Cluster Decomposition.Vanishing of the connected part of the S matrix.

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SUMMARY

The discussion centers on the vanishing of the connected part of the S matrix, specifically equation 4.3.8 from Weinberg's QFT Volume I, as the spatial distance between particle clusters increases. The key point is that the S matrix element, represented as S_{p_1'p_2'... , p_1 p_2}^C, contains a delta function ensuring 3-momentum conservation, which leads to the integral vanishing in the limit of large distances. The participants debate the implications of translational invariance and the Riemann-Lebesgue theorem on the behavior of the integral as coordinates approach infinity.

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  • Understanding of Quantum Field Theory (QFT) principles
  • Familiarity with S matrix formalism
  • Knowledge of Fourier transforms and their properties
  • Concept of translational invariance in physics
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Alex Cros
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Im following Weinberg's QFT volume I and I am tying to show that the following equation vanishes at large spatial distance of the possible particle clusters (pg 181 eq 4.3.8):

S_{x_1'x_2'... , x_1 x_2}^C = \int d^3p_1' d^3p_2'...d^3p_1d^3p_2...S_{p_1'p_2'... , p_1 p_2}^C \times e^{i p_1' . x_1'}...

(i.e the Fourier of the connected part of the corresponding S matrix element) vanishes when the distance between any two particles (states) is great.
He says that this only happens when this matrix element contains precisely one single delta function that ensures 3-momentum conservation:

S_{p_1'p_2'... , p_1 p_2}^C = \delta^3(p_1'+p_2'+... - p_1 - p_2)\times \delta(energy \ conservation) \times C_{p_1'p_2'... , p_1 p_2}

Can somebody tell my how this makes the first equation vanish when the relative distance of some states is large?

PD: sorry for the poor explanation but I am not going to explain it better than Weinberg, I am just pointing out where I am having troubles.
 
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Suppose that one x under the integral (4.3.8) goes to infinity. This means that the integral over the corresponding p is a high-frequency Fourier transform of a smooth function. In the infinite frequency limit, such a Fourier transform goes to zero. The same is true when all x's go to infinity in different directions, i.e., all particles separate.

Eugene.
 
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meopemuk said:
Suppose that one x under the integral (4.3.8) goes to infinity. This means that the integral over the corresponding p is a high-frequency Fourier transform of a smooth function. In the infinite frequency limit, such a Fourier transform goes to zero. The same is true when all x's go to infinity in different directions, i.e., all particles separate.

Eugene.

I disagree, because translational invariance tells us that if all coordinates go to infinity together the S matrix should be invariant.
I think I solved the problem yesterday, when you integrate expression (with this 3-spacial delta) 4.3.8 the exponentials become "coupled" i.e. if you integrate say p_1 then the exponential for exp(i p_1 x_1) becomes exp(ix_1(-all other momenta)) then by the Riemann-Lebesgue theorem the integral vanishes when some of the coordinates go to infinity BUT NOT IF ALL go to infinity together since they would cancel and the theorem wouldn't apply.

Correct me if I am wrong!
 

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