaps the state of $|\Psi_1\rangle and |Psi_2> only if the control bit is one. Finally, a Hadamard gate is applied to the control qubit again. So, the outcome of the circuit is defined as: $$|\Phi\rangle= (H \otimes I \otimes I) CSWAP (H \otimes I \otimes I)(|0\rangle|\Psi_1\rangle|\Psi_2\rangle)$$ CSWAP is defined as the following transformation matrix: $$CSWAP = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ |$\Phi$> is the joint state of the three qubits at the end of the circuit. I’ve written this final state in terms of the control bit, |$\Psi_1$> and |$\Psi_2$>: $$|\Phi\rangle = \frac{1}{2}(|0>|\Psi_1\rangle|\Psi_2\rangle + |1\rangle|\Psi_1>|\Psi_2\rangle + |0>|\Psi_2\rangle|\Psi_1\rangle - |1\rangle|\Psi_2\rangle|\Psi_1\rangle)$$ Now, in my assignment I have to calculate the chance of measuring zero for the control bit using the following formula: $$P_0 = \langle\Phi| |0\rangle\langle0| \otimes I \otimes I |\Phi\rangle$$ The formula above can be used to show that: $$P_0 = \frac{1}{2}+\frac{|\langle\Psi_1|\Psi_2\rangle|^2}{2}$$ This is where I am stuck. Using the formula I gave above, I calculated$ |0><0| \otimes I \otimes I $: $$|0\rangle\langle0| \otimes I \otimes I = V = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$ Also, I defined |$\Psi_1$> as$ \begin{bmatrix} \alpha_1 \\ \beta_1 \end{bmatrix} $and |$\Psi_2$\rangle as$ \begin{bmatrix} \alpha_2 \\ \beta_2 \end{bmatrix} $. Next, I tried to calculate$\langle\Phi| V |\Phi\rangle$and this gives me the following formula: $$\alpha^2_1 \alpha^2_2 + 2\frac{(\alpha_1 \beta_2 + \alpha_2 \beta_1)^2}{2} + \beta^2_1 \beta^2_2$$ But I’m unable to write this in the required form, namely$ P_0 = \frac{1}{2}+\frac{|\langle\Psi_1|\Psi_2\rangle|^2}{2} \$. Is there anyone out here who can help me to solve this problem? Any help would be greatly appreciated.

• This question should be asked on the Physics SE. – zoli Sep 24 '15 at 15:58