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

Question

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.

Answer 1

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.

Answer 2

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.