Disconnected diagram Feynman rules momentum space

In summary, a disconnected diagram in Feynman rules is a diagram that contains separate components not connected by lines or vertices. Feynman rules in momentum space are used to analyze and calculate particle behavior. The momentum space representation of a Feynman diagram tells us about the particles involved and their interactions. Rules for a disconnected diagram in momentum space are determined by considering each component separately. Disconnected diagrams in momentum space are significant in representing multiple independent interactions and are important in perturbation theory.
  • #1
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Hi!

I wonder if the Feynman rules in momentum space can also be applied to disconnected diagrams. So aussume I have a disconnected Feynman diagram 1 , i.e. the product of two connected Feynman diagrams 2 and 3.
I can translate my diagrams with the position space Feynman rules to explicit mathematical expressions:
1->D
2->C_2
3->C_3

D=C_2 * C_3

I can compute the Fourier transforms F(C_2) and F(C_3) by using the Feynman rules in momentum space.

Now if the FR in moentum space would be applicable for disconnected diagrams, this would mean
F(D)=F(C_2)*F(C_3)

This means the inverse Fourier transform yields
D= F^-1 [ F(C_2)*F(C_3) ]

=> C_2 * C_3 = F^-1 [ F(C_2)*F(C_3) ]

How can that be?? This rule is not generally true for Fourier transformations...

Martin
 
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  • #2


Hi Martin,

Thank you for your question. The Feynman rules in momentum space can indeed be applied to disconnected diagrams, but there are some important considerations to keep in mind.

Firstly, the Feynman rules in momentum space are derived from the Feynman rules in position space, so they should yield the same results. However, when dealing with disconnected diagrams, we need to be careful about how we interpret the results. In your example, you correctly translated the disconnected diagram 1 into the product of two connected diagrams 2 and 3. However, when taking the Fourier transform, we need to remember that the momentum conservation rules still apply. This means that the momenta of the particles in diagrams 2 and 3 must add up to the total momentum of the particles in diagram 1. So the Fourier transform of diagram 1 will not simply be the product of the Fourier transforms of diagrams 2 and 3, but rather a more complicated expression that takes into account the momentum conservation rules.

Secondly, the Fourier transform is a linear operation, which means that it can be applied to each term in a sum separately. This is why we can write the Fourier transform of a disconnected diagram as the product of the Fourier transforms of its connected components. However, this does not mean that we can simply take the inverse Fourier transform of the product of the Fourier transforms of connected diagrams to get the disconnected diagram. As you pointed out, this is not generally true for Fourier transformations. In fact, the inverse Fourier transform of the product of two functions is not equal to the product of their inverse Fourier transforms in most cases.

In summary, while the Feynman rules in momentum space can be applied to disconnected diagrams, we need to carefully consider the momentum conservation rules and the limitations of the Fourier transform when interpreting the results. I hope this helps clarify things for you.

 
  • #3


Hi Martin,

Thank you for your question. The Feynman rules in momentum space can indeed be applied to disconnected diagrams. In fact, the Feynman rules in momentum space are often preferred over those in position space because they simplify calculations and make certain properties of the diagrams more apparent.

In your example, the product of two connected diagrams in position space can be translated into a single disconnected diagram in momentum space, as you have correctly shown. However, the inverse Fourier transform of the product of two functions is not necessarily equal to the product of the inverse Fourier transforms of each individual function. This is because the Fourier transform is a linear operator, but not necessarily a multiplicative one.

In other words, the inverse Fourier transform of F(C_2)*F(C_3) is not equal to F^-1(F(C_2))*F^-1(F(C_3)). Instead, it can be written as a convolution of the inverse Fourier transforms of each individual function. This is why your final equation does not hold true.

I hope this helps clarify the application of Feynman rules in momentum space to disconnected diagrams. Keep exploring and asking questions!


 

1. What is a disconnected diagram in Feynman rules?

A disconnected diagram in Feynman rules is a type of diagram that contains two or more separate components that are not connected by any lines or vertices. This means that the components do not interact with each other and can be treated as independent parts of the overall diagram.

2. How are Feynman rules used in momentum space?

Feynman rules in momentum space are used to analyze and calculate the behavior of particles and their interactions in terms of their momenta and energies. This allows for a more intuitive understanding of the physical processes involved and can simplify complex calculations.

3. What does the momentum space representation of a Feynman diagram tell us?

The momentum space representation of a Feynman diagram provides information about the particles involved, their momenta and energies, and the interactions between them. It can also reveal important symmetries and conservation laws that govern the physical processes described by the diagram.

4. How do we determine the rules for a disconnected diagram in momentum space?

The rules for a disconnected diagram in momentum space are determined by considering each component separately and following the usual Feynman rules for those components. The overall diagram is then obtained by multiplying the individual components together.

5. What is the significance of disconnected diagrams in momentum space?

Disconnected diagrams in momentum space can represent physical processes that involve multiple independent interactions, such as two particles scattering off each other without exchanging any momentum. They also play an important role in perturbation theory and can help us understand the behavior of particles at high energies.

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