# Position Representation in QFT?

Gold Member
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?

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.

Eugene.

tom.stoer
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.

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.

Demystifier
Gold Member
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]

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 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.

Last edited: