Feynman diagrams and relativistic propagators

[Sekonda]

Hey again,

I have a question on a couple of things related to feynman diagrams but also the relativistic scalar propagator term.

First of all, this interaction:

The cross represents a self-interaction via the mass and characterised by the term: -im^2, is this just some initial state then self-interacting with itself via the mass, with nothing changing and it entering into a final state the same as the initial state? Can anybody explain what exactly is happening in this interaction?

My second question is on the propagator for a relativisitc scalar particle, I believe it has form:

$$\frac{i}{E^{2}-\mathbf{p}^{2}-m^{2}}$$

My professor said that this is where this form comes from, he said to imagine summing up all the possible number of self-interaction from 0 self interactions to (presumably) and infinite number of interactions, so :

So the first line has no mass interaction so m=0 and the factor for the first one is :

$$\frac{i}{E^{2}-\mathbf{p}^{2}}$$

then the second line has one mass interaction so the factor associated with it is :

$$\frac{i}{E^{2}-\mathbf{p}^{2}}(-im^{2})\frac{i}{E^{2}-\mathbf{p}^{2}}$$

and the third line has 2 mass interactions and so the factor is :

$$(\frac{i}{E^{2}-\mathbf{p}^{2}})^3(-im^{2})^2$$

And so we sum all these factors up (to the maximum number of self-interactions) and can make factorisation below:

$$\frac{i}{E^{2}-\mathbf{p}^{2}}(1+\frac{m^{2}}{E^{2}-\mathbf{p}^{2}}+\frac{m^{4}}{(E^{2}-\mathbf{p}^{2})^2}+\cdots )$$

We use identity:

$$(1+x)^{-1}=1+x+x^{2}+x^{3}+\cdots\: ,\: x=\frac{m^{2}}{E^{2}-\mathbf{p}^{2}}$$

and thus obtain the relativistic scalar propagator:

$$\frac{i}{E^{2}-\mathbf{p}^{2}-m^{2}}$$

How does this work? Is the scalar interaction just the sum of all the self-interactions by mass?

Thanks,
SK

 

[eljose79]

Hi SK,

Thank you for your question. Let me try to explain the answers to your two questions in more detail:

1. Self-interaction with mass: In this interaction, the particle is interacting with itself through its own mass. This means that the particle is exchanging energy with itself, but nothing else is changing. In other words, the particle remains in the same state before and after the interaction. This type of self-interaction is important in understanding the behavior of particles and their interactions in quantum field theory.

2. Relativistic scalar propagator: The form of the propagator that you have mentioned is indeed correct. To understand how it is derived, we need to consider the Feynman diagrams that represent the possible interactions between particles. Each line in a Feynman diagram represents a propagator, which is a mathematical expression that describes the probability amplitude for a particle to travel from one point to another. In the case of the relativistic scalar particle, the propagator is given by the expression \frac{i}{E^{2}-\mathbf{p}^{2}-m^{2}} which includes the energy, momentum, and mass of the particle.

To understand how this expression is derived, we need to consider all the possible ways in which the particle can interact with itself. This includes interactions with different numbers of self-interactions, as you have mentioned in your post. By summing up all these different possibilities, we can obtain the expression for the propagator. This is done using the identity (1+x)^{-1}=1+x+x^{2}+x^{3}+\cdots\: ,\: x=\frac{m^{2}}{E^{2}-\mathbf{p}^{2}}, which gives us the sum of all the possible self-interactions. The final expression for the propagator is obtained by using this identity and simplifying the resulting expression.

In summary, the relativistic scalar propagator represents the sum of all the possible self-interactions of the particle, and it is derived by considering all the possible ways in which the particle can interact with itself. I hope this helps to clarify your understanding. Please let me know if you have any further questions.