Take the 2-minute tour ×

In Quantum Electrodynamics by Landau and Lifshiz there is the following:

The correspondence between the spinor $\zeta^{\alpha \dot{\beta}}$ and the 4-vector is a particular case of a general rule: any symmetrical spinor of rank $(k,k)$ is equivalent to a symmetrical 4-tensor of rank $k$ which is irreducible (i.e. which gives zero upon contraction with respect to any pair of indices).

L&L also writes out this for a 4-vector $$ \zeta^{1\dot{1}}=\zeta_{2\dot{2}}=a^3+a^0 ,\quad \zeta^{2\dot{2}}=\zeta_{1\dot{1}}=a^0-a^3, $$ $$ \zeta^{1\dot{2}}=-\zeta_{2\dot{1}}=a^1-ia^2 ,\quad \zeta^{2\dot{1}}=-\zeta_{1\dot{2}}=a^1+ia^2, $$

Surely there must be an established method to do this in general just like it says above. I would like to know this method, so if someone would be kind enough to show me or refer me to a reference I would be grateful, (e.g. suppose I would like to know the components for $\zeta^{\alpha\beta\dot{\gamma}\dot{\delta}}$ in terms of the symmetric traceless rank-2, 4-tensor).


share|improve this question
You might want to look into Penrose and Rindler's Spinors and Space-Time which looks into this in great detail (it's what the two volumes are entirely about!). –  Alex Nelson Jul 9 '12 at 4:51

1 Answer 1

up vote 2 down vote accepted

The method is to contract every index with the $\sigma$ four vector:

$$ \sigma^\mu_{\alpha\dot{\beta}}$$

Where $\sigma^0$ is the identity, and $\sigma^i$ for i=1,2,3 is the Pauli spin matrix.

If you have a symmetric tensor with all lower indices, you contract each $\mu$ index with a sigma, and you get the dotted-undotted form.

$$ M_{\mu\nu} \sigma^\mu_{\alpha\dot{\beta}}\sigma^\nu_{\alpha'\dot{\beta}'} = M_{\alpha\alpha'\dot{\beta}\dot{\beta}'} $$

If M is symmetric on $\mu$ and $\nu$, this is symmetric under permutations of the pairs of corresponding dotted and undotted indices simultaneously. It is not symmetric under separate permutations of the dotted and undotted indices.

$$ M_{\alpha\alpha'\dot{\beta}\dot{\beta}'} = M_{\alpha'\alpha\dot{\beta}'\dot{\beta}}$$

But this is because I haven't implemented tracelessness. The inner product of two vectors in spinor form is given by using $\epsilon$ tensors:

$$ V_{\alpha\dot{\alpha}} V_{\beta\dot{\beta}} \epsilon^{\alpha\beta}\epsilon^{\dot{\alpha}\dot{\beta}}$$

If you make the contraction of M with g zero, in spinor form, this tells you that

$$ M_{\alpha\beta\dot{\alpha}\dot{\beta}} - M_{\beta\alpha\dot{\alpha}\dot{\beta}} = 0$$

Which together with the previous condition tells you that the M in spinor form is symmetric under separate permutations of the two indices. The higher rank proof is exactly the same.

share|improve this answer
Thanks Ron. Should the $\beta$s in the second line be dotted? –  kηives Jul 9 '12 at 19:03
@kηives: Yes, I'll fix it. I was suspended and couldn't do it. –  Ron Maimon Jul 10 '12 at 19:33

Your Answer


