Spinorial Field Representation

[Breo]

What does mean the next (why we write it like this, why is a sum, why first a 0 and secondly a 1/2 and viceversa):

$$ (\frac{1}{2} , 0 ) \oplus (0, \frac{1}{2}) $$

?

 

[The_Duck]

Do you know how the representations of the Lorentz group are classified? The (1/2, 0) and (0, 1/2) are two irreducible representations of the Lorentz group. These representations are also called the "left-handed Weyl spinor" and the "right-handed Weyl spinor." The $\oplus$ is the symbol for a direct sum of group representations.

 

[Breo]

I asked this to eventually understand the next decomposition. If we take the tensor product of 2 spinors $\Psi\Psi*$:

$$[(\frac{1}{2}, 0 ) \oplus (0, \frac{1}{2})] \otimes [(\frac{1}{2}, 0 ) \oplus (0, \frac{1}{2})] = 2(\frac{1}{2},\frac{1}{2}) \oplus (1,0) \oplus 2(0, 0 ) \oplus (0, 1)$$

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 $\bar{\Psi}\gamma^5 \gamma^{\mu}\Psi$ ?

 

[dextercioby]

May I suggest supplimenting whatever you're reading with Wu Ki Tung's group theory for physicists book? It explains things nicely. Miller's book is more detailed (especially on the Clebsch-Gordan theorem), but more intricate. Why are there 16 terms in a basis ? The representation space of the product representation is the tensor product of the 2 spaces. How do you compute dimensions?

 

[Breo]

$dim ( D \otimes D) = dim (D) .dim (D)$ ?

 

[dextercioby]

Bingo, so if the direct sum of 2 elementary Weyl spinors has dimension 4, what's the dimension of the product of 2 such direct sums?

 

[Breo]

16. I missunderstood the dimensions and terms... Thank you for clarification.

Do you know how to descompose the $2(1/2, 1/2)$ as I stated in the official post?

 

[dextercioby]

You're looking for the Fierz identities for the Dirac spinors in regular (1+3) Minkowski spacetime. http://arxiv.org/abs/hep-ph/0412245v4 equations 31 to 33.

 

[samalkhaiat]

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
$$\psi \otimes \bar{ \psi } = \left( \begin{array} {c} \psi_{1} \\ \psi_{2} \\ \psi_{3} \\ \psi_{4} \end{array} \right) \otimes \left( \psi_{ 1 }^{ * } , \ \psi_{ 2 }^{ * } , \ - \psi_{ 3 }^{ * } , \ - \psi_{ 4 }^{ * } \right) .$$
Or
$$\psi \otimes \bar{ \psi } = \left( \begin {array} {c c c c}<br /> ( \psi_{ 1 } \psi_{ 1 }^{ * } ) & ( \psi_{ 1 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 1 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 1 } \psi_{ 4 }^{ * } ) \\<br /> ( \psi_{ 2 } \psi_{ 1 }^{ * } ) & ( \psi_{ 2 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 2 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 2 } \psi_{ 4 }^{ * } ) \\<br /> ( \psi_{ 3 } \psi_{ 1 }^{ * } ) & ( \psi_{ 3 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 3 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 3 } \psi_{ 4 }^{ * } ) \\<br /> ( \psi_{ 4 } \psi_{ 1 }^{ * } ) & ( \psi_{ 4 } \psi_{ 2 }^{ * } ) & ( - \psi_{ 4 } \psi_{ 3 }^{ * } ) & ( - \psi_{ 4 } \psi_{ 4 }^{ * } )<br /> \end {array} \right) .$$
But any $4 \times 4$ matrix can be expanded in the standard (complete) basis of Clifford algebra $\mathcal{ C }_{ ( 1 , 3 ) } ( \mathbb{ C } )$:
$$\Gamma^{ A } = \{ I_{4 \times 4} , \gamma^{ \mu } , \gamma^{ \mu } \gamma_{ 5 } , \sigma^{ \mu \nu } , \gamma_{ 5 } \} ,$$
where $A = 1 , 2 , \cdots , 16$. So, we can write
$$\psi \otimes \bar{ \psi } = \sum_{ A = 1 }^{ 16 } a_{ A } \ \Gamma^{ A } . \ \ \ \ (1)$$
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
$$\mbox{ Tr } ( \Gamma_{ B } \psi \otimes \bar{ \psi } ) = a_{ I } \mbox{ Tr } ( \Gamma_{ B } ) + \sum_{ \Gamma^{ R } \neq I } a_{ R } \mbox{ Tr } ( \Gamma_{ B } \Gamma^{ R } ) . \ \ \ (2)$$
For $\Gamma_{ B } = I$, we find
$$a_{ I } = \frac{ 1 }{ 4 } \mbox{ Tr } ( \psi \otimes \bar{ \psi } ) = \frac{ 1 }{ 4 } \bar{ \psi } \psi .$$
For $\Gamma_{ B } \neq I$, Eq(2) gives
$$\mbox{ Tr } ( \Gamma_{ B } \psi \otimes \bar{ \psi } ) = 0 + \sum_{ \Gamma^{ R } \neq I } a_{ R } \mbox{ Tr } ( \Gamma_{ B } \Gamma^{ R } ) \sim 2^{ 2 } a_{ B } .$$
Thus
$$a_{ B } \sim \frac{ 1 }{ 4 } ( \Gamma_{ B } )_{ a b } ( \psi \otimes \bar{ \psi } )_{ b a } = \frac{ 1 }{ 4 } \bar{ \psi }_{ a } ( \Gamma_{ B } )_{ a b } \psi_{ b } = \frac{ 1 }{ 4 } \bar{ \psi }_{ a } \Gamma_{ B } \psi .$$
So, our expansion Eq(1) becomes
$$\psi \otimes \bar{ \psi } = \frac{ 1 }{ 4 } ( \bar{ \psi } \psi ) \ I + \frac{ c }{ 4 } \sum_{ \Gamma^{ B } \neq I } ( \bar{ \psi } \Gamma^{ B } \psi ) \Gamma_{ B } ,$$
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
$$\bar{ \psi } \psi + \bar{ \psi } \gamma_{ 5 } \psi \ \in ( 0 , 0 ) \oplus ( 0 , 0 ) , \mbox{ dim } : \ [1] \oplus [1] ,$$
$$\bar{ \psi } \gamma^{ \mu } \psi + \bar{ \psi } \gamma^{ \mu } \gamma_{ 5 } \psi \ \in ( 1/2 , 1/2 ) \oplus ( 1/2 , 1/2 ) \sim [4] \oplus [4] ,$$
and
$$\bar{ \psi } \sigma^{ \mu \nu } \psi \ \in ( 1 , 0 ) \oplus ( 0 , 1 ) \sim [6] .$$
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

 

[Breo]

Thank you very much.

Yes, I have started to studying theoretical physics this year. I have a lack of many "basic" things but I am trying to solving them :).

Indeed, I have one when reading you: where can I learn the properties of the traces? I know is the summ of the diagonal terms of a matrix but I do not figure out when I see some of his uses. Maybe if you could tell me why do you can used it in your derivation...

 

[ChrisVer]

Which traces derivation?
The $\gamma^\mu$ are traceless matrices, so are the $\gamma_5$
Then $\gamma^\mu \gamma_5$ are also traceless, take eg at some particular representation (Weyl):
$\gamma^\mu= \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{pmatrix}, ~~ \gamma_5 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$
with $\sigma^\mu = (1, \vec{\sigma}), ~~\bar{\sigma}^\mu = (1, -\vec{\sigma})$

Take the product:

$\gamma^\mu \gamma_5= \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}= \begin{pmatrix} 0 & \sigma^\mu \\ -\bar{\sigma}^\mu & 0 \end{pmatrix}$

Is still traceless. Of course using the anticommutation relations can also give you this answer (so it's a result of the algebra).
$Tr(\gamma^\mu \gamma_5)= Tr (\gamma_5 \gamma^\mu)= -Tr(\gamma^\mu \gamma_5) \Rightarrow Tr(\gamma^\mu \gamma_5)=0$

Where I used at first $Tr(AB)=Tr(BA)$ and then $\left \{ \gamma_5, \gamma^\mu \right \}=0$.

For the $\sigma^{\mu \nu} \equiv \frac{i}{2}[\gamma^\mu,\gamma^\nu]$
Well, since it's a commutor it's traceless...

$Tr(\sigma^{\mu \nu}) \sim Tr(\gamma^\mu \gamma^\nu)-Tr(\gamma^\nu \gamma^\mu) =Tr(\gamma^\mu \gamma^\nu)-Tr(\gamma^\mu \gamma^\nu)=0$

Where I used the trace property $Tr(AB)=Tr(BA)$

The rest is in a complete analogy with what you do for the 2x2 complex matrices with a basis $\Gamma^A = \left \{ 1, \sigma^i \right \}$ and when you write a general matrix:

$M= \sum_B a_B \Gamma^B$

and want to find the coefficients $a_B$.
http://en.wikipedia.org/wiki/Pauli_matrices#Completeness_relation

The even more simple form is what you do to find the coeffecients for a vector in some basis.
$\vec{v}= a_i x^i$
where you take the inner product with the basis vectors (the analogy to this is seen after you realize that the inner product is a scalar, as is the trace of a matrix)

 

[dextercioby]

Just as expected, samalkhaiat writes a stunning reply. :) Bravo!

 

[samalkhaiat]

Just as expected, samalkhaiat writes a stunning reply. :) Bravo!

Thanks, it is always a pleasure to help someone understand something. Also, it is fun to do.