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If $z \mapsto \mathrm{e}^{\mathrm{i}\delta}$, then $z^p \mapsto (\mathrm{e}^{\mathrm{i}\delta} z)^p = \mathrm{e}^{\mathrm{i}\delta p} z^p$. What we are actually looking at is that the "transformation of the $p$-th power of $z$" is a way to speak of the representation of the circle group $\mathrm{U}(1)$ labeled by $p$.


3

As I had wrote here, the Lorentz group can be represented as the direct product of two $SU(2)$ or $SO(3)$ groups. The $SU(2)$-realization of the irrep, $\left( \frac{n}{2}, \frac{m}{2}\right)$, refers to the spinor tensor $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$, where the sum $n + m$ represents the spin value of representation: $\frac{n + m}{2} = s$. ...



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