# Tag Info

5

Here's my two cents worth. Why Lie Algebras? First I'm just going to talk about Lie algebras. These capture almost all information about the underlying group. The only information omitted is the discrete symmetries of the theory. But in quantum mechanics we usually deal with these separately, so that's fine. The Lorentz Lie Algebra It turns out that the ...

2

Firstly, what book is this? It will help greatly if I can reference it myself. It is highly likely that when he says $\mbox{SO}(1,3)$ [or $\mbox{SO}(3,1)$!] that he means $\mbox{SO}(1,3)-\uparrow$, which is absolutely not the same! But most people are very lazy about this. Here you're picking out the simply-connected region of $\mbox{O}(1,3)$ ...

2

I don't think that Sasha made a mistake. I'll use the dotted/undotted notation which may clarify the possible SL(2,C) invariants. Let $\xi^{A}$, $\theta^{A}$, $\eta^{\dot{A}}$ and $\phi^{\dot{A}}$ be Weyl spinors. The Levi-Civita tensors $\epsilon_{AB}$ and $\epsilon_{\dot{A}\dot{B}}$ transform trivially under SL(2,C) so they can be used to lower indices. ...

3

You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant. Helicity defined $$\hat h = \vec\Sigma \cdot \hat p,$$ commutes with the Hamiltonian, $$[\hat h, H] = 0,$$ but is clearly not Lorentz invariant, because ...

0

There is a lot to be said about Majorana and Dirac neutrino masses, but I will try to address only the specific point you raise. When you write a Majorana mass term, for example, $$\mathcal{L}_{\text{Majorana}} = -\frac{m_M}{2} \left[\bar\nu_L(\nu_L)^C + \overline{\nu^C_L}\nu_L\right]$$ you are not assuming that $(\nu_L)^C = \nu_R$, but you can write that ...

0

III) A spinor is indeed an element of the vector space on which the representation of the Spin group acts. IV) I don't know V) The fact that the norm of the vector is null does not imply that has no nonvanishing entries. Remember that in physics applications one has a metric of indefinite signature, so contributions to the norm can (and most of the time ...

2

This question should probably be split up into several questions. Really quickly, I can answer the bit about the Newman-Penrose forumulation and tetrads. I think the best modern approach to the tetrad formalism is to treat the set of tetrads as a map from an internal orthonormal frame to an ordinary coordinate frame. So, the tetrad is a set of matrices ...

2

You need to work out the tensor product and will find a direct sum of different contributions \begin{multline} [(1/2, 0) \oplus (0, 1/2)] \otimes [(1/2, 0) \oplus (0, 1/2)] =\\ \big((1/2, 0) \otimes (1/2, 0)\big) \oplus \big((1/2, 0) \otimes (0, 1/2) \big)\oplus \quad \\\big((0, 1/2) \otimes (1/2, 0)\big) \oplus \big((0, 1/2) \otimes (0, 1/2)\big) = \\ (0, ...

0

short answer if $\hat {S}^{-1} S = \mathbb{I}$ I can give you a general example of $\psi^\dagger\psi$ not being invariant. because for Dirac spinor $\psi$ whe have the following transformation rules $$\psi(x) \rightarrow S[\Lambda] \psi(\Lambda^{-1}x)=S[\Lambda] \psi(x^\prime) \\ \psi^\dagger(x) \rightarrow \psi^\dagger(\Lambda^{-1}x) ... 1 If we assume that$$\Psi {'} = \hat {S}\Psi$$and$${\bar{\Psi}}{'} = \bar {\Psi} \hat {S}^{-1},$$it follows that the product of the two transforms as$$(\bar{\Psi}\Psi)'={\bar{\Psi}}{'}\Psi {'}=\bar {\Psi} \hat {S}^{-1}\hat {S}\Psi=\bar{\Psi}\Psi,$$which is a consequence of$$\hat {S}^{-1}\hat {S}=\mathbb{1}.$$1 In QFT, the Dirac spinor will also be promoted to a field, whose oscillation mode coefficients are creation and annihilation operators. BUT: For the Dirac spinor it is possible to well-define a probablility density and current:$$\rho^\mu \propto \bar\psi (\partial^\mu \psi) - (\partial^\mu \bar \psi) \psi  This current's zero component is positive ...

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