Show $e^{-a\sigma_3}\sigma_1e^{a\sigma_3} = \sigma_1e^{2a\sigma_3}$ How do you show that 
$$e^{-a\sigma_3}\sigma_1e^{a\sigma_3} = \sigma_1e^{2a\sigma_3}$$
where $\sigma_i$ are the Pauli matrices.
 A: In general, if $A,B$ are matrices and $A$ is invertible, then $$Ae^{B} = e^{ABA^{-1}} A$$
(That follow pretty directly from the definition of $e^M$ and the fact that $(ABA^{-1})^n = AB^nA^{-1}$.)
Setting $A=\sigma_1$, $B=a\sigma_3$, we get $A^{-1}=\sigma_1$ and $ABA^{-1}=a\sigma_1\sigma_3\sigma_1=-a\sigma_3$, so:
$$\sigma_1 e^{a\sigma_3} = e^{-a\sigma_3}\sigma_1$$
Multiply on the right by $e^{a\sigma_3}$ an you get:
$$\sigma_1 e^{2a\sigma_3} = e^{-a\sigma_3}\sigma_1e^{a\sigma_3}$$
[ In general, $e^{M}e^{N}\neq e^{M+N}$, but $e^{M}e^M=e^{2M}$ in general. ]
A: Try expanding the exponentials in a power series.  Since $\sigma_3 \sigma_3 = 1$ (the identity, if you insist on a matrix representation), you should get $\exp(-a \sigma_3) = \cosh a - \sigma_3 \sinh a$.
Because $\sigma_1 \sigma_3 = - \sigma_3 \sigma_1$, you can see that $\exp(-a \sigma_3) \sigma_1 = \sigma_1 \exp(a \sigma_3)$--if you're having trouble seeing this, try using the hyperbolic trig function form above.  After that, just combine the two exponentials, which is valid because $\sigma_3$ commutes with itself.
