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Physics
Quantum Physics
Commuting observables for Fermion fields?
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[QUOTE="maline, post: 6006697, member: 506251"] In nonrelativistic QM, we usually describe the Hilbert space by choosing a complete set of commuting observables, so that the set of states that are eigenstates of all the observables can be used as a basis. For instance, the "wavefunction" is the state as expressed in terms of "states" with definite positions for all particles (not normalizable, but that can be dealt with), which works because the positions (3 for each particle) are a complete set of commuting observables. Another classic example is the [I]n,l,m,s [/I]basis for the states of the hydrogen atom. For free boson fields, this concept seems to carry over well into QFT. For instance, in the case of a complex scalar field, the real and imaginary parts of the field at each point (on a spacelike hypersurface) are Hermitian and commute, so we can think of the well-defined field configurations as "basis states", and then a general state will be a functional giving a quantum amplitude to each configuration. This "simple" setup obviously does not work for fermion fields. There, we are given anticommutation relations, which don't seem very helpful for defining basis states. Even the real and imaginary parts of the field at the same point do not commute. What would a complete set of commuting observables look like for a fermion field, say a free Dirac field, expressed in terms of the field operators? [/QUOTE]
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Quantum Physics
Commuting observables for Fermion fields?
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