Does QFT specify particle propagation?

[friend]

I need to clear up my (mis)understanding about QFT.

Does QFT show how a particle propagates through spacetime? (Or maybe this is the realm of QM) Or does QFT only specify how a particle propagates as a particle through time without reference to where in space it is?

But... if QFT specifies how a particle propagated through spacetime, then does it account for the kinetic energy in terms of these virtual particle processes of QFT? Thanks.

 

[friend]

As I recall, QFT was only about how particles pop into existence out of nothing or interaction. There one was only concerned about particle number in Fock space. And this made no reference to exactly where those particles were. Or am I mistaken here? There were Feynman diagrams that plotted space on one axis and time on the other. But this is only a schematic to show that particles appeared sooner or later in spacetime. I don't think it is meant to diagram a particles trajectory through spacetime. Is this right?

 

[mfb]

QFT works with spacetime positions.
Often an integral over spacetime is performed for intermediate events, but apart from that: sure, you can use QFT to describe how a particle moves from A to B.

 

[FieldTheorist]

Does QFT show how a particle propagates through spacetime? (Or maybe this is the realm of QM) Or does QFT only specify how a particle propagates as a particle through time without reference to where in space it is?

Well, I guess my first question is, what do you mean by "propagate"? Do you mean the time evolution of a wave? If so, then yes, QFT does this. I'll give a very schematic overview of that concept here for a scalar quantum field.

In QFT, particles are the (wave-like) fluctuations of the (approximately free) quantum field. Quantum fields have a Hilbert space that's the infinite direct product of the vacuum (0-particle state, ## | \Omega \rangle ## ), 1-particle state with definite momenta k (e.g. ## | k \rangle ## ), the 2-particle state with definite momenta {k₁, k₂}, etc.

[itex] \mathcal{H} = \{ | \Omega \rangle \} \oplus \{ | k \rangle \} \oplus \{ | k_1 \rangle , | k_2 \rangle \} \oplus \cdots [/itex]

So the fluctuations in an (approximately free) quantum field are characterized by these particle states. So we can start a system out with an N particle state (or superpositions of particle states), and evolve it forward in time (If there's no interactions or source, you'll just evolve from an N-particle state to the same N-particle state with 100% probability). You can then evolve the prepared state forward with the dynamics of the Heisenberg equation,

[itex] \partial_t | \psi \rangle = i H | \psi \rangle [/itex](There's a related quantity called the "Feynman propagator," which is used in constructing S-matrix elements via Feynman scattering diagrams, but that's in the incoming-outgoing scattering picture. You can also setup a QFT in the Schwinger-Keldysh formalism, which allows you to evolve a specific quantum state forward in time just like QM, e.g. with a retarded propagator.)

 

[Vanadium 50]

As I recall, QFT was only about how particles pop into existence out of nothing or interaction.

This is a misconception, you keep spreading it, and I really really wish you would stop.

A consequence of this is that this is not an A-level thread.

 

[friend]

This is a misconception, you keep spreading it, and I really really wish you would stop.

A consequence of this is that this is not an A-level thread.

Virtual particles are not germane to the issue. What I'm trying to get at is whether QFT addresses the kinetic energy of a traveling particle. Some here say that QFT does tell us how a particle propagates from A to B. But I don't know how that QFT description would change with the velocity from A to B. I want to think in terms of Feyman diagrams for that propagation and how that description would change. For example, would there be more interaction with the surronding quantum fluctuation for fast moving particle than for slow particles? Thanks.

 

[FieldTheorist]

Virtual particles are not germane to the issue. What I'm trying to get at is whether QFT addresses the kinetic energy of a traveling particle. Some here say that QFT does tell us how a particle propagates from A to B. But I don't know how that QFT description would change with the velocity from A to B. I want to think in terms of Feyman diagrams for that propagation and how that description would change. For example, would there be more interaction with the surronding quantum fluctuation for fast moving particle than for slow particles? Thanks.

Velocity is a meaningless concept for a QFT particle if you're also trying to localize the fluctuation to points A and B. It's a wave, so it has a definite momentum or a definite position (assuming it's free), or a mixture of the two. But you cannot talk about moving from point A to point B with a definite momentum. The only time you can make meaningful statements about such things is if you're discussing classical point particles.

 

[friend]

Velocity is a meaningless concept for a QFT particle if you're also trying to localize the fluctuation to points A and B. It's a wave, so it has a definite momentum or a definite position (assuming it's free), or a mixture of the two. But you cannot talk about moving from point A to point B with a definite momentum. The only time you can make meaningful statements about such things is if you're discussing classical point particles.

Yes, of course. I'm sure I didn't mean to imply velocity with an exact momentum. But I did mean to imply a momentum with enough accuracy to talk about the particle's kinetic energy (not necessarily exact energy). Bottom line, is kinetic energy of particle (wave) motion at all considered by QFT, or is that a QM thing? Thanks.

 

[FieldTheorist]

Yes, of course. I'm sure I didn't mean to imply velocity with an exact momentum. But I did mean to imply a momentum with enough accuracy to talk about the particle's kinetic energy (not necessarily exact energy). Bottom line, is kinetic energy of particle (wave) motion at all considered by QFT, or is that a QM thing? Thanks.

Yes, more or less. (The details can be found in section 3.1.2 of Itzykson and Zuber, for example.) A single quantum fluctuation does have a definite energy (in a given frame), but it's not localized (just like QM, although in QFT it's much worse with interactions). However, obviously, you can prepare a mixture of states that are localized (e.g. Gaussian wavepacket), so:

[itex] | \psi \rangle = N \int d^3 k e^{ -L^2 k^2/2} | k \rangle \, , [/itex]

where N is just a normalization factor. Then, if you looked at the field value of this state, it would be localized as a wavepacket. So for the field operator, ## \hat \varphi(x)##, you'll obviously discover that your initial state is localized about the origin with a smeared scale L (up to factors of root two pi that I'm too lazy to calculate):

[itex] \langle \hat \psi | \varphi(x) | \psi \rangle \sim e^{-x^2/L^2} \, .[/itex]

You may then evolve this state forward in time by evolving the field operator, ## \hat{\varphi}(x,t) = e^{i \hat{H}t} \hat{\varphi}(x,0) e^{-i\hat{H}t} ##. And it will slowly smear out more like it does in QM. If you want kinematics, you'll need to do a Lorentz boost, which is certainly possible but somewhat obnoxious, but the rough change will be that the Gaussian distribution will look like ## \sim e^{(x - vt)^2/L^2} ##, and thus appear to be moving WRT your frame.

 

[mfb]

I want to think in terms of Feyman diagrams

Feynman diagrams are a tool to visualize calculations in perturbation theory. The diagrams are not the actual calculations, and the calculations in perturbation theory are not the actual quantum field theory.

 

[friend]

Feynman diagrams are a tool to visualize calculations in perturbation theory. The diagrams are not the actual calculations, and the calculations in perturbation theory are not the actual quantum field theory.

Yes, I'm aware of that. We are looking for a non-perturbative QFT. But it helps us think in terms of Feynman diagrams as a visualization tool. I'm sure the Feynman diagrams in the perturbation expansion for a traveling particle will be different from those of a particle that is moving faster. What I'm trying to get at is how those diagrams would change (to help me visualize a particle's kinetic energy in terms of those diagrams, if possible).

 

[mfb]

They would not change, because the 4-vectors of the particles are not part of the diagrams. They are part of the calculations, of course.

 

[friend]

They would not change, because the 4-vectors of the particles are not part of the diagrams. They are part of the calculations, of course.

Well, perhaps the diagrams themselves would not change with different velocities. But perhaps something else changes with speed to account for kinetic energy of the particle, such as frequency of occurrence.

 

[friend]

You may then evolve this state forward in time by evolving the field operator, ## \hat{\varphi}(x,t) = e^{i \hat{H}t} \hat{\varphi}(x,0) e^{-i\hat{H}t} ##. And it will slowly smear out more like it does in QM. If you want kinematics, you'll need to do a Lorentz boost, which is certainly possible but somewhat obnoxious, but the rough change will be that the Gaussian distribution will look like ## \sim e^{(x - vt)^2/L^2} ##, and thus appear to be moving WRT your frame.

Are you saying that the Hamiltonian for the field, ## \hat{H}, ## has a kinetic term of a particle?

 

[mfb]

Well, perhaps the diagrams themselves would not change with different velocities. But perhaps something else changes with speed to account for kinetic energy of the particle, such as frequency of occurrence.

That does not make sense.

 

[friend]

That does not make sense.

Do people draw a light cone on a Feynman diagram? That would at least be more specific about a particles velocity. If not, then that would seem to indicate that QFT does not take into account a particle's velocity.

I thought I read somewhere, or in a lecture I watched, that Feynman diagrams (and thus QFT itself) were only in and out states in the rest frame of the interaction. Does this sound right?

 

[Vanadium 50]

Do people draw a light cone on a Feynman diagram?

Why would anybody want to do that? Why is that better than drawing anything else on a Feynman diagram?

 

[friend]

Why would anybody want to do that? Why is that better than drawing anything else on a Feynman diagram?

What do you mean, Why? Feynman diagrams have space on one axis and time on the other. And so does a plot for a light cone. If QFT take into account a particle's velocity, then it seems natural to plot one WRT another. How would you avoid it in that case. But since it seem like nonsense to draw a light cone on a Feynman diagram, then the two must be unrelated and QFT does not take into account velocity.

 

[FieldTheorist]

Are you saying that the Hamiltonian for the field, ## \hat{H}, ## has a kinetic term of a particle?

The Hamiltonian evolves the state of a system --this is true classically as well as quantum mechanically. You can read up on the Louiville's theorem and Hamiltonian flows (time evolution of a system from IC to a later state) fro the classical case.

The quantum case for unitary time evolution of quantum operators is part of Heiseberg's equation and the Heisenberg picture, where operators have time dependence instead of the states. (But it's a choice, you can choose either.)

Yes, I'm aware of that. We are looking for a non-perturbative QFT. But it helps us think in terms of Feynman diagrams as a visualization tool. I'm sure the Feynman diagrams in the perturbation expansion for a traveling particle will be different from those of a particle that is moving faster. What I'm trying to get at is how those diagrams would change (to help me visualize a particle's kinetic energy in terms of those diagrams, if possible).

Let me state this cleanly: Feynman diagrams are not what you think that they are. What they actually represent is very mathematical and very subtle. They don't even represent a single transition probability amplitude from N particles to M particles. There's no notion of local time, you're exclusively looking at things in the scattering picture, so an initial state (infinitely far in the past) of incoming particles and, an infinite time later, what are the outgoing particles? That's the question the Feynman diagrams (and the S-matrix) answers. Do not try to extract any classical intuitions from Feynman diagrams unless you really actually understand QFT. It's just not a good place to start.

 

[friend]

Let me state this cleanly: Feynman diagrams are not what you think that they are. What they actually represent is very mathematical and very subtle. They don't even represent a single transition probability amplitude from N particles to M particles. There's no notion of local time, you're exclusively looking at things in the scattering picture, so an initial state (infinitely far in the past) of incoming particles and, an infinite time later, what are the outgoing particles? That's the question the Feynman diagrams (and the S-matrix) answers. Do not try to extract any classical intuitions from Feynman diagrams unless you really actually understand QFT. It's just not a good place to start.

Thank you for that, I appreciate it. So am I to understand that I remembered correctly when I recalled:

I thought I read somewhere, or in a lecture I watched, that Feynman diagrams (and thus QFT itself) were only in and out states in the rest frame of the interaction. Does this sound right?

Thanks.

 

[mfb]

Do people draw a light cone on a Feynman diagram?

No, and it would not make sense to do so.

If not, then that would seem to indicate that QFT does not take into account a particle's velocity.

That is not true. As I said, the diagrams are not the calculations. As long as you keep that misconception, this thread leads nowhere.

Feynman diagrams have space on one axis and time on the other.

Feynman diagrams do not have a space axis. They have a time axis, but even that is not a strict time ordering of events, its only purpose is to distinguish between incoming and outgoing particles.

 

[friend]

No, and it would not make sense to do so.That is not true. As I said, the diagrams are not the calculations. As long as you keep that misconception, this thread leads nowhere.Feynman diagrams do not have a space axis. They have a time axis, but even that is not a strict time ordering of events, its only purpose is to distinguish between incoming and outgoing particles.

My understanding (if you don't mind) is that QM involves position and conjugate momentum operators which explicitly address a particle trajectory. But QFT involves a field operator's amplitude and conjugate momentum of that amplitude at any given point. The field momentum addresses how the amplitude at a point changes, and not how the field is propagating through space. I don't see how that can address a particles trajectory and thus its kinetic energy. Does kinetic energy only enter QFT as a property of an incoming particle (put in as IC) which allows that particle to decay into other particles? For example, an incoming photon having enough energy to become a positron and electron. And so QFT does not address where in space or time a particle will decay, etc. only that it will decay. Is this right? We only put those vertices at different places on the Feynman graph to distinguish them from other events, and not to indicate any kind of propagation through actual spacetime. Right?

 

[Vanadium 50]

What do you mean, Why? Feynman diagrams have space on one axis and time on the other

No, they don't. They really don't. I don't expect you to believe me because you history shows that you never believe anyone (which makes me wonder why you bother to ask questions) but it is nonetheless the truth. Feynman diagrams are an alternative set of symbols - a mnemonic, if you will - for describing calculations, not cartoons on how the particles move in space and time. The only reason that they have an orientation at all in space and time is to specify which particles are incoming (and averaged over) and which are outgoing (and summed over).

Drawing a light cone on them is exactly like drawing a light cone on a "7" in a calculation. You can do it, but it won't do any good,

As I said, the diagrams are not the calculations. As long as you keep that misconception, this thread leads nowhere.

Exactly.

 

[friend]

No, they don't. They really don't. I don't expect you to believe me because you history shows that you never believe anyone (which makes me wonder why you bother to ask questions) but it is nonetheless the truth.

Don't be so negative. The history of this thread alone shows that I'm learning something.

Feynman diagrams are an alternative set of symbols - a mnemonic, if you will - for describing calculations, not cartoons on how the particles move in space and time. The only reason that they have an orientation at all in space and time is to specify which particles are incoming (and averaged over) and which are outgoing (and summed over).

As I show in post #22, that's what I'm understanding with your help. Thank you.

The reason I keep nagging is that I'm wondering if the kinetic energy is accounted for by the way that a particle interacts with the vacuum fluctuations as it travels through it. For it seems that would be the only difference for various velocities. But now I'm starting to think, no, that kinetic energy is not accounted for by the way particles propagate through the vacuum fluctuations. For if it were, then we would have energy defined in terms of spacetime and QFT. But as I understand it energy is a prior given in the development of QM and QFT. So energy can not be derive from QFT. But then again, we found mass as a being derived from QFT, the Higgs mechanism. So I remain open to the idea.

 

[FieldTheorist]

Thank you for that, I appreciate it. So am I to understand that I remembered correctly when I recalled:

I thought I read somewhere, or in a lecture I watched, that Feynman diagrams (and thus QFT itself) were only in and out states in the rest frame of the interaction. Does this sound right?

Thanks.

1.) I'm not sure what that means. There is a center of momentum frame for the incoming states (and by conservation, also the outgoing states), but it's not evolving in time. It's just the total momentum of the diagram.

2.) QFT is way more than Feynman diagrams. For instance, the calculation of a propagating particle I did above included no Feynman diagrams. There's path integrals, which go beyond perturbation theory (e.g. they include instanton corrections, which cannot be seen at any order in a perturbation theory/Feynman diagrams; this is relevant for QCD), and there's quantum field theories that have no Lagrangian (and thus no path integral definition). These are typically conformal field theories.

Another way of saying this is that Feynman diagrams are a useful tool in several cases, but it is not the central idea in QFT, it's not the defining feature of QFT, etc. It's just a tool that sometimes is helpful. That's more or less like saying Newtonian physics is just conservation of momentum. That's not true because other theories have conservation of momentum, and it hardly is the defining feature of Newtonian mechanics.

 

[friend]

1.) I'm not sure what that means. There is a center of momentum frame for the incoming states (and by conservation, also the outgoing states), but it's not evolving in time. It's just the total momentum of the diagram...

Perhaps vacuum fluctuations look the same at any constant speed. I know acceleration does funny things to the vacuum fluctuations. But if every inertial observer sees every other inertial observer's vacuum fluctuations to be the same as his, then I have my answer. Just as a fast moving yard stick appears shorter to those not moving, must the vacuum fluctuations of a fast mover be different than for a still mover? If not, then no, QFT cannot explain kinetic energy; that's just put in by hand as the energy of the input states.

A fast moving yard stick appears shorter to those not moving. So must the vacuum fluctuations seen by a fast moving observe be different than the vacuum fluctuations seen by a motionless observer? Would we say that the vacuum fluctuations seen by a fast moving observer are squeezed WRT us? I take it that there is so many quantum fluctuation per meter for any observer, and the fast moving meter is squeezed. So wouldn't we say that the fast moving observer must be experiencing a squeezed version of the vacuum fluctuations. Does this make any sense?

 

[mfb]

You can construct amplitude distributions in QFT that correspond to a particle with a given kinetic energy (within some uncertainty). This is done to represent incoming particles in collisions, for example. Those particles move through space in every way they do so in other theories.

For example, an incoming photon having enough energy to become a positron and electron.

A photon can never do that.

So energy can not be derive from QFT.

Can 7 derive from an apple?

Perhaps vacuum fluctuations look the same at any constant speed.

Constant speed relative to what? There are no absolute velocities.

But if the vacuum fluctuations always appear the same to all observers at any velocity, then I have my answer - no QFT cannot explain kinetic energy; that's just put in by hand as the energy of the input states.

What does "explain" even mean? Which theory explains 7? QFT states can have a property that we call "kinetic energy" because this quantity has some useful applications.

 

[friend]

I think I'm getting closer to the right language. A photon on average will encounter so many vacuum fluctuations per meter. The meter is defined by the propagation of light in a vacuum. A meter stick traveling close to the speed of light will appear shorter because ALL processes are squeezed in the direction of motion. Does this mean we still observers see their vacuum fluctuations squeezed as well in the direction of their motion? Since all processes are defined by propagation through the vacuum, And their meter sticks appear shorter, does that mean that their vacuum appears shorter as well?

 

[Vanadium 50]

Can 7 derive from an apple?

Colorless green ideas sleep furiously.

 

[mfb]

I think I'm getting closer to the right language. A photon on average will encounter so many vacuum fluctuations per meter.

It does not. You cannot count vacuum fluctuations.

Does this mean we still observers see their vacuum fluctuations squeezed as well in the direction of their motion?

That would violate special relativity. No, even if "squeezed vacuum fluctuations" would make sense (it does not).

 

[friend]

It does not. You cannot count vacuum fluctuations.

Just a moment. Isn't that what the cosmological constant is all about - the amount of quantum fluctuations in a certain volume of space?

If so, then consider that an empty box in a boosted frame is shrunk in the direction of motion. So what are we to think of the space inside the box. Does the box shrink but the space inside it does not? Then if the space inside the boosted box feels the same cosmological constant in a smaller space, then what can that mean except that a still observer would see the same amount of vacuum fluctuations inside a smaller volume?

 

[mfb]

Isn't that what the cosmological constant is all about - the amount of quantum fluctuations in a certain volume of space?

No.

If so, then consider that an empty box in a boosted frame is shrunk in the direction of motion. So what are we to think of the space inside the box. Does the box shrink but the space inside it does not? Then if the space inside the boosted box feels the same cosmological constant in a smaller space, then what can that mean except that a still observer would see the same amount of vacuum fluctuations inside a smaller volume?

Those questions do not make sense.

 

[friend]

No. Those questions do not make sense.

Anything and everything inside the empty boosted box would appear squeezed in the direction of motion, including the subatomic particles, and I assume any vacuum fluctuations that affect those squeezed subatomic particles. I recall an animation of particles colliding in the LHC. It showed the (what, point particles?) flattened out like two pancakes as they approached each other. If quantum fluctuations had any effect at all on these moving particles, then shouldn't we assume they'd be flattened out as well?

 

[mfb]

A vacuum in a box does not have a velocity. It is a vacuum.

If quantum fluctuations had any effect at all on these moving particles, then shouldn't we assume they'd be flattened out as well?

No, that does not make sense. How often do I have to repeat that?

 

[FieldTheorist]

Perhaps vacuum fluctuations look the same at any constant speed. I know acceleration does funny things to the vacuum fluctuations. But if every inertial observer sees every other inertial observer's vacuum fluctuations to be the same as his, then I have my answer. Just as a fast moving yard stick appears shorter to those not moving, must the vacuum fluctuations of a fast mover be different than for a still mover? If not, then no, QFT cannot explain kinetic energy; that's just put in by hand as the energy of the input states.

I would honestly suggest that you stick to the conceptual objects of "quantum particle" and "quantum field," before you traipse off into discussions using (very antiquated) terminology that you almost certainly don't understand. Vacuum fluctuations (like "holes" in the "Dirac sea") are a way of giving a picture to certain subcalculations of an observable, but the picture is pretty meaningless and just words to dress around a calculation.

If you want to take QFT seriously, you need to start by treating the quantum field as fundamental, learn the language you need to describe it, see how the Fock space structure of free QFTs leads to an infinite tower of states of N identical particles, and discover what the fundamental, physical observables of the theory are.

A fast moving yard stick appears shorter to those not moving. So must the vacuum fluctuations seen by a fast moving observe be different than the vacuum fluctuations seen by a motionless observer? Would we say that the vacuum fluctuations seen by a fast moving observer are squeezed WRT us? I take it that there is so many quantum fluctuation per meter for any observer, and the fast moving meter is squeezed. So wouldn't we say that the fast moving observer must be experiencing a squeezed version of the vacuum fluctuations. Does this make any sense?

Let's take a step back. The vacuum is the state of zero particles. (If you accelerate, you'll see a bath of particles, but it's a coordinate artifact. Let's assume we're in a Lorentz frame.) When I say "fluctuations in the quantum field" what I mean, to be clear, is exciting a particle state. If I turn on a source, that will excite the vacuum, and generate particles. Meaning you will no longer be in the vacuum state.

I think I'm getting closer to the right language. A photon on average will encounter so many vacuum fluctuations per meter. The meter is defined by the propagation of light in a vacuum. A meter stick traveling close to the speed of light will appear shorter because ALL processes are squeezed in the direction of motion. Does this mean we still observers see their vacuum fluctuations squeezed as well in the direction of their motion? Since all processes are defined by propagation through the vacuum, And their meter sticks appear shorter, does that mean that their vacuum appears shorter as well?

No, I don't see the statement here. You seem like you need to spend more time reading about QFT before you can ask full, coherent questions on QFT. I recommend reading Tong's intro to QFT, it's free and you can find it with a google search. If you make it through the first two chapters, I think you'll be able to answer many of your own conceptual confusions. Zee's QFT in a Nut Shell might also be a useful read.