# Hermitian operators in QM and QFT

• I

## Summary:

"Apparatus"? Or not?
I have always learnt 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.

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PeterDonis
Mentor
2019 Award
A second-quantised EM field for instance, has field operators associated with each spatial point
No, associated with each spacetime point. See below.

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.

• bhobba and WWCY
atyy
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.

DarMM
Gold Member
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.

• bhobba, vanhees71 and dextercioby
Thanks for the replies

##\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.

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.

Demystifier
Gold Member
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.

DarMM
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