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

Questions on chapter 95 in Srednickis QFT book, Supersymmetry

  1. Jun 20, 2009 #1

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Hello all

    Since there seems that quite many here have followed Professor Srednickis QFT book, I want to ask a couple of question I have from his chapter on Supersymmetry.

    i) on page 617 he defines the kinetic term of the Chiral Superfields as:
    [tex]L_{\text{kin}} = \Phi^{\dagger} \exp(-2gV) \Phi[/tex]

    but he evaluates [tex] \Phi^{\dagger} \Phi[/tex] in eq. 95.63

    But why? the relation is not [tex] \exp(-2gV) \Phi^{\dagger} \Phi[/tex]

    Can one pull that exponential like this, even though it contains anticommuting numbers (95.62)


    ii) On page 618, he says that "From eq 95.24, we see that [tex]D^*_{\dot{a}} = - \partial ^*_{\dot{a}} [/tex] acts on a function of y, theta and theta^*

    Has anyone confirmed this? I am lost, and what is the significance of this?

    iii) How does he starts off with equation 95.68, and why is there a factor of 2, comparing with 95.16??

    If anyone has any questions on this chapter, maybe we can start a study circle and solve the problems and derivations together?

    cheers
     
  2. jcsd
  3. Jun 20, 2009 #2

    Avodyne

    User Avatar
    Science Advisor

    Yes, because V and Phi are commuting; every term in them contains an even number of Grassmann variables. I believe he writes it that way because then it generalizes to the nonabelian case, where V is a matrix.

    As for your other questions, I don't have the book handy at the moment, will try to check back later.
     
  4. Jun 20, 2009 #3
    (i)

    The only field that is not commuting, Srednicki introduces on equation (95.65). You can tell because the field has a spinor index, (95.73) makes it concrete.

    (ii)

    This is correct. (95.24) proves two of the statements. The third is simple just use the definition (95.17).

    (iii)

    95.68 is true too. It's like a chain rule. The 2 comes out when you actually do the calculation of D_a on y^mu.
     
  5. Jun 20, 2009 #4

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Thank you all for input, will get my hands more dirty tomorrow with this hehe
     
  6. Jun 20, 2009 #5
    I just looked at your second question again (ii), and it's somewhat tricky.

    The supercovariant derivatives (D_a) was initially defined with respect to the spacetime variable x, I think on the 2nd page of the chapter. But Srednicki is writing the field W_a in terms of a new variable y now. So the question becomes how does D_a operate on a function of y now? 95.24 says D_a (where D_a operates on a field with argument x), operating on y (which is a function of x), is zero. So D_a operated on y is zero. However, you can write that as [tex]-\partial^{*}_{a} [/tex] (where there is a dot on the 'a') because operated on y, this is zero. So basically Srednicki is rewriting D_a to operate on a space with a y coordinate instead of an x one.
     
  7. Jun 20, 2009 #6

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [tex]D^*[/tex] on y and on [tex] \theta [/tex] gives zero , so the only action that it can have comes from its first term in 95.17, which is [tex]- \partial ^*_{\dot{a}} [/tex]. I am probably missing the point of your question.

    EDIT: Sorry, just realized that RedX had answered that question.
     
    Last edited: Jun 20, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Questions on chapter 95 in Srednickis QFT book, Supersymmetry
  1. Qft question (Srednicki) (Replies: 14)

Loading...