How Does QFT Describe or Predict the Position of a Particle?

[LarryS]

From what little I have read about QFT, apparently the position of a particle is not a observable - it is more like an index for a collection of quantum harmonic oscillators. Thus there is no QFT equivalent to the position probability density in QM.

So, how does QFT predict or describe a particle's position? What is the QFT equivalent of position?

As always, thank you in advance.

 

[martinbn]

There are no particles in QFT.

 

[Bill_K]

There are no particles in QFT.

Now I've heard everything. A quantum field theory is based on a Hilbert space, typically presented as a Fock space in which each state is labeled by the occupation numbers of a set of elementary excitations. The excitations are the particles.

referframe, There's a simple answer to your question and a more complicated one. We don't usually focus on the position operator because we work in momentum space in which the basis for the particle states are plane waves of momentum p. The simple answer is that the position operator is related to the momentum operator in QFT just the same way it is in regular quantum mechanics, namely x = iħ ∇_p.

The more complicated answer is that x as defined above does not have exactly the right properties. For one thing it is not Hermitian. It needs to be replaced by a Hermitian operator called the Newton-Wigner position operator X. The form of this differs for each kind of particle, but for a spin 0 particle, X = iħ ∇_p - ½iħ p/(p² + m²).

One can show that X does have the right properties, for example [X_i, X_j] = 0, [X_i, p_j] = iħ δ_ij and i[H, X] = p/E.

 

[vanhees71]

However, there are fundamental problems to define a position operator for massless particles with spin $\geq 1$. Particularly, there is no position operator for photons. See Arnold Neumaier's theory FAQ:

http://arnold-neumaier.at/physfaq/topics/position.html

 

[martinbn]

Now I've heard everything. A quantum field theory is based on a Hilbert space, typically presented as a Fock space in which each state is labeled by the occupation numbers of a set of elementary excitations. The excitations are the particles.

Yes, hence particles are convenient labels for states of the quantum field, those in a specific basis. And all this is defined only for free fields. There are no non-interacting fields in nature. So where are the particles?!

 

[questionpost]

Aren't particles themselves waves? So wouldn't quantum field theory be fine with that?

 

[Demystifier]

From what little I have read about QFT, apparently the position of a particle is not a observable - it is more like an index for a collection of quantum harmonic oscillators. Thus there is no QFT equivalent to the position probability density in QM.

So, how does QFT predict or describe a particle's position? What is the QFT equivalent of position?

As always, thank you in advance.

See
http://xxx.lanl.gov/abs/0904.2287 [ Int. J. Mod. Phys. A25:1477-1505, 2010]
especially
Sec. 3.2 "Free scalar QFT in the particle-position picture",
Sec. 4.1 "Probabilistic interpretation".

 

[the_pulp]

Hi, I wrote another thread a couple of days ago (Path Integral Formalism and Integral Limits or something like that), but this paper you mentioned solves much of my doubts. Nevertheless I have some of them remaining and it would be very helpful if some of you could check them out!
Here are my conclusions on this paper just to check if I am right (to simplify I'll take in Klein Gordon non Interacting theory)

1) The Hilbert space that describe the states in QFT, instead of QM where the Hilbert space is the quadratic convergent 3 dimention functions, has inside all the quadratic convergent functions builded with infinite products of 4 dimention function. Am I ok?
2) The wave function used to calculate the probability of finding 1 particle in (x,t) position is delta(x,t) properly simetryzed and normalized. Am I ok?
3) If I want to know the probability amplitude to go from a 1 particle with 4-momentum k1 (t=0) to 2 particles with 4-momentum k2a and k2b (t=T) using Path Integral Formalism (I know that it is 0 because no interaction implies that number of particles does not change, but I want to understand this formalism) I should take the following steps
3a) express the initial state as a superposition of delta(x,0) properly simetryzed and normalized (or the correction you make of point 2)
3b) express the final state as a superposition of products of delta(x1,T) and delta(x2,T) properly simetryzed and normalized (or the correction you make of point 2)
3c) perform the K-G integration over all paths that starts at 3a and finish in 3b
Am I ok with point 3?
Is there something I am missing?
Ps: Point 1 and 2 are completely related to this thread. Point 3 a little less but has something to do

 

[bhobba]

I think you may be referring to a well known issue in QFT. Since it is a relativistic theory time and space should be treated on the same footing. But in standard QM time is a parameter and position is an observable - if they are to be treated on the same footing they should really be the same ie both parameters or both observables. QFT treats them both as parameters ie position now is a parameter giving the value of a quantum field at that point. The second solution is evidently possible and gives exactly the same predictions but is supposedly more difficult so the usual relativistic treatment is QFT.

I am only really just getting into QFT - the math is both dazzling - and hard. I read Zee's book ages ago but am now moving onto harder stuff.

Thanks
Bill

 

[vanhees71]

Time cannot be an operator in relativistic QT (as it cannot be in nonrelativistic QT for the same reason) since then it would be conjugate to the Hamiltonian and thus would fulfill the corresponding commutation relation. As you know from the position operator-momentum operator commutation relation, it follows then that both time and energy must have $\mathbb{R}$ as spectrum and this contradicts the conjectured boundedness of energy from below). This conjecture one must fulfill in order to have a stable ground state. If one wouldn't have this, there wouldn't be stable matter in the theory, and this for sure is against any observation (let alone the fact of our very existence ;-)).

On the other hand, there's no first principle forbidding position and momentum operators, and indeed one can construct from Noether's theorem, applied to the Poincare group, a momentum operator as the generator of translations. It's given by the corresponding integrals over field operators (to be taken with a grain of salt since here we deal with composite operators which must be given an appropriate meaning; within perturbation theory you can use normal ordering and appropriate renormalization order by order in radiative corrections).

For massive particles there's also no trouble to define position operators, obeying all commutation relations of the Heisenberg algebra. There's trouble with such a definition, however, for massless particles of spin $\geq 1$ as detailed in Arnold Neumaier's Theoretical Physics FAQ (see link given in my posting yesterday).

 

[juanrga]

From what little I have read about QFT, apparently the position of a particle is not a observable - it is more like an index for a collection of quantum harmonic oscillators. Thus there is no QFT equivalent to the position probability density in QM.

So, how does QFT predict or describe a particle's position? What is the QFT equivalent of position?

As always, thank you in advance.

Effectively, in QFT position is not an observable but a parameter without physical meaning.

Therefore QFT does not deal with such issues as what is the probability that particle is here or there. In fact, the application of QFT in high energy experiments as those in CERN are done in an energy-momentum basis, no positions here.

Also QFT does not deal with gravitation neither with bound states.

 

[Bill_K]

Position certainly is an observable in quantum field theory.

Scattering amplitudes can be calculated either in position space or momentum space.

Quantum field theory does indeed deal with bound states.

 

[juanrga]

Position certainly is an observable in quantum field theory.

Scattering amplitudes can be calculated either in position space or momentum space.

Quantum field theory does indeed deal with bound states.

Fortunately textbooks say the contrary.

 

[Avodyne]

I recommend section 12.11 ("The problem of localizing photons") in the textbook on Quantum Optics by Mandel and Wolf.

 

[Bill_K]

And I would recommend Weinberg's QFT, Vol I. Feynman diagrams in position space are discussed in Sect 6.1, and bound states take up all of Chapter 14. A good reference on the position operator is harder to find, since it's a rather old topic. Here's one.

"The problem of localizing photons"

This is a separate issue. Massless particles like photons can't be localized.

 

[juanrga]

And I would recommend Weinberg's QFT, Vol I. Feynman diagrams in position space are discussed in Sect 6.1, and bound states take up all of Chapter 14.

Textbooks (Mand & Shaw, Sakurai, Landau & Peirls...) emphasize that x and t are dummy variables, not related to physical space and time. E.g. Bacry writes:

Every physicist would easily convince himself that all quantum calculations are made in the energy-momentum space and that the Minkowski x^μ are just dummy variables without physical meaning (although almost all textbooks insist on the fact that these variables are not related with position, they use them to express locality of interactions!)

Textbooks emphasize that Feynman diagrams are not spacetime diagrams but 'graph' representations of several terms in the QFT interaction. If my memory does not fail even the wikipedia remarks that Feynman diagrams would not be confused with spacetime diagrams...

Weinberg in his chapter 2 explains that physical states are described in a four-momentum basis and uses the notation ψ_p,σ where σ denotes other variables as spin.

Weinberg also explains in chapter 2 that physical states are only valid for non-bound states. And in chapter 13 he writes:

It must be said that the theory of relativistic effects and radiative corrections in bound states is not yet in entirely satisfactory shape.

He discusses why is not possible to study the two-body bound state in QFT unless both particles are non-relativistic in whose case we can just use the Schrödinger equation of QM.

The chapter 14 titled "Bound states in External Fields" deals with trivial cases as that of electrons in the external field of a heavy particle as atomic nuclei. This simple problem is solved with the one-particle Dirac equation in an external potential (known in RQM), where radiative corrections from QFT can be considered.

This is not a genuine bound state because nuclei is treated classically (physical states θ_N in 14.1.22 are only for the fermions not for the nuclei) and only externally, stationary regime...; moreover the treatment in chapter 14 is open to many objections.

 

[bhobba]

Time cannot be an operator in relativistic QT (as it cannot be in nonrelativistic QT for the same reason) since then it would be conjugate to the Hamiltonian and thus would fulfill the corresponding commutation relation.

In posting what I did I am thinking what I read in Srednicki page 10 which says it can be done - but is difficult - in fact he states:

'it turns out that any relativistic quantum physics that can be treated in one formalism can be treated in the other. Which we use is a matter of convenience and taste. And Quantum Field Theory, the formalism in which both position and time are both labels on operators, is much more convenient for most problems'

Perhaps you can clarify what is going on?

Thanks
Bill

 

[juanrga]

... And Quantum Field Theory, the formalism in which both position and time are both labels on operators, is much more convenient for most problems

"Labels on operators" is not the same than time and space itself being operators... In QFT space and time are not operators. vanhees71 is right

 

[bhobba]

"Labels on operators" is not the same than time and space itself being operators... In QFT space and time are not operators. vanhees71 is right

Sorry for any confusion but the alternative the quote was referring to that was more difficult to work with is one in which time is an operator.

I managed to find a copy of Srednicki on the web:
http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

The relevant quote is from page 25:

'We can solve our problem, but we must put space and time on an equal footing at the outset. There are two ways to do this. One is to demote position from its status as an operator, and render it as an extra label, like time. The other is to promote time to an operator. Let us discuss the second option first. If time becomes an operator, what do we use as the time parameter in the Schrodinger equation? Happily, in relativistic theories, there is more than one notion of time. We can use the proper time τ of the particle (the time measured by a clock that moves with it) as the time parameter. The coordinate time T (the time measured by a stationary clock in an inertial frame) is then promoted to an operator. In the Heisenberg picture (where the state of the system is fixed, but the operators are functions of time that obey the classical equations of motion), we would have operators Xµ(τ), where X0 = T. Relativistic quantum mechanics can indeed be developed along these lines, but it is surprisingly complicated to do so. (The many times are the problem; any monotonic function of τ is just as good a candidate as τ itself for the proper time, and this infinite redundancy of descriptions must be understood and accounted for.)'

Am I missing something?

Thanks
Bill

 

[juanrga]

Sorry for any confusion but the alternative the quote was referring to that was more difficult to work with is one in which time is an operator.

I managed to find a copy of Srednicki on the web:
http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf

The relevant quote is from page 25:

'We can solve our problem, but we must put space and time on an equal footing at the outset. There are two ways to do this. One is to demote position from its status as an operator, and render it as an extra label, like time. The other is to promote time to an operator. Let us discuss the second option first. If time becomes an operator, what do we use as the time parameter in the Schrodinger equation? Happily, in relativistic theories, there is more than one notion of time. We can use the proper time τ of the particle (the time measured by a clock that moves with it) as the time parameter. The coordinate time T (the time measured by a stationary clock in an inertial frame) is then promoted to an operator. In the Heisenberg picture (where the state of the system is fixed, but the operators are functions of time that obey the classical equations of motion), we would have operators Xµ(τ), where X0 = T. Relativistic quantum mechanics can indeed be developed along these lines, but it is surprisingly complicated to do so. (The many times are the problem; any monotonic function of τ is just as good a candidate as τ itself for the proper time, and this infinite redundancy of descriptions must be understood and accounted for.)'

Am I missing something?

Thanks
Bill

In QM time is evolution parameter and well-known theorems explain why cannot be a operator. Moreover, in RQM position cannot be an observable (its operator is non-Hermitian). By this and other inconsistencies RQM was abandoned...

In QFT, the inconsistencies of RQM are partially solved (really ignored) by downgrading x to unobservable parameter, there is not x operator and no localization problems. Wavefunctions are reinterpreted as field operators and the whole theory is formulated in energy-momentum space.

In SHP theory x is maintained as an operator and time t is introduced as another operator, but this t is not the time of QM and this x is not the operator of RQM.

Lacking an adequate time, another concept of time tau is introduced as an evolution parameter. SHP theory has two times, tau plays the role of QM time (and is not an operator) and t is an operator associated to x^0.

SHP is not QFT, the Hamiltonian of SHP is not QFT, but a quadratic Hamiltonian and in general tau is not proper time as Srednicki says, because in general a Hamiltonian using time t as operator cannot be on-shell. The whole theory is very complex, redundant (multiple times), and full of inconsistencies.

 

[bhobba]

The whole theory is very complex, redundant (multiple times), and full of inconsistencies.

Without going through the math, or seeing a paper that does it for me I am in a bit of a pickle. But thinking about it in a superficial way I tend to side with what you are saying. It is however a bit of a concern when a standard textbook doesn't get it quite right -it leaves students like me in a rather awkward position.

Does anyone know a paper or other source that discusses this stuff?

I do however understand the issues of RQM and why you need QFT - tons of books explain the negative probabilities and other problems.

BTW thanks for going to the trouble of explaining what's going on.

Thanks
Bill

 

[jimgraber]

What is SHP theory, please?
Thanks

 

[kith]

I recommend section 12.11 ("The problem of localizing photons") in the textbook on Quantum Optics by Mandel and Wolf.

Maybe you can answer one of my long-term questions. ;) A laser beam has clearly some kind of localization. How is this accounted for in quantum optics? How can position not be an observable, when I can clearly "observe" the spatial extent of a laser beam or detect photons in a small detector volume?

I don't say it is wrong, I just find it very hard to get the notion of photons together with what I know from non-relativistic QM. As far as I can tell, the experimental methods for detecting electrons and photons are quite similar. So it is kind of odd that a simple wavefunction interpretation is possible in one case and is not in the other.

 

[juanrga]

What is SHP theory, please?
Thanks

Stuckelberg, Horwitz, & Piron theory.

 

[sheaf]

Bialnycki-Birula gives a good overview of the problems of defining position operators for (and single particle wavefunctions for) photons.

 

[kith]

Bialnycki-Birula gives a good overview of the problems of defining position operators for (and single particle wavefunctions for) photons.

Thanks sheaf! This article explains quite a few things, but I don't know yet if it will answer one of my big questions.

 

[atyy]

Maybe relevant is a derivation of the LSZ formula for scattering that uses wave packets, which are very approximately localized.

http://www.phys.psu.edu/~collins/563/LSZ.pdf : "I will show a simple construction of suitable coordinate-space wave function ..."

 

[the_pulp]

Ok, the arguments for not considering time and position as operators sound convincing, but in the 4th post of this thread someone wrote a link to a paper which describes what it seems to be QFT from a point of view where time and position are things measurables in labs (what it shows that could be measurable is the probability to find certain amount of particles in certain positions during certain time intervals). This paper also sounds very convincing but yet, opposite to the idea that time and position is not an operator.
So who is right, the one who is here saying that position plays no role (dont remember your name, sorry), that paper in the 4th post, both? What am I missing?

 

[strangerep]

So who is right, the one who is here saying that position plays no role (dont remember your name, sorry), that paper in the 4th post, both? What am I missing?

Do you mean Hendrik Van Hees' reference to the section on position on Arnold Neumaier's FAQ? Arnold mentions that, although a position operator is problematic for the photon, one can (apparently) define satisfactory POVMs (Positive Operator-Valued Measures) instead:

Arnold Neumaier[/URL]'s FAQ]
If the concept of an observable is not tied to that of a Hermitian operator but rather to that of a POVM (positive operator-valued measure), there is more flexibility, and covariant POVMs for position measurements can be meaningfully defined, even for photons. See, e.g.,

* A. Peres and D.R. Terno, Quantum Information and Relativity Theory,
Rev. Mod. Phys. 76 (2004), 93. [see, in particular, (52)]

* K. Kraus, Position observable of the photon, in:
The Uncertainty Principle and Foundations of Quantum Mechanics,
Eds. W. C. Price and S. S. Chissick,
John Wiley & Sons, New York, pp. 293-320, 1976.

* M. Toller, Localization of events in space-time,
Phys. Rev. A 59, 960 (1999).

* P. Busch, M. Grabowski, P. J. Lahti, Operational Quantum Physics,
Springer-Verlag, Berlin Heidelberg 1995, pp.92-94.

I haven't yet had time to chase down these references, but there seems to be good reason to consider POVMs to be more important in general quantum theory than one might think, compared to the usual Hermitian operators that one usually insists on for observables. In the finite-dim case, in turns out that every POVM arises from a set of generalized coherent states, which in turn arise from quite general considerations of dynamical groups.

If a covariant POVM for photon position is indeed satisfactory, then perhaps the whole puzzle about nonexistence of a photon position operator is moot.

(Kith: this might also be relevant to what you were asking.)

 

[the_pulp]

No, sorry, I was meaning this link:

http://xxx.lanl.gov/abs/0904.2287

This sounded very convincing in relation to measure the probability of find certain particles in certain positions during certain times.

So, after reading this paper I don't get why I should keep on saying that QFT and position measurements are not compatible. What am I missing?

Thanks for your help!

 

[Avodyne]

In posting what I did I am thinking what I read in Srednicki page 10 which says it can be done - but is difficult - in fact he states:

'it turns out that any relativistic quantum physics that can be treated in one formalism can be treated in the other. Which we use is a matter of convenience and taste. And Quantum Field Theory, the formalism in which both position and time are both labels on operators, is much more convenient for most problems'

Perhaps you can clarify what is going on?

The formalism in which both space and time are operators is explained in string-theory books, and is summarized here:

http://www.physics.thetangentbundle.net/wiki/String_theory/relativistic_point_particle/action

 

[martinbn]

The formalism in which both space and time are operators is explained in string-theory books, and is summarized here:

http://www.physics.thetangentbundle.net/wiki/String_theory/relativistic_point_particle/action

The link doesn't explain that, it only talks about the classical action, there isn't even quantization yet. And the question was how it is in QFT, no?

 

[Demystifier]

The link doesn't explain that, it only talks about the classical action, there isn't even quantization yet.

True, but see
http://xxx.lanl.gov/abs/hep-th/0702060 [Found.Phys.39:1109-1138,2009]
especially Sec. 5.2.

 

[juanrga]

The formalism in which both space and time are operators is explained in string-theory books, and is summarized here:

http://www.physics.thetangentbundle.net/wiki/String_theory/relativistic_point_particle/action

In the first place this link is not about anything discussed here.

In the second place string theory explains nothing, that is the reason which is named TON (Theory Of Nothing) these days...