# Matrix product with Pauli matrices

I have to show that if $\hat{n} = (n_1,n_2,n_3) \in \mathbb{R}^3$ and $\vec{\sigma} = (\sigma_1, \sigma_2, \sigma_3)$, where $$\sigma_1=\left(\begin{matrix}0&1\\1&0\end{matrix}\right) \sigma_2 = \left(\begin{matrix}0&-i\\i&0\end{matrix}\right), \sigma_3 = \left(\begin{matrix}1&0\\0&-1\end{matrix}\right)$$ then $$(\hat{n}\cdot\vec{\sigma})^{2n} = \mathbb{1}$$ and I have to calculate $$(\hat{n}\cdot\vec{\sigma})^{2n+1}$$ I know that $\hat{n}\cdot\vec{\sigma} = n_1\sigma_1 + n_2\sigma_2 + n_3\sigma_3$ and that $\sum_i n_i^2 = 1$ but I can't get a good result. Can anyone help me?

• Try calculating $(\hat{n} \cdot \vec{\sigma})^2$ to start with. – Chappers May 9 '18 at 21:34
• I find $\left(\begin{matrix}n_3^2&n_1^2 - n_2^2 -2in_1n_2 \\ n_1^2 - n_2^2 +2in_1n_2 & n_3^2\end{matrix}\right)$ – shot22 May 9 '18 at 21:54

Personally, I find it helpful to refer to (or rederive) this multiplication identity for Pauli matrices: $$\sigma_a \sigma_b = \delta_{ab} I + i \epsilon_{abc} \sigma_c.$$ [Here, $\delta_{ab}$ is the Kronecker delta, $\epsilon_{abc}$ is the antisymmetric 3-tensor, and the summation convention is assumed.]
From this, it's possible to show that, for any 3-vectors $\vec a$ and $\vec b$, $$(\vec a . \vec \sigma) (\vec b . \vec \sigma) = (\vec a . \vec b)I + i (\vec a \times \vec b).\vec \sigma.$$
In your case, $\vec a = \vec b = \vec{\hat n}$ with $|\vec{ \hat{ n}} | = 1$, so $\vec a . \vec b = 1$ and $\vec a \times \vec b = \vec 0$, and hence, $(\vec {\hat n} . \vec \sigma)^2 = 1$.
• thank you. But what about $(\hat{n}\cdot\vec{\sigma})^{2n+1}$? – shot22 May 9 '18 at 23:23
• Well, that will just be equal to $\hat n . \vec \sigma$. – Kenny Wong May 10 '18 at 0:57