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I want to show that the Lorentz representation $(1/2,1/2)$ corresponds to the normal vectorial representation $A^\mu$. For this I need to show that the double spinors $A_{ij}=(A_\mu\sigma^\mu\sigma^2)_{ij}$ that transforms like $$ A_{ij}\rightarrow(\Lambda_R)_{ik}(\Lambda_L)_{jl}A_{kl} $$ implies that the $A_\mu$ transforms like $$ A^\mu\rightarrow\Lambda^\mu_{\>\>\nu} A^\nu. $$

But I really don't know how to even start this. Maybe I have issues understanding what is this representation. Here is what I did: $$ A'_\mu\sigma^\mu\sigma^2=\Lambda_R\Lambda_LA_\mu\sigma^\mu\sigma^2 $$ Then multiply by $\sigma^2\sigma_\nu$ on the left and taking the trace $$ Tr(\sigma^2\sigma_\nu A'_\mu{\sigma^\mu\sigma^2)=Tr(\sigma^2\sigma_\nu\Lambda_R\Lambda_LA_\mu\sigma^\mu\sigma^2)\\ 2A'_\mu\delta_\nu^\mu=Tr(\sigma^2\Lambda_R\sigma^2\Lambda_LA_\mu\delta^\mu_\nu)\\ 2A'_\nu=Tr(\Lambda_L^\dagger\Lambda_LA_\nu)\\ 2A'_\nu=Tr(\sigma^2\Lambda_L^{-1}\sigma^2\Lambda_LA_\nu)\\ 2A'_\nu=A_\nu} $$which can't be good.

EDIT:

I came up with a solution: $$ A_{ij}'=(\Lambda_R)_{ik}(\Lambda_L)_{jl}A_{kl}\\ (A_\mu'\sigma^\mu\sigma^2)_{ij}=(\Lambda_R)_{ik}(A_\mu\sigma^\mu\sigma^2)_{kl}(\Lambda_L)_{lj}^T\\ \text{Tr}(\sigma^2\sigma_\nu A_\mu'\sigma^\mu\sigma^2)=\text{Tr}(\sigma^2\sigma_\nu\Lambda_RA_\mu\sigma^\mu\sigma^2\Lambda_L^T)\nonumber\\ \text{Tr}(\sigma_\nu A_\mu'\sigma^\mu)=\text{Tr}(\sigma_\nu\Lambda_RA_\mu\sigma^\mu\sigma^2\Lambda_L^T\sigma^2)\nonumber\\ A_\mu'\text{Tr}(\sigma_\nu\sigma^\mu)=\text{Tr}(\sigma_\nu\Lambda_R\sigma^\mu\Lambda_R^*)A_\mu\nonumber\\ 2A_\mu'{\delta}{_\nu^\mu}=\text{Tr}(\sigma_\nu\Lambda_R\sigma^\mu\Lambda_R^*)A_\mu\nonumber\\ A_\nu'=\frac{1}{2}\text{Tr}(\sigma_\nu\Lambda_R\sigma^\mu\Lambda_R^*)A_\mu\nonumber\\ A_\nu'={\Lambda}{_\nu^{\>\>\mu}}A_\mu. $$ but I'm not sure why my indices are not good (compared to what I'm supposed to get ${A^{\mu}}'=\Lambda^\mu_{\>\>\nu}A^\nu$).

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/321292/2451 and links therein. $\endgroup$ – Qmechanic Oct 31 '17 at 16:55
  • $\begingroup$ It is not really what I want to do, but clearly it is related. $\endgroup$ – Soap312 Oct 31 '17 at 18:40
  • $\begingroup$ What is $\sigma^2$ matrix doing there and how come there are no dotted vs. undotted indices? $\endgroup$ – DanielC Nov 1 '17 at 11:50
  • $\begingroup$ @DanielC I really don't know. But I think the $\sigma^2$ is there because of the symmetries of the $\Lambda_R$ and $\Lambda_L$. $\endgroup$ – Soap312 Nov 1 '17 at 14:19
  • $\begingroup$ What book or lecture note do you use? $\endgroup$ – DanielC Nov 1 '17 at 14:56
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There is a one-to-one correspondence between a 2x2-matrix which transforms like a product of a undotted spinor and a dotted spinor (for more see below). As far as I've seen your post, you tried to combine 2 spinors of the same type which will not work to make up a 4-vector.

First of all extend the 3 Pauli-matrices to 4 matrices: $\sigma_i =(\bf{1},\sigma_1,\sigma_2,\sigma_3)$.

Furthermore we define $\tilde{\sigma}^i=\sigma_i$. (Note $\tilde{\sigma}^i\neq g^{ik}\sigma_i$, $g$ representing the Minkowski metric).

Construct the 2x2 matrix (summation of double indices which run from 0 to 3.) $X = x^i\sigma_i= \left( \begin{array}{cc} x^0+x^3 & x^1-ix^2\\ x^1+ix^2 & x^0-x^3 \end{array} \right)$. This relation can be inverted: $x^i = \frac{1}{2}Tr(X\tilde{\sigma}^i)$.

Important property of the matrix: $X$ is hermitian $X=X^\dagger$ (the vector $x^i$ is of course real).

We can transform $X$ with any complex unimodular matrix $A \in SL(2,\mathbb{C})$

$X' = A X A^{\dagger}$

The transformed matrix is again hermitian because $X'^{\dagger}= (AXA^{\dagger})^{\dagger} = A X^\dagger A^\dagger = A X A^\dagger = X'$. We can get the corresponding 4-vector from it: $x'^i = \frac{1}{2}Tr(X'\tilde{\sigma}^i)$.

We are almost done, but we have to check an important property. To $x^i \rightarrow x'^i$ be a Lorentz-transformation the length of both vectors has to be the same, so we have to compute $x^ix_i= (x^0)^2-\vec{x}^2 = det X$ (easy to check). This is of course also true for $x'^ix'_i = det X'$.

Now as the transformation matrix $A \in SL(2,\mathbb{C}), det(A)=1$. Therefore we can conclude $x'^ix'_i = det X'=det X=x^ix_i$.

There is just the rule missing how to build from the $A$-matrix the Lorentz-transformation: $x'^i =\frac{1}{2}Tr(X'\tilde{\sigma}^i) = \frac{1}{2}Tr( AXA^\dagger\tilde{\sigma}^i)= \frac{1}{2}Tr( A x^k\sigma_k A^\dagger \tilde{\sigma}^i) = \frac{1}{2}Tr( A \sigma_k A^\dagger \tilde{\sigma}^i) x^k =: L^i_k x^k$

A final remark, the matrix transformation $X' = A X A^{\dagger}$ has to be written with indices in the following form:

$X'^{A\dot{Y}} = A^A_B A^{\dot{Y}}_{\dot{Z}} X^{B\dot{Z}}$.

We can observe that this transformation law corresponds to a representation of the Lorentz-group $D(\frac{1}{2},\frac{1}{2})$ (the first entry $\frac{1}{2}$ in the $D$ corresponds to the undotted spinor, the second to the dotted spinor). The coefficients $A$ with the dotted indices are actually complex conjugated to the coefficients with undotted indices (They are defined like this). This explains why the transform law of $X$ is $X' = A X A^\dagger$.

If the coefficients had (what follows is purely hypothetical) only one type of indices ( $A^A_B=A^{\dot{A}}_{\dot{B}}$ ), the transformation law would be $X' = A X A^T$. But this such a law the correspondence between $x^i$ and $X$ would not be possible.

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