The Exercise

Two photons, a pump and a probe, each carry a qubit. They meet in a Kerr medium, which multiplies the state $|1\rangle$ of each photon by a phase of its own and gives the state $|11\rangle$ of the pair an extra Kerr phase on top of the two. The exercise is to build from this interaction a controlled not (CNOT) gate, with the pump as control and the probe as target, using only gates on one photon at a time. With the Kerr phase at $\pi$ the extra phase is a sign.

Two Qubits

A quantum system $S$ is a physical system described by its state space, a finite dimensional complex Hilbert space.^✦ The qubit is the two level system whose state space is $\mathbb{C}^2$.^✦ A pure state $|\psi\rangle$ of $S$ is a unit vector in the state space of $S$, $\lVert \psi \rVert = 1$.^✦ The basis states $|0\rangle$ and $|1\rangle$ of the qubit are the pure states whose vectors are the standard basis vectors $e_0$ and $e_1$ of $\mathbb{C}^2$; they form the computational basis.^✦ The composite system $S \otimes T$ of two systems $S$ and $T$ is the system whose state space is the tensor product of their state spaces.^✦ The pump and the probe form the composite $\mathrm{qubit} \otimes \mathrm{qubit}$, the pump first, with basis states $|jk\rangle = |j\rangle \otimes |k\rangle$.

A unitary $U$ of a system $S$ is an operator on the state space of $S$ with $U^\dagger U = U U^\dagger = 1$, where $1$ is the identity operator on that space; it is a closed system evolution of $S$.^✦ The identity evolution of $S$ is the unitary $1$.^✦ Given two unitaries $U$ and $V$ of $S$, $U$ after $V$ is the unitary $UV$, first $V$ and then $U$.^✦ The product unitary $U \otimes V$ of a unitary $U$ of $S$ and a unitary $V$ of $T$ is the unitary of $S \otimes T$ whose operator is the tensor product of the two operators.^✦ A unitary $U$ of $S$ applied to the first factor of $S \otimes T$, the system $S$ alone, is the gate $U \otimes 1$, with the identity evolution on $T$, and a unitary $V$ of $T$ applied to the last factor, the system $T$ alone, is the gate $1 \otimes V$.

The Gates

For a real angle $\alpha$, the phase shift gate $P(\alpha)$ is the unitary of the qubit that is diagonal in the computational basis, $$P(\alpha) = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\alpha} \end{pmatrix}.$$(1)^✦ The Hadamard gate $H$ and the Pauli $X$ gate, the quantum not, are the unitaries of the qubit whose operators on $\mathbb{C}^2$ have the matrices $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.$$(2)

Let $U$ be a unitary of a system $S$. The controlled gate $C(U)$ is the unitary of $\mathrm{qubit} \otimes S$, a control qubit and a target $S$, whose operator is built from the projectors onto the basis states of the control: $$C(U) = |0\rangle\langle 0| \otimes 1 + |1\rangle\langle 1| \otimes U.$$(3)^✦ The CNOT gate is the controlled gate $C(X)$ on two qubits.^✦ For a real angle $\Delta$, the controlled phase shift of the pump and the probe is the unitary $C(P(\Delta))$ of $\mathrm{qubit} \otimes \mathrm{qubit}$.^✦

For real phases $\varphi_a$, $\varphi_b$ and $\Delta$, the optical Kerr gate $K(\varphi_a, \varphi_b, \Delta)$ is the unitary of the pump and the probe that applies the controlled phase shift by the Kerr phase $\Delta$, then $P(\varphi_a)$ to the probe and $P(\varphi_b)$ to the pump.^✦ In the basis $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$ it is diagonal: $$K(\varphi_a, \varphi_b, \Delta) = \bigl(P(\varphi_b) \otimes 1\bigr)\bigl(1 \otimes P(\varphi_a)\bigr)\, C(P(\Delta)) = \mathrm{diag}\bigl(1,\ e^{i\varphi_a},\ e^{i\varphi_b},\ e^{i(\varphi_a + \varphi_b + \Delta)}\bigr).$$(4)^✦

The Claim

What Is Checked

The claim is stated over the pair of qubits formed by the pump and the probe and over its unitaries: the CNOT gate $C(X)$, the optical Kerr gate (4) with its controlled phase shift, and the Hadamard and phase shift gates on one qubit of the pair, composed one after another. For every unitary $U$ of a system $S$ the controlled gate $C(U)$ is a unitary of $\mathrm{qubit} \otimes S$, so $C(X)$ and the controlled phase shift are unitaries of the pair of qubits.^✦ The tensor product of two unitaries is again a unitary, and so is the composite of two unitaries, hence the Kerr gate and both sides of (5) are unitaries.