Undergrad Hermitian operators in QM and QFT

Click For Summary
SUMMARY

This discussion centers on the interpretation of Hermitian operators in Quantum Mechanics (QM) and Quantum Field Theory (QFT). Participants express skepticism about treating field operators as "experimental apparatus" at every point in space, particularly in the context of a second-quantized electromagnetic (EM) field. The conversation highlights that while operators in QFT correspond to measurements at specific spacetime points, the concept of smeared field operators provides a more practical approach by associating measurements with regions rather than individual points. The discussion concludes that self-adjoint smeared field operators can be viewed as measurement devices.

PREREQUISITES
  • Understanding of Hermitian operators in Quantum Mechanics
  • Familiarity with Quantum Field Theory concepts
  • Knowledge of second quantization and Fock space
  • Basic grasp of smeared field operators and their mathematical representation
NEXT STEPS
  • Research the mathematical formulation of smeared field operators in QFT
  • Study the role of self-adjoint operators in quantum measurements
  • Explore the implications of Fock space in particle creation and annihilation
  • Examine the differences between non-relativistic QM and QFT regarding operator interpretation
USEFUL FOR

Physicists, quantum field theorists, and students of theoretical physics who are interested in the foundational aspects of measurement and operator theory in quantum mechanics and quantum field theory.

WWCY
Messages
476
Reaction score
15
TL;DR
"Apparatus"? Or not?
I have always learned that a Hermitian operator in non-relativistic QM can be treated as an "experimental apparatus" ie unitary transformation, measurement, etc.

However this makes less sense to me in QFT. A second-quantised EM field for instance, has field operators associated with each spatial point; to think that there is some "apparatus" at every point in space seems slightly absurd (to me at least).

How I think of these field operators? Can I think of them as "fundamental" objects that (as part of a Hamiltonian) govern the evolution of Fock-space states by creating and annihilating particles (if so, how then does one think of the trajectory of a relativistic particle)? Are there any cases in which these field operators actually correspond to experimental apparatus?

Cheers.
 
Physics news on Phys.org
WWCY said:
A second-quantised EM field for instance, has field operators associated with each spatial point

No, associated with each spacetime point. See below.

WWCY said:
to think that there is some "apparatus" at every point in space seems slightly absurd (to me at least)

The fact that an operator appears in the math does not mean that operator has to be physically realized. A given quantum EM field operator corresponds to measuring the field at the particular spacetime point (some point in space at some instant of time) the operator is attached to. So to model measuring the quantum EM field at a particular point in space at a particular instant of time, you just pick the particular operator that is attached to that spacetime point and use that one to make predictions.
 
  • Like
Likes bhobba and WWCY
WWCY said:
However this makes less sense to me in QFT. A second-quantised EM field for instance, has field operators associated with each spatial point; to think that there is some "apparatus" at every point in space seems slightly absurd (to me at least).

In non-relativistic QM, using the Heisenberg picture, the position and momentum operators evolve simultaneously in time. However, the idea of simultaneous position and momentum measurements is absurd. So the operator evolving in time does not correspond to the measurement being made.
 
WWCY said:
Summary: "Apparatus"? Or not?

However this makes less sense to me in QFT. A second-quantised EM field for instance, has field operators associated with each spatial point; to think that there is some "apparatus" at every point in space seems slightly absurd (to me at least)
##\phi(x)## is not an operator, it's smeared version is:
$$\phi\left(f\right) = \int{\phi\left(x\right)f\left(x\right)d^{4}x}$$
where ##f## is a function that decays quickly (I can be more precise if you want), possibly vanishing outside of a compact region. Thus there are operators associated with weighted regions, not points.
 
  • Like
Likes bhobba, vanhees71 and dextercioby
Thanks for the replies

DarMM said:
##\phi(x)## is not an operator, it's smeared version is:
$$\phi\left(f\right) = \int{\phi\left(x\right)f\left(x\right)d^{4}x}$$
where ##f## is a function that decays quickly (I can be more precise if you want), possibly vanishing outside of a compact region. Thus there are operators associated with weighted regions, not points.
Demystifier said:
For that purpose you can us smeared field operators.

Can these smeared field-operators then be treated as "measurement devices" that are associated with a certain region rather than each point in spacetime?

Cheers.
 
WWCY said:
Can these smeared field-operators then be treated as "measurement devices" that are associated with a certain region rather than each point in spacetime?
Loosely speaking, yes.
 
WWCY said:
Can these smeared field-operators then be treated as "measurement devices" that are associated with a certain region rather than each point in spacetime?
If they are self-adjoint then loosely yes. Otherwise not, such as in the case of Fermion fields where only bilinears like ##\bar{\psi}\psi## are observable.
 
  • Like
Likes vanhees71

Similar threads

  • · Replies 15 ·
Replies
15
Views
819
  • · Replies 8 ·
Replies
8
Views
845
Replies
18
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
9K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 60 ·
3
Replies
60
Views
7K
  • · Replies 38 ·
2
Replies
38
Views
5K