General Lorentz boost for a spinor in Hyperbolic form with Pauli matrices I am having trouble deriving the general Lorentz boost for a spinor with rapidity $\rho $ using the hyperbolic functions. I know that the matrix: $$\exp [-\rho/2 \mathbf{n}\cdot \mathbf{\sigma }]=\cosh (\rho /2) I -\sinh (\rho /2) \mathbf n \cdot \mathbf \sigma. $$
What I have so far is that the above equality holds by reordering the Taylor series. I have also checked the action on the individual components $(x,y,z)$, for example $\Lambda =\exp (-\rho /2 \sigma _z)$. I have $$X=\left[\begin{matrix}t+z & x-iy\\ x+iy & t-z\end{matrix}\right]=tI+x\sigma _x+y\sigma _y+z\sigma _z$$ is the representation for a 4-vector $\mathbf x=(t,x,y,z)$. Then, the action of multiplication by $X'=\Lambda .X.\Lambda ^\dagger $ is the same as performing that hyperbolic rotation along the $z$-axis.
I have a general rotation-matrix by just multiplying the three coordinate rotations, but I cannot figure out how prove that this product equivalent to action $\Lambda .X.\Lambda ^\dagger $, where $\Lambda =\exp [-\rho/2 \mathbf{n}\cdot \mathbf{\sigma }]$
 A: To convert your calculations for $\exp(-\rho\,\sigma_z/2)$ to corresponding calculations for a boost in a general direction, first recall that in $\mathrm{SO}(1,\,3)$ a boost in a general direction has the matrix:
$$\Lambda = R\,\Lambda\,R^{-1}\tag{1}$$
where $R$ is a rotation that rotates a frame whose $z$-axis corresponds with the boost direction to the frame you wish to compute in. Since the Adjoint representation, which maps the double cover $\mathrm{SL}(2,\,\mathbb{C})$ to $\mathrm{SO}(1,\,3)$, is a homomorphism, you can use the same equation as (1) to find the preimage (or one of the two preimages) in $\mathrm{SL}(2,\,\mathbb{C})$ that maps to your general boost in $\mathrm{SO}(1,\,3)$.
So now you need to know the what a preimage of a rotation operator looks  in $\mathrm{SU}(2)\subset\mathrm{SL}(2,\,\mathbb{C})$; unsurprisingly, a rotation in $\mathrm{SU}(2)$ is given by:
$$R(\theta, \hat{n}) = \exp\left(-i\,\frac{\theta}{2}\,\hat{n}\cdot\sigma\right) = \cos\left(\frac{\theta}{2}\right)-\sin\left(\frac{\theta}{2}\right)\,\hat{n}\cdot\sigma\tag{2}$$
So that your calculation yields:
$$\Lambda = \left(\cos\left(\frac{\theta}{2}\right)-\sin\left(\frac{\theta}{2}\right)\,\hat{n}\cdot\sigma\right)\,\left(\cosh\left(\frac{\rho}{2}\right)-\sinh\left(\frac{\rho}{2}\right)\,\hat{n}\cdot\sigma_z\right)\,\left(\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)\,\hat{n}\cdot\sigma\right)\tag{3}$$
which I'll leave you to multiply out.
But take heed that the calculation's result naturally wipes out the rotational degree of freedom one subtly introduces - any frame that is a rotation about the $z$ axis of any other frame wherein the boost is along the $z$ axis is itself a frame wherein the boost is along the $z$-axis. This is because if we replace the rotation in (2) by any rotation of the form $R(\theta, \hat{n})\,\exp(-\phi\,\sigma_z/2)$ for $\phi\in\mathbb{R}$, the $\exp(-\phi\,\sigma_z/2)$ and $\exp(+\phi\,\sigma_z/2)$ that are introduced into (3) both commute with the central $z$-direction boost in (3) and the calculation's result is unchanged. Thus you can apply these ideas with any rotation that aligns the $z$ axis as you need, and the calculation's result will be independent of this choice.
