$(\psi_L^\dagger \psi_R)^\dagger \neq (\psi_R^\dagger \psi_L)^\dagger$ ? What is the transpose for spinors?

Question

The dirac mass term in terms of Weyl spinors is

$$\psi_L^\dagger \psi_R + \psi_R^\dagger \psi_L.$$

My understanding is that both terms are necessary to form a hermitian term. Naively, if you take the conjugate, you get:

$$(\psi_L^\dagger \psi_R)^\dagger=\psi_R^\dagger \psi_L$$

using the ordinary rules of "daggering".

However according to my lecture, grassmann variables (spinor components) have the special property

$$(\xi_1 \xi_2 )^* = \xi_2^*\xi_1^*$$

If we try to "be careful" about "daggering", by splitting it into transpose and conjugate in order to make use of this special spinor property, we get:

$$(\psi_L^\dagger \psi_R)^\dagger$$ $$=((\psi_L^\dagger \psi_R)^T )^*$$ $$=(\psi_R^T \psi_L^*)^*.$$

Now using the conjugate property for grassman variables on the components gives

$$...=\psi_L^T \psi_R^*.$$

This is not the expected result. I suspect my lecture just didn't mention a special negative sign which is gotten if the transpose involves commuting grassman variables. That would fix the expression.

$\implies$ Do we have a negative sign after transposing grassmann variables?

Answer 1

There should be no special "minus" sign for transposing a single spinor, yet here the minus sign must appear when you transpose the quantity $\psi_L^\dagger \psi_R$ because when you transpose it an ordinary way, the order of components is changed ($\psi_R$ becomes first and $\psi_L$ second).

Spinors consist of anticommuting Grassmann components. In components, if you write $\psi_L^\dagger = (\psi_1^*\,\psi_2^*)$, $\psi_R = (\chi_1\, \chi_2)^T$, then $$ \psi_L^\dagger \psi_R = \psi_1^* \chi_1 + \psi_2^* \chi_2 $$ and $$ \psi_R^T \psi_L^* = \chi_1 \psi_1^* + \chi_2 \psi_2^* $$ As all psi's and chi's are taken to be anticommuting, these two quantities indeed have different sign.