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2Point Correlator in CFTs

  1. Apr 9, 2012 #1
    Hi guys,

    i'm studying Conformal Field Theory using the big yellow book by Senechal et al. So far everything has been a smooth ride. I'm a bit stuck at the point where they derive the 2- and 3-point correlator for spinless fields.

    Based on invariance under rotations and translations the correlator should depend only on the relative coords of the quasi primary fields and moreover - because of scaling invariance - this dependence should be of the type

    [tex] f(|x_1-x_2|)\sim \lambda^{\Delta_1+\Delta_2}f(\lambda|x_1-x_2|)[/tex] where λ is the scaling and Δ the conformal weight.

    But then those guys say that this is nothing but

    [tex]\langle \phi(x_1)\phi(x_2)\rangle \sim \frac{1}{|x_1-x_2|^{\Delta_1+\Delta_2}} [/tex]

    which is cannot follow. How do they know that the dependence is in the denominator and where does the exponent come from explicitely?
    Any help is appreciated!
    Thanks,
    earth2
     
  2. jcsd
  3. Apr 9, 2012 #2
    Hello! I don't know whether I have understood what you have written; but in the case in which [itex]\langle \phi(x_1)\phi(x_2)\rangle = f(|x_1-x_2|)[/itex] and the behavior you have written is not just a behavior but an equality, then in my opinion you can try the following mathematical trick: the equation
    [itex]\lambda^{-\Delta_1-\Delta_2}f(|x_1-x_2|)=f(\lambda |x_1-x_2|)[/itex]
    is valid for every λ; you can, furthermore, subtract [itex]f(|x_1-x_2|)[/itex] and then divide by [itex]\lambda -1[/itex] both sides of the equation. In the limit [itex]\lambda\rightarrow 1[/itex] you can find a differential equation: if I didn't make any mistake it has the following form
    [itex]-\frac{(\Delta_1+\Delta_2)}{|x_1-x_2|}f(|x_1-x_2|)=f'(|x_1-x_2|)[/itex]
    The solution of this differential equation is the solution you have written with up to the multiplication of an unknown constant coefficient which depends on the border conditions.
    I hope I have been clear.
     
  4. Apr 9, 2012 #3
    Thank you for your answer! That is a nice way to understand this! From the book however I have the impression that the conclusion is much simpler to get and follows 'for free' from the invariance statements. But thanks a lot anyways!

    earth2

    P.s. you were right, the [tex] \sim [/tex] should be an equality in the first equation
     
  5. Apr 9, 2012 #4

    samalkhaiat

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    Science Advisor

    Poincare invariance implies
    [tex]
    \langle \Phi_{\Delta_{1}}(x_{1})\Phi_{\Delta_{2}}(x_{2}) \rangle = F(|x_{1}- x_{2}|).
    [/tex]
    Scale invariance;
    [tex]
    \Phi_{\Delta_{i}}(x_{i}) \rightarrow \lambda^{\Delta_{i}} \ \Phi_{\Delta_{i}}(\lambda x_{i}), \ \ i = 1,2 ,
    [/tex]
    leads to
    [tex]
    F(|x|) = \lambda^{\Delta}F(\lambda |x|), \ \ (1)
    [/tex]
    where
    [tex]|x| = |x_{1} - x_{2}| \ \ \mbox{and} \ \Delta = \Delta_{1} + \Delta_{2}.[/tex]
    Eq(1) tells you that [itex]F(|x|)[/itex] does not depend on [itex]\lambda[/itex] and it admits the following (most general) solution,
    [tex]F(|x|) = \frac{C(\Delta_{1}, \Delta_{2})}{|x|^{\Delta_{1} + \Delta_{2}}} \ \ (2).[/tex]
    (Put [itex]F(|x|) \propto |x|^{N}[/itex] in eq(1), you find [itex]N = -\Delta[/itex])

    Finally, demanding invariance under special conformal transformation, we find
    [tex]
    F(|x|) = \frac{C \ \delta_{\Delta_{1}, \Delta_{2}}}{|x|^{\Delta_{1} + \Delta_{2}}}
    [/tex]
    Where C is a constant depends on the type of the field. Thus, in order to have a non-vanishing two point function, the fields must have the same scaling dimension.
    If it is not obvious to you that eq(2) is the most general solution to eq(1), then do the following; write
    [tex]\lambda = 1 + \epsilon , \ \ |\epsilon| \ll 1,[/tex]
    then, expanding to first order in [itex]\epsilon[/itex], eq(1) gives you
    [tex]|x| \frac{dF(|x|)}{d|x|} = - \Delta F(|x|)[/tex]
    This you can solve to find eq(2).

    Sam
     
    Last edited: Apr 9, 2012
  6. Apr 10, 2012 #5
    Yupp, thank you for your answer!
     
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