In order to describe invariant forms on Dirac spinors $S$ one can find trivial subrepresentations in $S \otimes S$. If we use $S \cong (1/2, 0) \oplus (0, 1/2)$ then

\begin{multline} [(1/2, 0) \oplus (0, 1/2)] \otimes [(1/2, 0) \oplus (0, 1/2)] =\\ (0, 0) \oplus (1, 0) \oplus (1/2, 1/2) \oplus (1/2, 1/2) \oplus (0, 1) \oplus (0, 0) \end{multline}

Therefore representation theory predicts existence of two invariant forms. It is usually claimed that this two forms are $$ D_1(\chi, \psi)=\bar{\chi}\psi=\chi^T\gamma_0\psi=\chi_R^T\psi_L+\chi_L^T\psi_R $$ and $$ D_2(\chi, \psi)=\bar{\chi}\gamma_5\psi=\chi^T\gamma_0 \gamma_5\psi=\chi_R^T\psi_L-\chi_L^T\psi_R $$

The form $D_1$ is symmetric and it's quadratic form (with complex conjugation on the first argument) is usually used to construct Dirac's Lagrangian.

From the other hand, it is known that on Weyl spinors one can also find an antisymmetric invariant forms given as $$ \chi^T_L\sigma_2\psi_L . $$ Let me use this to construct one more anti-symmetric invariant form on Dirac spinors as a sum of two forms on Weyl spinors $$ D_3(\chi, \psi)=\chi^T_L\sigma_2\psi_L+\chi^T_R\sigma_2\psi_R $$

Form $D_3$ is not a linear combination of $D_1$ and $D_2$ and thus I get a contradiction with representation theory prediction. Where I made a mistake?


I don't think that Sasha made a mistake. I'll use the dotted/undotted notation which may clarify the possible SL(2,C) invariants. Let $\xi^{A}$, $\theta^{A}$, $\eta^{\dot{A}}$ and $\phi^{\dot{A}}$ be Weyl spinors. The Levi-Civita tensors $\epsilon_{AB}$ and $\epsilon_{\dot{A}\dot{B}}$ transform trivially under SL(2,C) so they can be used to lower indices. The consistent rules are, $$ \xi_{A}=\xi^{B}\epsilon_{BA} $$ and, $$ \eta_{\dot{A}}=\epsilon_{\dot{A}\dot{B}}\eta^{\dot{B}} $$ The SL(2,C) invariant Levi-Civita tensors are just similarity transformations which connect equivalent SL(2,C) irreps. Using upstairs and downstairs indices and complex conjugation $(^{*})$, one can make four SL(2,C) invariants, $\xi^{A}\theta_{A}$, $\eta^{\dot{A}}\phi_{\dot{A}}$, $$ (\xi^{A})^{*}\eta^{\dot{A}}=[\xi^{*}]_{\dot{A}}\eta^{\dot{A}} $$ and $$ \xi^{A}(\eta^{\dot{A}})^{*}=\xi^{A}[\eta^{*}]_{A} $$ The first and second are invariant under parity. The third and fourth are not invariant under parity. By adding and subtracting the third and fourth SL(2,C) invariants, one can make Sasha's bilinear forms $D_{1}$ and $D_{2}$. $D_{1}$ transforms trivially under parity whilst $D_{2}$ changes sign under parity. Thus $D_{1}$ is an $O(3,1)$ scalar and $D_{2}$ is an $O(3,1)$ pseudoscalar. Sasha's invariant $\chi^T_L\sigma_2\psi_L$ is my $\eta^{\dot{A}}\phi_{\dot{A}}$ modulo a factor of $i$ so Sasha's O(3,1) invariant $D_{3}$ is made by summing my first and second invariants.

Edit: My earlier draft said, "I don't see any contradiction with representation theory here because I don't see any reason for the expansion of Dirac spinors $$ [(1/2, 0) \oplus (0, 1/2)] \otimes [(1/2, 0) \oplus (0, 1/2)] $$ to exhaust the SL(2,C) invariants of the Weyl spinors." On reflection, my words were wrong. All I've done here is to list the bilinear O(3,1) invariants . I guess Sasha wants to see the decomposition of a general rank 2 Dirac tensor into O(3,1) irreps, I haven't done this part.

| cite | improve this answer | |

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.