1.3 Quantum Computation

Single and Multiple Qubtis

In order to tell what happens to s

Measurements in bases other than the computational basis

Quantum Circuits

Changes occurring to a quantum state can be described using the language of quantum computation.

The circuit is to be read from left-to-right. Each line in the circuit represents a wire in the quantum circuit. This wire does not necessarily correspond to a physical wire; it may correspond instead to the passage of time, or perhaps to a physical particle such as a photon.

It's also conventional to assum that the state input to the circuit is a computational basis state, usually the state consisting of $\lvert 0 \rangle$ s.

The figure accomplishes a simple but usaful task - it swaps the states of the two qubits.

$\begin{array}{ll} \lvert a, b \rangle \rightarrow \lvert a, a\oplus b \rangle \\ \lvert a, b \rangle \rightarrow \lvert a \oplus (a \oplus b), a\oplus b \rangle = \lvert b, a \oplus b \rangle \\ \lvert a, b \rangle \rightarrow \lvert b, (a\oplus b) \oplus b \rangle = \lvert b,a \rangle \end{array}$

remember XOR properties are:

$\begin{array}{ll} a\oplus b = b\oplus a \ (Commutative)\\ a\oplus ( b \oplus c) = (a \oplus b) \oplus c \ (Associative)\\ a\oplus 0 = a \ (Identity)\\ a\oplus a = 0 \ (Self-Inverse) \end{array}$

where all additions are done modulo 2. The effect of the circuit, therefore, is to interchange the state of two qubits. To review logic gate, please see Logic gate and more about Quantum logic gate (Wiki) and Quantum logic gate.

Figure 1: Circuit swapping two qubits and equivalent schematic symbol notation for this common and useful circuit.

There are a few features allowed in classical circuits that are not usually present in quantum circuits.

We don’t allow ‘loops’, that is, feedback from one part of the quantum circuit to another; we say the circuit is acyclic.
Classical circuits allow wires to be ‘joined’ together, an operation known as FANIN, with the resulting single wire containing the bitwise OR of the inputs. Obviously this operation is not reversible and therefore not unitary, so we don’t allow FANIN in our quantum circuits.
The inverse operation, FANOUT , whereby several copies of a bit are produced is also not allowed in quantum circuits. This is impossible in quantum mechanics.

Thus, we introduce a so called controlled-U gate, which is a natural extension of the controlled-NOT gate. Such a single single control qubit, indicated by the line with the black dot, and n target qubits, indicated by the boxed $U$ . If the control qubit is set to 0 then nothing happens to thetarget qubits. If the control qubit is set to 1 then the gate $U$ is applied to the target qubits.

Figure 2: Controlled- $U$ gate

Figure 3: Two different representations for the controlled- $NOT$ gate

Another important gate is the measurement. see figure 4 below.

Figure 4: quantum circuit symbol for measurement

Quantum copying circuit

Example: Bell states

Let's considering a more complicated circuit like image showed belowed The Bell state compose two state 1. Hadamard gate followed by a CNOT. The Hadamard gate takes the input $\lvert 00 \rangle$ to $\frac{(\lvert 0 \rangle + \lvert 1 \rangle)\lvert 0 \rangle)}{\sqrt{2}}$ and then the CNOT gives the output state $\frac{(\lvert 00 \rangle + \lvert 11 \rangle)}{\sqrt{2}}$ . first the Hadamard transform puts the top qubis in a superposition; theis then acts as a control input to the CNOT, and the target gets inverted only when the control is $1$ .

From Tensor Product, The initial state is $\lvert 00 \rangle = \lvert 0 \rangle \otimes \lvert 0 \rangle$ ,

Apply $H$ to the first qubit, $$ H\lvert 0 \rangle = \frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}} $$

Thus, the state of the system becomes: $$ (H\lvert 0 \rangle) \otimes \lvert 0 \rangle = (\frac{\lvert 0 \rangle + \lvert 1 \rangle}{\sqrt{2}}) \otimes \lvert 0 \rangle $$

This can be expanded as: $$ \frac{(\lvert 0 \rangle\lvert 0 \rangle + \lvert 1 \rangle\lvert 0 \rangle)}{\sqrt{2}} = \frac{(\lvert 00 \rangle + \lvert 10 \rangle)}{\sqrt{2}} $$

By the effect of the CNOT, which inverted only whe the control is $1$ , we have: $$ \frac{(\lvert 00 \rangle + \lvert 11 \rangle)}{\sqrt{2}} = \lvert \beta_{00} \rangle = \lvert \Phi^+\rangle $$

we can get all the following truth table,

$\begin{array}{|c|c|c|c|} \hline \text{Input Qubit 1 ($ A $)} & \text{Input Qubit 2 ($ B $)} & \text{Hadamard on $ A $} & \text{CNOT Output ($ \lvert\Phi^+\rangle $)} \\ \hline 0 & 0 & \frac{\lvert0\rangle + \lvert1\rangle}{\sqrt{2}} & \frac{\lvert00\rangle + \lvert11\rangle}{\sqrt{2}} \\ 0 & 1 & \frac{\lvert0\rangle + \lvert1\rangle}{\sqrt{2}} & \frac{\lvert01\rangle + \lvert10\rangle}{\sqrt{2}} \\ 1 & 0 & \frac{\lvert0\rangle - \lvert1\rangle}{\sqrt{2}} & \frac{\lvert00\rangle - \lvert11\rangle}{\sqrt{2}} \\ 1 & 1 & \frac{\lvert0\rangle - \lvert1\rangle}{\sqrt{2}} & \frac{\lvert01\rangle - \lvert10\rangle}{\sqrt{2}} \\ \hline \end{array}$

Bell States:

Below are the four possible Bell states:

$\begin{aligned} \lvert\beta_{00}\rangle = \lvert\Phi^+\rangle & = \frac{\lvert00\rangle + \lvert11\rangle}{\sqrt{2}} \\ \lvert\beta_{01}\rangle = \lvert\Psi^+\rangle & = \frac{\lvert01\rangle + \lvert10\rangle}{\sqrt{2}} \\ \lvert\beta_{10}\rangle = \lvert\Phi^-\rangle & = \frac{\lvert00\rangle - \lvert11\rangle}{\sqrt{2}} \\ \lvert\beta_{11}\rangle = \lvert\Psi^-\rangle & = \frac{\lvert01\rangle - \lvert10\rangle}{\sqrt{2}} \end{aligned}$