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I have been given the Lorentz generator $$J_{ab}=\lambda_a \frac{\partial}{\partial \lambda^b}+a\leftrightarrow b$$

I know $\frac{\partial}{\partial \lambda^b}\lambda_a=\epsilon_{ba}$ and $A_a=A^b\epsilon_{ab}$ (? not sure about these two). I want to show that this acting on a spinor bracket gives zero as this is supposed to be Lorentz invariant. Say I have a bracket $\langle A B \rangle = A_a B_b \epsilon^{ab}$ and now I act with $J_{ab}$ on it w.r.t $A$, i.e.

$$\begin{split}J^A_{ab}\langle A B \rangle &=(A_a \frac{\partial}{\partial A_b}+a\leftrightarrow b) A_cB_d\epsilon^{cd}\\&=A_a\epsilon_{bc}B_d\epsilon^{cd} + A_b\epsilon_{ac}B_d\epsilon^{cd}\\&=A_a\epsilon_{bc}B_{c}+A_b\epsilon_{ac}B_c\\&=?\end{split}$$

Where do I go wrong? Could someone please help me out? Why doesnt it sum to zero?

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1 Answer 1

up vote 4 down vote accepted

Firstly, I don't think your convention for the derivative is consistent with how you lower the index on $\lambda^b$. Let us start with $$ \frac{\partial}{\partial \lambda^b} \lambda^a = \delta_b^a $$ If you lower the index on $\lambda^b$ you get $$ \lambda_a = \epsilon_{ac} \lambda^c $$ Putting this together gives $$ \frac{\partial}{\partial \lambda^b} \lambda_a = \epsilon_{ac} \frac{\partial}{\partial \lambda^b} \lambda^c = \epsilon_{ab} = -\epsilon_{ba} $$ which has a different sign from what you have.

Furthermore, the spinor bracket you write is Lorentz invariant if you transform both spinors at the same time. The actions of $J^A$ and $J^B$ on $\langle AB \rangle = \epsilon_{cd} A^c B^d$ is given by \begin{align} J_{ab}^A\, \epsilon_{cd} A^c B^d &= \left(A_a \frac{\partial}{\partial A^b} + A_b \frac{\partial}{\partial A^a} \right) \, \epsilon_{cd} A^c B^d \\ &= \epsilon_{cd} \left( A_a \delta_b^c + A_b \delta_a^c \right) B^d = +(A_a B_b + B_b A_a) \\ % J_{ab}^B\, \epsilon_{cd} A^c B^d &= \left(B_a \frac{\partial}{\partial B^b} + B_b \frac{\partial}{\partial B^a} \right) \, \epsilon_{cd} A^c B^d \\ &= \epsilon_{cd} A^c \left( B_a \delta_b^d + B_b \delta_a^d \right) = -(A_a B_b + B_b A_a) \end{align} so if you sum them the transformation of $A$ is cancelled by the transformation of $B$.

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Thanks! Now i get it :) –  Sanji Jun 7 '13 at 9:51

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