Does a boson have a position operator?
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Creation and annihilation operators are just the field operators in disguise (in a different basis). It's just like a Fourier transform. Anything that can be expressed in terms of the fields, therefore, can be re-expressed in terms of creation and annihilation operators. If a QFT is local, then the only operators you'll get are fields and their derivatives.
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Does a boson have a position operator?
The Higgs boson seems to be a particule difficult to detect (elusive particle), in this context doesthe uncertainty principle apply as well as it applies to other less elusive …
The Higgs boson seems to be a particule difficult to detect (elusive particle), in this context doesthe uncertainty principle apply as well as it applies to other less elusive …
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