### Quantum bit (Ⅱ): Operators.

Having seen what a

is, let's look at the basic operations that can be applied to it.The operators used in quantum computation are all unitary (with the exception of measurements, which mathematically correspond to orthogonal projectors), and therefore in particular reversible. As a reminder, a unitary linear application in a Hilbert space is characterized by: \[ UU^{\dagger}= U^{\dagger}U=I \]

These are the first operators to remember, expressed in the computational basis \((\ket{0},\ket{1})\): \[ \begin{alignat*}{1} X&= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\\ Y&= \begin{bmatrix} 0 & -i\mkern1mu \\ i\mkern1mu & 0 \end{bmatrix}\\\ Z&= \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\ \end{alignat*} \] We call these matrices Pauli matrices.

Another essential operator is the Hadamard operator. It can be interpreted, among other things, as a transition matrix, from the (computational) basis \((\ket{0},\ket{1})\) to the (Hadamard) basis \((\ket{+},\ket{-})\): \[ \begin{alignat*}{1} H&= \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}\ \end{alignat*} \] with \[ \begin{alignat*}{1} \ket{+}&= \frac{1}{\sqrt{2}}(\ket{0}+\ket{1})\\ \ket{-}&= \frac{1}{\sqrt{2}}(\ket{0}-\ket{1})\\ \end{alignat*} \]

To finish this list of must-haves, here are \(S\) and \(T\), respectively *phase* and *\(\frac{\pi}{8}\)*:
\[ \begin{alignat*}{1} S&= \begin{bmatrix} 1 & 0 \\ 0 & i\mkern1mu \end{bmatrix}\\ T&= \begin{bmatrix} 1 & 0 \\ 0 & e^{i\mkern1mu \frac{\pi}{4} } \end{bmatrix}\\ \end{alignat*} \]

Remembering them is easy: note that: \[ \begin{alignat*}{1} S^2&= Z \\ T^2&= S \end{alignat*} \]

Why the name *\(\frac{\pi}{8}\)*? You can write \[ \begin{alignat*}{1} T&= e^{i\mkern1mu \frac{\pi}{8} }\begin{bmatrix} e^{-i\mkern1mu \frac{\pi}{8} } & 0 \\ 0 & e^{i\mkern1mu \frac{\pi}{8} } \end{bmatrix}\ \end{alignat*} \] Now we know that the factor \(e^{i\mkern1mu \frac{\pi}{8}}\) (global phase) has no physical significance, so we might as well remove it.

To remember \(Y\), we can think that \(-i\mkern1mu\) is *less* (minus) heavy than \(i\mkern1mu\), so floats to the top right. On the other hand, the \(i\mkern1mu\)s can't be on the diagonal, otherwise the operator would be just the same as \(Z\).

We have the relations: \[ \begin{alignat*}{1} X^2&=Y^2=Z^2=I \\ XY &= i\mkern1mu Z \\ YZ &= i\mkern1mu X \\ ZX &= i\mkern1mu Y \end{alignat*} \]

Now we need to familiarize ourselves with the action of these operators, in terms of Bloch's Sphere transformation. Let's look at what happens if we apply \(X\) to the qubit \(\ket\psi=\cos\frac{\theta}{2}\ket0 + \sin\frac{\theta}{2}e^{i\mkern1mu\varphi}\ket1\) \[ \begin{alignat*}{1} X\ket\psi &= \cos\frac{\theta}{2}X\ket0 + \sin\frac{\theta}{2}e^{i\mkern1mu\varphi}X\ket1 \\ &= \cos\frac{\theta}{2}\ket1 + \sin\frac{\theta}{2}e^{i\mkern1mu\varphi}\ket0 \\ &= \sin\frac{\theta}{2}e^{i\mkern1mu\varphi}\ket0 + \cos\frac{\theta}{2}\ket1 \\ &= e^{i\mkern1mu\varphi}(\sin\frac{\theta}{2}\ket0 + \cos\frac{\theta}{2}e^{-i\mkern1mu\varphi}\ket1) \end{alignat*} \] As usual, we get rid of the global phase, \[ \begin{alignat*}{1} X\ket\psi &= \sin\frac{\theta}{2}\ket0 + \cos\frac{\theta}{2}e^{-i\mkern1mu\varphi}\ket1 \\ &= \cos(\frac{\pi}{2}-\frac{\theta}{2})\ket0 + \sin(\frac{\pi}{2}-\frac{\theta}{2})e^{-i\mkern1mu\varphi}\ket1 \\ &= \cos(\frac{\theta}{2})\ket0 + \sin(\frac{\theta}{2})e^{-i\mkern1mu\varphi}\ket1 \end{alignat*} \]

So two things happened:

- \(\varphi\) has been transformed into \(-\varphi\): In \(\mathbb{R}^3\), this corresponds to a symmetry with respect to the plane \(y=0\).
- \(\theta\) has been transformed into \(\pi-\theta\): In \(\mathbb{R}^3\), this corresponds to symmetry with respect to the plane \(z=0\).

Now, we know that the compound of two symmetries is a rotation, whose axis is the intersection of the two planes (the x-axis), whose angle is twice the angle between the two planes (in this case \(2\frac{\pi}{2}=\pi\)).

So we've rotated the qubit around the x axis by an angle \(\pi\), or in other words, we have performed an axial symmetry. In images, it looks like this:

Unsurprisingly, we can easily verify that Y corresponds to a rotation of angle \(\pi\) around the y-axis and Z corresponds to a rotation of angle \(\pi\) around the z-axis.

It's also enlightening to determine the eigenvectors and eigenvalues of Pauli matrices:

Pauli Operator | Eigenvalue \(\lambda=1\) | Eigenvalue \(\lambda=-1\) |
---|---|---|

X | \(\frac{1}{\sqrt{2}}(\ket0+\ket1)\) | \(\frac{1}{\sqrt{2}}(\ket0-\ket1)\) |

Y | \(\frac{1}{\sqrt{2}}(\ket0+i\mkern1mu\ket1)\) | \(\frac{1}{\sqrt{2}}(\ket0-i\mkern1mu\ket1)\) |

Z | \(\ket0\) | \(\ket1\) |

More generally, rotation through an angle \(\theta\) about the unit vector \(n=\begin{bmatrix}n_x&n_y&n_z\end{bmatrix}\) is given by: \[ \begin{alignat*}{1} R_{\vec{n}}(\theta) &= e^{-i\mkern1mu \frac{\theta}{2}\vec{n}\cdot \vec{\sigma} } \\ &= e^{-i\mkern1mu \frac{\theta}{2}(n_xX+n_yY+n_zZ) } \\ &= \cos\frac{\theta}{2}I-i\mkern1mu\sin \frac{\theta}{2}(n_xX+n_yY+n_zZ) \end{alignat*} \]

All this is another way of looking at the body of quaternions: cf. this XENS

.