Tag Info

New answers tagged

2

This answer is mostly rephrasing Trimok's correct answer in other words. The super-Poincare group is supposed to be an extension of the Poincare group, which contains the Lorentz group and translations. We will complexify the Lorentz group. The Lie group $G:=SL(2,\mathbb{C})\times SL(2,\mathbb{C})$ is (isomorhic to the double cover of) the complexified ...


2

It is maybe simpler to consider all the generators as representations of $SL(2,C)$, so, using spinor indices, you will have : $M^{\alpha \dot \alpha \beta \dot \beta}, P^{\beta \dot \beta}, Q_\alpha, \bar Q^\dot\beta$ Indices are raised and lowered with the Levi-Civita symbols $\epsilon_{\alpha \beta}, \epsilon^{\alpha \beta},\epsilon_{\dot \alpha \dot ...


3

Short intro to ladders As you say, they're ladder operators. Let's get rid of the annoying $\hbar$ by setting it to one, and call them more systematically $L_{-1},L_0,L_1$ instead of $L_-,L_z,L_+$. Then, the commutation relations take the uniform form $$[L_n,L_m] = (m-n)L_{m+n}$$ If we had countably many of these, we'd have a Witt algebra, if there was ...



Top 50 recent answers are included