QFT: Prove this relationship In my notes from QFT appears this relationship between exponential and sinh and cosh, my question is: How demonstrate this?
$$\exp{(\pm\frac{1}{2} \eta \sigma^3)}=\cosh(\frac{1}{2}\eta) \mathbb{I}_{2x2}\pm \sinh(\frac{1}{2}\eta)\sigma^3,$$
where $\sigma^3$  is a Pauli matrix, $\eta$ is the "rapidity" and $\mathbb{I}_{2x2}$ is the 2x2 identity matix. Thank you for your attention.
 A: The first step is to expand the exponential as a Taylor series:
$$ \exp \left( \pm \frac{1}{2} \eta \sigma \right) = I \pm \left( \eta \sigma/2 \right) + \frac{1}{2} \left( \eta \sigma / 2 \right)^2 \pm \frac{1}{3!} \left( \eta \sigma / 2 \right)^3 \dots$$
Noting that $\sigma^2 = I$, this series can now be split into even and odd terms: the even terms sum to $\cosh \left( \eta/2 \right) \ I$, and the odd terms to $\pm \sinh \left( \eta/2 \right) \ \sigma$, which gives the result you want.
A: You just need to use the series expansion and observe that 
$$(\sigma^3)^{2n} = \mathbb I_{2}$$
and
$$(\sigma^3)^{2n+1} = \sigma^3$$
A: The shortest way is the following one. Define
$$f(\eta):= \exp{(\pm\frac{1}{2} \eta \sigma^3)}\:,\quad g(\eta):=\cosh(\frac{1}{2}\eta) \mathbb{I}_{2x2}\pm \sinh(\frac{1}{2}\eta)\sigma^3.$$
Now notice that
$$f'(\eta) = \pm\frac{1}{2}  f(\eta)\sigma_3$$
and
$$g'(\eta) = \frac{1}{2} \sinh(\frac{1}{2}\eta) \mathbb{I}_{2x2}\pm\cosh(\frac{1}{2}\eta)\sigma^3 =  \pm\frac{1}{2}  g(\eta)\sigma_3\:.$$
We conclude that both functions satisfy the same first order system of  differential equations 
$$y'= \pm \frac{1}{2} y\sigma_3$$
and the same initial condition 
$$y(0)=f(0)=g(0) = \mathbb{I}_{2x2}$$
Since the functions are smooth, we can safely apply the uniqueness theorem for first order differential equations concluding that:
$$f(\eta)=g(\eta) \quad \eta \in \mathbb{R}\:.$$
