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Why sometimes we remember that "classical" lagrangians of fermions are constructed from grassmann numbers, while sometimes don't?

For example, for Majorana's field in terms of 2-component spinors (you can read about this representation here in details), $$ L = \bar{\psi}(\alpha^{\mu}\partial_{\nu} - m)\psi , \quad \alpha^{\mu} = (\sigma_{2}K, \sigma_{1}, \sigma_{2}, \sigma_{3}), \quad K\psi = \psi^{*}, \quad \bar{\psi} = -\psi^{\dagger}\alpha_{0}, $$ and the mass term become $$ \bar{\psi} \psi = -\psi^{\dagger}\sigma_{2}\psi^{*} = -\begin{pmatrix}\psi_{1}^{*} & \psi_{2}^{*}\end{pmatrix}\sigma_{2}\begin{pmatrix} \psi_{1}^{*} \\ \psi_{2}^{*}\end{pmatrix} = 0 $$ if the fields are "classical". So we must assume their grassmanian nature before making all calculations (there this expression won't be equal to zero).

But when we assume the mass matrix for see-saw type 1 model, we introduce the term $$ -M_{ij}(\psi_{R}^{T})^{i}\hat{C}\psi_{R}^{j} + h.c., \quad M_{ij} = M_{ji}, \quad \hat{C} = i\gamma_{2}\gamma_{0}. $$ If the fields are grassmann numbers, this expression is equal to zero (don't forget about symmetric mass matrix). So we forget about grassmannian nature of the fields even in "classical" level.

Here "classical" means that we set all of anticommutators to zero. If we set them not to zero, we will have new problems, but it's the other story.

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    $\begingroup$ Is $\Psi_R$ a two-component spinor like you say in the first paragraph, or is it a four-component one, as implied by how your $C$ is represented in terms of 4-dimensional gamma matrices? $\endgroup$ – QuantumDot Aug 9 '14 at 21:05
  • $\begingroup$ @QuantumDot : the mass term in the second paragraph is built from 4-spinors, but this is only one of ways to write Majorana's mass term through 4-spinors: $$ M\bar{\Psi}_{M}\Psi_{M} = M\Psi_{R}^{T}\hat{C}\Psi_{R} + h.c., $$ where $$ \Psi_{M} = \Psi_{R} + \hat{C}\bar{\Psi}_{R}^{T}. $$ We may also write it in terms of 2-spinors. $\endgroup$ – Andrew McAddams Aug 9 '14 at 21:12
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Classical lagrangians of fermions are always constructed out of Grassmann numbers. No exception.

In both of OP's cases, the mass term is nonvanishing:

In the first case the mass term is proportional to $\psi_1^*\psi_2^*-\psi_2^*\psi_1^* = 2\psi_1^*\psi_2^* \neq 0.$

In the second case, I write in the chiral basis: $$\gamma^0=\begin{pmatrix}& \sigma^0 \\ \sigma^0 \end{pmatrix}\qquad \gamma^0=\begin{pmatrix} &\sigma^2 \\ -\sigma^2 \end{pmatrix} \enspace \Rightarrow \hat{C}=\begin{pmatrix}-i\sigma^0\sigma^2 \\ & i\sigma^0\sigma^2\end{pmatrix}$$

So that (picking the diagonal element of the mass matrix $i=j$ as the off-diagonal terms are obviously non-vanishing.)

$$ \begin{aligned}\psi_R^T \hat{C} \psi_R &= \begin{pmatrix}0&0&\psi_3&\psi_4\end{pmatrix} \begin{pmatrix}-i\sigma^0\sigma^2 \\ & i\sigma^0\sigma^2\end{pmatrix} \begin{pmatrix}0\\0\\\psi_3\\\psi_4\end{pmatrix} \\ &= \begin{pmatrix}\psi_3 & \psi_4 \end{pmatrix}i\sigma^0\sigma^2\begin{pmatrix}\psi_3\\\psi_4\end{pmatrix}\\ &=\psi_3\psi_4-\psi_4\psi_3 = 2\,\psi_3\psi_4 \neq 0 \end{aligned}$$ where in the last step I used the Grassmann property of anticommutation.

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  • $\begingroup$ Excuse me, it's my fault: I have thinked that symmetric mass matrix leads to the $\sum_{i, j}(\psi_{i}\psi_{j} + \psi_{j}\psi_{i}$ expression, which is zero. $\endgroup$ – Andrew McAddams Aug 10 '14 at 1:09

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