A Particle production in 1+1D QFT

[QFT1995]

I am currently studying the Massive Thirring Model (MTM) with the Lagrangian
$$
\mathcal{L} = \imath {\bar{\Psi}} (\gamma^\mu {\partial}_\mu - m_0 )\Psi - \frac{1}{2}g: \left( \bar{\Psi} \gamma_\mu \Psi \right)\left( \bar{\Psi} \gamma^\mu \Psi \right): .
$$
and Hamiltonian
$$
\int \mathrm{d}x \imath \Psi^\dagger \sigma_z \partial_x \Psi + m_0 \Psi^\dagger \Psi + 2g \Psi^\dagger_1 \Psi^\dagger_2 \Psi_2\Psi_1\\
$$
Due to the infinite set of conservation laws, particle production is said to be absent from this theory. However why isn't it sufficient to show that particle production is absent if the number operator $$N=\int \mathrm{d}x \Psi^\dagger \Psi$$ commutes the the Hamiltonian? Also, by particle production being absent, is that just a statement that all Feynman diagrams with self energy insertions evaluate to 0 but all other Feynman diagrams are possible?

 

[A. Neumaier]

Particle production is a nonperturbative phenomenon, and cannot be explained in terms of Feynman diagrams! The number operator is usually undefinable in an interacting theory - it doesn's survive renormalization; it is certainly not given by your expression.

 

[Demystifier]

Particle production is a nonperturbative phenomenon, and cannot be explained in terms of Feynman diagrams!

What? In the Standard Model, particle production is definitely a perturbative phenomenon described by Feynman diagrams.

 

[QFT1995]

Why isn't something like the following classed as particle production?

 

[PeterDonis]

Why isn't something like the following classed as particle production?

Because only one line comes in and only one line goes out. "Particle production" would mean more lines go out than come in.

 

[QFT1995]

Because only one line comes in and only one line goes out. "Particle production" would mean more lines go out than come in.

Okay thank you. Would you be able to explain to me (or provide a reference) of how particle production arrises in non-perturbative QFT?

 

[Demystifier]

Okay thank you. Would you be able to explain to me (or provide a reference) of how particle production arrises in non-perturbative QFT?

A nice class of examples are particle production from classical sources. See e.g. Itzykson and Zuber, Quantum Field Theory, Sec. 4-1-1.

 

[A. Neumaier]

What? In the Standard Model, particle production is definitely a perturbative phenomenon described by Feynman diagrams.

No. Feynman diagrams describe (usually infinite) contributions to S-matrix elements. Only the S-matrix as a whole describes scattering processes - among others those involving particle production.

 

[A. Neumaier]

Okay thank you. Would you be able to explain to me (or provide a reference) of how particle production arrises in non-perturbative QFT?

Since this is an A thread, you should read about Haag-Ruelle scattering theory, which is nonperturbative.

 

[Demystifier]

No. Feynman diagrams describe (usually infinite) contributions to S-matrix elements. Only the S-matrix as a whole describes scattering processes - among others those involving particle production.

Are you saying that only exact calculations describe physical processes, while approximations don't?

 

[A. Neumaier]

Are you saying that only exact calculations describe physical processes, while approximations don't?

No. But approximations should be at least finite. The only Feynman diagrams leading to finite contributions are the tree diagrams, and these give poor approximations.

Already to get masses for the particles one needs to resum and renormalize infinite series of diagrams with loops. This resummation is already regarded as a nonperturbative step.

 

[Demystifier]

No. But approximations should be at least finite. The only Feynman diagrams leading to finite contributions are the tree diagrams, and these give poor approximations.

In some cases those are good approximations, and are certainly perturbative.

Already to get masses for the particles one needs to resum and renormalize infinite series of diagrams with loops. This resummation is already regarded as a nonperturbative step.

But in many cases one can renormalize a finite number of diagrams, typically a one-loop integral. This is a perturbative operation.

 

[A. Neumaier]

In some cases those are good approximations, and are certainly perturbative.But in many cases one can renormalize a finite number of diagrams, typically a one-loop integral. This is a perturbative operation.

But these cases do not determine particle masses and particle production - which is the topic of the thread.

 

[vanhees71]

Why isn't something like the following classed as particle production?
View attachment 248324

Simply, because no new particle is produced. You start with one particle and end with one particle.

This can be two cases: If the in and out state refer to the same particle species, it's called a self-energy diagram. If they refer to different species, it's describing "particle oscillation". It's an interesting phenomenon (CP violation was discovered first in $K_0 \bar{K}_0$ mixing, resolving the "$\vartheta$-$\tau$ puzzle"; neutrino oscillations, demonstrating that at least 2 neutrino sorts must have a non-zero mass and solving the "solar-neutrino puzzle").