Tutorial

#### Quantum Gates

The essential objective in quantum computing is to apply physical effects — such as a precisely timed burst of a magnetic field — to tilt the odds that favor the quantum bit collapsing to one or the other state. A quantum program is a sequence of these physical effects that takes qubits from some initial state to one that solves the computational problem.

Fortunately, you don't need to know quantum mechanics to implement these physical effects. A standard set of devices, called quantum gates, encapsulate the quantum principles. These principles essentially amount to rotating the pentagon $\ket{0}$ and triangle $\ket{1}$ qubelets.

We'll look at a few quantum gates to see how they manipulate quantum states. Or more precisely, how they affect the pentagon $\ket{0}$ and triangle $\ket{1}$ qubelets.

##### NOT Gate

A quantum NOT gate, like its classical counterpart, toggles the qubelet. For example, when a NOT gate acts on a pentagon $\ket{0}$ qubelet, it switches it to a triangle $\ket{1}$ qubelet: Mathematically, we represent the action of the NOT gate on the quantum states as follows: \begin{eqnarray*} NOT: & \ket{0} \rightarrow \ket{1} \\ NOT: & \ket{1} \rightarrow \ket{0} \\ \end{eqnarray*} The rest of the gates you'll see in this section are "pure" quantum gates and have no classical equivalents.

##### Z Gate

The Z gate only acts on the $\ket{1}$ qubelets and flips their orientation $180^{\circ}$. The Z gate does not act on the pentagon $\ket{0}$ qubelets. For example, when the Z gates on a quantum state containing a pentagon $\ket{0}$ qubelet and a triangle $\ket{1}$ qubelet shown on the left in the figure below, it only turns the triangle $\ket{1}$ upside down: (On the right qubit, the faded triangle $\ket{1}$ behind the triangle $\ket{1}$ in the foreground shows its original position.)

Mathematically, we represent the action by the Z gate as follows: $$Z: \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1} \rightarrow \frac{1}{\sqrt{2}} \ket{0} - \frac{1}{\sqrt{2}} \ket{1}$$

##### T Gate

A T gate also only acts on the triangle $\ket{1}$ qubelet but rotates the triangle $\ket{1}$ qubelet by $45^{\circ}$ anti-clockwise: (On the right qubit, the faded triangle $\ket{1}$ behind the triangle $\ket{1}$ in the foreground shows its original position.)

Mathematically, we represent the action by the T gate as follows: $$T: \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1} \rightarrow$$

Other quantum gates, such as the $S$, $S^\dagger$ and $T^\dagger$ gates, rotate the triangle $\ket{1}$ by other amounts.

##### H Gate

Quantum gates do more than just rotate qubits though. A signature operation is to take a pure> qubit, such as one having only a pentagon $\ket{0}$ qubelet, and splits it so that the qubit ends up with both pentagons and triangles. For example, when the $H$, or Hadamard gate, acts on a pentagon $\ket{0}$ qubit, it splits it into a pentagon $\ket{0}$ and triangle $\ket{1}$:
image Mathematically, the action of the H gate on a $\ket{0}$ qubit is: $$H: \ket{0} \rightarrow \frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}$$
But, it acts differently on a $\ket{0}$ qubit than a $\ket{1}$ qubit. When it splits a $\ket{1}$ qubit, the triangle $\ket{1}$ qubelet is inverted: Mathematically, the action of the H gate on a $\ket{1}$ qubit is: $$H: \ket{1} \rightarrow \frac{1}{\sqrt{2}}\ket{0} - \frac{1}{\sqrt{2}}\ket{1}$$
Because the H gate can take "pure" $\ket{0}$ or $\ket{1}$ qubit and create a quantum state with both qubits simultaneously, the H gate is frequently used to kickstart a quantum program.

Like classical logic gates, these form the building blocks of quantum circuits that manipulate qubits to perform computational tasks. A quantum program encodes this circuit so that it runs on a quantum computer. In the next section, you'll see how chaining quantum gates lets you control how to evolve a quantum state in your programs.

##### Why Do Most Quantum Gates Act on $\ket{1}$ Qubelets Only?

There's nothing special about triangle $\ket{1}$ qubelets. The quantum gates could equally well have rotated pentagon $\ket{0}$ qubelets. In quantum computing, it's only the relative difference between the rotations of a $\ket{0}$ qubelet versus that of the $\ket{1}$ qubelet. So, by convention, the quantum gates rotate the $\ket{1}$ qubelet.