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referframe

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Then how is a position measurement represented/modeled in QFT?

As always, thanks in advance.

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referframe

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Then how is a position measurement represented/modeled in QFT?

As always, thanks in advance.

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E. V. Stefanovich, "Is Minkowski space-time compatible with quantum mechanics?", Found. Phys., 32 (2002), 673.

before forming your own opinion.

Eugene.

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tom.stoer

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I agree completely, the field argument 'x' in QFT has no relationship to the position observable and its eigenvalues. The true position operator should be built by the Newton-Wigner recipe.

Eugene.

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Even though it is true in the standard formulation of QFT, I think there is a way to reformulate QFT such that x becomes an operator just as in many-particle QM. See Sec. 3 of

http://xxx.lanl.gov/abs/0904.2287 [Int. J. Mod. Phys. A25:1477-1505, 2010]

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As you well say, position is not an observable in QFT and cannot be measured/modeled. As a consequence, any initial position dependence must be eliminated from the equations before going to the lab, and this is the reason for which the only physically relevant QFT object is the S-matrix, which does

Then how is a position measurement represented/modeled in QFT?

As always, thanks in advance.

As Mandl and Shaw note in his classic QFT textbook, this is a fundamental difference between QFT and QM.

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