The Exercise

The circuit of Figure 10.25 applies the $\pi/8$ gate $T$ to a data qubit $|\psi\rangle = a|0\rangle + b|1\rangle$ without ever acting on that qubit with $T$ . An ancilla is prepared in the state $|\Theta\rangle$ that a Hadamard gate $H$ followed by $T$ makes from $|0\rangle$ . A controlled NOT gate (CNOT) then acts with the ancilla as control and the data qubit as target, and the data qubit is measured in the computational basis. On outcome $0$ the ancilla is left in $T|\psi\rangle$ .^✦ On outcome $1$ the correction $SX$ , the bit flip $X$ followed by the phase gate $S$ , brings it to $T|\psi\rangle$ up to a global phase.

Gates and States

Each gate is a matrix in the computational basis $|0\rangle = (1,0)$ , $|1\rangle = (0,1)$ of $\mathbb{C}^2$ . The $\pi/8$ gate $T$ , the phase gate $S$ , the Pauli matrix $X$ and the Hadamard gate $H$ are the complex $2 \times 2$ matrices $T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix}, \qquad S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}, \qquad X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$ (1) The correction gate $SX$ is the product of the phase gate and the Pauli matrix $X$ , $SX = S\,X = \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix}.$ (2)^✦ The ancilla state $|\Theta\rangle$ is the qubit state $|\Theta\rangle = \frac{1}{\sqrt{2}} \bigl( |0\rangle + e^{i\pi/4} |1\rangle \bigr).$ (3)^✦

A vector of two qubits has four amplitudes, on $|00\rangle$ , $|01\rangle$ , $|10\rangle$ and $|11\rangle$ in this order. The tensor product of qubit vectors $|u\rangle = u_0|0\rangle + u_1|1\rangle$ and $|v\rangle = v_0|0\rangle + v_1|1\rangle$ is the vector of two qubits $|u\rangle \otimes |v\rangle = (u_0 v_0,\ u_0 v_1,\ u_1 v_0,\ u_1 v_1).$ (4)^✦ The CNOT with the first qubit as control is the $4 \times 4$ matrix that exchanges $|10\rangle$ and $|11\rangle$ and fixes $|00\rangle$ and $|01\rangle$ . The gate $T$ on the control qubit is $T \otimes I$ , where $I$ is the identity on the second qubit; it is the diagonal matrix $T \otimes I = \operatorname{diag}\bigl(1,\ 1,\ e^{i\pi/4},\ e^{i\pi/4}\bigr).$ (5)^✦

For complex numbers $a$ and $b$ , the state of the circuit after the CNOT is the vector of two qubits $|\chi\rangle = \mathrm{CNOT}\,\bigl(|\Theta\rangle \otimes |\psi\rangle\bigr) = \chi_0|00\rangle + \chi_1|01\rangle + \chi_2|10\rangle + \chi_3|11\rangle,$ (6)^✦ with the ancilla as the first qubit and the data qubit $|\psi\rangle = a|0\rangle + b|1\rangle$ as the second. The data qubit reads $0$ on the amplitudes $\chi_0$ and $\chi_2$ , and $1$ on $\chi_1$ and $\chi_3$ .

The Claim

Theorem 1. Let $a$ and $b$ be complex numbers, $|\psi\rangle = a|0\rangle + b|1\rangle$ , and $|\chi\rangle$ the state (6). The vector the ancilla is left with on outcome $0$ of the data qubit is $\chi_0|0\rangle + \chi_2|1\rangle = \frac{1}{\sqrt{2}}\, T|\psi\rangle,$ (7)^✦ and the correction $SX$ takes the vector it is left with on outcome $1$ to $SX \bigl( \chi_1|0\rangle + \chi_3|1\rangle \bigr) = \frac{e^{i\pi/4}}{\sqrt{2}}\, T|\psi\rangle.$ (8)^✦

Theorem 2. The Hadamard gate and $T$ prepare the ancilla state from $|0\rangle$ : $TH|0\rangle = |\Theta\rangle.$ (9)^✦ The gate $T$ on the control qubit commutes with the CNOT: $(T \otimes I)\,\mathrm{CNOT} = \mathrm{CNOT}\,(T \otimes I).$ (10)^✦ Conjugation by $T$ takes $X$ to a multiple of the correction gate, $T X T^\dagger = e^{-i\pi/4}\, SX,$ (11)^✦ while the product $TX$ differs from that multiple: $TX \neq e^{-i\pi/4}\, SX.$ (12)^✦

What Is Checked

The first theorem is stated over the state (6) after the CNOT and over the gates $T$ and $SX$ . What the library proves about these constructions is the second theorem, (9) to (12).