I'm having a trouble solving a problem 11.9 from Schwartz's QFT and SM. Here's a brief problem statement:

Given following Lagrangian (and assuming that only left neutrino $\nu_L$ couples to bosons): $$ {\cal{L}} = i \nu_L^\dagger \overline{\sigma}^\mu \partial_\mu \nu_L + \ i \nu_R^\dagger \sigma^\mu \partial_\mu \nu_R - \ m \left( \nu_L^\dagger \nu_R + \nu_R^\dagger \nu_L \right) + \ i \frac{M}{2} \left( \nu_R^T \sigma_2 \nu_R - \nu_R^\dagger \sigma_2 \nu_R^\star \right) $$ 1) Show that $\chi_L \equiv i \sigma_2 \nu_R^\star$ transforms as a left-handed spinor under the Lorentz group, so that it can mix with $\nu_L$.
2) Find the mass eigenstates? That is, find linear combinations $\psi_1$ and $\psi_2$ of $\chi_L$ and $\nu_L$ that satisfy the Klein-Gordon equation $\left( \square + m_i^2 \right) \psi_i = 0$

It is easy to prove the first statement, but the second question took a lot of fruitless effort so far. How am I supposed to approach it?

Here's how far I've been able to get so far.

$$ \frac{\partial{\cal{L}}}{\partial \nu_L^\star} = 0 = \ i \overline{\sigma}^\mu \partial_\mu \nu_L - m \nu_R $$

$$ \frac{\partial{\cal{L}}}{\partial \nu_R^\star} = 0 = \ i \sigma^\mu \partial_\mu \nu_R - m \nu_L - i M \sigma_2 \nu_R^\star $$

Multiplying first equation by $i \sigma^\nu \partial_\nu$:

$$ -\square \nu_L = m \left( i \sigma^\mu \partial_\mu \nu_R \right) = \ m^2 \nu_L + m M \chi_L $$

Which is equivalent to:

$$ \left( \square + m^2 \right) \nu_L = - mM \chi_L $$

Doing the same trick with second equation (after some algebra) gives me:

$$ \left( \square + m^2 + M^2 \right) \chi_L = - m M \nu_L $$

In my understanding, given that $\psi_i = \alpha_i \nu_L + \beta_I \chi_L$, I should be able to multiply above two equations by $\alpha_i$ and $\beta_i$ respectively and sum to get the Klein-Gordon equation from the problem above. However, my attempts at this so far have been futile.


You are almost there! Starting from where you stopped: $$ \begin{align} (\Box + m^2) \nu_L +mM\chi_L=0 \\ (\Box + m^2+M^2) \chi_L +mM\nu_L=0 \end{align} $$ we can re-write these equations into a matrix form: $$ [\;\Box + \begin{pmatrix} m^2 & mM \\ mM & m^2+M^2 \end{pmatrix}\;] \begin{pmatrix}\nu_L\\\chi_L \end{pmatrix} = (\Box +\tilde{M}^2)\begin{pmatrix}\nu_L\\\chi_L \end{pmatrix}= 0 $$ This is not quite the Klein-Gordon equation because $\tilde{M}^2$ is not diagonal ($\Box \cdot \mathbb{1}$ is already diagonal and any change of basis doesn't affect it). So all we need to do is to diagonalise $\tilde{M}^2$. The eigenvalues are found to be: $$ \begin{align} \tilde{m}^2_1 = \frac{1}{2} (M^2+2m^2-\sqrt{4m^2M^2+m^4})\tag{1a}\\ \tilde{m}^2_2 = \frac{1}{2} (M^2+2m^2+\sqrt{4m^2M^2+m^4})\tag{1b} \end{align} $$ From the eigenvectors of the mass matrix $\tilde{M}^2$ you can obtain the normalized mass eigenstates you are looking for: $$ \begin{align} \tilde{v}_1 &= \frac{[(M+\sqrt{4m^2+M^2})\;\nu_L -2m\;\chi_L]}{\sqrt{(M+\sqrt{4m^2+M^2})^2+4m^2}} \tag{2a}\\ \tilde{v}_2 &= \frac{[(M-\sqrt{4m^2+M^2})\;\nu_L -2m\;\chi_L]}{\sqrt{(M-\sqrt{4m^2+M^2})^2+4m^2}} \tag{2b} \end{align} $$ Performing this change of basis, you can obtain two separate pieces of Klein-Gordon equations: $$ \begin{align} (\Box+\tilde{m}_1^2)\;\tilde{\nu}_1 =0 \tag{3a}\\ (\Box+\tilde{m}_2^2)\;\tilde{\nu}_2 =0 \tag{3b} \end{align} $$

One interesting thing to notice is the case when $M \gg m$. In that case the mass eigenstate $\tilde{\nu}_{1}$ has a heavy overlap with the original $\nu_L$ (c.f. Eq.(2a)), while mass eigenstate $\tilde{\nu}_2$ has a heavy overlap with $\chi_L$ (c.f. Eq.(2b)). Also, in this limit: $$ \begin{align} \tilde{m}_1 = \sqrt{m^2+\frac{M^2}{4}}-\frac{M}{2} \approx \frac{m^2}{M} \\ \tilde{m}_2 = \sqrt{m^2+\frac{M^2}{4}}+\frac{M}{2} \approx M \end{align} $$ Hence, as the Majorana mass $M$ increases, the mass of $\tilde{\nu}_1$ (which looks like the left-handed neutrino $\nu_L$) decreases, while the mass of $\tilde{\nu}_2$ (which looks like $\chi_L$ and thus related to $\nu_R$) increases. This is the see-saw mechanism that Schwartz wants us to see.


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