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

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?

• @Qmechanic Can I ask why not the quantum-field-theory tag? That is the context in which these objects appear & though you could consider them abstractly instead, it is rare to talk about grassman outside of QFT. I suppose you could consider anything in QFT abstractly, anyhow. My concern is just that people interested in learning more about QFT would not find this question if they search by tag, despite it being relevant almost exclusively to them. Commented Jan 28, 2020 at 12:11
• Hi @doublefelix: The QFT tag is one of the most overused tags -- I think the question is more about supermathematics -- but put it in if you think it is better. Commented Jan 28, 2020 at 12:26

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.