Given a 4-vector $p^\mu$ the Lorentz group acts on it in the vector representation:
$$ \tag{1} p^\mu \longrightarrow (J_V[\Lambda])^\mu_{\,\,\nu} p^\nu\equiv \Lambda^\mu_{\,\,\nu} p^\nu. $$
However, I can always represent a 4-vector $p^\mu$ using left and right handed spinor indices, writing
$$ \tag{2} p_{\alpha \dot{\alpha}} \equiv \sigma^\mu_{\alpha \dot{\alpha}} p_\mu.$$
So the question is: **in what representation does the Lorentz group act on $p_{\alpha \dot{\alpha}}$?**

There are *a lot* of questions about this and related topics around physics.se, with a lot of excellent answers, so let me clear up more specifically what I am asking for.

I already know that the answer to this question is that the transformation law is
$$ \tag{3} p_{\alpha \dot{\alpha}} \rightarrow (A p A^\dagger)_{\alpha\dot{\alpha}}$$
with $A \in SL(2,\mathbb{C})$ (how is mentioned for example in this answer by Andrew McAddams).
I also understand that
$$ \tag{4} \mathfrak{so}(1,3) \cong \mathfrak{sl}(2,\mathbb{C}),$$
(which is explained for example here by Edward Hughes, here by joshphysics, here by Qmechanic).

So what is missing? Not much really. Two things:

How do I obtain **(3)** and what is the specific form of $A$, i.e. its relation with the vector representation $\Lambda^\mu_{\,\,\nu}$?
Defining the following
$$ (\tilde p) \equiv p^\mu, \qquad \Lambda \equiv \Lambda^\mu_{\,\,\nu},$$
$$ \sigma \equiv \sigma_{\alpha \dot \alpha}, \qquad \hat p \equiv p_{\alpha \dot \alpha},$$
we can rewrite **(1)** and **(2)** in matrix form as
$$ \tag{5} \hat p \equiv \sigma \tilde p
\rightarrow \sigma \Lambda \tilde p
= ( \sigma \Lambda \sigma^{-1}) \hat p,$$
however, *this disagrees with ***(3)** which I know to be right, so what is wrong with my reasoning?

Why does the transformation law **(3)** has a form
$$\tag{6} A \rightarrow U^{-1} A U,$$
while the usual vector transformation **(1)** has a form $V \rightarrow \Lambda V$?
I suspect this comes from a similar reason to that explained here by Prahar, but I would appreciate a confirmation about this.

This post imported from StackExchange Physics at 2015-01-13 11:51 (UTC), posted by SE-user glance