# How to interpret the field function Φ in QFT?

1. Oct 28, 2012

### yicong2011

Hi,

I have a question that how to interpret the field function Φ in Quantum Field Theory.

As I can see, it is an operator through second quantization and the co-ordinate representation no long exist after second quantization.

So we cannot regard it as wave function any more.

Could Quantum Field Theory still allow Born's Statistical Interpretation which exists in QM? I think not, since Statistical Interpretation is on wave function in QM.

What is your opinion? Thanks a lot!

2. Oct 28, 2012

### tom.stoer

I think that you cannot interpret the field operator directly; you should better look at Hilbert space states.

3. Oct 28, 2012

### Jazzdude

No, the Born rule was never about the wave function but quite more abstractly on rays in the Hilbert space which correspond to quantum states. If you want you can carry the probability interpretation directly to the hilbert space of QFT, using inner products just like in non-relativistic QT. But I'd recommend to avoid any statistical interpretation of the state.

4. Oct 28, 2012

### andrien

In qft ,there is no wavefunction representation because the number of particles are not fixed.Field operators are used to create and destroy particles and they obey certain kind of commutation or anticommutation relations.

5. Oct 29, 2012

### Ilja

Instead of the wave function, you have in QFT a wave functional $\Psi$defined on the classical configuration space of the field theory, that means, the space of classical field configurations.

So, there is a classical field $\varphi(x) \in C(\mathbb{R}^3)$, defined on the classical space, which is comparable to other classical fields like the EM field. And then this space of functions $Q\cong C(\mathbb{R}^3)$ can be used as the configuration space for a quantum theory, with the wave functional $\Psi(q)\in\mathbb{C}, \Psi \in L^2(Q,\mathbb{C})$.

This wave functional is seldom used. But this does not mean that there is no wave function in QFT. There is also some ideological reason that it is usually not even teached: The wave functional is obviously not Lorentz-covariant, and many people dislike everything which is not. The non-covariant parts of the quantum formalism are much less visible in the operator formalism.

6. Oct 29, 2012

### tom.stoer

the results are Lorentz covariant if the underlying theory is; it's harder to see or to prove that, but there's no principal problem;

anyway, interpreting the field operator is rather difficult

Last edited: Oct 29, 2012
7. Oct 29, 2012

### mpv_plate

I have only very basic awareness about QFT, so I'm not sure if I understand. Can QFT give probability that a particle will be found in a given location? Or this is completely out of scope of QFT?

8. Oct 30, 2012

### tom.stoer

You can ask this question in QFT - and it was one the first questions that have been asked. The result is scattering theory with the S matrix. You start with an initial state, e.g. a proton at rest plus an incoming photon (both with definite momentum,i.e. something like plane waves, which makes the direct physical interpretation impossible ;-); you identify possible final states - and b/c we are in the QFT context there is a plethora of allowed final states with different particle species, e.g. proton + photon, proton + electron-positron pair, proton + pion-antipion pair, break up of proton with hadronization in complex final state, ... For each particle in the final state you can ask for its scattering angle and momentum and therefore where it will be observed. This is the calculation of S-matrix elements via Feynman diagrams. This approach does not solve the problems with interpretation, it does not solve the "collapse" or whatever, it does not explain why using plane waves results in physically correct results, ... but it works! It gives you the differential cross section, the number of particles with momenta, spin, other quantum numbers like isospin etc. observed in your experiment.

9. Oct 30, 2012

### andrien

I would like to know where this concept is used because if it is not lorentz covariant then how one can receive it.

10. Oct 31, 2012

### Ilja

Sorry but I don't remember the places where I have seen it long ago. A natural place to look would be Bohmian field theory, and indeed Bohm uses some wave functional for the EM field.

Anyway quantum theory itself is not Lorentz covariant, only the predictions about observables are - in a weak sense, which ignores violations of Bell's inequalities and is satisfied if there is no possibility of explicit information transfer.

11. Oct 31, 2012

### tom.stoer

A wave functional approach is used in geometro-dynamics / Wheeler-deWitt equation

12. Nov 1, 2012

### andrien

it seems it is lorentz covariant. Something similar to action functional of feynman.

13. Nov 1, 2012

### tom.stoer

It is Lorentz-covariant, but not explicitly. The reason is that you discuss only one equation of constraint which corresponds to H ~ 0 implemented as an operator with functional derivatives acting on the wave functional (Wheeler-deWitt eq). But in principle you have to implement all generators of the (local) Lorentz group plus the diffeomorphism group in this way, and you have to ensure the correct algebra of these operators after regularizion. If you succeed with this program you have proved covariance.

Last edited: Nov 1, 2012
14. Nov 3, 2012

### A. Neumaier

In general in QM and QFT one can find a wave function representation given any complete set of commuting operators.

In QM, this is conventionally the set of positions or momenta (augmented y spins if necessary).

In scalar QFT, such a set is given by the field operators Phi(x) for x on a complete spacelike surface, called a multifingered time.
Changing the spacelike surface is the QFT analogue of changing time in QM. The change is given by the Tomonaga-Schwinger equations, which are the QFT analogue of the Schroedinger equation.
Everything is fully covariant though somewhat unwieldy because of the multifingered time.

15. Nov 3, 2012

### samalkhaiat

16. Nov 4, 2012

### Ilja

Multifingered time is, in my opinion, a clear case of multiplying entities without necessity.

A local measurement, then, probably causes a multifingered collapse. And all this only because of the need to obtain a Lorentz symmetry - something which was, initially, considered as a simplification.

17. Nov 4, 2012

### A. Neumaier

It exists automatically in each covariant field theory, even classically. Thus it doesn't involve any additional assumption that could be removed by Ochham's rasor.
It is reversible, hence there are no collapses. The latter only appear in approximate treatments, where the observer introduces a preferred frame of reference, and hence defines what the correct multifingered time is.

18. Nov 4, 2012

### jfy4

Is there a good QFT text that deal primarily in the wave functional approach, versus the fock space approach? I would be interested in reading more about using the wave functional.

19. Nov 4, 2012

### A. Neumaier

I haven't seen it in a textbook (though I haven't looked at many). But search for terms such as ''multifingered time'', ''Tomonaga-Schwinger'', ''functional Schrodinger equation'', and you'll find plenty of journal references.

Tomonaga and Schwinger developed this approach to QED at the same time as Feynman developed the path integral approach. All three were shown to be equivalent by Dyson in 1948. But Feynamn's approach turned out to be easier to teach and was aslo computationally less demanding. This is why textbooks usually concentrate on the latter.

In classical general relativity, multifingered time is just an expression of the fact that one can prescribe initial conditions on any maximal spacelike hypersurface, solve the field equations, and look at the result at any ''later'' spacelike hypersurface. See, e.g., Section 21.3 (p.497) in the well-known book Gravitation by Misner, Thorne and Wheeler.

But the same is possible in any Lorentz invariant classical theory, though much less useful in the flat case.

For multifingered time in quantum gravity, see, e.g.,
http://arxiv.org/pdf/1209.0065, Section 2B

20. Nov 4, 2012

### jfy4

Thanks Arnold.

21. Nov 5, 2012

### mpv_plate

When reading about the basics of QFT I found there is a so called "density operator" which gives the particle density at a given location. It is the combination of annihilation and creation operator in the position space.

Can the density operator be understood as another possible answer? It basically tells where the particles are in the space. When I use the scattering theory it seems I get similar answer: where the particles go (spatially) after they interact (if I understand that correctly). Is the density operator used in the scattering theory?

22. Nov 5, 2012

### tom.stoer

Can you please provide a defintion or a reference for this density operator? it seems that I have something in mind which does not allow for such an interpretation

23. Nov 6, 2012

### Sonderval

The wave functuional is very clearly explained in the book by Hatfield "Quantum Field Theory of particles and strings" - was quite a revelation to me this summer, when I got hold of it.
If you can read german, you can find a rather simple explanation of what the concpet behind it are on my blog, where I have a series on QFT. The wave functional is explained here:
http://scienceblogs.de/hier-wohnen-drachen/2012/06/05/qft-fur-alle-wir-verstehen-nichts/

24. Nov 6, 2012

### Demystifier

I think the book by Hatfield "QFT of point particles and strings", mentioned also by Sonderval above, is exactly what you need. In this book many problems are solved in parallel by 3 methods: wave functional, Fock space, and path integrals.

25. Nov 6, 2012

### Demystifier

One general comment.

Many people seem to mix three logically different concepts:
1. QFT per se
2. Interacting QFT
3. Relativistic QFT

For example, they often say that "the number of particles is not constant in QFT", but this is true only for interacting QFT.

Or they say that "the particle position operator is not well defined in QFT", but such a claim makes certain sense only in relativistic QFT, because the particle space-position operator turns out to be not Lorentz invariant.

Or sometimes they say that "the number of particles is not well defined in QFT", but it may have at least three different meanings.

- One possible meaning is that the number of particles changes with time, which, as I already said, is true only in interacting QFT.

- Another possible meaning is that the number of particles may be uncertain, i.e., the quantum state does not need to be a particle-number eigenstate. This is a genuine property of QFT per se, valid even for free and/or non-relativistic QFT. But it should be stressed that it does NOT imply that the OPERATOR of the number of particles is not well defined.

- Yet another meaning is that the OPERATOR of the number of particles may not be well defined, but this occurs only in attempts to combine QFT with GENERAL relativity. Namely, the operator of the number of particles may not be invariant under general coordinate transformations, even though it is invariant under Lorentz transformations.

I hope these notes will reduce confusion stemming from fails to distinguish the different concepts above.