I am not sure I understand what the short-hand anti-symmetrization means. I.e. I know that $$\delta_{cd}^{[ab]} ~=~ \frac{1}{2}(\delta_{c}^{a}\delta_{d}^{b} - \delta_{c}^{b}\delta_{d}^{a})$$ but how do we expand "bigger" analogues? I.e. $$\delta_{b_1 b_2 b_3 b_4}^{[a_1 a_2 a_3 a_4]} ~=~ ?$$

What is the order of permutation in this case?


$$ \delta^{[\mu_1\mu_2\ldots \mu_n]}_{\nu_1\nu_2\ldots \nu_n}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} \prod_{i=1}^n \delta^{\mu_{\pi(i)}}_{\nu_i}. $$

More generally,

$$ T^{[\mu_1\mu_2\ldots \mu_n]}~=~ \frac{1}{n!}\sum_{\pi\in S_n}{\rm sgn(\pi)} T^{\mu_{\pi(1)}\mu_{\pi(2)}\ldots \mu_{\pi(n)}}. $$

Here $S_n$ is the symmetric group of permutations, and $\pi\in S_n$ is a permutation with signature ${\rm sgn(\pi)}$. The order of $S_n$ is $$|S_n|~=~n!~.$$

See also this and this related Phys.SE posts.

  • $\begingroup$ Ehm, could you give me an example to understand this better? $\endgroup$ – Marion Dec 16 '14 at 18:37
  • $\begingroup$ Updated the answer. $\endgroup$ – Qmechanic Dec 16 '14 at 20:22
  • 2
    $\begingroup$ @Qmechanic - I suggest you write out an explicit example maybe for $n=3$ or $n=4$. $\endgroup$ – Prahar Mitra Dec 16 '14 at 20:31
  • 1
    $\begingroup$ I got it. It is rather long to write for $n=4$ or even for $n=3$. I have done it already and thanks for the answers. $\endgroup$ – Marion Dec 16 '14 at 23:21
  • 1
    $\begingroup$ @Marion: Perfect, that saves me the trouble of typing it out. For $n=3$, see also Prahar's answer. $\endgroup$ – Qmechanic Dec 16 '14 at 23:23

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.