Field operators and the uncertainty principle

[phoenix95]

Hi,
I am reading QFT by Lancaster and Blundell. In chapter 4 of the book the field operators are introduced:
"Now, by making appropriate linear combinations of operators, specifically using Fourier sums, we can construct operators, called field operators, that create and annihilate particles, but this time they don’t create/annihilate particles in particular momentum states but instead they create/annihilate particles localized at particular spatial locations. Thus the operator defined by
ψ^†(x)=I/√ν Σ_p a^†_pe^-ip.x"

Won't this violate the uncertainty principle? I thought maybe it just creates the particle, after which the particle obeys the relation, but in the very next example(4.1) they go ahead to find the position and momentum of the particle(which they do accurately it seems?). What am I missing here?

Thank you for your time

 

[Iliody]

The exp(-ipx) means that it is a wave, it hasn't a sharp position.

 

[phoenix95]

Oh so what does the operator mean then? Sum over all such waves with momentum p? What momentum state would that particle be in then?

EDIT: I thought Σe^-ip.x was Fourier tranform, since even a bunch of coupled oscillators behave as uncoupled oscillators in Fourier space

 

[Iliody]

The operator [itex] a^\dagger_{p,\lambda}[/itex] is the square root of the light intensity, the amplitude (in the classical sense, not in the wavefunction sense) of the given mode. The exponential also is fourier, because the reason you ell. The operators [itex] a^\dagger_{p,\lambda}[/itex] and [itex] a_{p,\lambda} [/itex] satisfy the commutation relations of harmonic oscillator amplitude (in the classical sense, not in the wavefunction sense), called creation-annihilation operators. Also, ψ⁺(x)=I/√ν Σp a^†pe-ip.x, but ψ^†(x)=I/√ν Σp a^†_pe^-ip.x+I/√ν Σp a_pe^ip.x.

 

[vanhees71]

Be careful with defining what's observable in relativistic QFT! E.g., I'd define "intensity" as in classical electrodynamics as the energy density of the em. field since it is (a) gauge invariant (nothing can be observed that is not gauge invariant) and (b) it is Lorentz covariant. The photon number isn't since there is no conserved current ti deine a Lorentz-invariant quantity of such a kind (indeed photons are strictly neutral). Usually the "number of photons" are defined as the total energy divided by a typical frequency. If there is no typical frequency, it's usually hard to define anything like a photon number that makes sense, but energy density always makes sense, no matter which photon state you are considering.

 

[PeterDonis]

Usually the "number of photons" are defined as the total energy divided by a typical frequency.

Why not the expectation value of the photon number operator? Of course most states are not eigenstates of this operator, and the expectation value in general won't be an integer, but that's no different from any other non-eigenstate in QM.

 

[vanhees71]

The photon-number operator is not Lorentz invariant. What's measured are in fact invariant momentum distributions of photons,
$$E \frac{\mathrm{d} N}{\mathrm{d}^3 \vec{q}}.$$
Of course, the expectation value of any "number" (in relativistic physics defined via conserved currents like electric charge, net-baryon number etc.) needs not necessarily by integer valued although the observable itself can only take integer numbers. That's indeed no problem at all.

 

[Iliody]

Maybe the photon number can be defined by an ideal detector in the next way: given the electromagnetic field mode that can excite the detector, given its energy levels of that mode, the photon number is the number of the energy levels below the energy level that the detector measure.

 

[Demystifier]

Sum over all such waves with momentum p? What momentum state would that particle be in then?

It would be in a state with a completely undetermined momentum, in agreement with the uncertainty principle.