# I Fields: Unobservable, Yet Physical?

1. Mar 11, 2017

### dm4b

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 clicked with me before: fields are un-observable. At least, if they have zero expectation value. Does this apply to all known fields in nature? Or, just scalar fields, where is where I am it in the book.

Also, I got caught up on the notion of how something can be un-observable, yet also physical. I guess the same may apply to virtual particles, though.

Any thoughts?

Thanks

2. Mar 11, 2017

### hilbert2

What do you mean by something being un-observable? All of us observe electromagnetic field excitations of certain wavelengths all the time with our eyes (except those who are blind). A gravitational field is observed by monitoring the trajectory of a mass or curvature of a light beam moving through it. There's no point in making a physical theory about something that can't be observed.

3. Mar 11, 2017

### dm4b

That was my initial thought!

Operators that reproduce the original state multiplied by an eigenvalue are observable with the eigenvalue being related to the probability of obtaining a particular eigenstate upon measurement.

Fields are operator-valued fields. These kinds of operators don't reproduce the original state, they change states from one state to another state and are generally not observable.

In addition, he also mentions how A_mu has zero expectation value and is also unobservable, or cannot be directly measured. This is the field behind EnM.

4. Mar 11, 2017

### dm4b

I can't help but wonder if we're getting into "virtual particles" here, or the virtual bosons involved in intermediating the force and found in the propagator, which are not directly observable, or cannot be measured.

5. Mar 11, 2017

### phinds

There is a FAQ here on PF about virtual particles, which are stated to be a mathematical fictions to (help with some computations) not real objects.

6. Mar 11, 2017

### dm4b

I've read that and been involved in the discussion. Damn good argument, but I didn't think it was %100 air tight, either. However, could the same thing be said about the unobservable fields in question?

7. Mar 11, 2017

### phinds

I'm not expert at all in this stuff, but I don't think so. It seems to me that virtual particles and fields are just not the same thing. I'm pretty sure I've never had a virtual particle hit my retina

8. Mar 11, 2017

### MrRobotoToo

It's somewhat subtle since all of the observables in a field theory (i.e., energy, momentum, angular momentum, etc.) can be expressed in terms of the fields and their derivatives, but in such a way that gauge transformations of the fields leave the observables unchanged.

9. Mar 11, 2017

### mikeyork

Non sequitur. What we observe is stimulation of the retina. Could just as equally well be described by a photon of a given energy.
No. We observe only a trajectory and we theorize about it's route.
Just about all theories in physics are built on stuff that cannot be observed, but predicts stuff that can.

10. Mar 11, 2017

### Staff: Mentor

Well the limit of QED is standard EM.

The fields of standard EM carry energy and momentum so physicists like to think of them as very real even though, as Feynman showed by devising a theory where they don't exist (the absorber theory is a direct action at a distance theory) they are obviously not directly observable. Feynman though was never able to come up with a quantum version but I read somewhere Penrose may have. It would be rather difficult to come up with a point in taking that limit where they all of a sudden change from real to not real - but I suppose you could hold to such a view.

That said they are fields of operators - how real and observable is a QM operator?

Not an easy question to answer - but I think of them as real for the reason I gave at the start.

Thanks
Bill

Last edited: Mar 11, 2017
11. Mar 11, 2017

### atyy

As far as I know, fields are not physical in any interpretation of quantum mechanics. In the Schroedinger picture, the wave function may be real. In QFT, one usually uses the Heisenberg picture with time evolution of field operators.

If you look at bhobba's post above, you see he says essentially the same thing.

A way to simplify the question is to go to non-relativistic quantum mechanics. Use the Heisenberg picture. There in the formalism, the position and momentum observables simultaneously exist and evolve with time. However, position and momentum do not commute.

12. Mar 12, 2017

### hilbert2

Difficult to say how much neural processing the visual stimulus has to go through before it becomes an "observation" that one can intellectually work on...

Something unobservable would be if I defined a field called "ghost field", denoted $G(x,y,z,t)$ and stated that it doesn't interact with anything else and has a field equation

$\frac{\partial^2 G}{\partial t^2}=a \nabla^2 G - b|G|^2$,

where a and b are some constants. Then someone else could say "no, you're wrong, its field equation is actually $\frac{\partial^2 G}{\partial t^2}=a \nabla^2 G - b|G|^4$", and there would be absolutely no way to show that either of us would be more correct in our claims.

13. Mar 12, 2017

### dm4b

I guess my next question would be ... if fields are indeed un-observable and non-physical, it seems like we can't say they are (more) fundamental, can we? This seems to make fields more akin to a convenient calculational model used to make predictions for (physical) things we really can observe/measure (like momentum, spin, etc), which again doesn't really seem to make fields more fundamental than the physical entities (particles and/or waves) that actually possess (and physically exhibit) these observable properties.

At the same time, I always found it pleasing how QFT makes sense of wave/particle duality by unifying what appeared to be separate entities (waves and particles) into a single, encompassing underlying (and more fundamental) field. The magic of this seems to fade away if fields are un-observable, un-physical and perhaps, by logical extension, un-real?

Seems like we're still missing something here ....

14. Mar 12, 2017

### Karolus

who established that virtual particles are "mathematical fictions" ?? What kind of nonsense! Then even a black hole is a "mathematical fiction"! Have you ever seen a black hole?

15. Mar 12, 2017

### phinds

Hey, look up the FAQ. I'm just saying there IS one, and telling you what it says.

16. Mar 12, 2017

### hilbert2

If I take a Hamiltonian operator $\mathbf{H} = \frac{\mathbf{p}^2}{2m} + \frac{1}{2}k\mathbf{x}^2 + \lambda\mathbf{x}^4$ and write its ground state as

$|\psi_0 > = |\phi_0 > + \lambda |\phi_1 > + \lambda^2 |\phi_2> + \dots$ ,

where $|\phi_0 >$ is the harmonic oscillator ground state, I can with good reason say that the terms $\lambda^n |\phi_n >$ are "mathematical fictions", but they can still be useful when calculating something.

17. Mar 12, 2017

### Karolus

you can apply similar reasoning to all physical. The black hole turns out to be purely theoretical calculations, not to mention the event horizon, and other abstract entities. The virtual particles are considered to explain the Casimir effect, that is anything but "virtual", or the Hawking radiation. If one day a hypothetical future, but not too hypothetical, could extract energy from the quantum vacuum, and the mechanism of virtual particles, they become quite "real", especially when you have to pay us taxes ...

18. Mar 12, 2017

### Staff: Mentor

What does "physical" mean? Be very careful of using terms that don't have precise definitions.

It seems like you're conflating models with observations. Virtual particles are models. So are black holes. So are quantum fields. The Casimir effect is an observation.

Any model can be considered a "mathematical fiction". But observations can't.

19. Mar 12, 2017

### mikeyork

Unobservable is not the same as unphysical. Something can be unobservable but if it accurately predicts the observable, then that would suggest it is at least a candidate as physical in the sense of reflecting reality.

20. Mar 12, 2017

### Staff: Mentor

We aren't missing anything.

Its purely a matter of what you consider real and physical which is a philosophical issue. They have energy and momentum - for me that makes them real - but its purely a matter of semantics. One of the silliest things you can argue about is semantics - it gets you no-where fast.

Here are the facts:

1. The fields themselves can never be observed - only their effects.
2, They do carry momentum and energy from Noether, but also from the simple observation since EM effects travel at the speed of light when an object radiates if momentum/energy conservation is to occur, and according again to Noether it must, then it must carry it away.
3, It can be formulated in a direct action at a distance way without fields - but no one really does it that way - it just seems a curiosity Feynman discovered.
4. In QED the field consists of a field of quantum operators.

These are the facts. Philosophers will argue if that implies they are real or not. We don't discuss philosophy here, so make up your own mind - it makes no difference. I think they are real, but as you can see from the above a decent argument could be made they are not. Its one of those things that you cant really answer because it depends on what you mean by the terms you are using like real etc. Observations are real - no VERY real - beyond that -.

Thanks
Bill

Last edited: Mar 12, 2017
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