Quantum Circuit

quantum-computing

Definition

Quantum Circuit

A quantum circuit is a finite sequence of quantum gates applied to a register of qubits.

Transformations

Let ${\mathcal{H}}_n$ be the state space of an $n$-qubit quantum register. A quantum circuit with gates

$$U_1,U_2,\dots ,U_T:{\mathcal{H}}_n\to {\mathcal{H}}_n$$

transforms an initial state $|{\psi }_0⟩\in {\mathcal{H}}_n$ by successive unitary operators:

$$|{\psi }_0⟩\stackrel{U_1}{\to }|{\psi }_1⟩\stackrel{U_2}{\to }|{\psi }_2⟩\stackrel{U_3}{\to }\cdots \stackrel{U_T}{\to }|{\psi }_T⟩.$$

At each time step,

$$|{\psi }_t⟩=U_t|{\psi }_{t-1}⟩.$$

Hence the whole circuit implements the composed unitary

$$U_{\mathrm{circ}}=U_T{U}_{T-1}\cdots U_2{U}_1,$$

so that

$$|{\psi }_T⟩=U_{\mathrm{circ}}|{\psi }_0⟩=U_T{U}_{T-1}\cdots U_1|{\psi }_0⟩.$$

The rightmost gate acts first. A gate drawn earlier in the circuit appears further to the right in the algebraic product.

Gates on Part of a Register

A gate may be drawn on only one wire of a circuit, but mathematically it must act on the whole quantum register. This section explains how a local one-qubit gate is embedded into the larger state space.

For a two-qubit register,

$${\mathcal{H}}_2={\mathbb{C}}^2\otimes {\mathbb{C}}^2.$$

The two copies of ${\mathbb{C}}^2$ are not two fixed vectors. They are two state spaces, one for each qubit subsystem:

$$\underset{\text{first qubit state space}}{\underbrace{{\mathbb{C}}^2}}\otimes \underset{\text{second qubit state space}}{\underbrace{{\mathbb{C}}^2}}.$$

A basis state such as $|00⟩$ abbreviates

$$|00⟩=|0⟩\otimes |0⟩.$$

Here the first $|0⟩$ is a state of the first qubit, and the second $|0⟩$ is a state of the second qubit. The qubits are the subsystems; the kets are their states.

A one-qubit gate

$$U:{\mathbb{C}}^2\to {\mathbb{C}}^2$$

acts on one qubit state space. To make it an operator on the whole two-qubit register, tensor it with the identity operator on the qubit that should remain unchanged.

To apply $U$ to the first qubit, use

$$U\otimes I.$$

To apply $U$ to the second qubit, use

$$I\otimes U.$$

On a product state, this means

$$(A\otimes B)(|a⟩\otimes |b⟩)=A|a⟩\otimes B|b⟩.$$

So $A$ acts on the first qubit state and $B$ acts on the second qubit state. For general superpositions, the same rule extends by linearity.

Why the identity appears

A one-qubit gate $U$ is too small for the whole two-qubit register: it only has type ${\mathbb{C}}^2\to {\mathbb{C}}^2$.

The full register has type

$${\mathbb{C}}^2\otimes {\mathbb{C}}^2.$$

Tensoring with $I$ fills the missing slot. The identity says: “leave this qubit state as it is, but keep it inside the full register.”

Flipping one qubit

With the Pauli-X gate,

$$\begin{array}{rl}(X\otimes I)|00⟩& =X|0⟩\otimes I|0⟩\\ & =|1⟩\otimes |0⟩\\ & =|10⟩.\end{array}$$

Thus $X\otimes I$ flips the state of the first qubit and leaves the state of the second qubit unchanged.

Similarly,

$$\begin{array}{rl}(I\otimes X)|00⟩& =I|0⟩\otimes X|0⟩\\ & =|0⟩\otimes |1⟩\\ & =|01⟩.\end{array}$$

Thus $I\otimes X$ leaves the state of the first qubit unchanged and flips the state of the second qubit.

For an $n$-qubit register, a gate on one qubit is embedded by placing $U$ in the corresponding tensor slot and identities in all other slots.