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While inspecting the $(\frac{1}{2},\frac{1}{2})$ representation of the Lorentz group and defining a right-handed spinor with upper dotted index and a left-handed spinor with lower undotted index and therefore

$$v^{\dot{a}}_b = v^{\nu} \sigma_{b \nu}^{\dot{a}}= \begin{pmatrix} v_1+iv_2&-(v_0+v_3) \\v_0-v_3&-(v_1-iv_2) \end{pmatrix}$$ and $$v^{ \dot{a} b} = \epsilon^{bc} v_c^{\dot{a}}$$

with spinor metric $$\epsilon^{bc}= \begin{pmatrix} 0&1 \\ -1&0 \end{pmatrix}$$

it can be seen that the rank 2 spinor $v^{ \dot{a} b}$ has exactly the same transformation properties as a 4-vector $v_{\mu}$. But in the same way it can seen that for example $v^{\dot{a}}_b$ transforms differently. Are there any physical interpretations of those objects? Do they describe specific particles (not vector particles?)? Any help or reading suggestion would be much appreciated.

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    $\begingroup$ It is better to use standard formalism, see for instance this paper, formulae $2.18 \to 2.32$. Now, the matrix $\epsilon^{bc}$ is clearly inverting the spin projection $m$ of the undotted (spinor) part, for instance, speaking of states, $( \pm \frac{1}{2}, +\frac{1}{2}) \to \mp ( \mp \frac{1}{2}, +\frac{1}{2})$. $\endgroup$ – Trimok Jan 15 '14 at 19:05
  • $\begingroup$ Thanks for your reading suggestion, the paper looks great. In eq. 2.32 of the paper the explicit connection between 4-vectors and bi-spinors is made. What is unclear to me is why this particular choice has to be made and what does a different choice like for example $v^{\dot{a}}_b = v^{\nu} \sigma_{b \nu}^{\dot{a}}= \begin{pmatrix} v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3 \end{pmatrix}$ describe? $\endgroup$ – jak Jan 15 '14 at 19:15
  • $\begingroup$ I think that the idea, for instance in looking at equation $2.30$, is that "matrices" have $2$ indices at the same level (up or down). Now, you could choose your own conventions, as long as the whole formalism is coherent, but it is better to follow the standard rules which are more or less used in scientific papers. $\endgroup$ – Trimok Jan 15 '14 at 19:41
  • $\begingroup$ This sound logically but as far as I can see the transformation matrices always have one index up and one index down and this somehow contradicts the convention. See for example in the line below 2.18. $\endgroup$ – jak Jan 15 '14 at 19:51
  • $\begingroup$ Yeah, but these matrices have space-time (vector) indices, not spinor indices, so there is no contradiction. $\endgroup$ – Trimok Jan 15 '14 at 19:53
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The Lorentz group is the set of matrices that preserve the four-vector dot product,

$v^2 = v_0^2 - v_1^2 - v _2 ^2 - v _3 ^2 $

Both $ v ^\mu $ and $ v ^{ \alpha \dot{\alpha} } \equiv v ^ {\mu} (\sigma_\mu) ^{ \alpha \dot{\alpha} } $ transform under the Lorentz group. They just transform under a different representation (but still under the same transformation!).

Furthermore, there is a one-to-one correspondence between every vector $ v ^\mu $ and $ v ^{ \alpha \dot{\alpha} } $. So you can always choose to work either in one representation or the other.

Note that the only important distinction between the two representations is that for every boost matrix for $ v ^\mu $, $ \Lambda ^{ \mu } _\nu $, there exist two equivalent boost matrices for $ v ^{ \alpha \dot{\alpha} } $, $N _{\alpha \beta}$. This turns out to fix some the inconveniences associated with the fundamental (four-vector) Lorentz group representation.

For example if you have four-vector and you want to boost it in a given direction. You can find the matrix $\Lambda ^\mu_\nu $ and multiply $v ^\mu $ as

$ v' ^\mu = \Lambda _\nu ^\mu v ^\nu $

or you could find the matrix $N _{\alpha \beta } $ and apply it to the rank 2 spinor:

$ v' _{\alpha \dot{\alpha}} = N _\alpha ^ \beta v _{\beta \dot{\gamma}} N ^{\ast \, \dot{\gamma}} _{\dot{\alpha}} $

Both methods are equivalent.

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  • $\begingroup$ Thanks for your answer. While i do agree with everything you say there is still something unclear to me. It makes a difference in the transformation behaviour of $v^{\nu}$ if i choose $v^{\dot{a}}_b = v^{\nu} \sigma_{b \nu}^{\dot{a}}= \begin{pmatrix} v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3 \end{pmatrix}$ or if i choose $v^{\dot{a}b} = v^{\nu} \sigma_{ \nu}^{\dot{a}b}= \begin{pmatrix} v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3 \end{pmatrix}$ as my first definition (i guess this corresponds in some way to choosing a basis). I don't understand what the different transforming object describes. $\endgroup$ – jak Jan 15 '14 at 18:25
  • $\begingroup$ We usually define the indices $\alpha, \dot{\alpha}$ such that $v ^{\alpha \dot{\alpha}} = (\sigma ^{\mu} v _\mu) ^{\alpha \dot{\alpha}}$. This leads to a particular form for the transformation matrices. However, If you have the transformation matrix for $v ^{\alpha \dot{\alpha}}$ you can easily calculate for $v ^{\alpha}_{ \dot{\alpha}}$ by applying the Levi Cevita tensor. The different transforming object describes a boost or rotation in space. $\endgroup$ – JeffDror Jan 15 '14 at 19:21
  • $\begingroup$ I made the same definitions for the indices, so $v ^{\alpha \dot{\alpha}} = (\sigma ^{\mu} v _\mu) ^{\alpha \dot{\alpha}}$ gives the correct transformations for the coefficents $v_{\nu}$. They transform like the coefficents of a 4-vector. Nevertheless choosing $v ^{\alpha \dot{\alpha}} = (\sigma ^{\mu} v _\mu) ^{\alpha \dot{\alpha}}$ appears to me somewhat unmotivated. With the same definitions for the indices I can define another object $v' ^{\alpha}_{ \dot{\alpha}} = (\sigma ^{\mu} v' _\mu) ^{\alpha}_{ \dot{\alpha}}$ that transforms completly different. What does this object describe? $\endgroup$ – jak Jan 15 '14 at 19:37
  • $\begingroup$ Sure. You can assign a different convention. But that will also change the convention of the your transformation matrices. So in the end if you switch back to four-vector form, you will still get the same transformed vector $v_ \mu '$. The position of the indices isn't sacred. All that is important is that you are consistent. $\endgroup$ – JeffDror Jan 15 '14 at 20:21
  • $\begingroup$ I don't mean a different convention. I mean staying in one convention and after looking at $v ^{\alpha \dot{\alpha}} = (\sigma ^{\mu} v _\mu) ^{\alpha \dot{\alpha}}$ I take a look at how $v' ^{\alpha}_{ \dot{\alpha}} = (\sigma ^{\mu} v' _\mu) ^{\alpha}_{ \dot{\alpha}}$ transforms. The coefficents $v' _\mu$ do not transform like the indices of a four vector. The definitions we make are as far as i understand for spinor indices, i.e. the position of right-handed and left-handed spinor indices. In other words: What fixes me to the first choice? $\endgroup$ – jak Jan 15 '14 at 20:36

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