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Grassmann integration

  1. Oct 30, 2014 #1
    Hi, everyone!

    I am trying to understand notation of this textbook http://arxiv.org/abs/hep-th/0108200

    page 8, formulas 2.1.4 and 2.1.5

    $$\int d \theta_\alpha \theta^\beta=\delta_\alpha^\beta$$

    this could be found in any textbook the weird that from the above formula follows

    $$\int d^2 \theta \; \theta^2=-1$$

    I know what θ2 means, but what is d2θ I could hardly guess. According to standard Berezin definition there should be $i$ in the r.h.s. of the last formula

    Please help to clarify this

    Best wishes
    Korybut Anatoly
     
  2. jcsd
  3. Oct 30, 2014 #2

    MathematicalPhysicist

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    Grassman integration is the same as Grassman's derivation, so unless I am mistaken it should be 2 and not -1.

    Think of it as: [tex]d^2(\theta^2)/d^2 \theta[/tex]

    I leave this to the experts.
     
  4. Oct 30, 2014 #3

    Avodyne

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    I'm not a fan of the notation in this text. But in general, for a Grassmann variable with two components like ##\theta##, we have ##d^2\theta=d\theta_1 d\theta_2##. This is like writing ##d^3x=dx_1 dx_2 dx_3## for a vector ##\vec x##.

    Then, if ##\theta^2 =\theta_1\theta_2## (which is true in everybody's convention up to some factor like ##-1## or ##i##), we have
    [tex]\int d^2\theta\,\theta^2 = \int d\theta_1d\theta_2\,\theta_1\theta_2=-\int d\theta_2d\theta_1\theta_1\theta_2=-\int d\theta_2\,\theta_2=-1.[/tex]
    Edit: I'm using the convention ##\int d\theta_\alpha\,\theta_\beta=\delta_{\alpha\beta}##, which differs from this text.
     
  5. Oct 31, 2014 #4
    I know that ##\theta## anticommute, but how one deduce that ##d\theta## obey the same rule
     
  6. Oct 31, 2014 #5

    julian

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    [itex]\int d \theta_1 d \theta_2 \theta_1 \theta_2 = - \int d \theta_1 d \theta_2 \theta_2 \theta_1 = - \int d \theta_1 \theta_1 = -1 = - \int d \theta_2 d \theta_1 \theta_1 \theta_2[/itex]

    and

    [itex]\int d \theta_1 d \theta_2 \theta_2 \theta_1 = 1 = \int d \theta_2 d \theta_1 \theta_1 \theta_2 = - \int d \theta_2 d \theta_1 \theta_2 \theta_1[/itex].
     
    Last edited: Oct 31, 2014
  7. Nov 1, 2014 #6

    julian

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    So I've proved

    [itex]\int d \theta_1 d \theta_2 ( \theta_1 \theta_2) = - \int d \theta_2 d \theta_1 (\theta_1 \theta_2)[/itex]

    This obviously implies

    [itex]\int d \theta_1 d \theta_2 ( \theta_2 \theta_1) = - \int d \theta_2 d \theta_1 (\theta_2 \theta_1)[/itex].

    So [itex]d \theta_1 d \theta_2 = - d \theta_2 d \theta_1[/itex]
     
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