Given a 4-vector, we can always define a 2x2 hermitian matrix:

$$X=x^\mu \sigma_\mu=\left(\matrix{x^0+x^3&x^1-ix^2\\x^1+ix^2&x^0-x^3} \right)$$

Where $\sigma_i$ are just the Pauli matrices. In this base, we can define the Lorentz transformations as $\Lambda(L)$, where $X'=LXL^\dagger$. This representation forms the basis of the linear group $\mathrm{SL}(2, \mathbb C)$.

However, I'm curious on the exact expression of the $2\times 2$ matrices that represent these Lorentz transformations (they don't appear in the literature).

I've read that they can be characterized by just 6 real parameters (which is reminiscent of the 6 parameters for the $\mathrm{SO}(3)$ Lorentz representation).


Most of the answer to the question can be found in the answer of the post Representation $(1/2,1/2)$ of Lorentz group,

I just add some additional information: The relationship

$$L^i_k =\frac{1}{2} Tr( A \sigma_k A^\dagger \tilde{\sigma}^i)$$

can be inverted and as a result we get the matrix $A$ expressed by a 4-vector Lorentz-transformation $L^i_k$:

$$ A =\frac{1}{N} L^i_k \sigma_i \tilde{\sigma}^k$$

with $N$ :

$$ N =\pm \sqrt{det(L^i_k \sigma_i \tilde{\sigma}^k)}$$

Spin representations as those of $SL(2,C)$ are double-valued which explains both signs.

  • $\begingroup$ Thank you very much. I didn't find that post before, but it gave me the information I was looking. Coupled with your answer, it was very useful to understand where the two signs come and thus why $\Lambda(L)=\Lambda(-L)$. $\endgroup$ – Charlie Aug 26 '18 at 15:40

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