Hermitian conjugate of Dirac field bilinear

[phypar]

In the standard QFT textbook, the Hermitian conjugate of a Dirac field bilinear
$\bar\psi_1\gamma^\mu \psi_2$ is $\bar\psi_2\gamma^\mu \psi_1$.

Here is the question, why there is not an extra minus sign coming from the anti-symmetry of fermion fields?

 

[Einj]

You don't need to anti-commute anything. The transposition automatically change the position of the fields:
$$
(\bar \psi_1\gamma_\mu\psi_2)^\dagger=\psi_2^\dagger\gamma_\mu^\dagger \gamma_0\psi_1=\psi_2^\dagger\gamma_0\gamma_\mu\psi_1=\bar \psi_2\gamma_\mu\psi_1
$$

 

[phypar]

This is exactly what i don't understand, so in the transposition there is a change of the postion of the fermions fields, but according to the anti-commutation rule of them, shouldn't there be a minus sign? I know there is something wrong in my understanding, but just cannot figure it out.

You don't need to anti-commute anything. The transposition automatically change the position of the fields:
$$
(\bar \psi_1\gamma_\mu\psi_2)^\dagger=\psi_2^\dagger\gamma_\mu^\dagger \gamma_0\psi_1=\psi_2^\dagger\gamma_0\gamma_\mu\psi_1=\bar \psi_2\gamma_\mu\psi_1
$$

 

[Bill_K]

In the standard QFT textbook, the Hermitian conjugate of a Dirac field bilinear
$\bar\psi_1\gamma^\mu \psi_2$ is $\bar\psi_2\gamma^\mu \psi_1$.

Here is the question, why there is not an extra minus sign coming from the anti-symmetry of fermion fields?

There IS an extra minus sign. $\gamma^\mu$ is Hermitian, but $\bar\psi\gamma^\mu \psi$ is anti-Hermitian. The Hermitian quantity is $i \bar\psi\gamma^\mu \psi$.

 

[Einj]

I'm not anti-commuting the fields. It's just the definition of transpose. Anti-commuting means, for example, to take $\bar \psi_1\gamma_\mu\psi_2$ and move $\psi_2$ on the other side, i.e. you are writing the same operator in a different way. When you take the transpose you are not rearranging the same operator, it's a new one (the transpose) and it is defined with the inverse order of operators.

 

[ChrisVer]

Is $\gamma^{\mu}$ hermitian?
I am not so sure...
$\gamma^{0}$ is
but $\gamma^{i}$ is antihermitian.
The conjugate of the gamma matrices, defined by Clifford Algebra $\left\{ \gamma^{\mu},\gamma^{\nu}\right\} = 2 n^{\mu \nu} I_{4}$ is given by:
$(\gamma^{\mu})^{\dagger} = \gamma^{0}\gamma^{\mu}\gamma^{0}$

In fact you have:
$(\bar{\psi_{1}} \gamma^{\mu} \psi_{2} )^{\dagger}=\psi_{2}^{\dagger} (\gamma^{\mu})^{\dagger} (\gamma^{0})^{\dagger} \psi_{1} = \psi_{2}^{\dagger} \gamma^{0}\gamma^{\mu}\gamma^{0}\gamma^{0} \psi_{1} = \bar{\psi_{2}} \gamma^{\mu} \psi_{1}$
The first is by definition of any "matrix", whether commuting or not, $(AB)^{\dagger}= B^{\dagger}A^{\dagger}$.
If you now want to anticommute the AB: $(AB)^{\dagger}=-(BA)^{\dagger}= -A^{\dagger}B^{\dagger}=B^{\dagger}A^{\dagger}$.

 

[Einj]

$\gamma_\mu$ is not Hermitian, but the additional $\gamma_0$ that you get from its conjugate goes together to $\psi^\dagger$ to form $\bar \psi$. I honestly don't think there should be any extra minus

 

[andrien]

One can chose all gamma matrices to be hermitian, it is merely a convention which one adopts and can be found in Sakurai, Mandl and Shaw. On the other hand one can chose only γ₀ to be hermitian which is most common used convention.