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In many text books and exercises, it is stated the the phase transformation is constant under differentiation iff the argument is real, which is necessary to show the invariance of the Lagrangian under such transformations. But why is this? That is, why does the following hold, and why only when the argument is real?

$$ \partial_\mu (e^{-i\alpha(x)} \phi(x)) = e^{-i\alpha(x)} \partial_\mu \phi(x) \leftrightarrow \alpha\in\mathbb{R}$$

a20
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1 Answers1

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It is only a constant, i.e commutes with $\partial_x$, if $\alpha$ does not depend on $x$. Being real has nothing to do with it. Where did you read this claim?

mike stone
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  • I cross-checked the text books again, but they said "if alpha is a real constant". In some lecture notes and exercises I have it only says real, but I take it that this is simply a typo then. Thank you for clarifying! – a20 Jun 02 '20 at 12:46