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!


2 Answers 2


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.

  • $\begingroup$ Just out of curiosity, how certain are you that the ($i\sigma_2$) invariant matrix lives in $V^* \otimes V^*$? $\endgroup$
    – Craig
    Commented Feb 26, 2022 at 9:27
  • $\begingroup$ 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 $V^*\otimes V^*$. $\endgroup$ Commented Feb 26, 2022 at 10:11
  • $\begingroup$ 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! $\endgroup$
    – Craig
    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.

  • $\begingroup$ 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. $\endgroup$
    – jak
    Commented Jan 13, 2014 at 9:08
  • $\begingroup$ 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. $\endgroup$
    – jak
    Commented Jan 13, 2014 at 9:13
  • $\begingroup$ @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? $\endgroup$ Commented Jan 13, 2014 at 9:57
  • $\begingroup$ 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? $\endgroup$
    – jak
    Commented Jan 13, 2014 at 10:16
  • 1
    $\begingroup$ 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: $\overline{L}$, $L^{t-1}$, $\overline{L^{t-1}}=L^{\dagger-1}$. $\endgroup$ Commented Jan 13, 2014 at 17:50

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