I am currently reading through Chapter 27 (Spinor helicity formalism) in Schwartz´s &quot;QFT and the SM&quot;. In this chapter it says that since 4-momenta transform in the (1/2,1/2) representation of the Lorentz group they are more naturally described as bispinors.  To represent momenta as bispinors, we use the $\sigma$-matrices: $p^{\alpha,\dot{\alpha}} \equiv \sigma^{\alpha,\dot{\alpha}}_{\mu}p^{\mu}$  Furthermore it says  In the special case that the momentum is lightlike we find $det(p^{\alpha,\dot{\alpha}}) = 0$.(...) As a result from linear algebra is that any 2x2 matrix with zero determinant can be written as an outer product $p^{\alpha,\dot{\alpha}} = \lambda^{\alpha}\lambda^{\dot{\alpha}}$ for two vectors $ \lambda^{\alpha}$ and $\tilde{\lambda}^{\dot{\alpha}}$.  I understand that a left-handed spinor (eg. $\lambda^{\alpha}$) transforms under the (1/2,0)-representation of the Lorentz group and that a right-handed spinor (eg. $\tilde{\lambda}_{\dot{\alpha}}$) transforms under the (0,1/2)-representation of the Lorentz group. On the other hand the bispinor has one dotted and one undotted index. Therefore I don´t know under which representation and how this object transforms.
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 ). I also understand that $$ \tag{4} \mathfrak{so}(1,3) \cong \mathfrak{sl}(2,\mathbb{C}),$$ (which is explained for example  by Edward Hughes,  by joshphysics,  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  by Prahar, but I would appreciate a confirmation about this.
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