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Invariant symbol with four spinor indices

  1. Apr 23, 2013 #1
    Hi,
    from Srednickis QFT textbook, we know the following coupling of Lorentz group representations: [tex](2,1)\otimes (2,1) = (1,1)_A \oplus (3,1)_S,[/tex] which yields [tex]\epsilon_{a b}[/tex] as an invariant symbol. Generalising, we can look at [tex](2,1)\otimes (2,1) \otimes (2,1) \otimes (2,1) = (1,1) \oplus (1,1)\oplus ...[/tex]
    which implies the existence of two invariant symbols with four spinor indices. One is just the trivial coupling [tex]\epsilon_{a b}\epsilon_{c d},[/tex] but what is the other one? It should be symmetric in the first two indices, and in the last two, and (I think) be further symmetric under the exchange of the two pairs. Could someone please confirm that it is just [tex](S_L^{\mu\nu})_{a b} (S_L^{\rho\sigma})_{c d} \epsilon_{\mu\nu\rho\sigma}? [/tex]
    (or in a different notation
    [tex](\sigma^{\mu\nu})_{a b} (\sigma_{\mu\nu})_{c d}.) [/tex]

    Thanks,
    torus
     
  2. jcsd
  3. Apr 23, 2013 #2

    fzero

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    The 2nd singlet here comes from reducing the symmetric tensor (in vector indices) into its trace and traceless part. The invariant symbol we use to do this is the Minkowski metric ##\eta_{\mu\nu}##.
     
  4. Apr 23, 2013 #3
    Hm, yes, but I am looking for a representation with spinor indices, i.e. the invariant symbol [tex]x_{a b c d}[/tex]
    corresponding to this singlet.
     
  5. Apr 23, 2013 #4

    Bill_K

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    He's taking the product of four undotted spinors. Don't you need to use dotted spinors also to form a vector/tensor quantity?
     
  6. Apr 23, 2013 #5

    fzero

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    I would try to express it as

    [tex]\sigma^{\mu})_{a b} (\sigma_{\mu})_{c d}. [/tex]

    I'm guessing that this can be reduced into some expression with ##\epsilon##s, but I don't have a good reference to steal standard results from at the moment.
     
  7. Apr 23, 2013 #6

    Bill_K

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    (2,1) ⊗ (2,1) = (3,1)S ⊕ (1,1)A. In terms of spinors this is φA χB reduced to tyhe symmetric part φ(A χB) and the trace φA χB εAB.

    Now take the product of two of these things, in each way:

    (3,1)S ⊗ (3,1)S = (5,1)S ⊕ (3,1) ⊕ (1,1)A
    (1,1)S ⊗ (3,1)A = (3,1)S
    (3,1)S ⊗ (1,1)A = (3,1)S
    (1,1)S ⊗ (1,1)S = (1,1)S

    (Total of 16 components)

    In spinor language, the first line:

    φ(A χB) ψ(C ωD) reduces to the totally symmetric part φ(A χB ψC ωD), a partial trace φ(A χX) ψ(Y ωD) εXY εBC and a full trace φ(W χX) ψ(Y ωZ) εXY εWZεBC εAD.

    The remaining three are φX χY ψC ωD εXY εAB, φA χB ψX ωY εXY εCD and φX χY ψW ωZ εXY εWZ εAB εCD.
     
  8. Apr 23, 2013 #7

    fzero

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    Ah, you're right that we need both left and right-handed spinors to get a Minkowski vector index. What confused me is that you can still view the ##\mathbf{3}## as an ##SO(3)## vector, in which case there is an invariant ##\delta_{ij}##. But it is not what I wrote down.
     
  9. Apr 24, 2013 #8
    @Bill_K: Thanks a lot, but isn't the singlet in 3x3 symmetric? Then you end up with the symmetry properties I proposed in the first post.
     
  10. Apr 26, 2013 #9
    Interesting problem. I used my semisimple-Lie-algebra package (My Science and Math Stuff) on it.

    I first created the algebra D2 ~ SO(4), and I then took a plethysm (power of rep with some symmetry) of various reps.

    torus, what you're working with is a Majorana spinor, (2,1) in multiplicity notation, (1,0) in highest-weight notation, and (1/2,0) in angular-momentum notation.

    It turns out that there are 2 mixed-symmetry combinations of 4 Majorana spinors (all nondotted or all dotted). Taking permutations of indices in torus's OP's one gives 3 possibilities. Only 2 of them are linearly independent, as I verified with a brute-force evaluation using Mathematica:
    [itex]\epsilon_{ab}\epsilon_{cd} - \epsilon_{ac}\epsilon_{bd} + \epsilon_{ad}\epsilon_{bc} = 0[/itex]

    Evaluating fzero's one with signature +---, I find
    [itex](\sigma^\mu)_{ab}(\sigma_\mu)_{cd} = 2\epsilon_{ac}\epsilon_{bd}[/itex]

    These looks like some Fierz identities (Wikipedia)

    I also tried it with a Dirac spinor, (2,1) + (1,2) / (1,0) + (0,1) / (1/2,0) + (0,1/2). For a square, the scalar products are all antisymmetric, and there are 2 of them. For a fourth power, the scalar products are 6 mixed-symmetry ones, with the same mixed symmetry as the Majorana one, and 1 completely antisymmetric one.
     
  11. Apr 28, 2013 #10
    Thanks for all the replies. I found a very nice reference:
    http://arxiv.org/abs/arXiv:0812.1594
    Eq. (2.95) together with (2.64) gives
    [tex] z_1 \sigma^{\mu\nu} z_2 z_3 \sigma_{\mu\nu} z_4 = - (z_1 z_4) (z_2 z_3) - (z_1 z_3) (z_4 z_2) ,[/tex]
    valid for both commuting and anticommuting spinors z_i, which has the symmetry properties proposed in the initial post.

    Regards,
    torus
     
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