Quantum Fredkin/CSWAP Gate

Definition¹

(From the definition of the classical Fredkin gate $(a, b, c) \mapsto \big(a, (\lnot a \land b) \lor (a \land c), (\lnot a \land c) \lor (a \land b) \big)$) For qubits $\ket{a, b, c} = \ket{a} \otimes \ket{b} \otimes \ket{c}$, the quantum Toffoli gate is defined as follows.

$$\begin{align*} F_{q} : (\mathbb{C}^{2})^{\otimes 3} &\to (\mathbb{C}^{2})^{\otimes 3} \\ \ket{a, b, c} &\mapsto \ket{a, (\lnot a \land b) \lor (a \land c), (\lnot a \land c) \lor (a \land b)},\quad \forall a,b,c \in \left\{ 0, 1 \right\} \end{align*}$$

$$F_{q}(\ket{a} \otimes \ket{b} \otimes \ket{c}) = \ket{a} \otimes \ket{(\lnot a \land b) \lor (a \land c)} \otimes \ket{(\lnot a \land c) \lor (a \land b)}$$

Here, $(\mathbb{C}^{2})^{\otimes 3}$ is the tensor product of vector spaces, $\ket{a} \otimes \ket{b} \otimes \ket{c}$ is the direct product, $\land$ is the logical AND, $\lor$ is the logical OR, and $\lnot$ is the logical NOT.

Description

This is the quantum computer version of the classical Fredkin gate. Whereas the classical Fredkin gate is a universal gate, the quantum Fredkin gate is not a universal gate. Not only the quantum Fredkin gate, but it is also impossible to find a universal gate in quantum computing.

The specific input and output of $F_{q}$ are as follows. The output changes only when the input is $\ket{101}, \ket{110}$.

$$F_{q} (\ket{000}) = \ket{0, (\lnot 0 \land 0) \lor (0 \land 0), (\lnot 0 \land 0) \lor (0 \land 0)} = \ket{000} \\[0.5em] F_{q} (\ket{001}) = \ket{0, (\lnot 0 \land 0) \lor (0 \land 1), (\lnot 0 \land 1) \lor (0 \land 0)} = \ket{001} \\[0.5em] F_{q} (\ket{010}) = \ket{0, (\lnot 0 \land 1) \lor (0 \land 0), (\lnot 0 \land 0) \lor (0 \land 1)} = \ket{010} \\[0.5em] F_{q} (\ket{011}) = \ket{0, (\lnot 0 \land 1) \lor (0 \land 1), (\lnot 0 \land 1) \lor (0 \land 1)} = \ket{011} \\[0.5em] F_{q} (\ket{100}) = \ket{1, (\lnot 1 \land 0) \lor (1 \land 0), (\lnot 1 \land 0) \lor (1 \land 0)} = \ket{100} \\[0.5em] F_{q} (\ket{101}) = \ket{1, (\lnot 1 \land 0) \lor (1 \land 1), (\lnot 1 \land 1) \lor (1 \land 0)} = \ket{110} \\[0.5em] F_{q} (\ket{110}) = \ket{1, (\lnot 1 \land 1) \lor (1 \land 0), (\lnot 1 \land 0) \lor (1 \land 1)} = \ket{101} \\[0.5em] F_{q} (\ket{111}) = \ket{1, (\lnot 1 \land 1) \lor (1 \land 1), (\lnot 1 \land 1) \lor (1 \land 1)} = \ket{111}$$

The matrix representation is as follows.

$$F_{q} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$