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Since hermitian conjugation and complex conjugation are serious issues in a QFT lagrangian with Grassmann variables, see here and here. Let us try to go to the bottom.

We start by accepting the Grassmann variable has the property $\eta \theta = -\theta \eta$. In Peskin & Schroeder eq 9.65, they wrote for complex Grassmann variable

It is convenient to define complex conjugation to reverse the order of products, just like the Hermitian conjugation of operators: $$\text{ P&S eq 9.65: }(\theta \eta)^* = \eta^* \theta^*= -\theta^* \eta^*. \tag{1}$$

question 1: why do we define complex conjugation $*$ this way? why $(\theta \eta)^* = \eta^* \theta^*= -\theta^* \eta^*$?

Note that this complex conjugation $*$ is not yet hermitian conjugate $\dagger$, see below.

Follow P&S below eq.9.56, the complex valued Grassmann variables $$ \theta =\theta_1 + i\theta_2, \quad \theta^* =\theta_1 - i\theta_2 $$ $$ \eta =\eta_1 + i\eta_2, \quad \eta^* =\eta_1 - i\eta_2 $$

Then we derive that $$ (\theta \eta)^*=((\theta_1 + i\theta_2)(\eta_1 + i\eta_2))^* =((\theta_1\eta_1 -\theta_2\eta_2) + i(\theta_1\eta_2 +\theta_2\eta_1))^* =(\theta_1\eta_1 -\theta_2\eta_2) - i(\theta_1\eta_2 +\theta_2\eta_1) $$ To make comparisons,
$$ \theta^* \eta^* =(\theta_1- i\theta_2)(\eta_1 - i\eta_2) =(\theta_1\eta_1 -\theta_2\eta_2) - i(\theta_1\eta_2 +\theta_2\eta_1). $$

Thus I showed that $$(\theta \eta)^* = +\theta^* \eta^*. \tag{2}$$

So P&S equation (1) contradicts with my (2). Why is that?

question 2: Does it make sense to define tranposed on Grassmann variable? do we have $(\theta \eta)^T = \pm \eta^T \theta^T= \pm \eta \theta= \mp \eta \theta$?

What is the plus or minus sign here?

question 3: Does it make sense to define hermitian conjugate on Grassmann variable? if we can combine the above to get $(\theta \eta)^\dagger =$?

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    $\begingroup$ "Since hermitian conjugation and complex conjugation are serious issues in a QFT lagrangian with Grassman variables" <--- it absolutely is not a serious issue in any way. You are having trouble with it and that's fine, but there's no reason to claim that any serious issue exists. $\endgroup$
    – Prahar
    Mar 29 at 17:06
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    $\begingroup$ There is a good answer to your first question on the math stackexchange here. The answers to the others follow a similar logic. $\endgroup$
    – cpollack
    Mar 29 at 17:07
  • $\begingroup$ I thought hermitian conjugation and complex conjugation are serious issues always for quantum mechanics? should they be something deep in quantum mechanics? thanks I voted up $\endgroup$ Mar 29 at 17:10
  • $\begingroup$ Related: physics.stackexchange.com/q/695933/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 29 at 17:12
  • $\begingroup$ @МаринаMarinaS it’s absolutely not. It maybe a difficult subject to learn for young students but it’s perfectly well defined. No issue exists at all! $\endgroup$
    – Prahar
    Mar 29 at 17:13

1 Answer 1

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You cannot take the transpose or hermitian conjugate of a (Grassmann) number. Those operations are only defined for matrices.

Complex conjugation for complex numbers satisfies $a^* a > 0$. We would like to choose our definition for complex conjugation so that $\eta^*\eta>0$. Let's assume this is true for $\eta$ and $\theta$. Then, suppose that $$ (\eta \theta)^* = \epsilon \eta^* \theta^* $$ Then, $$ (\eta \theta)^* (\eta\theta) = \epsilon \eta^* \theta^* \eta \theta = - \epsilon ( \eta^* \eta ) ( \theta^* \theta) $$ It follows that this is positive if and only if $\epsilon=-1$. Thus, the correct definition for complex conjugation is $$ (\eta\theta)^* = \theta^*\eta^* $$ The mistake you made in doing your P&S calculation is that you assumed $(\theta_1\eta_2)^* = + \theta_1\eta_2$ but this sign is not true.

Now, let's talk about transpose and hermitian conjugate of Grassmann valued matrices. These are matrices such that each element $A_{ij}$ of the matrix is a Grassmann number. Then, $$ [(AB)^T]_{ij} = (AB)_{ji} = A_{jk} B_{ki} = (A^T)_{kj} (B^T)_{ik} $$ At this stage, we have to exchange $A$ and $B$. But these are Grassmann valued, so we have a sign $$ [(AB)^T]_{ij} = - (B^T)_{ik} (A^T)_{kj} = - (B^TA^T)_{ij} \qquad \implies \qquad (AB)^T = - B^T A^T $$

The conjugation properties for matrices follows similarly $$ [(AB)^*]_{ij} = (A_{ik} B_{kj})^* = B_{kj}^* A_{ik}^* $$ To bring it back into matrix multiplication form, we need to exchange the elements. So $$ [(AB)^*]_{ij} = - A_{ik}^* B_{kj}^* = - (A^*B^*)_{ij} \quad \implies \quad (AB)^* = - A^*B^*. $$

Hermitian conjugation is a complex conjugation + transpose and you can use the results above to derive $(AB)^\dagger = + B^\dagger A^\dagger$.

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  • $\begingroup$ This trick $(𝜂𝜃)∗(𝜂𝜃)=𝜖𝜂∗𝜃∗𝜂𝜃=−𝜖(𝜂∗𝜂)(𝜃∗𝜃)>0$ is very neat, where did you learn this from? +1 $\endgroup$ Mar 29 at 17:19
  • $\begingroup$ @МаринаMarinaS I don’t recall anymore. It was more than 15 years ago. It’s pretty standard. It’s also described in the math stackexchange link that someone put in the comments. $\endgroup$
    – Prahar
    Mar 29 at 17:23

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