A collection of gates and wires.
NOT: x \mapsto 1 -
xAND (x, y) \mapsto x \cdot
yCOPY: x \mapsto (x,
x)OR: (x, y) \mapsto x + y -
xyCNOT: (x, y) \mapsto (x, x
\oplus y) where \oplus is
XORCCNOT (TOFFELI): (x, y, z) \mapsto (x, y, z \oplus xy)circuit: \{0, 1\}^{n_{\text{IN}}} \to \{0, 1\}^{n_{\text{OUT}}}
Theorem: any function f: \{0, 1\}^n \to \{0, 1\}^m has a circuit and vice versa
A gate G where G is its inverse. Or
actually, a gate for which we can use itself to restore \bar x from G(\bar
x)
Theorem: any function f: \{0,
1\}^n \to \{0, 1\}^m has a circuit made of reversible gates
{CNOT, CCNOT, COPY,
NOT}
The quantum state is a unit vector in a Hilbert space. The state of n qubits is | \psi \rangle \in \mathbf{C}^{2^n} with ||| \psi \rangle||^2 = 1
Equivalently, we can say that an operation is a unitary matrix U \in \mathbf{C}^{2 \times 2}: f(| \psi \rangle) = U| \psi \rangle
We observe x \in \{0, 1\}^n with probability |\alpha_x|^2
If quantum system 1 is in state | \psi_1 \rangle \in \mathbf{C}^{2^{n_1}} and quantum system 2 is in state| \psi_2 \rangle \in \mathbf{C}^{2^{n_2}} then the join state is | \psi_{12} \rangle = | \psi_1 \rangle \otimes | \psi_2 \rangle
An entangled state is such that cannot be decomposed into a product of two other ones. There is no simple way to know if a state is entangled or not.
\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}
\begin{pmatrix} 1 & 0 \\ 0 & e^{i\frac{\pi}{4}} \\ \end{pmatrix}
\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \\ \end{pmatrix}
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix}
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}
Turns any function into an invertible one.
For f: \{0, 1\}^n \to \{0, 1\}^m we define \tilde f: \{0, 1\}^{n+m} \to \{0, 1\}^{n+m} as (x, y) \mapsto (x, y \oplus f(x))
Theorem: For all unitary matrices U \in \mathbf{C}^{2 \times 2} and all \delta > 0 there exists \tilde U such that
Theorem: For all unitary matrices U \in \mathbf{C}^{2^n \times 2^n} and all \delta > 0 there exists \tilde U such that