Position Representation in QFT?

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  • #1
referframe
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I have read that, in QFT, unlike QM, there is no position probability density function because position is not considered an observable.

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

As always, thanks in advance.
 

Answers and Replies

  • #2
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Position operator and position representation can be defined in relativistic QFT just as well as in ordinary quantum mechanics. You may want to check

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

before forming your own opinion.

Eugene.
 
  • #3
tom.stoer
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Just a second: there is a big difference between the 'x' in QM and the 'x' in QFT. 'x' in QM is an operator, whereas 'x' in QFT is a continuous index.
 
  • #4
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Just a second: there is a big difference between the 'x' in QM and the 'x' in QFT. 'x' in QM is an operator, whereas 'x' in QFT is a continuous index.
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.
 
  • #5
Demystifier
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Just a second: there is a big difference between the 'x' in QM and the 'x' in QFT. 'x' in QM is an operator, whereas 'x' in QFT is a continuous index.
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]
 
  • #6
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I have read that, in QFT, unlike QM, there is no position probability density function because position is not considered an observable.

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

As always, thanks in advance.
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 not depend of position (spatial coordinates are integrated out and eliminated from the matrix).

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