Breo said:
I asked this to eventually understand the next decomposition. If we take the tensor product of 2 spinors ## \Psi\Psi* ##:
$$[(\frac{1}{2}, 0 ) \oplus (0, \frac{1}{2})] \otimes [(\frac{1}{2}, 0 ) \oplus (0, \frac{1}{2})] = 2(\frac{1}{2},\frac{1}{2}) \oplus (1,0) \oplus 2(0, 0 ) \oplus (0, 1)$$
The LHS of this equation is the vector space S \otimes \bar{ S } of all 16-component vectors of the form \psi \otimes \bar{ \psi }. We can also take this “vector” to be the 4 \times 4 matrix (of fields on M^{ ( 1 , 3 ) }) \psi_{ a } \bar{ \psi }_{ b } formed by the following direct product
<br />
\psi \otimes \bar{ \psi } = \left( \begin{array} {c} \psi_{1} \\ \psi_{2} \\ \psi_{3} \\ \psi_{4} \end{array} \right) \otimes \left( \psi_{ 1 }^{ * } , \ \psi_{ 2 }^{ * } , \ - \psi_{ 3 }^{ * } , \ - \psi_{ 4 }^{ * } \right) .<br />
Or
<br />
\psi \otimes \bar{ \psi } = \left( \begin {array} {c c c c}<br />
( \psi_{ 1 } \psi_{ 1 }^{ * } ) & ( \psi_{ 1 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 1 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 1 } \psi_{ 4 }^{ * } ) \\<br />
( \psi_{ 2 } \psi_{ 1 }^{ * } ) & ( \psi_{ 2 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 2 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 2 } \psi_{ 4 }^{ * } ) \\<br />
( \psi_{ 3 } \psi_{ 1 }^{ * } ) & ( \psi_{ 3 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 3 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 3 } \psi_{ 4 }^{ * } ) \\<br />
( \psi_{ 4 } \psi_{ 1 }^{ * } ) & ( \psi_{ 4 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 4 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 4 } \psi_{ 4 }^{ * } )<br />
\end {array} \right) .<br />
But any 4 \times 4 matrix can be expanded in the standard (complete) basis of Clifford algebra \mathcal{ C }_{ ( 1 , 3 ) } ( \mathbb{ C } ):
<br />
\Gamma^{ A } = \{ I_{4 \times 4} , \gamma^{ \mu } , \gamma^{ \mu } \gamma_{ 5 } , \sigma^{ \mu \nu } , \gamma_{ 5 } \} ,<br />
where A = 1 , 2 , \cdots , 16. So, we can write
<br />
\psi \otimes \bar{ \psi } = \sum_{ A = 1 }^{ 16 } a_{ A } \ \Gamma^{ A } . \ \ \ \ (1)<br />
To find the coefficients a_{ A } we use the trace properties of \Gamma^{ A }: \mbox{ Tr } ( \Gamma^{ A } ) = 0 , \ \mbox{for} \ \Gamma^{ A } \neq I ,
\mbox{ Tr } ( \Gamma^{ A } ) = 4 , \mbox{for} \ \Gamma^{ A } = I ,
and
\mbox{ Tr } ( \Gamma_{ B } \Gamma^{ A } ) \sim 2^{ 2 } \delta^{ A }_{ B } .
So, multiplying Eq(1) with \Gamma_{ B } and taking the trace, we find
<br />
\mbox{ Tr } ( \Gamma_{ B } \psi \otimes \bar{ \psi } ) = a_{ I } \mbox{ Tr } ( \Gamma_{ B } ) + \sum_{ \Gamma^{ R } \neq I } a_{ R } \mbox{ Tr } ( \Gamma_{ B } \Gamma^{ R } ) . \ \ \ (2)<br />
For \Gamma_{ B } = I, we find
<br />
a_{ I } = \frac{ 1 }{ 4 } \mbox{ Tr } ( \psi \otimes \bar{ \psi } ) = \frac{ 1 }{ 4 } \bar{ \psi } \psi .
For \Gamma_{ B } \neq I, Eq(2) gives
<br />
\mbox{ Tr } ( \Gamma_{ B } \psi \otimes \bar{ \psi } ) = 0 + \sum_{ \Gamma^{ R } \neq I } a_{ R } \mbox{ Tr } ( \Gamma_{ B } \Gamma^{ R } ) \sim 2^{ 2 } a_{ B } .<br />
Thus
<br />
a_{ B } \sim \frac{ 1 }{ 4 } ( \Gamma_{ B } )_{ a b } ( \psi \otimes \bar{ \psi } )_{ b a } = \frac{ 1 }{ 4 } \bar{ \psi }_{ a } ( \Gamma_{ B } )_{ a b } \psi_{ b } = \frac{ 1 }{ 4 } \bar{ \psi }_{ a } \Gamma_{ B } \psi .<br />
So, our expansion Eq(1) becomes
<br />
\psi \otimes \bar{ \psi } = \frac{ 1 }{ 4 } ( \bar{ \psi } \psi ) \ I + \frac{ c }{ 4 } \sum_{ \Gamma^{ B } \neq I } ( \bar{ \psi } \Gamma^{ B } \psi ) \Gamma_{ B } ,<br />
where c is a constant whose value depends on the definitions of \gamma_{5} and \sigma^{ \mu \nu } and on the signature of the spacetime metric.
From this it follows that
<br />
\bar{ \psi } \psi + \bar{ \psi } \gamma_{ 5 } \psi \ \in ( 0 , 0 ) \oplus ( 0 , 0 ) , \mbox{ dim } : \ [1] \oplus [1] ,<br />
<br />
\bar{ \psi } \gamma^{ \mu } \psi + \bar{ \psi } \gamma^{ \mu } \gamma_{ 5 } \psi \ \in ( 1/2 , 1/2 ) \oplus ( 1/2 , 1/2 ) \sim [4] \oplus [4] ,<br />
and
<br />
\bar{ \psi } \sigma^{ \mu \nu } \psi \ \in ( 1 , 0 ) \oplus ( 0 , 1 ) \sim [6] .<br />
So, going back to the vector space S \otimes \bar{ S }, we can say that the 16 \times 16 matrices which act naturally on the 16-component “vector” \psi \otimes \bar{ \psi } \in S \otimes \bar{ S } can be brought into block diagonal form with matrices of dimensions [1] , [1] , [4] , [4] and [6] on the diagonal.
Why we say these are 16 terms and how to descompose ##2(\frac{1}{2},\frac{1}{2})## in the vector ## \bar{\Psi}\gamma^{\mu}\Psi ## and pseudovector ## \bar{\Psi}\gamma^5 \gamma^{\mu}\Psi ## ?
You need to spend more time on group theory in general and the representation theory of Lorentz algebra in particular.
Sam
