# Recovering 4-vector Lorentz transformation from spinor formalism

I'm trying to recover the 4-vector transformation laws using spinors. I have defined

$$v^{\dot{a}b} = v^{\nu} \sigma_{\nu}^{\dot{a}b}$$

as usual, with $\sigma_0=1$.

Now with the rules for dotted and undotted spinor indices i get the transformed v for a boost in the z-direction

$$v^{\dot{a}b} \rightarrow v^{\dot{a'}b'}= \left({\mathrm{e }}^{ - \theta \frac{\sigma_3}{2}} \right)^{\dot{a'}}_{\dot{a}} \left({\mathrm{e }}^{ \theta \frac{\sigma_3}{2}} \right)^{b'}_{b} v^{\dot{a}b} = \begin{pmatrix} {\mathrm{e }}^{- \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{\theta}{2}} \end{pmatrix} \begin{pmatrix} v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3 \end{pmatrix} \begin{pmatrix} {\mathrm{e }}^{ \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ - \frac{\theta}{2}} \end{pmatrix}$$

where i have used the fact that $\sigma_3$ is diagonal and that ${\mathrm{e }}^A= \begin{pmatrix} {\mathrm{e }}^{A_{11}}&0\\0&{\mathrm{e }}^{A_{22}} \end{pmatrix}$ holds for every diagonal matrix A. This gives me the wrong transformation!

It would give me the correct transformation if I had

$$v^{\dot{a}b} \rightarrow v^{\dot{a'}b'}= \left({\mathrm{e }}^{ - \theta \frac{\sigma_3}{2}} \right)^{\dot{a'}}_{\dot{a}} \left({\mathrm{e }}^{ \theta \frac{\sigma_3}{2}} \right)^{b'}_{b} v^{\dot{a}b} = \begin{pmatrix} {\mathrm{e }}^{- \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{\theta}{2}} \end{pmatrix} \begin{pmatrix} v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3 \end{pmatrix} \begin{pmatrix} {\mathrm{e }}^{ - \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{\theta}{2}} \end{pmatrix}$$

But I can't figure out why this should be the case. The only possibility I can think about would be if a relation like $\ A'_{\nu\mu} = M_{\mu}^{\ \rho}(M_{\nu}^{\ \theta})^{-1}A_{\rho\theta }$ $\rightarrow$ $A'=MAM$ would hold, but i can't find a formula like this. Doing the summation by hand i get the same result as with normal matrix multiplication without using the inverse matrix on the right hand side. Any tip or help would be much much appreciated!

There are four complex 2-dimensional spaces of "spinors with two components":

$V$ space of right spinors, with the abstract index notation: $\xi^a$,

$\overline{V}$ space of conjugate spinors, $η^{\dot{a}}$, (the complex linear space of linear maps $\overline{V}^*\to C$)

$V^∗$ space of dual spinors, $\xi_a$, (the complex linear space of linear maps $V\to C$)

$\overline{V}^∗$ space of dual-conjugate spinors or left spinors, $\eta_{\dot{a}}$ (the complex linear space of anti-linear maps $V\to C$)

There is a non-degenerate skew-symmetric form $\epsilon: V\otimes V \to C$ and it fixes canonical bases in $V$: all those where $\epsilon$ is represented by the matrix $i\sigma_2$.

If $\{e_i\}_{i=1,2}$ is such a canonical basis, every other canonical basis is obtained as $\{e'_i\}_{i=1,2}$, with:

$$e_i = {L^j}_i e'_j \quad L \in SL(2,C)\:,$$

so that

$\xi = \xi^a e_a = \xi'^b e'_b$ verifies:

$$\xi'^b = {L^b}_a \xi^a\:.$$

A canonical basis $\{e_a\}_{a=1,2}\subset V$ induces analogous associated canonical bases in $V^*$, $\overline{V}$ and $\overline{V}^*$, respectively indicated by $\{e^{*a}\}_{a=1,2}\subset V^*$, $\{\overline{e}_{\dot{a}}\}_{a=1,2}\subset \overline{V}$, $\{\overline{e}^{\dot{a}}\}_{a=1,2}\subset \overline{V}^*$, by the requirments: $$e^{*a}(e_b)= \delta^a_b\:, \quad \overline{e}_{\dot{a}}(\overline{e}^{*\dot{b}}) = \delta^{\dot{b}}_{\dot{a}}\:,\quad \overline{e}^{*\dot{a}}(e_b) = \delta^{\dot{a}}_b$$

Referring to these bases, if $\xi^a \in V$ then $\overline{\xi^a} \in \overline{V}$, therefore components of tensors in $\overline{V}$ transforms with $\overline{L}$, when changing canonical basis. Similarly in $V^*$ one has to use $L^{t-1}$ and $L^{\dagger -1}$ in $\overline{V}^*$.

Finally, with the given definitions, there is a canonical isomorphism iduced by the metrical spinor $V \to V^*$, in components of canonical bases (there is a sign to be fixed depending on preferred conventions)

$$\xi^a \to \eta_{b}:= i \sigma_{2 ab} \xi^a\:.$$

The relation between spinors and $4$-vectors is based on the following theorem connecting real 4-vectors with Hermitean tensors in $V\otimes \overline{V}$.

THEOREM. Let $\pi: SL(2,C) \to SO(1,3)\uparrow$ be the covering Lie-group homomorphism ($SL(2,C)$ being the universal covering of $SO(1,3)\uparrow$). Let $\{e_a\}_{a=1,2}$ and $\{\overline{e}_{\dot{a}}\}_{a=1,2}$ be associated canonical bases of $V$ and $\overline{V}$ respectively and $\{f_\mu\}_{\mu=0,1,2,3}$ a pseudo-orthonormal basis in Minkowski spacetime. If $v:= v^\mu f_\mu$ is a real $4$-vector and $\Xi_v := v^\mu \sigma_\mu^{a\dot{b}} e_a\otimes \overline{e}_{\dot{b}}$, then:

$$\Xi_{\pi(L)v} = L \Xi_v L^{\dagger}\:, \quad \forall L \in SL(2,C)\:.$$

If $L$ is such that $\pi(L)$ is a boost, then, as is well-known $L=L^\dagger$, so your final supposition is true.

• Just out of curiosity, how certain are you that the ($i\sigma_2$) invariant matrix lives in $V^* \otimes V^*$? Commented Feb 26, 2022 at 9:27
• I am not sure to understand well your question. We are referring to a linear map from $V\otimes V$ to $\mathbb{C}$ (equivalently, bilinear from $V\times V$ to $\mathbb{C}$), so this map stays in the dual space of the above tensor product. That dual space is $V^*\otimes V^*$. Commented Feb 26, 2022 at 10:11
• Yes sorry it wasnt clear. Mostly was unsure if the bilinear form took two undotted spinors or one of each type. But after some reflecting it's clear. Thanks! Commented Feb 26, 2022 at 18:05

First thing is that you are missing the imaginary unit in the exponential. The correct transformation matrix should be $$M=e^{i\frac{\theta}{2}}.$$ Up to this small misprint, the first expression you wrote actually gives the desired transformation of the components $v_1$ and $v_2$. If we do the computation \begin{aligned} &\begin{pmatrix} {\mathrm{e }}^{- \frac{i\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{i\theta}{2}} \end{pmatrix} \begin{pmatrix} v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3 \end{pmatrix} \begin{pmatrix} {\mathrm{e }}^{ \frac{i\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ - \frac{i\theta}{2}} \end{pmatrix}=\\ &\begin{pmatrix} v_0+v_3 & e^{-i\theta }(v_1-iv_2) \\ e^{i\theta }(v_1+i v_2) & v_0 + v_3 \end{pmatrix}. \end{aligned} This implies \begin{aligned} v_1'-iv_2'&=e^{-i\theta}(v_1-iv_2),\\ v_1'+iv_2'&=e^{+i\theta}(v_1+iv_2), \end{aligned} that finally gives the correct transformation.

I'm not sure, may be you made a mistake in multiplying of matrices.

• Thanks for you answer. What you consider is a rotation, i'm looking at a boost, therefore there is no "i" in the exponent. Furthermore this is exactly my problem: I get a rotation as a result if i take a look at a boost and a boost if i'm considering a rotation. Dotted and undotted indices transform the same under rotations but differ under boost by a minus sign. This gives rise to the problem described above. For a boost i need the same exponents on both sides, for a rotation i need different signs in the exponent.
– jak
Commented Jan 13, 2014 at 9:08
• I derived the transformation laws for dotted and undotted indices and checked them with a book: $v^{\dot{a'}b}= \left({\mathrm{e }}^{ \vec{ i \phi} \vec{\frac{\sigma}{2}} - \vec{ \theta} \vec{\frac{\sigma}{2}}} \right)^{\dot{a'}}_{\dot{a}} v^{\dot{a}b}$ and $v^{\dot{a}b'}= \left({\mathrm{e }}^{ \vec{ i \phi} \vec{\frac{\sigma}{2}} + \vec{ \theta} \vec{\frac{\sigma}{2}}} \right)^{b'}_{b} v^{\dot{a}b}$ But if there is anything wrong with them, this would explain a lot.
– jak
Commented Jan 13, 2014 at 9:13
• @JakobH It seem to me that there is nothing wrong in your final suggestion, whereas I do not understand what is the general rule you assume to transform spinors. Since I adopt different conventions I have to ask you to make explicit yours. Referring to yours what is the transformation rule for dotted spinors? I mean, if $\xi'^a = {L^a}_b \xi^b$ then $\eta^{\dot{a}} = {M^{\dot a}}_b \eta^{\dot{b}}$. What is the relation between $L$ and $M$ referring to your conventions? Commented Jan 13, 2014 at 9:57
• I can transform a dotted (LH) into a undotted (RH) spinor by complex conjugating and multiplication with $i \sigma^2$. To be explicit $i \sigma^2 (\Psi_L)^{\star} = \Psi_R$ . I don't know any rule relating the transformation laws L and M as you indicated explicitly because i have never done it, but a first guess would be that $i \sigma^2$ does this. Do i have to transform the indices of the transformation acting on the undotted indice first to dotted before i can write it in matrix form? If yes, why?
– jak
Commented Jan 13, 2014 at 10:16
• the bar is the complex conjugation. You have 4 spaces: $V$ spinors $\xi^a$, $\overline{V}$ conjugate spinors $\eta^{\dot{a}}$, $V^*$ dual spinors $\xi_{a}$, $\overline{V}^*$ dual conjugated spinors $\eta_{\dot{a}}$. When you change spinors in $V$ with some $L \in SL(2,C)$, spinors in the other spaces change under the action of, respectively: $\overline{L}$, $L^{t-1}$, $\overline{L^{t-1}}=L^{\dagger-1}$. Commented Jan 13, 2014 at 17:50