Question about Propagators in QFT

[masteralien]

TL;DR Summary

In QFT the propagator is an important object. I want to know if it really describes the propagation of a particle the same way as in Quantum Mechanics

In Quantum Field Theory the Propagator is computed as the time ordered expectation value of products of fields.

$$\langle 0|\hat{T}\{\phi(x)\phi(y))\}|0\rangle$$

I want to know does it describe the amplitude for a particle to propagate from one point to another like in QM where the wavefunction can be computed with the Propagator through this formula
$$\Psi({\bf{r},t})=\int{G({\bf{r}},{\bf{r’}}},t)\Psi_0 ({\bf{r’}})d^3{\bf{r’}}$$
I want to know if this same method applies in QFT this time the “wavefunction” is this matrix element
$$\langle 0|\phi(x)|1\rangle$$
Where
$$|1\rangle = \int{\psi(k) |k\rangle}d^3k$$

Presumably one would write the evolution of this matrix element in this form

$$\Psi(x)=\int{G(x-y)\Psi_0(y)}d^3y$$

Does it make sense to think of the Propagator in the same way as regular QM as evolving an initial state in time.

 

[Adrian59]

In quantum field theory (QFT) a propagator is as you correctly gave defined as <0| T{φ(x) φ(y)} |0>. Using Feynman's derivation this is computed to ∫ d^4p i exp(-ip.(x-y).
(p^2 - m^2 + iε)

The reason I showed this expression is that I think this explains the idea of of propagator well. The internal lines of a Feynman diagram are indicative of virtual particles and not real particles like in non-QFT quantum mechanics. Of course the rather archaic terminology is that when p^2 = m^2 the situation is called 'on shell'.

Of course, exactly 'on shell' means we have a divergence as the integral goes to infinity, and this is called an infrared (IR) divergence as opposed to an ultraviolet divergence when the momentum tends to infinity.

However, one can see that as p^2 approaches m^2 the integral gets larger which to me represents the increasing likelihood of a virtual particle being formed.

So the answer is that this represents an amplitude for a virtual particle.

 

[Adrian59]

Apologies the posting has destroyed my carefully arranged response.

The equation should read ∫ d^4p i /(p^2 - m^2 + iε) exp(-ip.(x-y)

NB the bar notation represents division by 2π.

 

[apostolosdt]

TL;DR Summary: In QFT the propagator is an important object. I want to know if it really describes the propagation of a particle the same way as in Quantum Mechanics

In Quantum Field Theory the Propagator is computed as the time ordered expectation value of products of fields.I want to know does it describe the amplitude for a particle to propagate from one point to another like in QM where the wavefunction can be computed with the Propagator through this formula
[…]

Does it make sense to think of the Propagator in the same way as regular QM as evolving an initial state in time.

Your first expression is known as correlation function or two point Green’s function; as such, it is related to your second expression. Calling it “propagator” is the physical interpretation we assign to it. Using that term in nonrelativistic quantum mechanics requires one’s attention. What propagates outside the QFT realm? The particle associated with the wavefunction?

Moving from nonrelativistic QM to QFT is not without extra physical concepts.

 

[masteralien]

Your first expression is known as correlation function or two point Green’s function; as such, it is related to your second expression. Calling it “propagator” is the physical interpretation we assign to it. Using that term in nonrelativistic quantum mechanics requires one’s attention. What propagates outside the QFT realm? The particle associated with the wavefunction?

Moving from nonrelativistic QM to QFT is not without extra physical concepts.

So is the formula for how the particle’s wavefunction evolves according to the propagator correct in the context of Relativistic QFT like does a Propagator have the same interpretation.

 

[vanhees71]

In relativistic QFT first of all there are no more wave functions. The 1st-quantization formalism is inadequate for relativistic theories, because it does not take into account the possibility for creation and annihilation of particles and you cannot formulate a causal theory with it. What's necessary for that is the use of QFT and the introduction of antiparticles. For a very nice introduction, see Coleman's famous lectures on relativistic QFT:

https://arxiv.org/abs/1110.5013

or more complete as a textbook:

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

 

[apostolosdt]

So is the formula for how the particle’s wavefunction evolves according to the propagator correct in the context of Relativistic QFT like does a Propagator have the same interpretation.

In addition to what vanhees71 just said, perhaps it is also important to emphasize that, in nonrelativistic QM, the nature of the potential function in, say, Schroedinger's equation, is used without any restrictions. In QFT, Lorentz invariance imposes very stringent restrictions upon interacting terms in the Hamiltonian.

 

[masteralien]

In relativistic QFT first of all there are no more wave functions. The 1st-quantization formalism is inadequate for relativistic theories, because it does not take into account the possibility for creation and annihilation of particles and you cannot formulate a causal theory with it. What's necessary for that is the use of QFT and the introduction of antiparticles. For a very nice introduction, see Coleman's famous lectures on relativistic QFT:

https://arxiv.org/abs/1110.5013

or more complete as a textbook:

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

But isnt the Propagator describing the amplitude for a particle to travel from one point to another in spacetime so is the interpretation different than in QM. Do the Propagators not actually describe Particle Propagation

 

[masteralien]

In relativistic QFT first of all there are no more wave functions. The 1st-quantization formalism is inadequate for relativistic theories, because it does not take into account the possibility for creation and annihilation of particles and you cannot formulate a causal theory with it. What's necessary for that is the use of QFT and the introduction of antiparticles. For a very nice introduction, see Coleman's famous lectures on relativistic QFT:

https://arxiv.org/abs/1110.5013

or more complete as a textbook:

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

by “Wavefunction” I didnt mean standard QM wavefunction rather amplitudes of this form and how they evolve

$$\int{\langle{0}|\hat{\phi}(x)|y\rangle}\psi(y)d^3y=\int\langle{0}|\hat{\phi}(x)\hat{\phi}(y)|0\rangle \psi(y)d^3y$$

$$=\int{G(x-y)\psi(y)d^3y}$$

 

[renormalize]

But isnt the Propagator describing the amplitude for a particle to travel from one point to another in spacetime so is the interpretation different than in QM. Do the Propagators not actually describe Particle Propagation

Yes, in ordinary QM the propagator is the amplitude you describe and it's used to evolve the ordinary wave function $\psi$. See, e.g., https://en.wikipedia.org/wiki/Propagator:

 

[masteralien]

Yes, in ordinary QM the propagator is the amplitude you describe and it's used to evolve the ordinary wave function $\psi$. See, e.g., https://en.wikipedia.org/wiki/Propagator:
View attachment 337842

I know the interpretation in normal QM does this interpretation still hold in QFT thats my question. Do Relativistic Propagators really describe particle propagation

 

[vanhees71]

In "vacuum QFT" what you calculate are S-matrix elements, which describe transition amplitudes from a given asymptotic in-state (usually two particles in a collision experiment) to an asymptotic out-state (usually many particles). This is done in perturbation theory, and what you calculate there are the $N$-point functions, i.e., time-ordered autocorrelation functions of fields. The $2$-point function is the propagator.

 

[masteralien]

In "vacuum QFT" what you calculate are S-matrix elements, which describe transition amplitudes from a given asymptotic in-state (usually two particles in a collision experiment) to an asymptotic out-state (usually many particles). This is done in perturbation theory, and what you calculate there are the $N$-point functions, i.e., time-ordered autocorrelation functions of fields. The $2$-point function is the propagator.

I am aware that you use Propagators and n point functions to get S Matrix elements my question was does the QFT propagator 2 point function have the same interpretation as in QM as a particle propagating in spacetime. Are the Propagators in QFT really describing spacetime propagation so one can write the time evolution of the one particle amplitude of this form .
$$\Psi(x)=\bra{0}\hat{\phi}(x)\ket{1}$$

 

[jzz]

I am aware that you use Propagators and n point functions to get S Matrix elements my question was does the QFT propagator 2 point function have the same interpretation as in QM as a particle propagating in spacetime. Are the Propagators in QFT really describing spacetime propagation so one can write the time evolution of the one particle amplitude of this form .
$$\Psi(x)=\bra{0}\hat{\phi}(x)\ket{1}$$

In quantum mechanics, the propagator typically refers to the probability amplitude for a particle to propagate from one position to another over a certain period of time. This concept is fundamental in understanding the time evolution of quantum systems described by Schrödinger's equation. In quantum field theory (QFT), however, the propagator can have a broader interpretation. It can represent the amplitude for the propagation of a particle or a quantum excitation in the underlying field. Can this interpretation be taken to mean that in QFT propagators can be explained in terms of creation and annihilation operators?