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

1. Jan 26, 2012

### referframe

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.

2. Jan 26, 2012

### martinbn

There are no particles in QFT.

3. Jan 26, 2012

### Bill_K

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 [Xi, Xj] = 0, [Xi, pj] = iħ δij and i[H, X] = p/E.

4. Jan 26, 2012

### vanhees71

5. Jan 26, 2012

### martinbn

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?!

6. Jan 26, 2012

### questionpost

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

7. Jan 27, 2012

### Demystifier

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".

8. Jan 27, 2012

### 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 Im 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 Im missing?
Ps: Point 1 and 2 are completely related to this thread. Point 3 a little less but has something to do

9. Jan 27, 2012

### 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

10. Jan 27, 2012

### 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).

11. Jan 27, 2012

### juanrga

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.

12. Jan 27, 2012

### 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.

13. Jan 27, 2012

### juanrga

Fortunately textbooks say the contrary.

14. Jan 27, 2012

### Avodyne

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

15. Jan 27, 2012

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

16. Jan 28, 2012

### juanrga

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:

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 ψp,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:

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 θN 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.

Last edited: Jan 28, 2012
17. Jan 28, 2012

### bhobba

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

18. Jan 28, 2012

### juanrga

"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

19. Jan 28, 2012

### bhobba

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

20. Jan 28, 2012

### juanrga

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.

Last edited: Jan 28, 2012