Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I think I missed a rule somewhere, because I can contract the following expression in multiple ways.

$\epsilon^{\alpha \beta} \sigma_{\dot{\alpha} \alpha} \epsilon^{\dot{\alpha} \dot{\beta}} = \sigma_{\dot{\alpha} \alpha} \epsilon^{\alpha \beta} \epsilon^{\dot{\alpha} \dot{\beta}}=\sigma_{\dot{\alpha}}^{~\beta}\epsilon^{\dot{\alpha} \dot{\beta}}=\sigma^{ \dot{\beta}\beta}$

I also know that $\epsilon^{\alpha \beta}=-\epsilon^{ \beta \alpha}$, so

$\epsilon^{\alpha \beta} \sigma_{\dot{\alpha} \alpha} \epsilon^{\dot{\alpha} \dot{\beta}} = -\epsilon^{ \dot{\beta} \dot{\alpha}}\sigma_{\dot{\alpha} \alpha} \epsilon^{\alpha \beta} =-\sigma^{\dot{\beta}}_{~\alpha}\epsilon^{\alpha \beta}=-\sigma^{ \dot{\beta}\beta}$

What is going wrong here?

share|improve this question
1  
What is $\epsilon$ and what is $\sigma$? What are the dots denoting? Why did you introduce the negative sign in the second computation? –  Qiaochu Yuan Jul 1 '12 at 17:55
    
$\epsilon$ is a completely antisymmetric tensor, the negative comes from the exchange of a pair of indices. The $\sigma$ is a Pauli matrix and the dots indicate that the index refers to a conjugated spinor. The dots shouldn't be important however, they can just be regarded to be different indices than their dotless counterparts. –  Ben Ruijl Jul 1 '12 at 18:26

1 Answer 1

up vote 3 down vote accepted

Both calculations are wrong, I'm afraid. Since when does $\epsilon^{\lambda\eta}\sigma_{\eta\nu} = \sigma^\lambda{}_\nu$? This is valid only for metric tensor, but $\epsilon^{\lambda\eta}$ ($\begin{pmatrix}0&1\\-1&0\end{pmatrix}$) is not a metric tensor, assuming $\epsilon^{\lambda\eta}$ is the two-dimensional Levi-Civita symbol.

Suppose $T^{\beta\delta} = \epsilon^{\alpha\beta}\sigma_{\gamma\alpha}\epsilon^{\gamma\delta}$, and the metric is identity. Since there are only 4 distinct elements, we could carry out the computation directly:

  • $T^{00} = \epsilon^{10}\sigma_{11}\epsilon^{10} = \sigma_{11} = -\sigma_{00}$
  • $T^{01} = \epsilon^{10}\sigma_{01}\epsilon^{01} = -\sigma_{01}$
  • $T^{10} = \epsilon^{01}\sigma_{10}\epsilon^{10} = -\sigma_{10}$
  • $T^{11} = \epsilon^{01}\sigma_{00}\epsilon^{01} = \sigma_{00} = -\sigma_{11}$

or use the identities $\epsilon_{ab}=\epsilon^{ab}$ and $\epsilon_{ab}\epsilon^{cd} = \delta_a^c\delta_b^d - \delta_a^d\delta_b^c$ to arrive at

$$ T^{\beta\delta} = (\delta^{\alpha\gamma}\delta^{\beta\delta} - \delta^{\alpha\delta}\delta^{\beta\gamma})\sigma_{\gamma\alpha} =\delta^{\beta\delta}\operatorname{tr}(\sigma) -\sigma^{\beta\delta} = -\sigma^{\beta\delta}. $$


Edit: If you define $\epsilon^{\mu\nu}$ to be the metric, then your first formula is wrong, because index raising is done by

$$\huge x^{\color{red}\mu} = g^{\color{red}\mu\color{green}\nu} x_{\color{green}\nu} \tag{Correct} $$

and not

$$\huge x^{\color{red}\mu} = g^{\color{#FF4000}\nu\color{#808000}\mu} x_{\color{green}\nu} \tag{Wrong} $$

especially when your metric $g^{\mu\nu}$ is asymmetric.

share|improve this answer
    
$\epsilon$ behaves like a metric for Weyl spinors and so it has the same behaviour for Pauli matrices. I should have clarified this, sorry for that. –  Ben Ruijl Jul 1 '12 at 20:53
    
@BenRuijl: OK. See update. –  KennyTM Jul 1 '12 at 21:17
    
How does this work with multiple indices, should I contract the $\nu$ first: $x^{\mu \alpha}=g^{\mu \nu} g^{\alpha \beta} x_{\nu \beta}$, or doesn't it matter? That's where my confusion arose. –  Ben Ruijl Jul 2 '12 at 8:56
    
@BenRuijl: It doesn't matter whether you contract $\nu$ and $\beta$ first. For summation with finite terms, $\sum_a\sum_b f(a,b)=\sum_b\sum_a f(a,b)$. –  KennyTM Jul 2 '12 at 10:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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