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Mass dimension of a scalar field in two dimensions?

  1. Jul 9, 2012 #1
    Which is the mass dimension of a scalar filed in 2 dimensions?
    In 4 dim I know that a scalar field has mass dimension 1, by imposing that the action has dim 0:
    [itex]S=\int d^4 x \partial_{\mu} A \partial^{\mu} A [/itex]
    where
    [itex]\left[S\right]=0[/itex]
    [itex]\left[d^4 x \right] =-4[/itex]
    [itex]\left[ \partial_{\mu} \right]=1[/itex]
    [itex]\Rightarrow \left[A\right]=1[/itex]
    Doing the same in 2 dim I found
    [itex]\left[A\right]=0[/itex]
    Is it right?
    I need it for a model in supersymmetry.
     
  2. jcsd
  3. Jul 9, 2012 #2

    haushofer

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    Yes, that's right, if you stick to the same action.
     
  4. Jul 9, 2012 #3
    Ok thanks!
    But now I have a problem in writing the action of the superfield
    [itex] \phi =A +i \bar{\theta} \psi + \frac{i}{2}\bar{\theta} \theta F [/itex]
    Phi has the same mass dimension of A, 0 in two dimension.
    In the kinetic part of the action, there must be a quadratic term, such as [itex] \phi \bar{\phi} [/itex], which would have mass dimension zero. But the invariants
    [itex] \int d^2 x d\theta [/itex] and [itex] \int d^2 x d^2 \theta [/itex]
    need after them something of dimension 3/2 or 1 respectively, assumed that the dimension of [itex] d\theta [/itex] is 1/2 .
    How can I resolve this?
     
  5. Jul 10, 2012 #4

    haushofer

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    That quadratic term you mention doesn't seem to be a kinetic term; for that you need derivatives. See e.g. chapter 4.1 of Green,Schwarz,Witten (vol.1).
     
  6. Jul 13, 2012 #5
    I'm studying on Paul West's book and at page 112 he says that [itex]\bar{\phi}\phi[/itex] is the kinetic term because if you resolve the integral in theta you find the kinetic action for the component of the superfield.
    I'd like to do the same thing in two dimension: writing down the action of the superfield with the kinetic term, the term with the mass and the cubic interaction term. Only in the case of cubic phi I have to resolve the integral over theta.
    But my problem is the dimensions as I've just said... So I don't know what to do! Maybe I'm doing some errors?
    In two dimensions theta has two components, and I don't have used the chiral formalism; in addition the components are real.
    In the superfield phi, which I wrote in a my previous post, appear both theta and bar theta.
    Do you have an idea of what to do? Thank you!
     
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