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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...

I see, thank you for your insight. I take that by the qcd scale you mean the scale at which the coupling constant diverges, and as such perturbation theory can't be used.

My background of some introductory courses in particle physics has left me with severe shortcomings. Say we start from a hadron, which decays purely to other hadrons. My question is this: through which interaction does the process take place? Is there a preferred interaction, and why/why not...

Thanks for the reply, yeah. Let ## a,b## be exclusive events. For classical physical things ## P(a\mathrm{ \, or \, }b)= P(a) + P(b)##. In quantum physics, however, we have ## A(a\mathrm{ \, or \, }b)= A(a) + A(b)## where ##A \in \mathbb{C}## is a 'probability amplitude' so that ## P(x)= A(X)...

I was looking at what Cohen-Tannudji has to say on compatibility of observables. Assumptions: ## A,B## are operators such that ##[A,B]=0 ## and we denote ## |a_i \,b_j\rangle## to be states for which ##A | a_i \, b_j \rangle= a_i | a_i \, b_j \rangle##, ##B | a_i \, b_j \rangle= b_j | a_i \...

I agree, now that you said it, my last example probably goes by the name 'semiclassical'. And one more note about your first question. The intuition I tried to give is more of a guideline, I think, to determine whether or not one should be aware of the commutation relations. I didn't really...

Let me try to answer your first question first. The essential thing, as you said, is to compare Planck's constant to other variables in your system. If other variables in your system are 'small enough', Planck's constant becomes larger in comparison. For example, let's look at the scale of the...

Possibly a last update. I think this problem is pretty much solved (or, well, solved in an alternative way). I still couldn't see ##\mathrm{d}/\mathrm{d}t \, \mathrm{d}\Gamma \, \rho= 0## from any equations. But, as the last messages show, I was confused about the definition of the probability...

I see. Thank you for explaining (and apologies for my slow replies). I see that energy and Hamiltonian coincide in the rotating frame, and are both conserved in that frame. About when ## H## is and isn't energy (in a classical system). One can write down the following. ## H= p_i \dot{q}_i - L##...

I do have some updates on this problem. I'm still not convinced with the above derivation. Maybe it works with some explanation, but I prefer something mathematically more straight-forward. However, I did find an alternative derivation, which, to my mind, goes more smoothly. As a reminder, we...

True, thanks for the correction (as you can see, at first I talked about quantum systems). I've heard this before, but I must admit I haven't thought about it that much. In your example the idea is to get the equation of motion for the ball in pipe's rest frame? The pipe doesn't rotate in its...

Hamiltonian gives the energy of a system. Let's discuss the case of pure states (where we have quantum states that can be written as vectors ## | \psi \rangle ##). Conservation of energy means that the (expectation value of) amount of energy does not change in time, i.e. ## \mathrm{d} \langle...

First, two definitions: let ## \varrho (M)## be the probability density of macro states ##M ## (which correspond to a subgroup of the phase space) and ## \mathrm{d} \Gamma ## be the volume element of a phase space. In my lecture notes, the derivation for continuity equation of probability...