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

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  • Thread starter Lapidus
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  • #1
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..where can be found:
Zee 1.PNG


What in the whole wide world does Zee mean with

zee 2.PNG
??

Thank you
 

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  • #2
DrDu
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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.
 
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  • #3
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But on what else could the partial derivative act on than on A?
 
  • #4
DrDu
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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.
 
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  • #5
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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
 
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  • #6
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Got it! Thanks DrDu and mfb!
 
  • #7
vanhees71
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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 :-(.
 
  • #8
ChrisVer
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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)
 
  • #9
nrqed
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A better way to write this would be
$$(\partial_i A_i + A_i \partial_i) X = (\partial_i A_i) + 2 A_i \partial_i X = 2 A_i \partial_i X$$
for some wave function X and ignoring indices there. Just the product rule applied to ##\partial_i A_i X##
Just to point out a small misprint (mfb just forgot): there is an X missing in the middle expression$$(\partial_i A_i + A_i \partial_i) X = (\partial_i A_i) X + 2 A_i \partial_i X = 2 A_i \partial_i X$$
 
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