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Interacting part of the $SU(2)_L$describing the Higgs and fermionic sector with one family

$$\mathcal{L}= \bar{l}_Li\not Dl_L+\bar{e}_Ri\not\partial e_R+ \bar{\nu}_Ri\not\partial\nu_R - (y_{\nu}\bar{l}_L\tilde{\Phi}\nu_R - y_{e}\bar{l}_L{\Phi}e_R+h.c)$$

Where $l_L=(\nu_L, e_L)^T$ and $D_{\mu}=\partial_{\mu}+ig\hat{W_{\mu}}$ , $\tilde{\Phi}=i\sigma_{2}\Phi^{*}$ , $\Phi = \Big(\frac{v+h}{\sqrt{2}}\Big)e^{i\Pi/v}(0,1)^T$

From the interaction terms in the Lagrangian only 3-point interactions are:

$$\mathcal{L_{int}}= -\frac{g}{2}\gamma^{\mu}[\bar{\nu_L} W^{3}_{\mu} \nu_L + \sqrt{2}\bar{\nu_L}W^{+}_{\mu} e_L + \sqrt{2}\bar{e_L}W^{-}_{\mu} \nu_L -\bar{e_L} W^{3}_{\mu} e_L] -y_{\nu}\frac{1}{\sqrt{2}}\bar{\nu}_L h \nu_R + iy_{\nu}\frac{1}{2\sqrt{2}}\bar{\nu}_L \Pi^{3} \nu_R + iy_{\nu}\frac{1}{2}\bar{e}_L \Pi^{-} \nu_R + \frac{y_e}{\sqrt{2}}\bar{e_L}h e_R + \frac{iy_e}{2}\bar{\nu_L}\Pi^{+}e_R - \frac{iy_e}{2\sqrt{2}}\bar{e_L}\Pi^3 e_R$$

And from these, Feynman graphic rules are as follows:

int

Since I want to calculate the processes contributing $\nu_e \bar{\nu_e} \rightarrow e^- e^+$ . I would expect to get the following scattering diagrams:

Scat

The way I proceeded is:

I matched the diagrams with the same propagators (and with their h.c. in case of charged propagators). However I don't know how to interpret $\nu_R$ and $\bar{\nu}_L$ in vertex diagrams.

Aren't they right-handed neutrino and left-handed anti-neutrino? Why do they even appear?

Just by looking at the vertex with Z boson it seems like something is off since L.H. neutrino annihilates with L.H. anti-neutrino (yet L.H. neutrino should annihilate with R.H. anti-neutrino). Same mismatch appears in the vertex with Higgs however this time there is a R.H. neutrino and L.H. anti-neutrino which is even more confusing.

Edit

At this point, if anyone could show how to correctly glue the vertex diagrams to get neutrino pair annihilation to electron positron pair, I will accept the answer.

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  • $\begingroup$ What's this freaky non-charge-conserving 2nd term in your Lagrangian??? $\endgroup$ Commented Sep 4, 2022 at 12:32
  • $\begingroup$ You do appreciate how fantastically small $y_\nu$ is, and why, right? $\endgroup$ Commented Sep 4, 2022 at 12:40
  • $\begingroup$ Thank you, it was a typo, I corrected it. Yes I do know that it's incredibly small but I don't understand why we have $\nu_R$ in the Lagrangian. From the first term in the interaction Lagrangian 'g' is larger compared to Yukawa term however I still see a left handed anti-neutrino. I've been trying to wrap my head around it for hours but I feel like I'm missing something... $\endgroup$
    – Monopole
    Commented Sep 4, 2022 at 12:49

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I'm really not sure what your trouble or question are. When in doubt, check your expressions versus (10.2) of the PDG review. In your expression, you need to intercalate the γ matrices in-between the two spinors.

For neutrinos, the neutral current is particularly (uniquely among fermions!) clean. Only left-chiral neutrinos and right-chiral antineutrinos couple to the Z bosons, since the neutrino fails to couple to the photon, and only couples to the Z. So right-chiral neutrinos and left-chiral antineutrinos do not couple to it: $$ {\sqrt{g^2+g'^2}\over 2} \overline {\nu_L}Z_\mu \gamma^\mu \nu_L. $$ This term destroys a Z and supplants it with a created pair of L neutrino and R antineutrino; or, vice versa, creates a Z out of a destroyed pair of L neutrino and R antineutrino. (It can also represent an L neutrino emitting a virtual Z in the t-channel and continuing on as an L neutrino.)

By contrast, the Higgs coupling, $\propto y_\nu h \bar\nu \nu= y_\nu h \bar\nu (L+R)\nu= y_\nu h (\overline{\nu_R} \nu_L+\overline{\nu_L} \nu_R )$, lacking a chirality flipping γ, creates an L neutrino and L antineutrino when a higgs is destroyed, and, in the same breath, with the same strength, a R neutrino and R antineutrino. But with a ridiculously small strength, as $$ y_\nu \sim m_v/v < 10^{-12}. $$ It of course also creates a higgs out of a destruction of a LL or RR neutrino-antineutrino pair. Recall $\overline {\nu_L}=\bar\nu R$.

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  • $\begingroup$ It's always incredibly helpful to read your answers, I completely agree with what you said about the Z boson. However I fail to see the part of Higgs, to me from $\bar{\nu_L} h{\nu_R} $ term in the Lagrangian I get from the destruction of Higgs right neutrino $\nu_R$ and right anti-neutrino $\bar{\nu_L}$ or higgs production from right neutrino $\nu_R$ and right anti-neutrino $\bar{\nu_L}$. I don't see how to get left neutrino and left anti-neutrino from Higgs. $\endgroup$
    – Monopole
    Commented Sep 5, 2022 at 8:37
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    $\begingroup$ But you skipped the relevant h.c. term, $h\overline{\nu_R} \nu_L$, which destroys a L neutrino and a L antineutrino... I supplemented my answer. $\endgroup$ Commented Sep 5, 2022 at 11:06
  • $\begingroup$ Thank you, now I do see where they come from, but then with the h.c. terms I'm gonna have $\bar{\nu_L} h \nu_R$ & $\bar{\nu_R} h \nu_L$ and from the electron part $\bar{e_L} h e_R$ , $\bar{e_R} h e_L$ so if I want to calculate $\bar{\nu_e} \nu_e \to e^- e^+$ scattering, only with Higgs propagator, I'm going to have 4 diagrams, is that correct? $\endgroup$
    – Monopole
    Commented Sep 5, 2022 at 12:49
  • $\begingroup$ As you see, the L and R projections are for conceptual convenience. The minuscule Higgs term has no need of them... they are like mass terms. So, as per your second blackboard, the 2nd diagram is one, with two different vertices, and no need for L, R subclasses... $\endgroup$ Commented Sep 5, 2022 at 12:55
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    $\begingroup$ They too are small. All I'm saying is just plug the full spinors in, and don't split them into L and R components in those higgs-mediated diagrams. $\endgroup$ Commented Sep 5, 2022 at 13:40

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