# How does this term in the Majorana mass not vanish?

This is classical field theory. In the Majorana mass term, we have the expression $$\nu_L^T\sigma_2\nu_L \tag{1}$$ where the left-handed spinor field $$\nu$$ has a Grassmann-valued amplitude, i.e., $$\nu = \mathcal{N} \text{(blah)}$$ where $$\mathcal{N}$$ is a Grassmann number. How does $$(1)$$ not vanish? For, we can write $$(1)$$ as $$\nu_L^T\sigma_2\nu_L = \mathcal{N}^2\text{(blah)}=0$$ by antisymmetry of $$\mathcal{N}$$ with itself.

Now, say that the above problem is resolved. Then, it seems like $$\nu_L^T\sigma_2\nu_L = (\nu_L^T\sigma_2\nu_L)^T = \nu_L^T\sigma_2^T\nu_L = -\nu_L^T\sigma_2\nu_L,$$ which would again suggest that this expression does equal $$0$$. How does this expression not vanish given these two observations?

Remember that $$\nu_L$$ is a two-component column vector with Grassmann entries $$\nu_L^1$$ and $$\nu_L^2$$, so with $$\sigma_2 =\left[\matrix{0&-i\cr i&0}\right],$$ we have $$\nu_L^T \sigma_2 \nu_L = \left[\matrix{\nu_L^1, &\nu_L^2}\right] \left[\matrix{0&-i\cr i&0}\right] \left[\matrix{\nu_L^1 \cr \nu_L^2}\right]= -i(\nu_L^1\nu_L^2- \nu_L^2 \nu_L^1) =-2i\nu_L^1\nu_L^2\ne0.$$
• Oh, so we should not assume that it is merely the amplitude that is Grassmann-valued. It is, in general, the components themselves that are Grassmann-valued? Also, are Grassmann numbers affected by transpose? I.e., is $(\mathcal{N}_1 \mathcal{N}_2)^T = \mathcal{N}_2 \mathcal{N}_1$? Commented Apr 3 at 23:30
• I guess I mean to ask a slightly separate question about taking the tranpose $(\nu_L^T \sigma_2 \nu_L)^T$, which I keep finding to be antisymmetric. However, I know this term should not vanish and so should not be antisymmetric. Commented Apr 3 at 23:38
• There is no meaning to $(\nu_L^T\sigma_2 \nu_L)^T$. How would you transpose a number? --- Grassman or not..... Now you can define an involution such that $(ab)^*=b^*a^*$, for Grassmanns, but that is not the same as a transpose, which is a matrix operation. Commented Apr 3 at 23:41
• The matrix in $(\nu^T\sigma \nu))$ has to be skew symmetric, and transposition interchanges the vectors being transposed, so both the interchange and the transpose gives a minus sign, making the overall sign unchanged. Commented Apr 4 at 11:53