I have an elementary question. In quantum field theory observables are operators. But the quantum fields are operators too. This means the quantum fields are observable? I read somewhere that quantum fields are not observable.
Asked
Active
Viewed 716 times
1
-
2Related: https://physics.stackexchange.com/q/54603 – Nihar Karve May 03 '21 at 08:19
-
1Also related: https://physics.stackexchange.com/questions/633170/are-observables-in-qft-actually-observable – joigus May 03 '21 at 10:44
-
2Not a comment about quantum field theory, but let me stress that density matrices in "standard" quantum mechanics are (also self-adjoint!) operators, but they are not observables. In this context the point is that one identifies states with the operators acting on them because finite-dimensional Hilbert spaces are reflexive, but density states and observables actually live in two different spaces. I honestly don't know how this translates in field theory, but I hope this helped nonetheless! – Fabio Di Nocera May 03 '21 at 11:24
-
1Fields are, in general, not measurable. – Prahar May 03 '21 at 12:26
-
3In QFT, the operators representing observables are expressed in terms of the field operators, but the field operators themselves are not necessarily observables. In some cases they are, but usually not. Expressing observables in terms of field operators is convenient partly because the time-dependence of the field operators is described by relatively simple equations of motion, like Maxwell's equations and the Dirac equation. States constructed using field operators certainly do have observable properties (probed using observables), though, so the answer depends on exactly what you're asking. – Chiral Anomaly May 03 '21 at 13:39
2 Answers
1
I am familiar with the quantum field theory of the standard model of physics, and my answer is within this frame.
The elementary particles in the axiomatic table of the standard model, the electrons, quarks, etc are mathematically represented as a field, an electron field, a quark field, etc, by the plane-wave wave functions of the corresponding quantum mechanical equation without a potential, Dirac for electron, quantized maxwell for photon etc. , at each point in space time. As is well known the wave-function $Ψ$ is not measurable (only $Ψ^*Ψ$), and afaik, these fields are not operators.
Creation and annihilation operators operate on these fields, creating a particle and annihilating it, and that creation would be observable, if it were not the difficulty of plane waves that can not be normalized to probability amplitudes. Real particles have to be represented by wavepackets of the plane waves.
Fortunately for experiments in particle physics it is by the use of Feynman diagrams that the observable crossections and decays are calculated, and the predictions are validated continually in new experiments.
By successfully predicting using Feynman diagrams ( see paragraph 4) a specific decay, or a crossection, using creation and annihilation operators on the relevant fields, one can say that the existence of the fields is validated, and in that sense the fields are observable, because if they were not there the creation and annihilation operators would not work. So it is the combined "creation and annihilation operators on the fields" that are really observable.
-
I'm not sure about the comment I'm about to make: Since creation operators are not self-adjoint, when you say that you can observe the creation do you mean that you can a-posteriori observe $N = a^\dagger a$ which is an observable?
Sorry for the "handwaving" explanation, but I'm not really into field theory, and my observation comes from "single-particle" quantum mechanics reasoning.
– Fabio Di Nocera May 03 '21 at 11:19 -
1@FabioDiNocera I am saying that since to calculate observable values, like crossections and decay times with Feynman diagrams one uses use the creation and annihilation operators,(see 4. here http://home.thep.lu.se/~bijnens/fytn04/feynmandiagrams.pdf ) and in that sense the set (operators + fields) is observable in the measurement sense..Of course single particles exist, propagating with continuous creation and annihilation, so the fact that the theory can describe particles also validates it. – anna v May 03 '21 at 12:23
-
-
1
Observables in QFT are generally constructed as products of field operators such as $\hat{\phi}^- \hat{\phi}^+$. Note that such combination are Hermitian as they need to be to qualify as observables. Individual fields operator may not always be Hermitian. Even in those cases where the field operator is Hermitian (such as the photon), the single field operator would then to give zero. For instance $\langle n|\hat{E}|n\rangle = 0$, where $\hat{E}$ is the field operator for the electric field and $|n\rangle$ is a number state for the photon.
flippiefanus
- 14,614