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On spinor representations and SL(2,C)

  1. Nov 12, 2011 #1
    Hi guys!

    I still have problem clearing once and for all my doubt on the spinor representation. Sorry, but i just cannot catch it.

    1)

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    Take a left handed spinor, [itex]\chi_L[/itex].
    Now, i know it transforms according to the Lorentz group, but why do i have to take the [itex]\Lambda_L[/itex] matrices belonging to [itex]SL(2,\mathbb C)[/itex],
    [tex]\chi'=\Lambda_L\chi[/tex]??

    Dimensionally it makes sense, it's like
    [tex]\left(\begin{align}\chi'_{L1}\\ \chi'_{L2}\end{align}\right)=\left(\begin{align}A &B\\C&D\end{align} \right) \left(\begin{align} \chi_{L1}\\ \chi_{L2} \end{align}\right)[/tex]

    but why exacly [itex]SL(2,\mathbb C)[/itex] matrices and not every other generic 2x2 complex matrix?


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    2)

    ----
    Is it right to say that [itex]\Lambda_L[/itex] is the representation of the lorentz group which acts on spinors?

    I have this doubt becaouse i read everywhere that the spinor is a [itex]\left(0,\frac{1}{2}\right)[/itex] representation of the lorentz group, but i'd say that the spinor os the thing on which the [itex]\Lambda_L[/itex] acts, and it is the [itex]\Lambda_L[/itex] itself to be a represetation of the group!!
     
    Last edited: Nov 12, 2011
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  3. Nov 12, 2011 #2

    vanhees71

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  4. Nov 12, 2011 #3

    Bill_K

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    Yes, that's correct.
     
  5. Nov 12, 2011 #4
    OK, that's reassuring, thanks Bill_K, but why then i do read everywhere that the the spinor is a (0,1/2) representation of the Lorentz group???


    (I'm reading the pages you suggested me vanhees71...)
     
  6. Nov 12, 2011 #5

    vanhees71

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    There are two two-dimensional inequivalent irreducible representations of the proper orthochronous Lorentz group, labeled (1/2,0) and (0,1/2). The pair of numbers are the spin values of the two pseudo-su(2) subalgebras according to the relation between the Lie algebras [itex]\text{sl}(2,\mathbb{C})[/itex] and [itex]\text{su}(2) \oplus \text{su}(2)[/itex].

    In my manuscript I describe two ways to describe the irreps of the Lorentz group, namely the formalism with dotted and undotted tensor indices (Theorem 8) and via the above mentioned Lie algebra equivalence (Theorem 10), which is more convenient for applications.
     
  7. Nov 12, 2011 #6

    dextercioby

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    If you're going to Lie algebras, then you needn't use [itex] \mathfrak{sl}(2,\mathbb{C}) [/itex], because of the isomorphism

    [tex] \mathfrak{so}(1,3)^{\text{C}} \simeq \mathfrak{su}(2) \oplus \mathfrak{su}(2) [/tex]

    The Lie algebra of SU(2) should be known from angular momentum theory.
     
  8. Nov 12, 2011 #7
    Ok, that seems to clarify most of my problem on the reason of using SL(2,C).

    But i'm still stuck on the representation thing.

    I just cannot solve that.

    For example, the group SO(3) can be represented by 3x3 orthgonal matrices, which act on 3-dimensional vectors belonging to a vector space.

    Well, i cannot transpose this mechanism on the lorentz (poincarè) group, becaouse everywhere i read that the spinor is a (0,1/2) (or (1/2,0) ) representation of the group; while i understand it as the equivalent of the 3-dimensional vector of the example above.

    Can you clarify that to me?
     
  9. Nov 12, 2011 #8

    vanhees71

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    A linear representation of a group is the realization of that group by linear operator on a vector space.

    More formally it's a homomorphism [itex]\Phi:G \rightarrow \text{GL}(V)[/itex], where [itex]G[/itex] is the group under consideration and [itex]\text{GL}(V)[/itex] is the group of all invertible linear operators on the vector space [itex]V[/itex]. The group product for this is the product of the linear operators (i.e., the composition of linear mappings of the vector space to itself).

    Further, a homomorphism between two groups obeys the rule

    [tex]\Phi(g_1 g_2)=\Phi(g_1) \Phi(g_2)[/tex]

    for all [itex]g_1,g_2 \in G[/itex].

    All these mathematical ideas can be applied also to Lie algebras, leading to the analogous notion of linear representations of Lie algebras.
     
  10. Nov 12, 2011 #9

    Bill_K

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    They're being casual with the language. As indicated above, the representation is the set of matrices that correspond to the group elements. A spinor is an element of a vector space whose components transform according to the (0,1/2) representation, but it's just too cumbersome to keep saying that.
     
  11. Nov 12, 2011 #10
    So it's on an improper use of the term representation i've been struggling upon!!


    But another thing then:

    When we compose two right (left) weyl spinor to get something that transforms as a 4-vector, namely by doing [tex]\left(0,\frac{1}{2}\right)\otimes\left(0,\frac{1}{2}\right)=(0,0)\oplus(0,1)[/tex]
    we are actually combining togheter the spinors themselves, but as far as i've understood the notation [itex]\left(0,\frac{1}{2}\right)[/itex] (as well as the others) stands for the linear operators, right?

    Is that another misleading symbology?

    thanks!
     
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