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4

Essentially you are asking why only scalars are allowed to develop a vacuum expectation value (VEV). A scalar (as the name suggests) does not point to any direction -it has spin 0- therefore it can have a VEV without breaking the Lorentz symmetry. On the other hand, a boson with higher spin, e.g. a vector (spin 1) would spontaneously break Lorentz by ...

3

Frankly I find this so-called pedagogical article quite unintelligible and fail to see what the author wanted to say about these two operations. I also can't make any sense of the "derivation" of 7.3 based on the chiral projection being a "numerical matrix" and therefore commuting with charge conjugation operator. Moreso the remark: Elaborate statements ...

3

After thinking about it some more, I think I have a resolution. Consider a single Weyl fermion, which has equation of motion $$\sigma^\mu \partial_\mu \psi = 0.$$ Just like the Dirac equation, the Weyl equation is linear in momenta and hence has negative energy solutions. We then perform the usual procedure to regard these as positive energy solutions ...

4

I think your problem is mostly a problem of notation. If you write two Weyl spinors inside a Dirac spinor, you should use different symbols to avoud confusion, i.e. $$\psi = \begin{pmatrix} \xi_L \\ i \sigma_2 \xi_L^* \end{pmatrix}.$$ Now, your object $\Psi$ has a left-chiral component $\xi_L$ and a right-chiral component $i \sigma_2 \xi_L^*$. (A Dirac ...

0

Multiplying matrices is not correct here. You need to work these things out using the definition, $(\bar{\sigma}^\mu)^{\dot{\alpha}\alpha}=\epsilon^{\alpha\beta}\epsilon^{\dot{\alpha}\dot{\beta}}(\sigma^\mu)_{\beta\dot{\beta}}$. We have, for example, $$(\bar{\sigma}^0)^{11}=\epsilon^{12}\epsilon^{12}(\sigma^0)_{22}=(\sigma^0)_{22}=1,$$ while ...

2

In the rest frame $\not p \rightarrow - m \gamma^0$ and $$(\not p - m)u(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I + \gamma^0) u(\mathbf{0},s) = 0 \\ {\bar u}(\mathbf{p},s)(\not p - m) = 0 \;\; \rightarrow \;\; {\bar u}(\mathbf{0},s) (I + \gamma^0) = 0 \\ (\not p + m)v(\mathbf{p},s) = 0 \;\; \rightarrow \;\; (I - \gamma^0) v(\mathbf{0},s) = 0 \\ {\bar ... 0 A scalar is, like other scalars, merely just a number. Think about their matrix representation:$$ \psi=(\phi_R\; \phi_L)^T$$and$$\bar{\psi}=\psi^\dagger\gamma^0 =(\phi^*_L \; \phi_R^*). It is clear that $\bar{\psi}\psi$ is a 1x1 matrix (scalar), and of course the operation is legitimate. Those other forms are also 1x1 matrices. However under Lorentz ...

0

For non-hermitian products of Dirac field operators the parity is not well defined and depends on the phase $\eta=\pm1,\,\pm i$ of the parity transformation $\eta \gamma^0$. For example, $I_P = -\eta^2 I$, where $I = \overline{\Psi_C}\Phi$. In the $S$-matrix elements, however, all phases go away eventually, because creation and annihilation operators come ...

0

(I) Assuming there are $N$ distinct fermions in your Lagrangian (e.g. quarks in the standard model), the kinetic term will have the $U(N)\times U(N)$ symmetry. This symmetry is broken by the mass term, however, which couples fermions with different handedness. (II) If $V$ is the fundamental representation of a Lie group (with dual/conjugate representation ...

1

Yes, the first part of your question is appreciated and answered soundly. The fermion kinetic term splits into two independent parts involving left and right Weyl spinors respectively, so each is independent under a separate U(N) as your wrote down. The second question is a matter of language. A generator is a matrix with one adjoint index, a here, ranging ...

-3

Its just a consequence of garlac experiment the particles on the have possibly two eigen states up |z>and down ie half of particles go up nd rest half down |_z> that's y they are called spin half. And note that spin is nothing to do with rotation it is purely intrinsic and quantum mechanical

1

First notice that $(\bar u\gamma^\mu v)$ is just a scalar, i.e. number. So taking a dagger would be the same as taking its complex conjugate. So, $(\bar u\gamma^\mu v)^\dagger=v^\dagger(\gamma^\mu)^\dagger\gamma^0u=v^\dagger\gamma^0\gamma^0(\gamma^\mu)^\dagger\gamma^0u=\bar v\gamma^\mu u.$ Notice that I have inserted $\gamma^0\gamma^0 = 1$, and used that ...

2

Another one of these issues where careless terminology is our downfall. Indeed, on the level of the Dirac equation, such a thing as a "left-handed electron" does not exist. Every pure electron and every pure positron state is an equal mixture of left- and right-handed components. I'll call this electron (positive-frequency solution electron) the "mass basis ...

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