Exercise 4.4 (Reviewing)

The idea here is that a unitary gate \(U\) can be decomposed in the following form,

(1)\[ U = e^{i\alpha}R_{x}(\beta)R_{z}(\gamma)R_{x}(\delta) \]

Therefore, we need to find the decomposition form of Hadmard gate with some specific rotational angle and \(\varphi\).

The Hadmard gate \(H\) is defined by

(2)\[\begin{split} H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{split}\]

The rotational-\(x\) gate is defined by

(3)\[\begin{split} R_{x}(\theta) = \begin{pmatrix} \cos\frac{\theta}{2} & -i\sin\frac{\theta}{2} \\ -i\sin\frac{\theta}{2} & \cos\frac{\theta}{2} \\ \end{pmatrix} \end{split}\]

The rotational-\(z\) gate is defined by

(4)\[\begin{split} R_{z}(\theta) = \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \\ \end{pmatrix} \end{split}\]

Notice that we could re-write the Hadmard gate into the sum of Pauli matrices, that is,

(5)\[\begin{split} H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \frac{X + Z}{\sqrt{2}} \end{split}\]

Since Hadmard gate is a sum of Pauli matrices, we could re-write rotational gate into the sum of Pauli matrices with some special angle \(\theta\), and use the combination of special rotational gates to construct the Hadmard gate.

We could also re-write the rotational gate into sum of Pauli matrices,

(6)\[\begin{split} R_{x}(-\pi/2) = \begin{pmatrix} \cos\frac{\pi}{4} & i\sin\frac{\pi}{4} \\ i\sin\frac{\pi}{4} & \cos\frac{\pi}{4} \\ \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & i \\ i & 1 \\ \end{pmatrix} = \frac{I + iX}{\sqrt{2}} \end{split}\]

and also we have for rotational-\(z\) gate,

(7)\[\begin{split} \begin{align} R_{z}(-\pi/2) &= \begin{pmatrix} e^{i\pi/4} & 0 \\ 0 & e^{-i\pi/4} \\ \end{pmatrix} \\ &= \begin{pmatrix} \cos\frac{\pi}{4}+i\sin\frac{\pi}{4} & 0 \\ 0 & \cos\frac{\pi}{4}-i\sin\frac{\pi}{4} \\ \end{pmatrix} \\ &= \frac{1}{\sqrt{2}}\begin{pmatrix} 1+i & 0 \\ 0 & 1-i\\ \end{pmatrix} = \frac{I + i Z}{\sqrt{2}} \end{align} \end{split}\]

Then we could have

(8)\[\begin{split} \begin{align} R_{x}(-\pi/2)R_{z}(-\pi/2)R_{x}(-\pi/2) &= \left(\frac{I + iX}{\sqrt{2}}\right)\left(\frac{I + iZ}{\sqrt{2}}\right)\left(\frac{I + iX}{\sqrt{2}}\right) \\ &= \frac{1}{2\sqrt{2}}( I+ iX + iZ + i^2ZX + iX + i^2X^2 + i^2XZ + i^3 XZX) \\ &= \frac{1}{2\sqrt{2}}( I+ iX + iZ -ZX + iX -X^2 -XZ - i XZX) \\ &= \frac{1}{2\sqrt{2}}( iX + iZ + iX + i Z) \\ &= i\frac{X+Z}{\sqrt{2}} = iH\\ \end{align} \end{split}\]

Note that we adopt \(XZX = -Z\) and \(XZ + ZX = \{X, Z\} = 0\) in the calculation. Then we could get

(9)\[ H = -iR_{x}(-\pi/2)R_{z}(-\pi/2)R_{x}(-\pi/2) = e^{-i\pi/2}R_{x}(-\pi/2)R_{z}(-\pi/2)R_{x}(-\pi/2) \]

In addition, we could also find that

(10)\[ R_{x}(\pi/2) = \frac{I - iX}{\sqrt{2}}, R_{z}(\pi/2)= \frac{I - iZ}{\sqrt{2}} \]

Then we could have

(11)\[\begin{split} \begin{align} R_{x}(\pi/2)R_{z}(\pi/2)R_{x}(\pi/2) &= \left(\frac{I - iX}{\sqrt{2}}\right)\left(\frac{I - iZ}{\sqrt{2}}\right)\left(\frac{I - iX}{\sqrt{2}}\right) \\ &= \frac{1}{2\sqrt{2}}( I- iX - iZ + i^2ZX - iX + i^2X^2 + i^2XZ - i^3 XZX) \\ &= \frac{1}{2\sqrt{2}}( I- iX - iZ -ZX - iX -I -XZ - i Z) \\ &= -\frac{1}{2\sqrt{2}}( iX + iZ + iX + i Z) = -iH\\ \end{align} \end{split}\]

Finally, we could get

(12)\[ H = i R_{x}(\pi/2)R_{z}(\pi/2)R_{x}(\pi/2) = e^{i\pi/2}R_{x}(\pi/2)R_{z}(\pi/2)R_{x}(\pi/2) \]