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Hermitian conjugate of Dirac field bilinear 
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#1
Apr2914, 05:36 AM

P: 11

In the standard QFT textbook, the Hermitian conjugate of a Dirac field bilinear
[itex]\bar\psi_1\gamma^\mu \psi_2[/itex] is [itex]\bar\psi_2\gamma^\mu \psi_1[/itex]. Here is the question, why there is not an extra minus sign coming from the antisymmetry of fermion fields? 


#2
Apr2914, 08:31 AM

P: 328

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


#3
Apr2914, 09:41 AM

P: 11

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 anticommutation 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.



#4
Apr2914, 11:27 AM

Sci Advisor
Thanks
P: 4,160

Hermitian conjugate of Dirac field bilinear



#5
Apr2914, 11:27 AM

P: 328

I'm not anticommuting the fields. It's just the definition of transpose. Anticommuting means, for example, to take [itex]\bar \psi_1\gamma_\mu\psi_2[/itex] and move [itex]\psi_2[/itex] 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.



#6
Apr2914, 11:50 AM

P: 1,058

Is [itex]\gamma^{\mu}[/itex] hermitian?
I am not so sure... [itex]\gamma^{0}[/itex] is but [itex]\gamma^{i}[/itex] is antihermitian. The conjugate of the gamma matrices, defined by Clifford Algebra [itex]\left\{ \gamma^{\mu},\gamma^{\nu}\right\} = 2 n^{\mu \nu} I_{4}[/itex] is given by: [itex] (\gamma^{\mu})^{\dagger} = \gamma^{0}\gamma^{\mu}\gamma^{0} [/itex] In fact you have: [itex] (\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} [/itex] The first is by definition of any "matrix", whether commuting or not, [itex] (AB)^{\dagger}= B^{\dagger}A^{\dagger}[/itex]. If you now want to anticommute the AB: [itex] (AB)^{\dagger}=(BA)^{\dagger}= A^{\dagger}B^{\dagger}=B^{\dagger}A^{\dagger}[/itex]. 


#7
Apr2914, 11:54 AM

P: 328

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



#8
Apr2914, 12:52 PM

P: 1,020

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 γ_{0} to be hermitian which is most common used convention.



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