2
$\begingroup$

I am learning (Weyl) spinor formalism from Müller-Kirsten and Wiedemann's Introduction to Supersymmetry (2nd Ed., WS, 2010, here). I am quite confused about the transformation between left-handed spinors and right-handed spinors.

On P31, left-handed spinors are defined to transform under self-representation $M\in SL(2,\mathbb{C})$: \begin{equation} \psi^{\prime}_A=M_A^B \psi_B.\tag{1.60} \end{equation}

On P32, right-handed spinors are defined to transform under complex conjugate self-representation $M^{*},M\in SL(2,\mathbb{C})$: \begin{equation} \bar{\psi}^{\prime}_\dot{A}=(M^*)_\dot{A}^\dot{B} \bar{\psi}_\dot{B} \qquad (1.62) \end{equation}

An equivalent representation to complex conjugate self-representation is discussed on P38: \begin{equation} \bar{\psi}^{\prime\dot{A}}=(M^{*-1T})^\dot{A}_\dot{B} \bar{\psi}^\dot{B} \qquad (1.81) \end{equation}

The authors claim that: \begin{equation} \psi^{A}=\bar{\psi}^{*}_{\dot{A}}\qquad (1.200) \end{equation} or equivalently:

\begin{equation} \psi^{A}=\bar{\psi}^{*}_{\dot{B}}(\bar{\sigma}^0)^{\dot{B}A}\qquad (1.199) \end{equation}

The right-hand sides in the two equations are the same since $(\bar{\sigma}^0)^{\dot{B}A}$ equals identity matrix in components.

Here is my question: it seems like (1.200) contradicts the definitions (1.60) and (1.81). To be precise, for a $M\in SL(2,\mathbb{C})$, we can write the following: \begin{align} &\psi^{\prime}=M\psi\\ &\bar{\psi^{\prime}}=(M^{*-1T})\bar{\psi}\\ &\psi^{*}=\bar{\psi} \end{align} where I regard the lower undotted index and upper dotted index as column indices to rewrite the above relations in matrix form (and leave indices implicit). We can deduce that \begin{equation} M=M^{-1T} \end{equation} which is NOT the property of matrices in $SL(2,\mathbb{C})$. Is there any way out?

$\endgroup$
2
  • $\begingroup$ @NiharKarve But why? To my understanding, $M^*{}_\dot\alpha{}^\dot\beta$ is defined to be $(M_{\alpha}{}^\beta)^*$. Since for a given $M\in SL(2,\mathbb{C})$, the components of $M$ are fixed, is there any reason why $M^*$ is not composed of the same elements? $\endgroup$
    – ZHC
    Mar 22 at 15:43
  • $\begingroup$ Maybe try writing the transformation matrices out explicitly. Your conclusion about $\mathcal M = \mathcal M^{-1T}$ is almost correct - $\mathcal M$ is not equal, but equivalent (in the representation-theoretic sense) to $\mathcal M^{-1T}$ by conjugation with $\epsilon_{\alpha\beta}$. I'll write an answer if I have time. $\endgroup$ Mar 24 at 12:02
0
$\begingroup$

First of all there are covariant and contravariant spinors which transform according to representations of $SL(2,\mathbb{C})$ like:

$$ \psi'_A = M^{\,B}_A \psi_B \quad\text{and}\quad \psi'^C = M_{\,D}^C \psi^D \quad\tag{1} $$.

The matrix $M^{C}_{\,D} = (M^{-1T})_{C}^{\,D}$ is the contragredient to $M$. So it is a different matrix, but actually it is equivalent to the matrix representation $M$ since a similarity transformation $S$ can be found so that

$$M^{-1T}\equiv M^{cg} =S M S^{-1}\quad \text{where}\quad S=\sigma_2$$

($\sigma_2$ is the second Pauli-matrix) or explicitly:

$$M^{-1T} = \sigma_2 M \sigma_2^{-1}$$

So basicly they are the same. From both we can now construct the complex-conjugate representations of $SL(2,\mathbb{C})$ by taking the complex-conjugate of equation (1):

$$ \psi'_\dot{A} = M^{\,\dot{B}}_\dot{A} \psi_\dot{B} \quad\text{and}\quad \psi'^\dot{C} = M_{\,\dot{D}}^\dot{C} \psi^\dot{D}$$.

where we understand $M_{A}^{\ast\,B} = M^{\,\dot{B}}_\dot{A} $ and $M_{\,\,D}^{\ast C} =M_{\,\,\dot{D}}^{\dot{C}}$ and $\psi^\ast_A =\psi_\dot{A}$ and $\psi^{\ast B} = \psi^\dot{B}$.

The equation $\overline{\psi'} = M^{\ast -1 T}\overline{\psi}$ in the post has to be understood with writing indices like:

$$ \psi'^\dot{C} = M_{\,\dot{D}}^\dot{C} \psi^\dot{D}$$.

So if we take off the complex-conjugation from it we get:

$$\psi'^C = M_{\,D}^C \psi^D \quad\text{or}\quad \psi' = M^{-1T}\psi $$

The transformation (matrix) is different from

$$\psi'_C = M_{C}^{\,D} \psi_D$$

but $M_{\,D}^C$ and $M_{C}^{\,D}$ are equivalent representations.

By the way I don't understand the equation (1.200). In particular the indices on both sides are not really equivalent. Also equation (1.199) is confusing. However, covariant and contravariant indices can be lifted up or down with the $\epsilon$-matrix :

$\psi^A =\epsilon^{AB} \psi_B \quad\text{where}\quad \epsilon = i\sigma_2$

again with $\sigma_2$ as second Pauli-matrix. Actually in supersymmetry if a Dirac-spinor $\binom{\psi_A}{\chi^\dot{B}}$ is considered then it is often shown that $i\sigma_2\chi^\ast$ transforms like $\psi$. In correct index writing it can be understood as (the dot is cancelled by the complex-conjugation and $\sim$ means "transforms like"):

$$ \epsilon_{BA} (\chi^\dot{A})^\ast \sim\psi_B$$

More details on the transformation of spinors transforming according to $SL(2,\mathbb{C})$ are given in my post How to prove these relations for Pauli matrices? .

$\endgroup$
1
  • $\begingroup$ You must be using different conventions to the OP: I haven't seen many (any?) sources say $(\bar\psi^\dot\alpha)^*=\psi^\alpha$ $\endgroup$ Mar 25 at 5:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.