Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spinors in various dimensions

  1. Dec 16, 2009 #1
    Dear guys,

    I wanna understand the spinors in various dimensions and Clifford algebra. I tried to read the appendix B of Polchinski's volume II of his string theory book. But it's hard for me to follow and I stuck in the very beginning. I will try to figure out the outline and post my questions later.

    For now, I wanna ask for very simple, introductory articles for the construction of gamma matrices and spinors in various dimensions. (Is the appendix B of Polchinski the simplest article among all?:blushing:)

    Thanks for your help!!

    Ismaili
     
  2. jcsd
  3. Dec 16, 2009 #2

    haushofer

    User Avatar
    Science Advisor

    Maybe "A menu of supergravities" of Van Proeyen can help you. I found it quite understandable.
     
  4. Dec 18, 2009 #3
    Excuse me,
    I searched for this book in libraries nearby and on google but I couldn't find it?
    Was this book published in english?

    ---
    In the following, I briefly present the content and one of my question by which I stuck.

    In the appendix B of Polchinski's string book.
    One starts from the Clifford algebra in SO(d-1,1)
    [tex] \{ \gamma^{\mu} , \gamma^\nu \} = 2\eta^{\mu\nu}
    [/tex]
    In the even dimension, [tex] d = 2k+2 [/tex], one can group the [tex]\gamma^\mu [/tex] into [tex] k+1 [/tex] sets of anticommuting creation and annihilation operators,
    [tex]
    \gamma^{0\pm} = \frac{1}{2} (\pm\gamma^0 + \gamma^1)
    \quad\quad \gamma^{a\pm} = \frac{1}{2}(\gamma^{2a} \pm i \gamma^{2a+1})
    [/tex]
    where [tex] a=1,2,\cdots, k[/tex].
    One then found that,
    [tex]
    \{ \gamma^{a+}, \gamma^{b-} \} = \delta^{ab}\quad\quad
    \{ \gamma^{a+} , \gamma^{b+} \} = \{ \gamma^{a-} , \gamma^{b-} \} = 0
    [/tex]
    That is, one finds that the gamma matrices can be grouped into the creation and annihilation operators of [tex]k[/tex] species of fermions. In particular, from
    [tex] (\gamma^{a-})^2 = 0 [/tex]
    one sees there exist a vacuum [tex] |\xi\rangle [/tex] annihilated by all [tex]\gamma^{a-}[/tex].
    Thus, by this observation, one constructed the representation of Clifford algebra in the following space,
    [tex]
    (\gamma^{k+})^{s_k+1/2}\cdots(\gamma^{0+})^{s_0+1/2} |xi\rangle
    [/tex]
    , i.e. a space of the tensor product of [tex]k[/tex] species fermions; so, the dimension of this representation is [tex]2^{k+1}[/tex].

    In [tex] d = 2 [/tex], one can easily work out the matrix form of the gamma matrices,
    [tex] \gamma^0 = \left(\begin{array}{cc}0 &1\\ -1 &0\end{array}\right) = i\sigma^2[/tex]
    [tex] \gamma^1 = \left(\begin{array}{cc}0 &1\\ 1 &0\end{array}\right) = \sigma^1[/tex]

    One can construct the representation in higher dimensional even space recursively, by [tex] d \rightarrow d+2 [/tex]. But now comes my question, for [tex] d = 6 [/tex]
    [tex]
    \gamma^0 = i\sigma^2\otimes\textcolor{red}{(-\sigma^3)}\otimes\textcolor{red}{(-\sigma^3)}
    [/tex]
    [tex]
    \gamma^1 = \sigma^1 \otimes \textcolor{red}{(-\sigma^3)} \otimes \textcolor{red}{(-\sigma^3)}
    [/tex]
    [tex]
    \quad\quad\quad\vdots
    [/tex]
    [tex]
    \gamma^4 = I \otimes I \otimes \sigma^1
    [/tex]
    [tex]
    \gamma^5 = I \otimes I \otimes \sigma^2
    [/tex]
    where [tex] I [/tex] is the 2 by 2 unit matrix.
    My question is that, why do we use [tex] \textcolor{red}{\sigma^3} [/tex]? I thought it should be the 2 by 2 identity matrix!

    Anybody guides me through this?
    Thank you so much for your help!
     
  5. Dec 20, 2009 #4
    I think I know the answer to the use of [tex]\sigma^3[/tex].
    The gamma matrices in d = 2 invole only [tex]\sigma^1, \sigma^2[/tex].
    When we add the spacetime dimension by 2,
    in order to get the correct anti-commutation relations,
    we have to tensor product the original gamma matrices by [tex]\sigma^3[/tex].

    -----

    After figuring out the construction of higher dimensional gamma matrices,
    I was confused by the suddenly born conjugation matrix [tex]B[/tex] and charge conjugation matrix [tex] C [/tex]
    ...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Spinors in various dimensions
  1. Spinors and Parity (Replies: 13)

  2. Spinor representations (Replies: 1)

Loading...