Recent content by terra

Ok, I think I got your point. Under an infinitesimal transformation ##\psi_{\nu,L} \to \psi_{\nu,L} \big( 1 + i \alpha_3 \big) + \psi_{e,L} \big( i \alpha_1 + \alpha_2 \big)##, ##\psi_{e,L} \to \psi_{e,L} \big( 1 - i \alpha_3 \big) + \psi_{\nu,L} \big( i \alpha_1 - \alpha_2 \big)## and I can...

Yes, but as you can see, I can't see how.

Ok. But for me, ##\ell^c_L## is just a symbol with components ##\ell_1## and ##\ell_2##, so I still can't see what tells me it doesn't transform appropriately.

Whoops. If ##\ell^c_L## is left-chiral, it should transform trivially, right? But I still don't see how does this help.

I had two suggestions. As I see it, my naive suggestion would transform exactly the same way, whereas the one multiplied with ##-i \sigma^2## would transform as ##\begin{pmatrix} -\psi^c_{e,L} \\ \psi^c_{\nu,L} \end{pmatrix} \to \begin{pmatrix} -\psi^c_{e,L} \big[ \cos(|\boldsymbol\alpha|) +...

Ok. I appreciate your help and sorry I can't follow your point. I'll go back to your previous reply. The components transform as ##\psi_{\nu,L} \to \psi_{\nu,L} \big[ \cos(|\boldsymbol\alpha|) + \sin(|\boldsymbol\alpha|) i \alpha_3 \big] + \psi_{e,L} \sin(|\boldsymbol\alpha|) (i \alpha_1 +...

My question was not about transforming ##\ell_L## under ##SU(2)## but under hermitian and charge conjugation and what is meant by the notation ##\bar\ell_L##. I'm sorry, but I still can't see how I see them from the transformation properties of ##\ell_L## under ##SU(2)##, as I don't know how is...

Assuming ##\psi_{\{\nu,e\}L}## are Dirac spinors for which ##\psi_L := P_L \psi## I have, in Weyl's representation defined by: \begin{align*} \gamma^0 &= \begin{pmatrix} 0 & \mathbb{1} \\ \mathbb{1} & 0 \end{pmatrix}, \gamma^i = \begin{pmatrix} 0 & \sigma^i \\ -\sigma^i & 0 \end{pmatrix}...

##\ell_L \to U \ell_L = \begin{pmatrix}\cos(|\boldsymbol\alpha|) \psi_{\nu,L} + \sin(|\boldsymbol\alpha|) \big[ i \alpha_3 \psi_{\nu,L} + (i \alpha_1 + \alpha_2) \psi_{e,L} \big] \\ \cos(|\boldsymbol\alpha|) \psi_{e,L} + \sin(|\boldsymbol\alpha|) \big[ (i \alpha_1 - \alpha_2) \psi_{\nu,L}- i...

Yes, that's what I meant. I also know the conjugation rules for Dirac spinors. I don't see where you are going, neither do I see how can I derive the transformation properties of these doublets from those of the Dirac spinors unless they follow 'trivially', that is ##\bar\ell = \begin{pmatrix}...

Which one? I'm sorry, but I can't follow. An infinitesimal ##SU(2)## transformation would read ##\mathbb{1} + i \alpha^a \sigma^a / 2 = \mathbb{1} + \begin{pmatrix} i \alpha_3 & i \alpha_1 + \alpha_2 \\ i \alpha_1 - \alpha_2 & - \alpha_3 \end{pmatrix}## (or something along those lines), right...

I have a left-handed ##SU(2)## lepton doublet: ## \ell_L = \begin{pmatrix} \psi_{\nu,L} \\ \psi_{e,L} \end{pmatrix}. ## I want to know its transformation properties under conjugation and similar 'basic' transformations: ##\ell^{\dagger}_L, \bar{\ell}_L, \ell^c_L, \bar{\ell}^c_L## and the general...

That's right, sorry, I was sloppy adding that. Weinberg shows that $$ U(\Lambda) \Psi_{p,\sigma} = \sum_{\rho} D^{(j)}_{\rho \sigma}(W(\Lambda,p)) \Psi_{\Lambda p, \rho} $$ where ##W(\Lambda,p)## is such that ##W^{\mu}_{\nu} k^{\nu} = k^{\mu}## so such transformations form a little group for...

Let's take a quantum state ##\Psi_p##, which is an eigenstate of momentum, i.e. ##\hat{P}^{\mu} \Psi_p = p^{\mu} \Psi_p##. Now, Weinberg states that if ##L(p')^{\mu}\,_{\nu}\, p^{\nu} = p'##, then ##\Psi_{p'} = N(p') U(L(p')) \Psi_{p}##, where ##N(p')## is a normalisation constant. How to...

I want to learn how to write down a particle state in some inertial coordinate frame starting from the state ##| j m \rangle ##, in which the particle is in a rest frame. I know how to rotate this state in the rest frame, but how does one write down a Lorentz boost for it? Note that I am not...