Let's assume seesaw 1 type of generation of left neutrino Majorana mass:

$$ L_{m} = -G_{ij}\begin{pmatrix} \bar{\nu}_{L}& \bar{l}_{L}\end{pmatrix}^{i}i\sigma_{2}\begin{pmatrix}\varphi_{1}^{*} \\ \varphi^{*}_{2} \end{pmatrix}\nu_{R}^{j} - M_{ij}(\nu_{R}^{T})^{i}\hat{C}\nu_{R}^{j} + h.c. $$ After using the unitary gauge and shifting the vacuum we can get the mass terms $$ \tag 1 L_{m} = -\frac{1}{2}\begin{pmatrix} \nu_{L} & \nu_{R}^{c}\end{pmatrix}^{T}\hat{C}^{-1}\begin{pmatrix} 0 & \hat{m}_{D}^{*} \\ \hat{m}_{D}^{\dagger} & \hat{M}^{\dagger} \end{pmatrix}\begin{pmatrix} \nu_{L}\\ \nu_{R}^{c}\end{pmatrix}+h.c., $$ where $\nu^{c} = \hat{C}\bar{\nu}^{T}, \quad \hat{m}^{ij}_{D} = \eta G^{ij}$,

and interaction terms with the Higgs field $\sigma $: $$ \tag 2 L_{int} = -\sigma G_{ij} \bar{\nu}_{L}^{i}\nu_{R}^{j} + h.c. $$ We can "diagonalize" $(1)$ if we assume that $||m_{D}|| << ||M||$: $$ L_{m} \approx \nu_{L}^{T}\hat{M}_{1}\nu_{L} - \nu_{R}^{T}\hat{M}_{2}\nu_{R} + h.c., \quad \hat{M}_{1}= -\hat{m}_{D}^{T}\hat{M}^{-1}\hat{m}_{D}, \quad \hat{M}_{2} = \hat{M}. $$ But how to "delete" interaction term $(2)$ (i.e., to integrate out heavy field $\nu_{R}$)?

Or this interaction is the prediction?


1 Answer 1


The interaction term disappears because when you integrate out the right handed neutrinos you insert in a linear combination of the other fields in their place. For example I expect that we would have something of the form,

\begin{equation} G _{ ij} \sigma \bar{\nu} _L ^i \nu _R ^j \rightarrow G _{ij} G _{jk}\sigma ^2 \bar{\nu} _L ^i \nu _L ^k \end{equation} This isn't an exact expression since you are doing the matching after diagonalization in which case you are integrating out the interaction basis right handed neutrinos as opposed to the mass basis right handed neutrinos.

To calculate exactly what this term is you can use tree level matching or integrate out the field using the equation of motion.


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