If the Dirac field $\psi(x)$ is to the electron as the Electromagnetic field is to the photon, why is it that we can measure the Electromagnetic field, whereas the Dirac field we cannot?

  • $\begingroup$ What does an observable always have to be? $\endgroup$ – ACuriousMind Dec 1 '15 at 17:08
  • $\begingroup$ Hermitian... hm, is the answer that simple? $\endgroup$ – DaYu1729 Dec 1 '15 at 17:10
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    $\begingroup$ Locality forbids you from measuring any fermionic operators. $\endgroup$ – Meng Cheng Dec 1 '15 at 17:21
  • $\begingroup$ @MengCheng, thanks for the response! Could you elaborate in an answer? That sounds interesting and I would like to learn more about the significance of locality in measurement, as well as the distinction between fermionic and bosonic operators. $\endgroup$ – DaYu1729 Dec 1 '15 at 17:24
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    $\begingroup$ Localized observables should commute when their supports are causally separated. This fact does not hold for fermionic fields in view of fermionic commutation relations...To obtain observables you should consider (bosonic) currents constructed out of fermionic fields... $\endgroup$ – Valter Moretti Dec 1 '15 at 19:53

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