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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:

  1. 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?

  2. 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.

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Your equation (3) comes from the following steps. First, a dotted index transforms in the complex conjugate representation of an undotted index. For a tensor product, each index transforms according to its own representation. Thus $$p_{a\dot a} \mapsto A_{ab} \bar A_{\dot a \dot b} p_{b\dot b} = A_{ab} p_{b\dot b} A^\dagger_{\dot b \dot a}$$ where on the left side of the equal sign we have elementwise complex conjugation. Putting the conjugate matrix on the right we have to take a transpose to get order of indices right.

In reasoning about (4) and (5) you are neglecting the transformation of $\sigma^\mu_{a\dot a}$. The correct description of the relation $p_{a\dot a} = \sigma^\mu_{a\dot a} p_\mu$ is that the 4-vector representation is equivalent to the $(\frac 1 2,0)\otimes(0,\frac 1 2)$ representation, by means of the linear transformation $$\sigma^\mu_{a \dot a} : V\to (\frac12,0)\otimes(0,\frac12)$$ meaning that $\sigma^\mu_{a\dot a}$ belongs to the space $(\frac 1 2,0)\otimes (0,\frac12) \otimes V^*$, on which the (double cover of the) Lorentz group acts. In fact, it acts like $$\sigma^\mu_{a\dot a} \mapsto A_{a\dot b} \sigma^\nu_{b\dot b} A^\dagger_{\dot b \dot a} (\Lambda^{-1})_\nu^\mu $$ so that $A^\mu_{a \dot a} p_\mu$ indeed has the correct transformation law.

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    $\begingroup$ Thank you very much, that definitely sorted it out. Just one thing: could you also provide some reference where I can find more on the subject (particulary where I can find an exposition of the transformation rules of $\sigma^\mu_{\,\,\alpha\dot \alpha}$ that you quoted)? $\endgroup$ – glS Jan 12 '15 at 15:38
  • $\begingroup$ That $\sigma^\mu_{a \dot a}$ transforms in that manner is really implicit in the indices it has, so I don't know if it's written out anywhere. My best guess is Penrose & Rindler, but I haven't checked. $\endgroup$ – Robin Ekman Jan 12 '15 at 16:06

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