Graduate In QFT, what is the momentum of a created particle?

[Christian Thom]

TL;DR

In QFT, a particle is created by the creation operator which is depending on space only, so how can its momentum be defined ?

This seems important to me, since in some interactions, particles are produced by pairs of opposite momentum in the rest frame of the interaction.

 

[Demystifier]

Can you write down the explicit formula, or refer to a specific equation in a textbook, so that we can see what exactly do you mean that it depends on space only?

 

[Christian Thom]

Can you write down the explicit formula, or refer to a specific equation in a textbook, so that we can see what exactly do you mean that it depends on space only?

Well, you got me there ! It is something that I recall from the Leonard Susskind course on youtube, but I can't recall the precise "episode".
More precisely, it seems to me that the creation operators exist either with a momentum or a space variable but not both, but interactions being localized, it seems strange to me.

 

[Gaussian97]

An electron is described by a quantum state. This quantum state has all the information about the electron, depending on the QS you can have an electron with some definite momentum or an electron with some definite position (although these two cases are usually taken as "ideal" and not physically possible).
Essentially, given an arbitrary quantum state $\left|\psi\right>$ there exists an operator that creates an electron in such state.
Of course, there are relations between the states, so you can find a relation between the different creation operators (for example if you have a state $\psi$ that is a linear combination of states $\psi_1$ and $\psi_2$, then the creation operators will also be linearly dependent).

So for me, the question of whether the electron has a definite momentum or not has nothing to do with the creation operators, but with its state. And it is exactly equivalent to asking whether a non-relativistic wave function (the usual you have probably studied in quantum physics) has a definite momentum or not.

 

[vanhees71]

It's really hard to answer the question without a definite definition of the physical problem. Are you talking about relativistic or non-relativistic QFT?

In the latter case you have annihilation and creation operators for momentum eigenstates of (asymptotic) free fields, and the field operators in position space always must consist of superpositions of creation and annihilation operators (due to the "Feynman Stueckelberg trick" to get the correct interpretation for negative-frequency modes as representing anti-particles with positive energy, or, in more modern terms, to fulfill the microcausality principle for local observables).

 

[Christian Thom]

An electron is described by a quantum state. This quantum state has all the information about the electron, depending on the QS you can have an electron with some definite momentum or an electron with some definite position (although these two cases are usually taken as "ideal" and not physically possible).
Essentially, given an arbitrary quantum state $\left|\psi\right>$ there exists an operator that creates an electron in such state.
Of course, there are relations between the states, so you can find a relation between the different creation operators (for example if you have a state $\psi$ that is a linear combination of states $\psi_1$ and $\psi_2$, then the creation operators will also be linearly dependent).

So for me, the question of whether the electron has a definite momentum or not has nothing to do with the creation operators, but with its state. And it is exactly equivalent to asking whether a non-relativistic wave function (the usual you have probably studied in quantum physics) has a definite momentum or not.

Thank you for your answer that makes a lot of sense. So my problem may lie more in the fact that during interactions, it seems that the position and the momentum of particles must be known, if only for the purpose of conserving the usual quantities...

 

[vanhees71]

During interactions there is no particle interpretation for relativistic QFT. Only (asymptotic) free Fock states have a (many-)particle interpretation.

 

[Christian Thom]

During interactions there is no particle interpretation for relativistic QFT. Only (asymptotic) free Fock states have a (many-)particle interpretation.

Is this still an active research subject ?

 

[vanhees71]

Sure, high-energy particle experiments are up and running. LHC has just started the next round of even more accurate experiments after a big upgrade of its detectors.

 

[Christian Thom]

I meant more the topic of studying the mechanisms at play during the interactions themselves than the prediction of the results.

 

[vanhees71]

That cannot be resolved. What is measured in particle-collision experiments are the probabilities for certain "scattering events". E.g., in elastic electron-electron scattering you take two asymptotic free electrons with some momenta (and maybe polarizations) and look at all cases where after the scattering you still have two electrons and investigate the probability that they have scattered to two other momenta (and maybe polarization states). The momenta are always in agreement with energy-momentum conservation and the energy-momentum relations $E^2=c^2 p^2+m^2 c^2$ of each asymptotic free electron. The probabilities are usually expressed in terms of scattering cross sections.

 

[mattt]

I meant more the topic of studying the mechanisms at play during the interactions themselves than the prediction of the results.

Neumaier's answer in this link is more detailed than what I was about to write, so read it there: https://physics.stackexchange.com/q...m-electrodynamics-at-finite-times/21687#21687

 

[Christian Thom]

Thank you to all of you. I see that it is a problem that there seem to be for the moment no tools to tackle.