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Zee question I.2.2

  1. Nov 16, 2007 #1
    I've decided to slowly work my way through Zee's quantum field theory in a nutshell over the vacation. I'm confused by the second question of the book which deals with matrix differentiation.

    1. The problem statement, all variables and given/known data

    Derive the equation [itex]\langle x_i x_j \cdots x_k x_l \rangle = \sum_{\textrm{wick}} (A^{-1})_{ab}\cdot (A^{-1})_{cd}[/itex] where {a,b,...,c,d} is a permutation of {i,j,...,k,l}


    2. Relevant equations

    [itex]\langle x_i x_j \cdots x_k x_l \rangle = \frac{\int d\mathbf{x} e^{1/2 \mathbf{x}^TA \mathbf{x}}x_i x_j \cdots x_k x_l}{\int d\mathbf{x} e^{1/2 \mathbf{x}^TA \mathbf{x}}}[/itex]

    3. The attempt at a solution

    I have no trouble deriving this equation for the 1-variable case. For the multi-variable case Zee suggests repeatedly differentiating the equation

    [itex]\int d\mathbf{x} e^{-1/2 \mathbf{x}^TA\mathbf{x} + J\mathbf{x}} = [(2\pi)^N /\det A]e^{(1/2) J A^{-1}J}[/itex]

    with respect to J. How does one differentiate wrt a matrix anyway?
     
    Last edited: Nov 16, 2007
  2. jcsd
  3. Nov 16, 2007 #2
    Doh! It just occured to me that J is a vector and therefore we can differentiate componentwise. I still don't quite get how to prove this, however. I can show that the claim is true for small n. Is some kind of inductive argument required here??
     
  4. Nov 16, 2007 #3
    I figured it out: For future knowledge, the trick is to use that the matrix A must be symmetric, and thus the derivative of the [itex]\frac{1}{2}\mathbf{x}^t A^{-1} \mathbf{x}[/itex] with respect to [itex]x_i[/itex], say can be written

    [itex]\sum_n (A^{-1})_{in} x_n[/itex].
     
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