In the "Quantum Computation and Quantum Information 10th Anniversary textbook by Nielsen & Chuang", they claim that Eqn(4.75) is a rotation about the axis along the direction ( ${cos(\pi/8)}$, ${sin (\pi/8)}$, ${cos(\pi/8)}$ ). They then defined an angle ${\theta}$ such that:

${cos(\theta/2)}$ = $cos^2$(${\pi/8}$)

and its also claimed to be an irrational multiple of 2${\pi}$.
We know that the rotational matrices about any arbitrary axis takes the form of

${cos(\theta/2)}$ ${I}$ - ${i}$ (${n_x}$${X}$ + ${n_y}$${Y}$ + ${n_z}$${Z}$) ${sin(\theta/2)}$ ,

but Eqn(4.75) gives:

${cos^2(\pi/8)}$ ${I}$ - ${i}$ [ ${cos(\pi/8)(X+Z)+sin(\pi/8)Y}$ ] ${sin(\pi/8)}$

My question is how does this ${\theta}$ be able to simultaneously satisfy ${sin(\pi/8})$? Why does ${\theta}$ be referenced from ${cos^2(\pi/8)}$ instead of ${sin(\pi/8})$?


1 Answer 1


In order to compare to the Pauli vector exponentiation formula, we need to write in terms a normalized unit vector:

$${\displaystyle R = \cos^2(\pi/8) I_2 -\frac{i}{\sqrt{1+\cos^2(\pi/8)} } \times \quad \times \left [\cos(\pi/8)(X+Z)+ \sin(\pi/8)Y \right ] \sqrt{1+\cos^2(\pi/8)}\sin(\pi/8)} $$

Now, the result can be seen by inspection from the comparison to the general formula:

$${\displaystyle e^{ia({\hat {n}}\cdot {\vec {\sigma }})}=I\cos {a}+i({\hat {n}}\cdot {\vec {\sigma }})\sin {a}}$$

  • $\begingroup$ Okay I've not thought about this before. But now there's a constant factor below the $i$. Do we bring in this factor into the Pauli matrices and define $\sigma$ as such? $\endgroup$
    – C.C.
    Commented Dec 5, 2018 at 1:12
  • $\begingroup$ No, the constant factor is just the normalization factor of the unit radius vector $\hat{n} = \frac{[ cos(\pi/8), sin(\pi/8), cos(\pi/8)] }{\sqrt{1+cos^2(\pi/8)}}$. The Pauli vector exponentiation formula is correct only if the vector $\hat{n}$ is a unit vector. Nielsen and Chuang wrote this vector without normalization; they used the vector notation $\vec{n}$ to emphasize this fact and left it for the reader to do the above manipulation. $\endgroup$ Commented Dec 5, 2018 at 8:25

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