# Is $\bar{\psi} \psi$ its own complex (*)? transpose ($T$)? hermitian conjugate ($\dagger$)?

Related to this earlier A common standard model Lagrangian mistake?

Here I am treating Dirac equation of Dirac field as QFT.

## You may want to consider the quantized version or the classical version.

Now my question is:

## Is $$\bar{\psi} \psi$$ its own complex (*)? its own transpose ($$T$$)? its own hermitian conjugate ($$\dagger$$)?

Naively in the Dirac lagrangian we have its mass term: $$m\bar{\psi} \psi =m \psi^\dagger \gamma^0 \psi$$ based on the standard notation in Peskin's QFT book (using the Weyl representation).

Since $$\bar{\psi} \psi$$ is a Lorentz scalar number (that depends on the spacetime point $$(t,x)$$) and integrated over the 4d spacetime, I believe it is a real number as a Lorentz scalar (yes), so $$\bar{\psi} \psi \in \mathbb{R} (?).$$ If this is true, we may check that $$T$$ as a transpose and $$*$$ as a complex conjugation, these should be true $$(\bar{\psi} \psi)^T=\bar{\psi} \psi ? \tag{(i)}$$ $$(\bar{\psi} \psi)^*=\bar{\psi} \psi ? \tag{(ii)}$$ However, a simple calculation shows that: $$(\bar{\psi} \psi)^T=(\psi^\dagger \gamma^0 \psi)^T={\psi}^T \gamma^0 \psi^*= \sum_{a,b}{\psi}_a \gamma^0_{ab} \psi^*_b= \sum_{a,b} \{{\psi}_a,\psi^*_b\} \gamma^0_{ab} - \sum_{a,b} \psi^*_b \gamma^0_{ab} {\psi}_a =-\psi^\dagger \gamma^0_{} {\psi} = -\bar{\psi} \psi$$ here $$a,b$$ are 4-component spinor indices. We use the fermi anticommutation relation $$\{{\psi}_a,\psi^*_b\}={\psi}_a\psi^*_b +\psi^*_b \gamma^0_{ab} {\psi}_a =\delta^3$$ which is up to a delta function for the equal time anticommutation relation. We derive the last equality based on $$\gamma^0_{ab}=\gamma^0_{ba}$$, so $$\sum_{a,b} \psi^*_b \gamma^0_{ba} {\psi}_a = -\bar{\psi} \psi$$. This means we derive that: $$(\bar{\psi} \psi)^T=-\bar{\psi} \psi. \tag{(i')}$$ Similarly, a simple calculation shows that: $$(\bar{\psi} \psi)^*=(\psi^\dagger \gamma^0 \psi)^*={\psi}^T \gamma^0 \psi^*= \sum_{a,b}{\psi}_a \gamma^0_{ab} \psi^*_b= \sum_{a,b} \{{\psi}_a,\psi^*_b\} \gamma^0_{ab} - \sum_{a,b} \psi^*_b \gamma^0_{ab} {\psi}_a =-\psi^\dagger \gamma^0_{} {\psi} = -\bar{\psi} \psi$$ here $$a,b$$ are 4-component spinor indices. This means we derive that: $$(\bar{\psi} \psi)^*=-\bar{\psi} \psi. \tag{(ii')}$$

## question 1:

I imagine that the Dirac path integral implies a $$e^{ -i m \bar{\psi} \psi}$$ phase with $$\bar{\psi} \psi$$ shall be a real number. But it turns out to lead to a conflict between the equations (i), (ii) to that of (i'), (ii'). Is there a simple explanation?

## question 2:

Is the delta function important here? the equality above is not precise since we use $$\{{\psi}_a,\psi^*_b\}={\psi}_a\psi^*_b +\psi^*_b \gamma^0_{ab} {\psi}_a =\delta^3$$.

## question 3:

However, regardless the conflict of its complex (*) and its transpose ($$T$$), we always have its hermitian conjugate ($$\dagger$$) to be itself $$(\bar{\psi} \psi)^\dagger=+\bar{\psi} \psi. \tag{(iii)}$$ Why is that?

There may be good physics intuition behind.

• Not sure who voted down -- what I asked is reasonable thing about taking "complex (*)? its transpose (𝑇)? its hermitian conjugate (†)?" in which coordinates/spinors/matrix/space? -- please leave an answer if you find what I said is trivial or obvious Mar 29, 2022 at 2:31
• $\psi$ is a Grassmannian variable. All your manipulations are valid only for complex-valued spinors. You have to be careful about the extra minus signs you get when you move fermions across each other (eg. In your transpose calculation) Mar 29, 2022 at 8:47
• To be clear, if $A$ and $B$ are Grassmann-valued, then $(AB)^T = - BA$ and $(AB)^* = - A^*B^*$. Hermitian conjugate is complex conjugation + transpose so $(AB)^\dagger = ((AB)^T)^* = - (B^TA^T)^* = + B^\dagger A^\dagger$ so the formula for Hermitian conjugation is exactly the same as for complex-valued matrices. Mar 29, 2022 at 8:54
• Thanks Prahar I agree with you -- your answer agrees with mine. I think the other answer by Gold only treats $\psi$ as a classical variable. Does Grassmann-valued count as a quantized variable? Or can Grassmann-valued be a classical variable? Mar 29, 2022 at 12:44
• How about the commutator $\{{\psi}_a,\psi^*_b\}={\psi}_a\psi^*_b +\psi^*_b \gamma^0_{ab} {\psi}_a =\delta^3$? Is the $\delta^3$ involved? Should it be equal time on a space $\delta^3$ or should it be a spacetime $\delta^4$? Do we need to worry about the $\delta$ function? could you be so kind to write an answer? (I think Gold answer may be partial only.) Mar 29, 2022 at 12:47

In classical field theory, $$\psi$$ is a Grassmann-valued field so you have to be super careful with complex conjugation and transpose. If $$A$$ and $$B$$ are Grassmann-valued matrices, then $$(AB)^T = - B^TA^T , \qquad (AB)^* = - A^* B^* , \qquad (AB)^\dagger = + B^\dagger A^\dagger$$ As a comparison, for complex-valued matrices, the corresponding formulae are $$(AB)^T = + B^TA^T , \qquad (AB)^* = + A^* B^* , \qquad (AB)^\dagger = + B^\dagger A^\dagger$$ If you keep track of these extra minus signs, you will find that $${\bar \psi} \psi$$ is a real-valued number.

• I think 𝜓¯𝜓 is a real valued Grassman number? thanks! Mar 29, 2022 at 13:55
• Product of two Grassmann numbers is a complex number so ${\bar \psi} \psi$ is real-valued. It is NOT Grassmann-valued. Mar 29, 2022 at 13:56
• But I thought that only $\int [D {\psi}][D \bar{\psi}]𝜓¯𝜓$ or $\int [D {\psi}][D \bar{\psi}] \exp(a 𝜓¯𝜓)$ gives a number (NOT Grassmann)? The $𝜓¯𝜓$ is still a real valued Grassman number? Mar 29, 2022 at 13:58
• Product of two Grassmann numbers commutes with ALL other numbers (complex or Grassmann). Therefore, it is not a Grassmann number (Grassmann numbers ANTICOMMUTE with other Grassmann numbers). $(\theta \eta) \psi = - \theta \psi \eta = + \psi ( \theta \eta)$. Thus, $\theta\eta$ commutes with $\psi$. This implies that $\theta\eta$ is a complex number. Mar 29, 2022 at 14:01
• The action has to be a real number, right? How can you have a term of the form ${\bar \psi} \psi$ in the action if it's not real?!?!? Mar 29, 2022 at 14:02

Well, by definition we have, $$\bar \psi = \psi^\dagger \gamma^0=(\psi^T)^\ast \gamma^0\tag{1}.$$

Now, $$\psi$$ is a $$4\times 1$$-matrix, so $$\psi^T$$ is a $$1\times 4$$-matrix. Clearly $$(\psi^T)^\ast$$ will also have the same dimensionality. Since $$\gamma^0$$ is a $$4\times 4$$-matrix it follows $$(\psi^T)^\ast\gamma^0$$ is again a $$1\times 4$$-matrix. So this establishes that $$\bar \psi$$ is just a row vector.

In that case $$\bar \psi \psi$$ is a row vector times a column vector. The result is a $$1\times 1$$-matrix, or just a number. This by itself already establishes that $$(\bar \psi \psi)^T = \bar \psi \psi$$.

So: Is $$\bar \psi \psi$$ its transpose? Yes.

Now even though $$\bar \psi \psi$$ is just a number it is expressed as a product of matrices and therefore we can use the following identity $$(\bar \psi \psi)^\dagger = \psi^\dagger \bar \psi^\dagger = \psi^\dagger (\psi^\dagger \gamma^0)^\dagger = \psi^\dagger(\gamma^0)^\dagger \psi = \psi^\dagger \gamma^0 \psi = \bar \psi \psi\tag{2}$$

where we have used that $$\gamma^0$$ is hermitian.

So: Is $$\bar \psi \psi$$ its hermitian conjugate? Yes.

Finally combining the two we answer the other question: Indeed we have that $$(\bar \psi \psi)^\ast = [(\bar \psi \psi)^T]^\ast = (\bar \psi \psi)^\dagger = \bar \psi \psi\tag{3}.$$

So: Is $$\bar \psi \psi$$ its complex conjugate? Yes.

• thanks so much for your answer, but there may be different meanings when we take complex conjugation and transpose. Mar 28, 2022 at 22:26
• I think you are doing the classical version (but there is an issue that the Dirac $\psi$ is not a normalized wavefunction see Peskin chap 3). But I am doing the quantized version where when I exchange the fermion operator, there are (-1) sign. Maybe we can compare our calculations carefully :) Mar 28, 2022 at 22:28
• My calculation follows closely a related one as eq. 3.147 of Peskin QFT chap 3. Mar 28, 2022 at 22:29
• Not sure who voted down -- what I asked is reasonable thing about taking "complex (*)? its transpose (𝑇)? its hermitian conjugate (†)?" in which coordinates/spinors/matrix/space? Mar 29, 2022 at 2:31
• thanks, I upvote anyway. Mar 29, 2022 at 12:47