Recent content by bayners123

OK, right I think I'm understanding this a little better. So what I should have written, for muon-electron scattering, is: \bar{\psi}_{\nu_\mu} \left[ g_R^{\nu_\mu} \frac{(1-\gamma^5)}{2} + g_L^{\nu_\mu} \frac{(1+\gamma^5)}{2} \right] \psi_{\nu_\mu} multiplied by the term for the electron...

The Z0 is a linear combination of W0 and B0 bosons, so unlike the charged current interaction it can interact with both handednesses (is that a word). In the search to quantify this mixing, people measured the cross section for muon (anti)neutrinos to scatter off electrons. This is a reaction...

Brilliant, thanks for both your help in understanding this!

Hmm ok. Is it possible to think of this in terms of combining quantum numbers? So adding two 1^{--} systems and obtaining a 0^{-+}? If not, what's special about photon which makes this possible? Also, does this imply that a two photon system can have either parity depending on what decayed?

Sorry, I don't quite understand your answer. Are you referring to the polarization of a single photon? If so, then yes I agree: that's why I put the Parity eigenvalue of the photon as -1, making the parity of a two photon system (-1) \times (-1)^L = (-1)^{L+1}

\pi^0s decay to two photons via the EM interaction. The J^{PC} of the pion is 0^{-+} and of a \gamma is 1^{--}. \gamma\gamma therefore has J^{PC} = 0^{++}, 1^{-+}, 2^{++}. This does not match the pion, so how can this decay occur?

I did, thanks! I wasn't aware that radial excitation was possible without changing the name of the particle.

The J/Psi is a state of charmonium with J=1, S=1, L=0. So J^{PC} = 1^{--}. It can be excited to states J^\prime \textrm{ and } J^{\prime\prime}, but these don't change any of these numbers. So what is changing?

I've got a question that asks what the maximum scattering angle in \nu_\mu e \rightarrow \nu_\mu e is. The electron is stationary in the lab frame and after the collision has E >> m_e. The answer given is \sqrt{\frac{2 m_e}{E_e}}, independent of the neutrino's energy. How can I get...

Brilliant, thanks. I think I understand that better now. The one that still confuses me is how the hadrons break down as \mathbf{3}\otimes\mathbf{3}\otimes\mathbf{3}=\mathbf{10}_S\oplus\mathbf{8}_M\oplus\mathbf{8}_M\oplus\mathbf{1}_A Where does this breakdown come from? That's rhetorical by...

Hmm ok thanks. I think I need to understand group theory better.

The pseudoscalar mesons have J^P = 0^- They form a nonet: for S = ±1, I (isospin) = 1/2 and so there are two particles for each value of strangeness. This account for 4 particles: the ground-state Kaons. For S=0, I can be 0 or 1. I=1 gives a triplet: \pi^\pm \mbox{ and } \pi^0. For...

Ah ok.. The reason this came up is that I was looking at latter operators on paired states. So with 2 atoms in a 1,1 state you can get \mid 2,2 \rangle = \mid 1,1 \rangle\mid 1,1 \rangle And then you can use ladder operators to go down: J_-\mid 2,2 \rangle = (J_-\mid 1,1 \rangle)\mid 1,1...

Hi, Just a little thing that's been puzzling me: Consider a state \mid \psi \rangle = \frac{1}{\sqrt{2}} \mid A \rangle + \frac{1}{\sqrt{2}} \mid B \rangle This is normalised since [\frac{1}{\sqrt{2}}]^2 + [\frac{1}{\sqrt{2}}]^2 = 1 Now let A = B: \mid \psi \rangle =...

Ah, thanks