- #1
Gleeson
- 30
- 4
Suppose ##\lambda_A## and ##\bar{\lambda}_A## are fermions (A goes from 1 to N) and ##\{ \lambda_{A \alpha}, \bar{\lambda}_B^{\beta}\} = \delta_{AB}\delta_{\alpha}^{\beta}##.
Let ##\sigma^i## denote the Pauli matrices.
Does it follow that ##[\bar{\lambda}_A \sigma^i \lambda_A, \bar{\lambda_B} \sigma^j \lambda_B] = 0##? Or should it be ##2i \epsilon_{ijk}\bar{\lambda}_C \sigma^k \lambda_C##?
I think it should be the former. ##\bar{\lambda_A} \sigma^i \lambda_A## is a sum over A of a (2 entry row, times a 2x2 matrix, times a 2 entry column), and so is no longer a matrix. And overall it is bosonic. So I think the commutator should be zero. If this is not the case, then why not?
In case it is not clear, ##\bar{\lambda}_A \sigma^i \lambda_A = \bar{\lambda}_A^{\alpha}\sigma^{i \beta}_{\alpha}\lambda_{A \beta}##. Repeated indices are summed.
Let ##\sigma^i## denote the Pauli matrices.
Does it follow that ##[\bar{\lambda}_A \sigma^i \lambda_A, \bar{\lambda_B} \sigma^j \lambda_B] = 0##? Or should it be ##2i \epsilon_{ijk}\bar{\lambda}_C \sigma^k \lambda_C##?
I think it should be the former. ##\bar{\lambda_A} \sigma^i \lambda_A## is a sum over A of a (2 entry row, times a 2x2 matrix, times a 2 entry column), and so is no longer a matrix. And overall it is bosonic. So I think the commutator should be zero. If this is not the case, then why not?
In case it is not clear, ##\bar{\lambda}_A \sigma^i \lambda_A = \bar{\lambda}_A^{\alpha}\sigma^{i \beta}_{\alpha}\lambda_{A \beta}##. Repeated indices are summed.
Last edited: