I'm taking up QFT for the first time. Basic question really. Given the free complex Klein-Gordon field [tex]\phi[/tex] under the global gauge transformation [tex]exp(\imath\alpha)[/tex] where [tex]\alpha[/tex] is some real parameter, the Noether current is given by(adsbygoogle = window.adsbygoogle || []).push({});

[tex] j^{\mu}=\imath \left( \phi \partial^{\mu} \phi\dagger - \phi\dagger \partial^{\mu} \phi \right) [/tex]

Now, the conserved charge is Q, given by

[tex] Q = \int d^3x j^0 = -\imath \int d^3x \left( \pi \phi - \pi\dagger\phi\dagger \right) [/tex]

where [tex]\pi[/tex] is the conjugate field momentum.

Question is: if I want to express Q in terms of the ladder operators

[tex] \hat{a}[/tex], [tex]\hat{a}\dagger [/tex], [tex] \hat{b}[/tex] and [tex]\hat{b}\dagger [/tex], do I need to symmetrize the field operators [tex] \hat{\phi} [/tex] and [tex] \hat{\pi} [/tex], since they do not commute?

I searched the usual books but there seems to be no comment on the subject. I performed the calculation for [tex]\hat{Q}[/tex] using symmetrized field operators and obtained the usual result, but perhaps it was unnecessary?

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# Taking up QFT for the first time

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