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I've seen this identity used several times but never seen a proof and have struggled to come up with it myself $$U\sigma_2U^T=\sigma_2\qquad \text{with}\qquad U\in SU(2)$$

Any hints, and is this identity also valid for the other Pauli matrices?

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Every element in $SU(2)$ can be parametrized as follows: $$\begin{bmatrix} a&-\overline{b}\\b&\overline{a} \end{bmatrix}$$ where $|a|^2 + |b|^2 = 1$ and the overline denotes complex conjugation.

Together with the explicit form of the second Pauli matrix you should be able to check the identity explicitly through matrix multiplication. The other cases can be checked in a similar way.

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    $\begingroup$ I think it is better written as $ (-i\sigma_2)^{-1} U (-i\sigma_2) =U^* $ which shows that $-i \sigma_2$ gives the equivalence of the fundamental representaion of ${\rm SU}(2)$ and its conjugate represenation. As NDewolf says, it is easy to check using his explicit form of the general $U$. $\endgroup$
    – mike stone
    Aug 30, 2018 at 15:39

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