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Notation question: partial derivative with arrow

  1. Mar 9, 2010 #1
    I'm reading thru Brown's QFT.

    He uses the notation of the gradient operator or a partial differentiation operater with an arrow over the operator. The arrow points either left or right.

    Can someone please tell me what this means?

    thanks
     
  2. jcsd
  3. Mar 10, 2010 #2
    It just denotes if the derivative is acting to the left or the right. You can usually change this direction by integration by parts.

    if the arrow is point to the left and u had [tex] f\overleftarrow{\nabla} g[/tex] it really just means [tex] (\nabla f) g [/tex].

    So integrating:

    [tex] \int d^4x f\overleftarrow{\nabla} g= [fg]-\int d^4x f\overrightarrow{\nabla} g[/tex], providing functions go to zero at infinity (compact support), you just throw away the [] term.

    Translating into conventional notation:

    [tex] \int d^4x (\nabla f) g= [fg]-\int d^4x f(\nabla g)[/tex]
     
    Last edited: Mar 10, 2010
  4. Mar 10, 2010 #3

    Ben Niehoff

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    Two more things to add:

    1. The reason one wants to do this is that when using the Feynman "slash" notation, the derivative operator also carries a gamma matrix, which can't be moved to the other side of the field being differentiated.

    2. A derivative with a double arrow on top means:

    [tex]\alpha \overleftrightarrow {\partial_\mu} \beta = (\partial_\mu \alpha) \beta - \alpha \partial_\mu \beta[/tex]

    But be careful, because sometimes this is defined to mean the other way around!
     
  5. Mar 10, 2010 #4
    Thank you very much for the clarification.
     
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