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Spinorial Field Representation

  1. Dec 7, 2014 #1
    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):

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

    ?
     
  2. jcsd
  3. Dec 7, 2014 #2
    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.
     
  4. Dec 7, 2014 #3
    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 ## ?
     
  5. Dec 8, 2014 #4

    dextercioby

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    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?
     
    Last edited: Dec 8, 2014
  6. Dec 8, 2014 #5
    ## dim ( D \otimes D) = dim (D) .dim (D) ## ?
     
  7. Dec 8, 2014 #6

    dextercioby

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    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?
     
  8. Dec 8, 2014 #7
    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?
     
  9. Dec 8, 2014 #8

    dextercioby

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  10. Dec 11, 2014 #9

    samalkhaiat

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

    Sam
     
    Last edited: Dec 11, 2014
  11. Dec 11, 2014 #10
    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...
     
  12. Dec 11, 2014 #11

    ChrisVer

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

    Take the product:

    [itex]\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} [/itex]

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

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

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

    [itex]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[/itex]

    Where I used the trace property [itex]Tr(AB)=Tr(BA)[/itex]

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

    [itex]M= \sum_B a_B \Gamma^B [/itex]

    and want to find the coefficients [itex]a_B[/itex].
    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.
    [itex]\vec{v}= a_i x^i [/itex]
    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)
     
    Last edited: Dec 11, 2014
  13. Dec 11, 2014 #12

    dextercioby

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    Just as expected, samalkhaiat writes a stunning reply. :) Bravo!
     
  14. Dec 14, 2014 #13

    samalkhaiat

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