I was asked to find that pauli matrix which satisfies the relation: $\left[U^{\dagger} \sigma_1 U = \sigma_3\right]$. Also, I got as a clue that $\left[U \right]$ should be the rotation matrix with respect to $y-$axis, and therefore expressible as: $\left[U=e^{i\theta \sigma_2} \right]$. So, here begun a painful comprehension process:

  1. First of all, how should i had to know that an unitary matrix acts on pauli matrices as rotation in the space? I mean, to me it was not obvious to search the solution among the unitary matrices that represent a rotation. So, are the pauli matrices linked by rotations? If yes, how?

  2. I was able to show (explicitely) that: $\left\{R(\theta)=\begin{bmatrix}\cos{\theta} & -\sin{\theta}\\\sin{\theta} & \cos{\theta}\\\end{bmatrix}= \cos({\theta})\mathbb{I}-i\sigma_2\sin({\theta})=e^{-i\theta \sigma_2}\right\}$. Note the minus sign at the exponent, which is missing in the suggested expression (???). Anyway, this refers to $2D-$space rotations, so i think it's not what i need; in fact, the clue speaks of a rotation with respect to the $y-$axis, which therefore refers to $3D-$space rotations! While googling, i got the formula: $\left[R_{\hat{n}}(\theta)=e^{-i\theta(\hat{n}\cdot\vec{\sigma})} \right]$ which is said to be referred to rotations of an angle $(\theta)$ with respect to the axis $(\hat{n})$. Is that true? How can it be shown? And especially, how can it be referred to three-dimensional rotations?

  • 1
    $\begingroup$ The relationship between Pauli matrices, quaternions and spatial rotations, the groups $SU(2)$ and $SO(3)$ is an important topic in geometry and quantum mechanics. P. Woit's book contains this an a lot more. $\endgroup$
    – Kurt G.
    Apr 24 at 16:46

1 Answer 1


You wnat $U$ so that

$$\sigma_1 U=U \sigma_3$$

Taking your hint, set $U= e^{i\theta \sigma_2}$. Then

$$\sigma_1 e^{i\theta \sigma_2}=\sigma_1\sum_{n=0}^\infty \frac{i^n \theta^n}{n!}\sigma_2^n$$

Separating this sum modulo $4$ gives

$$\sum_{n=0}^\infty \frac{i^n \theta^n}{n!}\sigma_2^n= \sum_{n\equiv 0\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}I +\sum_{n\equiv 1\ (\text{mod }4)}^\infty \frac{i\theta^n}{n!}\sigma_2 -\sum_{n\equiv 2\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}I -\sum_{n\equiv 3\ (\text{mod }4)}^\infty \frac{i\theta^n}{n!}\sigma_2$$

Then left multiplying by $\sigma_1$ gives

$$\sigma_1\sum_{n=0}^\infty \frac{i^n \theta^n}{n!}\sigma_2^n= \sum_{n\equiv 0\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}\sigma_1 -\sum_{n\equiv 1\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}\sigma_3 -\sum_{n\equiv 2\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}\sigma_1 +\sum_{n\equiv 3\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}\sigma_3$$


In the same way, we find that

$$e^{i\theta \sigma_2}\sigma_3=\sum_{n\equiv 0\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}\sigma_3 -\sum_{n\equiv 1\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}\sigma_1 -\sum_{n\equiv 2\ (\text{mod }4)}^\infty \frac{\theta^n}{n!}\sigma_3 +\sum_{n\equiv 3\ (\text{mod }4)}^\infty \frac{i\theta^n}{n!}\sigma_1$$


So we need $\theta$ such that $\cos(\theta)=-\sin(\theta)$. One such angle is $\theta=\frac{3\pi}{4}$. This gives the matrix $U$

$$U=e^{i\frac{3\pi}{4}\sigma_2}=-\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} $$

Then it is easy to check that

$$-\sqrt{2}\sigma_1U=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$

$$-\sqrt{2}U\sigma_3= \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}=\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$

  • $\begingroup$ Thanks for your answer, but maybe it's not what exactly i was looking for.. i mean, i was able to find the final solution given the hint.. but what about the hint itself? This is what i'm struggling with... $\endgroup$
    – TheRvela
    Apr 24 at 16:50
  • $\begingroup$ I'd start with the Bloch Sphere en.wikipedia.org/wiki/… $\endgroup$
    – QC_QAOA
    Apr 24 at 16:56

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