Now that we have some intuition about what these rotations are, we can begin to express $RX$ and $RY$ in terms of matrices and describe how they act on the basis states. The matrix representation of an $X$ rotation is $$ RX(\theta) = \begin{pmatrix} \cos \left(\frac{\theta}{2} \right) & -i \sin \left(\frac{\theta}{2} \right) \\ -i\sin \left(\frac{\theta}{2} \right)& \cos \left(\frac{\theta}{2} \right) \end{pmatrix}. \tag{3} $$ ___ ***Exercise I.6.1.*** Determine $\theta$ for which $RX(\theta) = X$ (up to a global phase).

Solution.

We need an angle such that $\cos\left(\theta/2\right) = 0,$ and $\sin(\theta/2) = 1.$ The angle $\pi$ satisfies these requirements; there is a global phase of $-i$ present, but as in the previous node, global phases have no effect and can be safely removed. $$ \begin{align*} RX(\pi) &= \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \\ &= -i X. \end{align*} $$

▢

___ The matrix representation of a $Y$ rotation looks very similar to that of $RX,$ however there is no complex component: $$ RY(\theta) = \begin{pmatrix} \cos \left(\frac{\theta}{2} \right) & - \sin \left(\frac{\theta}{2} \right) \\ \sin \left(\frac{\theta}{2} \right)& \cos \left(\frac{\theta}{2} \right) \end{pmatrix}. \tag{4} $$ ___ ***Exercise I.6.2.*** Just like how Pauli $X$ and $Z$ are special cases of $RX$ and $RZ,$ Pauli $Y$ is the special case of $RY(\theta)$ when $\theta = \pi.$ Typically, it is written as $$ Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}. $$ Furthermore, there is a nice relationship between $X,$ $Y,$ and $Z.$ Show that $Y = iXZ = iRY(\pi).$

Solution.

For this exercise, let's use matrix multiplication to prove the identity. First, $$ \begin{align*} i XZ &= i \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\ &= i \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \\ &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \\ \end{align*}. $$ Now if we use instead the $RY$ rotation, $$ \begin{align*} i RY(\pi) &= i \begin{pmatrix} \cos \left(\frac{\pi}{2} \right) & - \sin \left(\frac{\pi}{2} \right) \\ \sin \left(\frac{\pi}{2} \right)& \cos \left(\frac{\pi}{2} \right) \end{pmatrix}. \\ &= i \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \\ &= \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}. \end{align*} $$

▢

___ ***Exercise I.6.3.*** Evaluate the action of $RX(\theta)$ and $RY(\theta)$ on $\vert 0 \rangle$ and $\vert 1 \rangle.$ Express your results as linear combinations of these computational basis states.

Solution.

This can be worked out easily enough using matrix multiplication. For $RX,$ we have $$ \begin{align*} RX(\theta)\vert 0\rangle &= \begin{pmatrix} \cos \left(\frac{\theta}{2} \right) & -i \sin \left(\frac{\theta}{2} \right) \\ -i\sin \left(\frac{\theta}{2} \right)& \cos \left(\frac{\theta}{2} \right) \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ &= \begin{pmatrix} \cos \left( \frac{\theta}{2} \right) \\ -i \sin \left( \frac{\theta}{2} \right) \end{pmatrix} \\ &= \cos \left( \frac{\theta}{2} \right) \vert 0 \rangle -i \sin \left( \frac{\theta}{2} \right) \vert 1 \rangle, \end{align*} $$ and $$ \begin{align*} RX(\theta) \vert 1\rangle &= \begin{pmatrix} \cos \left(\frac{\theta}{2} \right) & -i \sin \left(\frac{\theta}{2} \right) \\ -i\sin \left(\frac{\theta}{2} \right)& \cos \left(\frac{\theta}{2} \right) \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} -i \sin \left( \frac{\theta}{2} \right) \\ \cos \left( \frac{\theta}{2} \right) \end{pmatrix} \\ &= -i \sin \left( \frac{\theta}{2} \right) \vert 0 \rangle + \cos \left( \frac{\theta}{2} \right) \vert 1 \rangle. \end{align*} $$ You can follow the same steps for $RY$ to obtain $$ \begin{align*} RY(\theta) \vert 0 \rangle &= \cos \left( \frac{\theta}{2} \right) \vert 0 \rangle + \sin \left( \frac{\theta}{2} \right) \vert 1 \rangle, \\ RY(\theta) \vert 1 \rangle &= -\sin \left( \frac{\theta}{2} \right) \vert 0 \rangle + \cos \left( \frac{\theta}{2} \right) \vert 1 \rangle. \\ \end{align*} $$

▢

___ ***Exercise I.6.4.*** There are many other relationships you can derive between these three types of rotation. Such relationships are often useful for simplifying sequences of quantum operations in circuits. For example, what is the result of applying $X$ both before and after an application of $RY(\theta)?$

Solution.

Let's work through the math in bra-ket notation; recall we need only see how the operations affect the computational basis states to know how they act in general. $$ \begin{align*} X RY(\theta) X \vert 0 \rangle &= X RY(\theta) \vert 1 \rangle \\ &= X \left( -\sin \left( \frac{\theta}{2} \right) \vert 0 \rangle + \cos \left( \frac{\theta}{2} \right) \vert 1 \rangle \right) \\ &= \cos \left( \frac{\theta}{2} \right) \vert 0 \rangle -\sin \left( \frac{\theta}{2} \right) \vert 1 \rangle. \end{align*} $$ This looks almost like $RY(\theta) \vert 0\rangle = \cos \left( \frac{\theta}{2} \right) \vert 0 \rangle + \sin \left( \frac{\theta}{2} \right) \vert 1 \rangle$ , but with a negative sign. Let's check what it does to $\vert 1 \rangle:$ $$ \begin{align*} X RY(\theta) X \vert 1 \rangle &= X RY(\theta) \vert 0 \rangle \\ &= X \left( \cos \left( \frac{\theta}{2} \right) \vert 0 \rangle -\sin \left( \frac{\theta}{2} \right) \vert 1 \rangle \right) \\ &= -\sin \left( \frac{\theta}{2} \right) \vert 0 \rangle + \cos \left(\frac{\theta}{2} \right)\vert 1 \rangle. \end{align*} $$ We can see that, again, that we're *close* to $RY(\theta) \vert 1\rangle$ but have a negative sign on the $\sin$ term. However, recall that $- \sin \left(\theta \right) = \sin(-\theta).$ We can simply absorb the negative sign into the argument, meaning that we obtain the "backwards" rotation by $-\theta:$ $$ X RY(\theta) X = RY(-\theta). $$ Or, expressed as a circuit,

Whenever we see such a pattern of three gates in a circuit, we can perform this small optimization to replace them by a single gate! You can show that similar relationships exist among the other $RX, RY, RZ$ and $X, Y, Z$ too. ▢

___ ***Exercise I.6.5.*** Show that $RX,$ $RY,$ and $RZ$ can be represented as follows: $$ \begin{align*} RX(\theta) &= e^{-i\theta X/2}, \\ RY(\theta) &= e^{-i\theta Y/2}, \\ RZ(\theta) &= e^{-i\theta Z/2}. \end{align*} $$ (While this may look unusual, this representation is actually very important in applications of quantum computing such as *Hamiltonian simulation*.)

Hint.

Start by taking the [matrix exponential](https://en.wikipedia.org/wiki/Matrix_exponential), and then consider what are some special properties of powers of $X,$ $Y,$ and $Z.S$

Solution.

We show the case of $RX;$ the other gates have similar proofs. We can exponentiate a matrix by expanding the exponential out as a series: $$ \begin{align*} e^{-i\theta X/2} &= I + \left(-i\frac{\theta}{2} X\right) + \frac{1}{2!}\left(-i \frac{\theta}{2} X\right)^2 + \frac{1}{3!}\left(-i \frac{\theta}{2} X\right)^3 + \frac{1}{4!}\left(-i \frac{\theta}{2} X\right)^4 + \cdots \\ &= I - i\frac{\theta}{2} X -\frac{1}{2} \left(\frac{\theta}{2}\right)^2 X^2 + \frac{1}{3!}i \left( \frac{\theta}{2}\right)^3 X^3+ \frac{1}{4!} \left(\frac{\theta}{2}\right)^4 X^4 + \cdots \end{align*} $$ However, consider what happens when we apply $X$ more than once - all it does is flip the bit, so applying it twice brings us right back to where we started! Therefore, all even powers in this expansion will reduce to $I,$ and all odd ones to $X:$ $$ \begin{align*} &= \left(1 - \frac{1}{2}\left(\frac{\theta}{2}\right)^2 + \frac{1}{4} \left(\frac{\theta}{2}\right)^4 \right)I - i\left(\left(\frac{\theta}{2}\right) - \frac{1}{3!}\left(\frac{\theta}{2}\right)^3 + \cdots \right) X . \end{align*} $$ These expressions in the parentheses are simply the expansions of cosine and sine, so we can write this in closed form, and recover the matrix expression. $$ \begin{align*} &= \cos \left(\frac{\theta}{2} \right) I - i \sin \left(\frac{\theta}{2} \right) X \\ &= \cos \left(\frac{\theta}{2} \right) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} -i \sin \left(\frac{\theta}{2} \right) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ &= \begin{pmatrix} \cos \left(\frac{\theta}{2} \right) & -i \sin \left(\frac{\theta}{2} \right) \\ -i\sin \left(\frac{\theta}{2} \right)& \cos \left(\frac{\theta}{2} \right) \end{pmatrix}. \end{align*} $$

▢

___

Codercise I.6.1 — Applying RX

Codercise I.6.2 — Plotting RX

Codercise I.6.3 — Plotting RY

The Bloch sphere

RX and RY