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JohnA.
JohnA.
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Following chapter 38 of Srednicki, "bar"-ing means (based on eq. 38.15) $$\bar{A} = \beta A^\dagger\beta$$ One can show for instance that $$\bar{\gamma^\mu} = \gamma^\mu$$ My question is, using the definition (eq. 36.39): $$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}$$ we try to find the bar so we get

$$\beta(\gamma^\mu)^\dagger\beta = \begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix}\begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}^\dagger\begin{pmatrix} 0 &I\\ I & 0 \end{pmatrix}$$ would the $\dagger$ in the $A^\dagger$ mean to transpose the matrix as a 4x4 or as a 2x2?I.e, is $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^* & C^* \\ B^* & D^* \end{pmatrix}$$ where $A,B,C,D$ are 2 by 2 matrices. Or $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^\dagger & B^\dagger\\ C^\dagger & D^\dagger \end{pmatrix}$$$$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^\dagger & C^\dagger\\ B^\dagger & D^\dagger \end{pmatrix}$$ Which of these 2 is what is meant by the author? For a general matrix, they are different because the component matrices may be not hermitian.

Following chapter 38 of Srednicki, "bar"-ing means (based on eq. 38.15) $$\bar{A} = \beta A^\dagger\beta$$ One can show for instance that $$\bar{\gamma^\mu} = \gamma^\mu$$ My question is, using the definition (eq. 36.39): $$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}$$ we try to find the bar so we get

$$\beta(\gamma^\mu)^\dagger\beta = \begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix}\begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}^\dagger\begin{pmatrix} 0 &I\\ I & 0 \end{pmatrix}$$ would the $\dagger$ in the $A^\dagger$ mean to transpose the matrix as a 4x4 or as a 2x2?I.e, is $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^* & C^* \\ B^* & D^* \end{pmatrix}$$ where $A,B,C,D$ are 2 by 2 matrices. Or $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^\dagger & B^\dagger\\ C^\dagger & D^\dagger \end{pmatrix}$$ Which of these 2 is what is meant by the author? For a general matrix, they are different because the component matrices may be not hermitian.

Following chapter 38 of Srednicki, "bar"-ing means (based on eq. 38.15) $$\bar{A} = \beta A^\dagger\beta$$ One can show for instance that $$\bar{\gamma^\mu} = \gamma^\mu$$ My question is, using the definition (eq. 36.39): $$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}$$ we try to find the bar so we get

$$\beta(\gamma^\mu)^\dagger\beta = \begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix}\begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}^\dagger\begin{pmatrix} 0 &I\\ I & 0 \end{pmatrix}$$ would the $\dagger$ in the $A^\dagger$ mean to transpose the matrix as a 4x4 or as a 2x2?I.e, is $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^* & C^* \\ B^* & D^* \end{pmatrix}$$ where $A,B,C,D$ are 2 by 2 matrices. Or $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^\dagger & C^\dagger\\ B^\dagger & D^\dagger \end{pmatrix}$$ Which of these 2 is what is meant by the author? For a general matrix, they are different because the component matrices may be not hermitian.

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Qmechanic ♦
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Question about meaning of "bar"-ing in the context of diracDirac fields

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JohnA.
JohnA.
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Question about meaning of "bar"-ing in the context of dirac fields

Following chapter 38 of Srednicki, "bar"-ing means (based on eq. 38.15) $$\bar{A} = \beta A^\dagger\beta$$ One can show for instance that $$\bar{\gamma^\mu} = \gamma^\mu$$ My question is, using the definition (eq. 36.39): $$\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}$$ we try to find the bar so we get

$$\beta(\gamma^\mu)^\dagger\beta = \begin{pmatrix} 0 & I\\ I & 0 \end{pmatrix}\begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma^\mu} & 0 \end{pmatrix}^\dagger\begin{pmatrix} 0 &I\\ I & 0 \end{pmatrix}$$ would the $\dagger$ in the $A^\dagger$ mean to transpose the matrix as a 4x4 or as a 2x2?I.e, is $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^* & C^* \\ B^* & D^* \end{pmatrix}$$ where $A,B,C,D$ are 2 by 2 matrices. Or $$\begin{pmatrix} A & B \\ C & D \end{pmatrix}^\dagger =\begin{pmatrix} A^\dagger & B^\dagger\\ C^\dagger & D^\dagger \end{pmatrix}$$ Which of these 2 is what is meant by the author? For a general matrix, they are different because the component matrices may be not hermitian.