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The double line notation and the adjoint representation

  1. Jan 12, 2015 #1
    hi!

    in the first page of the attached pdf, after the title " 't hooft double line notation", he says that we have to consider the gluon as NxN traceless hermitian matrices to convince ourselves about the double line notation.

    there is my question: if you want the indices a,b to run from 1 to N, you need the T^A generators to be NxN matrices, and that means that you are in the fundamental representation, right?

    but when you look for a demonstration of the 't hooft's planar limit, it seems that you need the gauge bosons to be in the adjoint representation of U(N)... that doesn't mean that the a,b indices in the pdf run from 1 to N^2 - 1 (number of generators)?

    what am I misunderstanding here?

    thanks!
     

    Attached Files:

  2. jcsd
  3. Jan 13, 2015 #2

    ChrisVer

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    are you sure?
    For SU(3) let's say, you know you have 3x3 matrices [the Gellmann matrices are 3x3]... same for SU(2) you have 2x2 matrices [the Pauli matrices are 2x2]....
    The N^2 -1 is the number of generators... so in SU(3) you have 9-1=8 such 3x3 matrices... and for SU(2) you have 4-1=3 such 2x2 matrices...

    so the a,b run from 1 to N... and A index runs from 1 to N^2-1
     
  4. Jan 13, 2015 #3
    yes, but that's in the fundamental representation.

    in the adjoint representation the dimension of the generators is equal to the number of them, for example in SU(2) you define the adjoint representation as:

    gif.gif

    then, for SU(2), in the adjoint representation you have three 3x3 matrices, for SU(3) eight 8x8 matrices, and so on...

    and that's my question, I would think that you need to use the fundamental representation so the gluon is a NxN matrix... but in the adjoint representation, the gluon is a (N^2-1)x(N^2-1) matrix, because it's a linear combination (with coefficients A_mu^A accordng to the pdf) of (N^2-1)x(N^2-1) matrices (the generators in the adjoint representation).

    thanks for the answer!
     
    Last edited: Jan 13, 2015
  5. Jan 13, 2015 #4

    samalkhaiat

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    Okay, the Lie algebra of [itex]SU(n)[/itex] is given by [tex][ t^{ a } , t^{ b } ] = i f^{ a b c } \ t^{ c } , \ \ a = 1 , 2 , \cdots , n^{ 2 } - 1 .[/tex] The dimension of the generator matrices [itex]( t^{ a } )_{ A B }[/itex] is equal to the dimension of the representation space on which they act. In the defining (fundamental) representation, they are given by [itex]n \times n[/itex] hermitian traceless matrices [itex]( t^{ a } )_{ \alpha \beta }[/itex] , [itex]\alpha = 1 , 2 , \cdots , n[/itex].
    The fact that the gauge fields transform in the adjoint representation (I assume you know why that is so!) means that, for any spacetime index [itex]\mu[/itex] and any point [itex]x[/itex], we have [itex]n^{ 2 } - 1[/itex] REAL NUMBERS [itex]A_{ \mu }^{ a } ( x )[/itex]. In other words, it is an element in [itex]( n^{ 2 } - 1 )[/itex]-dimensional vector space. However, it is more convenient to represent the gauge field (equivalently) by a traceless hermitian [itex]n \times n[/itex] matrix formed by the following linear combination [tex]\mathbb{ A }_{ \mu } = \frac{ 1 }{ 2 } A^{ a }_{ \mu } \ t^{ a } , \ \ \ \ \ (1)[/tex] or, in the fundamental representation, [tex]( \mathbb{ A }_{ \mu } )_{ \alpha \beta }= \frac{ 1 }{ 2 } A^{ a }_{ \mu } \ ( t^{ a } )_{ \alpha \beta } . \ \ \ \ (2)[/tex] Using the trace normalization [itex]\mbox{ tr } ( t^{ a } t^{ b } ) = 2 \delta^{ a b }[/itex] we can invert (1) to obtain [tex]A_{ \mu }^{ b } = \mbox{ tr } ( \mathbb{ A }_{ \mu } \ t^{ b } ) = ( \mathbb{ A }_{ \mu } )_{ \alpha \beta } ( t^{ b } )_{ \beta \alpha } . \ \ \ \ (3)[/tex] The consistency between (3) and (2) follows from the following tensor identity [tex]\frac{ 1 }{ 2 } ( t^{ a } )_{ \alpha \beta } ( t^{ a } )_{ \gamma \eta } = \delta_{ \alpha \eta } \ \delta_{ \beta \gamma } - \frac{ 1 }{ n } \delta_{ \alpha \beta } \ \delta_{ \gamma \eta } . \ \ (4)[/tex] Finally, I am going to give you some homework to do, and trust me if you manage to do them you will never get confuse about these stuff again. If you struggle with them you can ask me for help.

    Ex(1) Drive the identity (4).

    Ex(2) Let [itex]U_{ \alpha }{}^{ \beta } = \delta_{ \alpha }{}^{ \beta } + i \epsilon_{ \alpha }{}^{ \beta }[/itex] be an infinitesimal [itex]SU(n)[/itex] transformation, where [itex]\epsilon[/itex] is hermitian [tex]\epsilon_{ \alpha }{}^{ \beta } = \epsilon^{ \beta }{}_{ \alpha } = ( \epsilon^{ \alpha }{}_{ \beta } )^{ * } .[/tex] A (Lorentz) scalar field [itex]\phi_{ \alpha }[/itex] (in the fundamental rep. [itex][n][/itex]) and its complex conjugate [itex]\phi^{ \alpha } = ( \phi_{ \alpha } )^{ * }[/itex] (in the conjugate rep. [itex][n^{ * }][/itex]) transform according to [tex]\phi_{ \alpha } \to \phi_{ \alpha } + i \epsilon_{ \alpha }{}^{ \beta } \ \phi_{ \beta } ,[/tex] [tex]\phi^{ \alpha } \to \phi^{ \alpha } - i \epsilon^{ \alpha }{}_{ \beta } \ \phi^{ \beta } .[/tex] Using these transformations we find, for (Lorentz) scalar fields in the adjoint representation [itex]\Phi_{ \alpha }{}^{ \beta } \equiv \phi_{ \alpha } \phi^{ \beta } - ( 1 / n ) \delta_{ \alpha }^{ \beta } \ \phi^{ 2 }[/itex], the following transformation [tex]\Phi_{ \alpha }{}^{ \beta } \to \Phi_{ \alpha }{}^{ \beta } + i \epsilon_{ \alpha }{}^{ \eta } \ \Phi_{ \eta }{}^{ \beta } - i \epsilon^{ \beta }{}_{ \eta } \ \Phi_{ \alpha }{}^{ \eta } .[/tex] The action of the covariant derivative on [itex]\phi_{ \alpha }[/itex] is given by [tex]( D_{ \mu } \phi )_{ \alpha } = \partial_{ \mu } \phi_{ \alpha } + i g ( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \beta } \ \phi_{ \beta } .[/tex] (i) Show that the gauge field transforms according to [tex]( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \beta } \to ( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \beta } + i \epsilon_{ \alpha }{}^{ \eta } ( \mathbb{ A }_{ \mu } )_{ \eta }{}^{ \beta } - i \epsilon^{ \beta }{}_{ \eta } ( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \eta } - \frac{ 1 }{ g } \partial_{ \mu } \epsilon_{ \alpha }{}^{ \beta } .[/tex] (ii) Show that the action of the covariant derivative on (Lorentz) scalars in the adjoint representation is given by [tex]( D_{ \mu } \Phi )_{ \alpha }{}^{ \beta } = \partial_{ \mu } \Phi_{ \alpha }{}^{ \beta } + i g \left( ( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \eta } \Phi_{ \eta }{}^{ \beta } - ( \mathbb{ A }_{ \mu } )_{ \eta }{}^{ \beta } \Phi_{ \alpha }{}^{ \eta } \right) .[/tex] (iii) Expand the matrices [itex]\mathbb{ A }_{ \mu }[/itex] and [itex]\Phi[/itex] in terms of the [itex]n \times n[/itex] hermitian traceless matrices [itex]t^{ a }[/itex] : [tex]( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \beta } = \frac{ 1 }{ 2 } A_{ \mu }^{ a } ( t^{ a } )_{ \alpha }{}^{ \beta } , \ \ \ \Phi_{ \alpha }{}^{ \beta } = \phi^{ b } ( t^{ b } )_{ \alpha }{}^{ \beta } ,[/tex] and show that [tex]D_{ \mu } \phi^{ a } = \partial_{ \mu } \phi^{ a } - \frac{ g }{ 2 } f^{ a b c } A_{ \mu }^{ b } \ \phi^{ c } .[/tex] Notice that, like the Lorentz vectors [itex]A_{ \mu }^{ a }[/itex], the Lorentz scalars [itex]\phi^{ a }[/itex] are components of a vector in [itex]( n^{ 2 } - 1 )[/itex]-dimensional vector space.

    Sam
     
  6. Jan 14, 2015 #5
    first of all, thanks for answering. I think I'm having a notation problem... I'll try to explain it:

    the fact that for any index [itex]\mu[/itex] you have [itex]n^{ 2 } - 1[/itex] real numbers [itex]A_{ \mu }^{ a } ( x )[/itex] is always true. it doesn't depend on which representation you decide to use, am I right? so the fact that you have [itex]n^{ 2 } - 1[/itex] real numbers and you can think about them as the independent parameters of a [itex] n \times n[/itex] hermitian traceless matrix is always true... my question has to do with your sentence between brackets: what does exactly mean that the gauge fields transform in the adjoint representation?

    because when you write your equation (2), you describe [itex]A_{ \mu }^{ a } ( x )[/itex] as a linear combination of generators of the fundamental representation.

    I agree with all you said after equation (2)... but I have that nomenclature problem... I would say that we are working in the fundamental representation, I don't really see what role is playing the adjoint representation here.

    put in another way: what do we have to change to your argument if the gauge fields transform in the fundamental representation (i.e: if we use the "single line" notation)? I mean, we also have [itex]n^{ 2 } - 1[/itex] real numbers there... why can't we think of them as the independent parameters of a [itex] n \times n[/itex] hermitian traceless matrix and then say that we have two color indices?

    thanks again!
     
  7. Jan 14, 2015 #6

    samalkhaiat

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    I know that, this is why I gave you exercises to do

    No, having [itex](n^{ 2 } - 1)[/itex] numbers means that you are dealing with an element (vector) in an [itex](n^{ 2 } - 1)[/itex]-dimensional vector space (THE ADJOINT REP. SPACE).

    Absolutely true.
    This should have become obvious to you if you have done exercise (2). It simply means that the gauge fields transform by the adjoint map. You obtain the adjoint representation space by subtracting all invariant subspaces from the direct product space [itex][n] \otimes [n^{*}][/itex]. I am sure you have seen this [tex][n] \otimes [n^{*}] = [n^{ 2 } - 1] \oplus [1] .[/tex] This an equation for (representation) vector spaces. The space [itex][n^{ 2 } - 1][/itex] is the adjoint representation space. You can translate the above equation into equation between vectors in those spaces. For example, if you take a set of [itex]n[/itex] Lorentz scalars [itex]\phi_{ \alpha } \in [n][/itex] and their complex conjugates [itex]\phi^{ \alpha } \in [n^{*}][/itex], then you can translate the above equation into the following equation (between [itex]SU(n)[/itex] tensors) [tex]\phi_{ \alpha } \phi^{ \beta } = ( \phi_{ \alpha } \phi^{ \beta } - \frac{ 1 }{ n } \delta^{ \beta }_{ \alpha } \phi^{ 2 } ) + \frac{ 1 }{ n } \delta^{ \beta }_{ \alpha } \phi^{ 2 } .[/tex] So, we have [itex]\Phi_{ \alpha }{}^{ \beta } \equiv ( \phi_{ \alpha } \phi^{ \beta } - (1 / n ) \delta^{ \beta }_{ \alpha } \phi^{ 2 } )[/itex] belonging to the adjoint representation space [itex][ n^{ 2 } - 1 ][/itex], and the invariant (singlet) [itex]\delta_{ \alpha }^{ \beta } \phi^{ 2 } \in [ 1 ][/itex] (the trivial representation space). In exercise (2), we worked out the transformation law for the fields [itex]\Phi_{ \alpha }{}^{ \beta }[/itex] in the adjoint representation (the adjoint map). You were also asked to derive the transformation law for the gauge fields [itex]( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \beta }[/itex], so you should have noticed that, apart from the inhomogeneous term [itex]\partial_{ \mu } \epsilon[/itex], this transformation law is identical to that of the adjoint representation [itex]\Phi_{ \alpha }{}^{ \beta }[/itex] of the group [itex]SU(n)[/itex].

    For more details see the PDF in
    https://www.physicsforums.com/threads/photons-and-qed.764368/page-2
    First, the expansion [itex]\mathbb{ A }_{ \mu } = A_{ \mu }^{ a } t^{ a }[/itex] is valid in any representation. Here, the generators [itex]t^{ a }[/itex] NEED NOT be the fundamental representation matrices. So, choosing [itex]t^{ a }[/itex] to be the fundamental generators DOES NOT change the fact that the gauge field transforms by the ADJOINT MAP.
    Do the exercises or study my PDF above.
    The gauge fields DO NOT transform in the fundamental representation. How can they? There are only [itex]n[/itex] numbers in the fundamental representation. The fundamental representation [itex][ n ][/itex] is the space of n-vectors NOT [itex]( n^{ 2 } - 1 )[/itex]-vector which what the gauge field is.
    Look, in the graphical representation, if you want to couple the gauge fields to the matter (Noether) current [itex]J_{ \mu }^{ a }[/itex] you use [itex]A_{ \mu }^{ a }[/itex]. If you want to couple the gauge fields to other gauge fields and coloured quarks it is more convenient to use [itex]( \mathbb{ A }_{ \mu } )_{ \alpha \beta }[/itex] where [itex]( \alpha , \beta )[/itex] are the n colour indices.

    Sam
     
    Last edited: Jan 14, 2015
  8. Jul 1, 2015 #7

    samalkhaiat

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    I was asked to derive the last equation, i.e. the action of the covariant derivative on scalar field in the adjoint representation:
    The adjoint representation [itex]\Phi_{\alpha}{}^{\beta}[/itex] transforms in the same way as the tensor product of the fundamental representations [itex]\phi_{\alpha} \phi^{\beta}[/itex]: [tex]\Phi_{\alpha}{}^{\beta} \to \Phi_{\alpha}{}^{\beta} + i \epsilon_{\alpha}{}^{\eta} \Phi_{\eta}{}^{\beta} - i \epsilon_{\eta}{}^{\beta} \Phi_{\alpha}{}^{\eta} .[/tex] So, the action of the covariant derivative on [itex]\Phi_{\alpha}{}^{\beta}[/itex] is obtained by replacing [itex]\epsilon_{\alpha}{}^{\beta}[/itex] by [itex]g (\mathbb{A}_{\mu})_{\alpha}{}^{\beta}[/itex]: [tex](D_{\mu}\Phi)_{\alpha}{}^{\beta} = \partial_{\mu}\Phi_{\alpha}{}^{\beta} + i g (\mathbb{A}_{\mu})_{\alpha}{}^{\eta} \Phi_{\eta}{}^{\beta} - i g \Phi_{\alpha}{}^{\eta} (\mathbb{A}_{\mu})_{\eta}{}^{\beta} .[/tex] This is because the combination of the second and the third terms (as you can work it out) transforms in the adjoint representation of the global group. In terms of matrix multiplication, we can rewrite the above equation as [tex]D_{\mu}\Phi_{\alpha}{}^{\beta} = \partial_{\mu}\Phi_{\alpha}{}^{\beta} + i g [ \mathbb{A}_{\mu} , \Phi ]_{\alpha}{}^{\beta} . \ \ \ \ (1)[/tex] Now using the relations [tex]\Phi = \phi^{b} t^{b} , \ \ \ \mathbb{A}_{\mu} = \frac{1}{2} A_{\mu}^{a} t^{a} ,[/tex] together with the algebra [tex][t^{a} , t^{b}]_{\alpha}{}^{\beta} = i f^{a b c} (t^{c})_{\alpha}{}^{\beta} ,[/tex] Eq(1) becomes [tex]D_{\mu}\phi^{c} (t^{c})_{\alpha}^{\beta} = \partial_{\mu}\phi^{c} (t^{c})_{\alpha}^{\beta} - \frac{g}{2} f^{a b c} A_{\mu}^{a}\phi^{b} (t^{c})_{\alpha}^{\beta} .[/tex] Or [tex]D_{\mu}\phi^{a} = \partial_{\mu}\phi^{a} - \frac{g}{2} f^{a b c} A_{\mu}^{b} \phi^{c} .[/tex]

    Sam
     
    Last edited: Jul 1, 2015
  9. Jul 1, 2015 #8
    Hi Sam,

    Thanks for you reply, but I see that we replaced ## \epsilon^\beta_\alpha ## by ##g~
    (\mathbb{A}_{\mu})_{\alpha}{}^{\beta} ## in the transformation of ##
    \Phi_{\alpha}{}^{\beta}
    ## adjoint representation, we will get the action of the covariant derivative on ## \Phi_{\alpha}{}^{\beta} ## as

    $$ ( D_{ \mu } \Phi )_{ \alpha }{}^{ \beta } = \partial_{ \mu } \Phi_{ \alpha }{}^{ \beta } + i g \left( ( \mathbb{ A }_{ \mu } )_{ \alpha }{}^{ \eta } \Phi_{ \eta }{}^{ \beta } - ( \mathbb{ A }_{ \mu } )_{ \eta }{}^{ \beta } \Phi_{ \alpha }{}^{ \eta } \right) ,$$

    ( as in (ii) in your first reply), and not by

    $$ (D_{\mu}\Phi)_{\alpha}{}^{\beta} = \partial_{\mu}\Phi_{\alpha}{}^{\beta} + i g (\mathbb{A}_{\mu})_{\alpha}{}^{\eta} \Phi_{\eta}{}^{\beta} - i g \Phi_{\alpha}{}^{\eta} (\mathbb{A}_{\mu})_{\eta}{}^{\beta} . $$

    Where I think the commutator in (1) ##
    " [ \mathbb{A}_{\mu} , \Phi ] "##comes from
    ## (\mathbb{A}_{\mu})_{\alpha}{}^{\eta} \Phi_{\eta}{}^{\beta} - \Phi_{\alpha}{}^{\eta} (\mathbb{A}_{\mu})_{\eta}{}^{\beta} ##.


    Edit: I think you mean that, in general for a second rank tensor:
    $$ \Phi ^{\alpha \beta } \to \Phi^{\alpha \beta } + i \epsilon^{ \alpha \eta } \ \Phi ^{ \eta \beta } - i \Phi^{\alpha \eta } \epsilon^{\eta \beta } .$$
    So when we make the replacement by## (\mathbb{A}_{\mu})_{\alpha}{}^{\beta} ## to get the covariant derivative:
    $$ (D_{\mu} \Phi) ^{\alpha \beta } = \partial_{\mu} \Phi^{\alpha \beta } + i g (\mathbb{A}_{\mu})^{ \alpha \eta } \ \Phi ^{ \eta \beta } - ig \Phi^{\alpha \eta } (\mathbb{A}_{\mu})^{\eta \beta } . $$

    Bests.
    S.
     
    Last edited: Jul 1, 2015
  10. Jul 2, 2015 #9

    samalkhaiat

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    There is no difference. [itex]\mathbb{A}_{\mu}[/itex] and [itex]\Phi[/itex] are the non-commuting matrices, but [itex](\mathbb{A}_{\mu})_{\eta}{}^{\beta}[/itex] and [itex]\Phi_{\alpha}{}^{\eta}[/itex] are commuting matrix elements, i.e. [tex](\mathbb{A}_{\mu})_{\eta}{}^{\beta} \Phi_{\alpha}{}^{\eta} = \Phi_{\alpha}{}^{\eta} (\mathbb{A}_{\mu})_{\eta}{}^{\beta} = ( \Phi \mathbb{A}_{\mu})_{\alpha}{}^{\beta} .[/tex]
     
  11. Jul 4, 2015 #10
    Hi Sam,

    Okay, I got it. May I ask for a reference notes for the equations you mentioned in your first reply, because i'd like to summarize all that ..

    Also let me ask about other point: it seems you don't approve that
    Rather you say in the adjoint representation we have ## n^2-1## generators which are ## n \times n ## hermitian matrices ## ( t^{ a } )_{ \alpha \beta },~ \alpha, \beta=1,..,3 ~## (for example in SU(3)), I'm I right ?

    But then, what dose it mean that in the adjoint representation the generators can be written in the form of:
    $$ (T_i)_{jk}= - f_{ijk} ~~ (1) $$
    ( Still in SU(3) ), where i,j,k =1,..8 ?
    That's are in all group notes as: http://astro.sunysb.edu/steinkirch/books/group.pdf

    I think that it is not arbitrary for a a gauge field ## A^{ a }_{ \mu } \ t^{ a } , ## or a scalar field in
    the adjoint representation ## S^{ a } \ t^{ a } ## to substitute by ## f_{ijk}## instead of ## t^{ a }##, because in that case the colour factor of the vertex of three gluons for example 'd be ## f_{ijk} f_{lmn} f_{qrs}## which is obviously wrong ..

    So when (1) can be indeed used?
    In Group's note as: http://www.physics.indiana.edu/~dermisek/QFT_08/qft-II-19-2p.pdf, I found it's used in the Jacobi identity to drive the Lie Algebra in case of the adjoint representation ..

    What do you think ?


    Cheers,
    S.
     
  12. Jul 4, 2015 #11

    ChrisVer

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    I guess the first t^a you wrote defines the generators of the infinitiseimal transformation of the fundamental repr, which is a 3 dimeensional space.
    The other (1) is a higher dimensional repr of SU(3): the adjoin, that is 8dimensional.
     
  13. Jul 5, 2015 #12

    samalkhaiat

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    That is good question, I don’t really know of particular reference for that material. I suppose all good textbook on QCD should have something to say about it. Also try to look up works on “Chiral Perturbation Theory”.


    Where did I say that? The [itex]n^{2} - 1[/itex] traceless hermitian [itex]n \times n[/itex] generator matrices ARE the defining (fundamental) representation of [itex]SU(n)[/itex]. Of course the generator matrices of the adjoint representation ARE [itex](n^{2} - 1) \times (n^{2} - 1)[/itex], and there are [itex]n^{2}-1[/itex] of them. This is exactly what the relation [itex]ad(t^{a})^{bc} = - i f^{abc}[/itex] tells you.
    Look I am busy at the moment; I will post long reply for you some times next week explaining almost everything you are confused about.
     
  14. Jul 7, 2015 #13
    Okay, thanks for your reply ..
     
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