Recall that the most general expression of a single-qubit unitary matrix looks like this: $$ U(\phi, \theta, \omega) = \begin{pmatrix} e^{-i(\phi + \omega)/2} \cos(\theta/2) & -e^{i(\phi - \omega)/2} \sin(\theta/2) \\ e^{-i(\phi - \omega)/2} \sin(\theta/2) & e^{i(\phi + \omega)/2} \cos(\theta/2) \end{pmatrix}. \tag{1} $$ It's possible to find a set of angles $\phi, \theta, \omega$ for all the gates we encountered in previous nodes, such that $U(\phi, \theta, \omega)$ is equivalent to the gate up to a global phase. However, $U$ can actually be expressed in a simpler way. ___ ***Exercise I.7.1.*** Prove that $$ U(\phi, \theta, \omega) = \begin{pmatrix} e^{-i(\phi + \omega)/2} \cos(\theta/2) & -e^{i(\phi - \omega)/2} \sin(\theta/2) \\ e^{-i(\phi - \omega)/2} \sin(\theta/2) & e^{i(\phi + \omega)/2} \cos(\theta/2) \end{pmatrix}. $$ can be expressed using only 3 gates from the set $\{RZ, RY\}.$

Hint.

Recall that their matrix forms are $$ RY(\theta) = \begin{pmatrix} \cos \left(\frac{\theta}{2} \right) & - \sin \left(\frac{\theta}{2} \right) \\ \sin \left(\frac{\theta}{2} \right)& \cos \left(\frac{\theta}{2} \right) \end{pmatrix}, \quad RZ(\theta) = \begin{pmatrix} e^{-i \frac{\theta}{2}} & 0 \\ 0 & e^{i \frac{\theta}{2}}. \end{pmatrix} $$

Solution.

First, knowing that we need 3 $RZ$ and $RY$ operations, consider how they can be combined. If we put two $RZ$ next to each other, that would be equivalent to applying an $RZ$ with the sum of the angles (the same goes for $RY$). Thus, if we are using only 3 gates, they must alternate, so our options are $RZ,$ $RY,$ $RZ,$ or $RY,$ $RZ,$ $RY.$ Looking at the form of the matrix representation above, we see that the portion of the matrix that depends on $\theta$ looks exactly like a $Y$ rotation. Furthermore, the phase in the exponential contains two of the angles. This suggests that we need $RY(\theta),$ and then $RZ(\omega)$ and $RZ(\phi).$ Indeed, if you work through the matrix multiplication, you will find that $$ U(\phi, \theta, \omega) = RZ(\omega) RY(\theta) RZ(\phi) $$ as desired. ▢

___ ***Exercise I.7.2.*** Prove that $$ U(\phi, \theta, \omega) = \begin{pmatrix} e^{-i(\phi + \omega)/2} \cos(\theta/2) & -e^{i(\phi - \omega)/2} \sin(\theta/2) \\ e^{-i(\phi - \omega)/2} \sin(\theta/2) & e^{i(\phi + \omega)/2} \cos(\theta/2) \end{pmatrix}. $$ can be expressed using only 3 gates from the set $\{RZ, RX\}.$

Hint.

You already know how to express this in terms of $RZ$ and $RY,$ so the question is whether we can turn the $RY$ into $RZ$ and $RX.$ There is a convenient circuit identity that relates $Y$ and $X:$ $Y = SXS^\dagger.$ Use this as starting point for expressing $RY(\theta) = e^{-i\frac{\theta}{2}Y}$ in terms of $RX$ and $RZ.$

Solution.

Let's take advantage of the hint. If $Y = S X S^\dagger,$ then $$ \begin{align*} RY(\theta) &= e^{-i\frac{\theta}{2}Y} \\ &= \cos \frac{\theta}{2} I - i \sin \frac{\theta}{2} Y \\ &= \cos \frac{\theta}{2} I + i \sin \frac{\theta}{2} SXS^\dagger \\ &= S \left( \cos \frac{\theta}{2} I + i \sin \frac{\theta}{2} X \right) S^\dagger \\ &= S RX(-\theta) S^\dagger. \end{align*} $$ Now, since $S = RZ(\frac{\pi}{2}),$ we can recover $U$ by writing $$ \begin{align*} U(\phi, \theta, \omega) &= RZ(\omega) RY(\theta) RZ(\phi) \\ &= RZ(\omega) RZ\left(\frac{\pi}{2} \right) RX(-\theta) RZ\left(-\frac{\pi}{2}\right) RZ(\phi) \\ &= RZ\left(\omega + \frac{\pi}{2}\right) RX(-\theta) RZ\left(\phi - \frac{\pi}{2}\right). \end{align*} $$

▢

*Tip*. If you have experience with 3D computer graphics, or flying planes, you might notice that there are a lot of similarities here to Euler angle decompositions for rotations in 3-dimensional space. There are 12 different ways to decompose an arbitrary single-qubit operation into a sequence of 3 $RX, RY, RZ$ gates; a combination of $RZ$ and $RY$ is the most commonly encountered.

___ These two exercises have led us to something very interesting: with just two types of rotations, we can implement *any* single-qubit unitary operation! This is great news, especially in terms of building actual quantum hardware: the hardware only needs to be able to do these two things well, rather than having to implement a large set of gates. This idea has a name: any two rotations of $\{RX, RY, RZ\}$ gates are a **universal gate set** for single-qubit operations. A gate set is universal if combinations of the gates therein can be used to approximate any unitary matrix up to arbitrary precision. More formally, for *any* $U,$ we can find $V$ composed only of gates from the universal set such that $$ || U - V || \leq \epsilon \tag{2} $$ for $\epsilon$ arbitrarily small. Finding the sequence of gates that make up $V$ is called **quantum circuit synthesis**. In the case of $RX$ and $RZ,$ or $RZ$ and $RY,$ the precision can be taken for granted, since we can rotate by arbitrary angles. However, there are also universal gate sets that do not contain any parametrized rotations: a particularly famous one is the set $\{H, T\}.$ For *any* single-qubit operation, you can approximate it up to arbitrary precision with just $H$ and $T.$ That sequence of $H$ and $T$ might be absolutely enormous, should you want to obtain a particularly precise approximation, but it can always be found (and for the single-qubit case, can be found quite efficiently).

Codercise I.7.1 — Universality of rotations

Codercise I.7.2 — Synthesizing a circuit

Codercise I.7.3 — Universality of H and T

Universal gate sets