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

• A
• Lapidus

#### Lapidus

..where can be found: What in the whole wide world does Zee mean with ??

Thank you

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.

• Lapidus
But on what else could the partial derivative act on than on A?

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.

• Lapidus
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##

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• Lapidus
Got it! Thanks DrDu and mfb!

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 :-(.

Well it's so and so; it seems like a bigger generalization of the derivation of $xp -px = i \hbar$. Without having to write $(xp - px ) f(x) = i \hbar f(x)$.
Although I think similar expressions were in Griffith's Intro to Electrodynamics.
$$(\partial_i A_i + A_i \partial_i) X = (\partial_i A_i) + 2 A_i \partial_i X = 2 A_i \partial_i X$$
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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