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Lets have the scalar Klein-Gordon field interacting with EM field:

$$ L = \partial_{\mu}\varphi \partial^{\mu}\varphi - m^2\varphi \varphi^{*} - j_{\mu}A^{\mu} + q^{2} A_{\mu}A^{\mu}\varphi \varphi^{*} - \frac{1}{4}F_{\mu \nu}F^{\mu \nu}. \qquad (1) $$ I heard that the normalization of Klein-Gordon field in a theory $(1)$ is invariant under gauge transformations. What normalization is meaned? Does it refer to the factor $\frac{1}{\sqrt{2(2 \pi)^{3} E_{\mathbf p}}}$? How to prove it?

An edit.

It was the invariance of condition $\int j^{0}d^{3}\mathbf r = q$ under $U(1)$ local gauge transformations. $j^{0} = \frac{q}{2m}(\psi^{*}\partial^{0}\psi - \psi \partial^{0}\psi^{*}) - \frac{q^2}{m}A^{0}|\Psi |^{2}$.

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I can't be sure what the source meant without seeing the context, however I suspect the author meant the following. A $ U(1) $ gauge transformation acting on a charged scalar field gives: \begin{equation} \phi (x) \rightarrow e ^{ i \alpha (x) } \phi (x) \end{equation} Under such a transformation the normalization is invariant since $\phi$ simply gains a phase. This is just the definition of a $U(1)$ rotation.

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  • $\begingroup$ It was the invariance of condition $\int j^{0}d^{3}\mathbf r = q$ under $U(1)$ local gauge transformations. $j^{0} = \frac{q}{2m}(\psi^{*}\partial^{0}\psi - \psi \partial^{0}\psi^{*}) - \frac{q^2}{m}A^{0}|\Psi |^{2}$. $\endgroup$ – Andrew McAddams Feb 21 '14 at 14:55

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