How to interpret the field function Φ in QFT?

In summary: QFT.In summary, in Quantum Field Theory, the field function Φ is an operator through second quantization and does not have a wave function representation. Instead, there is a wave functional defined on the classical configuration space of the field theory. This wave functional is not Lorentz-covariant and is seldom used in practice. However, it is possible to use it to calculate probabilities for scattering events in QFT. The field operator can be seen as the position operator in QFT and can be used to define a complete set of commuting operators.
  • #1
yicong2011
75
0
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!
 
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  • #2
I think that you cannot interpret the field operator directly; you should better look at Hilbert space states.
 
  • #3
yicong2011 said:
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.

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
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
andrien said:
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.

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

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

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
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
 
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  • #7
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
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
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.
I would like to know where this concept is used because if it is not lorentz covariant then how one can receive it.
 
  • #10
andrien said:
I would like to know where this concept is used because if it is not lorentz covariant then how one can receive it.
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
A wave functional approach is used in geometro-dynamics / Wheeler-deWitt equation
 
  • #12
it seems it is lorentz covariant. Something similar to action functional of feynman.
 
  • #13
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.
 
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  • #14
Ilja said:
This wave functional is seldom used. But this does not mean that there is no wave function in QFT. [...] The wave functional is obviously not Lorentz-covariant
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.
 
  • #16
A. Neumaier said:
Everything is fully covariant though somewhat unwieldy because of the multifingered time.
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
Ilja said:
Multifingered time is, in my opinion, a clear case of multiplying entities without necessity.
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.
Ilja said:
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.
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
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
jfy4 said:
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.

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
Thanks Arnold.
 
  • #21
mpv_plate said:
Can QFT give probability that a particle will be found in a given location?

tom.stoer said:
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.

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
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
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
jfy4 said:
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.
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
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.
 
  • #26
Thank you Demyistifier, I'll check it out.
 
  • #27
Sonderval said:
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.

Hatfield's book is actually called ''Quantum Field Theory of Point Particles and Strings''.

For a commentary on Hatfield's book, see Forrester's Winter 08 Lecture Notes http://aforrester.bol.ucla.edu/educate.php ,
starting with week 4.
 
  • #28
tom.stoer said:
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

The density operator is described for example http://www.colorado.edu/physics/phys7450/phys7450_sp10/notes/2nd_quantization.pdf. See the expression (54) on page 6.

It seems to describe how particles are distributed in the space.
 
  • #29
OK, I see. You are right, interpreting this expression in QM it corresponds to a density (like a charge density) in space.

I would never call this density a "density operator" b/c a density operator is already defined in non-rel. QM and its meaning is something totally different
 
  • #30
they often say that "the number of particles is not constant in QFT", but this is true only for interacting QFT.
Where else can it be used.
 
  • #31
tom.stoer said:
OK, I see. You are right, interpreting this expression in QM it corresponds to a density (like a charge density) in space.

I would never call this density a "density operator" b/c a density operator is already defined in non-rel. QM and its meaning is something totally different

The correct name is (operator-valued) density field.
 
  • #32
mpv_plate said:
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?

Yes, the expectation of the density field multiplied by the mass is what becomes in the case of macroscopically many particles the thermodynamical mass density, and tells about where the mass is concentrated.
 

1. What is the field function Φ in QFT?

The field function Φ in QFT (Quantum Field Theory) is a mathematical representation of a quantum field, which is a physical quantity that exists throughout space and time. It describes the behavior of particles and their interactions, and is a fundamental concept in understanding the behavior of matter at the subatomic level.

2. How is the field function Φ used in QFT?

The field function Φ is used in QFT to calculate the probability amplitudes of different particle interactions. It is also used to determine the energy and momentum of particles, as well as to describe how particles behave and interact in different physical situations.

3. What is the significance of the field function Φ in QFT?

The field function Φ is significant in QFT because it allows us to describe and understand the behavior of particles at the subatomic level. It is a fundamental concept in modern physics and is essential in many areas of research, including particle physics, cosmology, and condensed matter physics.

4. How is the field function Φ related to other concepts in QFT?

The field function Φ is closely related to other important concepts in QFT, such as the Hamiltonian, Lagrangian, and Feynman diagrams. These concepts all work together to describe the behavior of particles and their interactions, and are essential in making predictions and calculations in QFT.

5. Is the field function Φ a measurable quantity in QFT?

No, the field function Φ is not a directly measurable quantity in QFT. It is a mathematical construct that helps us understand and make predictions about the behavior of particles. However, the effects of the field function Φ can be observed and measured through experiments and observations of particle interactions.

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