I'm currently working through Robert Klauber's Student Friendly Quantum Field Theory, which by the way is much more accessible than other texts like, say, Peskin and Schroeder, for others also coming into QFT via the self-study path.

Anyhow, he mentioned something that never really jumped out at me before, like it is now: Fields are un-observable.

Also, I got caught up on the notion of how something can be un-observable, yet also physical. And, how we can call fields fundamental, while at the same time fields are undetectable, even in principle?

Curious what others make of this.

  • 1
    $\begingroup$ If fields were unobservable, even in principle, then all field theories would be equivalent. This is not the case. $\endgroup$ – tfb Mar 28 '17 at 19:15
  • 2
    $\begingroup$ It might help if you provide a quote with some context around "fields are un-observable", so we can better understand what the author meant. $\endgroup$ – Conifold Mar 28 '17 at 20:28
  • $\begingroup$ Why have you concluded that fields are unobservable? Can we not measure amplitudes i.e. VEVs of combinations of fields? There are superselection rules though... $\endgroup$ – innisfree Mar 28 '17 at 20:38
  • 1
    $\begingroup$ Here's a simpler example. There's no way to measure energy. One can measure location, mass, and time, and from those calculate gravitational potential energy and kinetic energy, but there are no direct energy measurements. So is energy real? The fact that it is conserved makes it seem pretty real. $\endgroup$ – Paul B Mar 28 '17 at 22:11

Thanks for the responses everyone, much appreciated. Anna, thanks for the detailed response, I have come to some of the conclusions you have stated, still, it leaves me feeling a bit uneasy as it's not a very satisfying conclusion.

Here is a quote of what is stated in the text I refer to:

"In QFT there are two kinds of operators. One kind is the usual one from NRQM and RQM representing dynamical variables of classical theory, such as the Hamiltonian (energy), the 3-momentum operator, charge, etc. The other kind comprises creation and destruction operators.

The first kind, when operating on an eigenstate, reproduces the original eigenstate, multiplied by an eigenvalue. The second kind changes the state to another state ... note that operators of this kind do not have eigenvalues, since their operation on a state changes that state, rather than reproducing it .. and hence they are generally not observable."

From another chapter:

"Recall ... fields such as phi, psi, A_mu are themselves not observable. They cannot be measured directly (we prove they have zero expectation value in Chapter 7). But properties of fields like energy, momentum and charge are measurable. Our dynamical variable operators, which include number operators, reflect this. They typically have non-zero expectation value

(phi is the scalar field of the Klein-Gordon equation, psi the spinor field of electrons and A_u the vector field of QED)

I guess we get into the whole, "what is real" debate. But, are fields "real", or are they as real as absolute space-time was in Newtonian Physics, in that they're just an approximation, or partial appearance, of some deeper underlying reality? Absolute space-time doesn't exist anywhere in nature, any more than a perfect circle does? Can the same be said for fields?

| cite | improve this answer | |

Also, I got caught up on the notion of how something can be un-observable, yet also physical. And, how we can call fields fundamental, while at the same time fields are undetectable, even in principle?

Quantum field theory is a mathematical model, with creation and annnihilation operators operating on quantum fields . These do not have to be the fields of fundamental particles but cover many problems and are called effective field theories. In this sense the "field" on which the creation and annihilation operators operate can be various ,( starting from the harmonic oscillator ground state).

In particle physics the ground state is the wave function of the free particle solution for the respective particle fields from the table of elementary particles.

Wave functions are not directly observable. It is only their complex conjugate squared which gives probability density distributions.

A concrete example is the harmonic oscillator , using its solutions one can define a field theory for creation and annihilation operators:


This is astonishingly convenient. It presages a form of operator algebra that proceeds without ever looking at the form of ψ(x) and does not require direct evaluation of integrals

The fields existence is implied by the creation and annihilation operators on these ground states, which will describe the particles, and particles are observable.

To sum up, mathematical models fitting observations have components which are not directly observable, but are necessary for calculating measurable distributions. In particle physics calculations are made with Feynman diagrams which describe the perturbative expansion of crossections etc.

Fundamental but not physical: The (x,y,z) coordinates mapping a terrain are unobservable, a mathematical concept, nevertheless the map is described by the functions of (x,y,z), and the (x,y,z) concept is the context for having a map. Particle fields provide the fundamental context on which the physical observables of particles can be modeled.

| cite | improve this answer | |
  • 1
    $\begingroup$ Creation and annihilation operators are the quantum fields which operate on Fock space $\endgroup$ – innisfree Mar 28 '17 at 20:28
  • $\begingroup$ The ground state is the state with no particles, not a wave function. $\endgroup$ – innisfree Mar 28 '17 at 20:30
  • $\begingroup$ @innisfree mathematically, so that one can calculate the feynman diagrams , it is the ground state wave function on which the creation and annihilation operators operate. ( the operators are differentials that have to operate on something, and give the expectations value of one if a creation operator or -1 if an annihilation one) $\endgroup$ – anna v Mar 29 '17 at 3:37
  • $\begingroup$ @innisfree look at the harmonic oscillator which started the whole field: ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2013/… $\endgroup$ – anna v Mar 29 '17 at 3:45
  • 1
    $\begingroup$ i have no idea what you are talking about. the creation/annihilation operators do indeed operate on something, and that something is a Fock space, not a wave function from QM. $\endgroup$ – innisfree Mar 29 '17 at 4:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.