# Show $\epsilon e^{{i}{\vec\phi\over2}\cdot \vec{\sigma}^*} (-\epsilon) =e^{-{\vec\phi\over2}\cdot \vec\sigma}$ for Pauli matrices

In pg. 76 of the Physics from Symmetry book, it was stated that the following relation is true:

$$\epsilon e^{{i}{\vec\phi\over2}\cdot \vec{\sigma}^*} (-\epsilon) =e^{-i{\vec\phi\over2}\cdot \vec\sigma}$$

where $$\epsilon = \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$$, $$\vec\phi$$ is a constant vector and $$\vec\sigma = (\sigma_x,\sigma_y,\sigma_z)$$ represent the Pauli matrices.

But for a function $$f(\epsilon \sigma_i \epsilon^{-1})$$

It was stated that this was shown true by using the fact that $$\epsilon\sigma_i^*(-\epsilon)=-\sigma_i.$$

I had been able to prove that the first relation is true up to a second order expansion in Pauli matrices explicitly. However, is there an easier way to see that it is true generally without having to explicitly expand to the $$n^\text{th}$$ order?

Edit: I had tried the methods the comments and answer had suggested but is still unsuccessful. The main challenge I face is that in the expansion of $$\epsilon e^{{i}{\vec\phi\over2}\cdot \vec{\sigma}^*} (-\epsilon)$$ there will be cross terms like $$\epsilon {\sigma_x^*}^n {\sigma_y^*}^m {\sigma_z^*}^l (-\epsilon)$$ where $$n,m,l$$ are integers. If it can be shown that $$\epsilon {\sigma_x^*}^n {\sigma_y^*}^m {\sigma_z^*}^l (-\epsilon) = (-1)^{n+m+l} {\sigma_x}^n {\sigma_y}^m {\sigma_z}^l$$ , then the first relation will be true. However, I don't know how to show it.

• The exponential of a Pauli matrix has a simple form. This probably follows from that. – G. Smith Nov 29 '20 at 17:41
• See the answer to this question for a hint in your calculations – Jeanbaptiste Roux Nov 29 '20 at 17:43
• Consider the $n$th power of your last equation. – G. Smith Nov 29 '20 at 18:31

$$\epsilon = i\sigma_2 = -\epsilon^{-1}, \\ \leadsto \epsilon \sigma_i \epsilon^ {-1}= -\sigma_i^*,$$ the conjugate representation.
As a similarity transformation it leads to $$\epsilon \sigma_i^n \epsilon^ {-1}= (-\sigma_i^*)^n, \leadsto \\ f(\epsilon \sigma_i \epsilon^ {-1})= f(-\sigma_i^*),$$ for any function f regular at the origin, including the exponential.