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How *exactly* do you get the generators (1/2,1/2)

  1. Aug 15, 2010 #1
    With the Lorentz group (SO(1,3) = SU(2)xSU(2)), if you're working with the "true" SU(2) generators N^+_i and N^-_i (rather than the generators J_i and K_i), it is straightforward to find the form of the generators in either spinor rep (0,1/2) or (1/2,0) by setting either N^+ or N^- to zero and setting the other to the Pauli matrices.
    Is there an analogous procedure to recover the normal J_i and K_i expressions if you start with (1/2,1/2)? If so, can someone outline how it's done? I'm looking for an explicit derivation of the standard forms of J_i and K_i, using only the fact that we're looking at a (1/2,1/2) representation of SU(2)xSU(2).
     
  2. jcsd
  3. Aug 15, 2010 #2
    Use the definitions

    [tex]J_{\pm}=(J\pm iK)/2[/tex]

    Here, [tex]J_{\pm}[/tex] are the generators in the (1/2,0) and (0,1/2) representations, which you said you can explicitly calculate. Now the generators of the Lorents group in the (1/2,1/2) representations are simply

    [tex]J=J_-+J_+[/tex]

    [tex]K=i(J_--J_+)[/tex]
     
  4. Aug 15, 2010 #3
    In (1/2,0), J_- = 0 and J_+ = (1/2)Pauli, which are 2x2 matrices. In (0,1/2), J_+ = 0 and J_- = (1/2)Pauli, which are again 2x2.
    So if you take both to have the 1/2, you would have
    J = J_- + J_+ = Pauli - again 2x2 matrices.

    I don't see how doing this gives you the canonical 4x4 vector generators.
     
  5. Aug 15, 2010 #4
    You have to combine the two 2x2 representations in the diagonal blocks of a 4x4 representation:

    [tex]J_+=\left(\begin{array}{cc}\sigma/2&0\\0&0\end{array}\right)[/tex]

    [tex]J_-=\left(\begin{array}{cc}0&0\\0&\sigma/2\end{array}\right)[/tex]

    [tex]J=\left(\begin{array}{cc}\sigma/2&0\\0&\sigma/2\end{array}\right)[/tex]

    [tex]K=\left(\begin{array}{cc}-i\sigma/2&0\\0&i\sigma/2\end{array}\right)[/tex]
     
  6. Aug 15, 2010 #5
    Just to make sure if i think it right.
    The generators constructed above should be the ones in the representation [tex](1/2,0)\oplus(0,1/2) [/tex] right?
    which is different from [tex](1/2,1/2)[/tex]
     
  7. Aug 15, 2010 #6
    Those definitions of J and K are not the canonical definitions for the vector representation. The canonical ones are the antisymmetric basis (times i). Is there a similarity transformation that gets you there from the definitions written here?
     
  8. Aug 15, 2010 #7
    Also, those definitions for J and K don't exponentiate to the standard Lorentz vector transformations.
     
  9. Aug 15, 2010 #8
    Yes, you are right, the two representations are different (the second one is an irreducible representation of the proper orthocronous Lorentz group, the first one is reducible, and becomes irreducible only if you include parity into the group). I didn't read well your question, I'm sorry. The fact is that I thought you were interested in the Dirac representation for the electron. What mislead me was the fact you mentioned the Pauli matrices, which enter in the Dirac representation, but not really in the (half,half).

    I give you a clue on how to solve your problem, (if I interpret well this time!):

    For each of the [tex]J^{(\pm)}[/tex], consider the z-component, and the creation and annihilation operators. They act like this (standard angular momentum theory):

    [tex]J^{(\pm)}_x\pm iJ^{(\pm)}_y}|m>=\sqrt{(1/2\mp m)(1/2\pm m+1)}|m\pm 1>[/tex]

    [tex]J^{(\pm)}_z|m>=m|m>[/tex]

    Now consider the tensor product ket

    [tex]|m^{+},m^{-}>[/tex]

    You know how every component of [tex]J^{(\pm)}[/tex] acts on theese four states, so you can find the representation matrices. And then you find the representation matrices of J and K with the same formula I mentioned in my first post.

    I don't know if it's clear. If it's not, I'll show you the explicit calculation. (I hope not, though!!)
     
    Last edited: Aug 15, 2010
  10. Aug 15, 2010 #9
    No problem, and thanks. That helps. When I work things out according to what you said, it seems to make sense to give the vectors |m+, m-> two indices, a + index and a - index, each taking two values (up or down), rather than one spacetime index. Then the generators each have four indices, two + indices and two - indices, rather than two spacetime indices. I get the following (upper case indices are + and lower case indices are -):
    [tex](N^+_1)_{ABcd} = {1 \over 2}\delta_{cd}(\sigma_1)_{AB}[/tex]
    [tex](N^+_2)_{ABcd} = {1 \over 2}\delta_{cd}(\sigma_2)_{AB}[/tex]
    [tex](N^+_3)_{ABcd} = {1 \over 2}\delta_{cd}(\sigma_3)_{AB}[/tex]
    [tex](N^-_1)_{ABcd} = {1 \over 2}\delta_{AB}(\sigma_1)_{cd}[/tex]
    [tex](N^-_2)_{ABcd} = {1 \over 2}\delta_{AB}(\sigma_2)_{cd}[/tex]
    [tex](N^-_3)_{ABcd} = {1 \over 2}\delta_{AB}(\sigma_3)_{cd}[/tex]

    Is there a way to convert from these indices to spacetime indices?
     
  11. Aug 15, 2010 #10
    Oh - one more thing.

    As I said, the states in this setup have one + index and - index, so they are
    [tex] v_{Ab} [/tex]
    Srednicki says to convert these spinor indices to a spacetime index using
    [tex] \sigma^{\mu}_{Ab} [/tex]
    where [tex] \sigma^{\mu}_{Ab} = (I, \sigma_1, \sigma_2, \sigma_3)[/tex],
    so that
    [tex] v_{Ab} = \sigma^{\mu}_{Ab} v_{\mu} [/tex]
    But it's not particularly clear to me why.
     
  12. Aug 16, 2010 #11
    Rather tha working with two indices that go from -half to half, you can work with just one index that goes from 0 to 3. For example you may choose this basis:

    [tex]C=\{|0>=|\downarrow\uparrow>,|1>=|\downarrow\downarrow>,|2>=|\uparrow\uparrow>,|3>=|\uparrow\downarrow>\}[/tex]

    and you can calculate the J^{\pm} with respect to this basis, and you get 4x4 matrices. (I'n not typing them here because it would take me 2 hours)

    That's right. With the matrices I just mentioned I made some linear algebra exercices, and found out that, if instead of the basis C you use the basis

    [tex]B=\{|0'>=|0>,|1'>=(|1>+i|2>)/\sqrt{2},|2'>=(-i|1>+|2>)/\sqrt{2},|3'>=|3>\}[/tex]

    then you get exactly the generators of the Lorentz group, in the standard 4-dimensional representation (the one with the lambda's). These generators are not antisymmetric because they are defined with the row index upstairs and the column index downstairs. If You use the metric to lower the upper index, you find antysimmetric matrices.
    The similarity transformation you asked is thus

    [tex]U=\mathscr{M}^B_C(Id)=\left(\begin{array}{cccc}1&0&0&0\\0&1/\sqrt{2}&-i\sqrt{2}&0\\0&-i/\sqrt{2}&1/\sqrt{2}&0\\0&0&0&1\end{array}\right)[/tex]

    I don't know that book. I suppose this is a more elegant way for doing the unitary transformation I mentioned.
     
  13. Aug 25, 2010 #12
    I read section 33, 34 of Srednicki these days too.
    He mentioned about something related to representation theory, which i'm not familiar to too.
    I also have a few questions:
    (1) start from (34.1),
    [tex]U(\Lambda)^{-1}\psi_a(x)U(\Lambda) = L_a{}^b(\Lambda)\psi_b(\Lambda^{-1}x) [/tex]
    , we have (34.6),
    [tex] [\psi_a(x),M^{\mu\nu}]
    = \mathcal{L}^{\mu\nu}\psi_a(x) + (S_\text{L}^{\mu\nu})_a{}^b \psi_b(x) [/tex], where [tex]\mathcal{L}^{\mu\nu} = -i(x^\mu\partial^\nu - x^\nu\partial^\mu)[/tex],
    by evaluating at [tex] x = 0[/tex], we arrive at (34.7)
    [tex]\epsilon^{ijk}[\psi_a(x) , J_k] = (S_\text{L}^{ij})_a{}^b \psi_b(0) [/tex]
    Then, I don't understand why he could deduce that [tex] (S_\text{L}^{ij})_a{}^b = \frac{1}{2}\epsilon^{ijk}\sigma_k [/tex] by just recalling the (2,1) representation of Lorentz group includes angular momentum 1/2 only?
    Another question about this is, he started from (34.1), where the operator [tex]U(\Lambda)[/tex] should be the representation of Lorentz group in some Hilbert space, but we haven't quantize the theory, how can we get the Hilber space representation?

    (2) He did explain the invariant symbol a little bit. I read from somewhere else that sometimes the metric which can pull up or lower down the index is an antisymmetric symbol, i.e. Levi Civita symbol. Srednicki gave some explanation here. He said, "whenever the product of a set of representations includes the singlet, there is a corresponding invariant symbol." He gave two examples, however, he didn't explain how he got those. Could anybody shed some light on this? Why he could know
    [tex] (2,2)\otimes(2,2) = (1,1)_s \oplus (1,3)_A \oplus (3,1)_A \oplus (3,3)_S [/tex]
    implies the existence of [tex]g_{\mu\nu} = g_{\nu\mu} [/tex].
    And, how do we determine the subscripts of summand of direct sums? (The symmetric or antisymmetric)

    (3) He showed how to map a 4D spacetime Lorentz representation to a (2,2) representation by the formula given by the OP: [tex] A_{a\dot{a}}(x) = \sigma^{\mu}_{a\dot{a}}A_\mu(x)[/tex] He also gave another example, which is the mapping from (3,1) representation to the antisymmetric rank two, self-dual tensor:
    [tex] G^{\mu\nu}(x) \equiv (S_\text{L}^{\mu\nu})^{ab}G_{ab}(x) [/tex]
    How did he find the mapping operator [tex] \sigma^\mu_{a\dot{a}}[/tex] and [tex] (S_\text{L}^{\mu\nu})^{ab}[/tex] ?

    Thanks very much!
     
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