# A Does QFT specify particle propagation?

1. Jan 28, 2016

### 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.

2. Jan 28, 2016

### 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?

3. Jan 28, 2016

### Staff: Mentor

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.

4. Jan 28, 2016

### FieldTheorist

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.

$\mathcal{H} = \{ | \Omega \rangle \} \oplus \{ | k \rangle \} \oplus \{ | k_1 \rangle , | k_2 \rangle \} \oplus \cdots$

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,

$\partial_t | \psi \rangle = i H | \psi \rangle$

(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.)

5. Jan 28, 2016

Staff Emeritus
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.

6. Jan 28, 2016

### friend

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.

7. Jan 28, 2016

### FieldTheorist

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.

8. Jan 28, 2016

### friend

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.

9. Jan 28, 2016

### FieldTheorist

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:

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

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):

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

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.

Last edited: Jan 28, 2016
10. Jan 29, 2016

### Staff: Mentor

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.

11. Jan 29, 2016

### friend

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).

12. Jan 29, 2016

### Staff: Mentor

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.

13. Jan 29, 2016

### friend

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.

14. Jan 29, 2016

### friend

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

15. Jan 29, 2016

### Staff: Mentor

That does not make sense.

16. Jan 29, 2016

### friend

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?

17. Jan 29, 2016

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

18. Jan 29, 2016

### friend

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.

19. Jan 29, 2016

### FieldTheorist

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.)

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.

20. Jan 29, 2016

### friend

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.