Tag Info

New answers tagged

1

If $S_\text{tot}^2$ is a symmetry i.e., commutes with the Hamiltonian, then one can choose a basis in where it is diagonal. This corresponds to going to the coupled basis, the triplet (total spin=1) and singlet (total spin=0) combinations of the two spins. Those are the four choices that you have listed for the `spin functions' -- the first three are the ...


6

The fundamental representation of a Lie group $G$, as commonly used in this context, is the smallest faithful (i.e. injective) representation of the group. We do not require fermions to belong to the fundamental rep, it is just the case that, in the Standard model, they always either belong to the fundamental or the trivial representation (as that is ...


1

Dirac fermions is only the direct sum of left- and right-handed Weyl representations (which leads to time inversion, charge inversion and spatial inversion invariance of the theory). Two Weyl representations are mixed by the mass term in the Dirac equation. If we set mass to zero, we will get two uncoupled equations, each of which describes Weyl fermion. But ...


1

Yes, the parity symmetry can be implemented only for the four-component Dirac fields and not for two-component Weyl fields. This fact, mathematically speaking, does not depend on the value of the mass. Physically speaking, however, I am not sure that a massless four component spinor makes much sense in standard theories (Sorry, I do not know anything about ...


0

The real reason is in following. Let's assume Majorana field: $$ \Psi_{M} = \Psi_{L} + \hat{C}\bar{\Psi}^{T}_{L}, \quad \hat{C} = i\gamma_{2}\gamma_{0}, \quad \Psi_{L} = \begin{pmatrix} \psi_{L} \\ 0 \end{pmatrix}. $$ By using this notation it's not hard to see that kinetic term is equal to $$ \bar{\Psi}_{M}\gamma^{\mu}\partial_{\mu}\Psi_{M} = ...


0

The short asnwer to your question is that the overall factor $\frac{1}{2}$ from the Lagrangian of a Majorana field (in the 4-component notation) $$\mathcal{L}=\frac{1}{2}(\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi -m\bar{\psi}\psi)$$ compared to the general Dirac Lagrangian is usual for self-conjugate fields and it is introduced to ensure a consistent ...



Top 50 recent answers are included