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

A Zee, QFT Nut, III.6, p. 194

  1. May 15, 2016 #1
    ..where can be found:
    Zee 1.PNG

    What in the whole wide world does Zee mean with

    zee 2.PNG ??

    Thank you
     
  2. jcsd
  3. May 15, 2016 #2

    DrDu

    User Avatar
    Science Advisor

    I guess that ##(\partial_i A_i)##is another way to say div A, which probably vanishes in the gauge Zee is considering. The bracket probably means that the action of the partial derivative is restricted to A.
     
  4. May 15, 2016 #3
    But on what else could the partial derivative act on than on A?
     
  5. May 15, 2016 #4

    DrDu

    User Avatar
    Science Advisor

    On what does it act in ##A_i \partial_i##? I mean, you are considering a momentum operator, so it will act on some wavefunction, besides A.
     
  6. May 15, 2016 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    A better way to write this would be
    $$(\partial_i A_i + A_i \partial_i) X = (\partial_i A_i)X + 2 A_i \partial_i X = 2 A_i \partial_i X$$
    for some wave function X and ignoring indices there. Just the product rule applied to ##\partial_i A_i X##

    Edit: Added missing X
     
    Last edited: May 18, 2016
  7. May 15, 2016 #6
    Got it! Thanks DrDu and mfb!
     
  8. May 16, 2016 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    This book by Zee, although apparently written in a quite colloquial way, confuses (at least me) more than it helps. This is an example. Instead of writing the simple matter in a simple way as mfb did in #5 he only gives some short-hand notation, which is more confusing than helpful :-(.
     
  9. May 17, 2016 #8

    ChrisVer

    User Avatar
    Gold Member

    Well it's so and so; it seems like a bigger generalization of the derivation of [itex] xp -px = i \hbar[/itex]. Without having to write [itex] (xp - px ) f(x) = i \hbar f(x)[/itex].
    Although I think similar expressions were in Griffith's Intro to Electrodynamics.
    (feels like Zee's advocate)
     
  10. May 18, 2016 #9

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Just to point out a small misprint (mfb just forgot): there is an X missing in the middle expression$$(\partial_i A_i + A_i \partial_i) X = (\partial_i A_i) X + 2 A_i \partial_i X = 2 A_i \partial_i X$$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Zee, QFT Nut, III.6, p. 194
  1. Equations of QFT (Replies: 13)

  2. Causality in QFT (Replies: 5)

  3. Nonlinerality of QFT (Replies: 6)

  4. Groups in QFT (Replies: 3)

Loading...