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 $(\theta \eta)^\dagger =$?

  • 1
    $\begingroup$ "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, 2022 at 17:06
  • 1
    $\begingroup$ 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, 2022 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, 2022 at 17:10
  • $\begingroup$ Related: physics.stackexchange.com/q/695933/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 29, 2022 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, 2022 at 17:13

1 Answer 1


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 $(AB)^\dagger = + B^\dagger A^\dagger$.

  • $\begingroup$ 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, 2022 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, 2022 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.