In QFT of a real Klein-Gordon-Field, the field operator(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\phi(x)[/itex] is an observable.

Mathematically, this is the case because it is a sum (over all k) of [itex]a[/itex] and [itex]a^\dagger[/itex] and this yields a Hermitian operator.

Physically, I can understand this because this equation would describe, for example, a membrane treated with QFT.

However, if the field is complex (charged particle) [itex]\phi(x)[/itex], the operator is the sum of particle lowering and anti-particle raising operators (or vice versa). Mathematically, I can see that it is not Hermitian and therefore it will not have Eigenvalues and does not correspond to an observable.

But I don't really understand this physically: Why can I (in principle) measure the uncharged pion field but not the charged pion field? I suspect this has to do with the scalar particle being its own anti-particle, but I don't really see how this affects the measurability of the field or the possibility to create a local field excitation (which is what operating with [itex]\phi(x)[/itex] should do)?

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# When is operator phi(x) an observable in QFT?

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