About two integrals in QCD textbook by muta

In summary: This is what allows you to simplify the equation for the bracket in eq.2.3.154:$$\begin{aligned}&\delta\left(q^2+k'^2-4m^2\right)\delta\left(q^2+k'^2-2q\cdot k'-4m^2\right)\\&=\delta\left(q^2+k'^2-2q\cdot k'-4m^2+2q\cdot k'^2+4m^2\right)\end{aligned}$$
  • #1
Thor Shen
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http://d.kankan3d.com/file/data/bcs/2014/0508/w65h1446064_1399517186_873.jpg

1.How to deal with the delta functions in eq.2.3.153 to obtain the eq.2.3.154 by integrating over q'?
2.How to caculate the integral from eq.2.3.154 to eq.2.3.156, especially the theta function?
 
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  • #2
You already have a [itex]\delta^4(q'-q)[/itex] so just put q=q' and remove the integral over dq'. Then in Eq. (2.3.154) you have [itex]\delta(k'\cdot q)[/itex] that, in the COM, becomes just [itex]\delta(k_0\sqrt{s})[/itex]. Therefore, taking care of the Jacobian coming out of the delta function, this just tells you that [itex]k'_0=0[/itex]. Hence you don't need to worry about the thetas since they only give you [itex]\theta(q_0)=\theta(\sqrt{s})=1[/itex] since its argument is positive.
 
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  • #3
Actually, when I try to simplify the eq.2.3.153 for obtain the bracket in eq.2.3.154, I find we must use the two delta functions in eq.2.3.154, but I wonder whether we can use δ(1/4(q'+k')^2-m^2)=δ(k'^2+s-4m^2)δ(k'*q) directly, and the δ(k*q) is equivalent the δ(k'*q) because of the k' is integrated over total space?
 
  • #4
Thor Shen said:
I wonder whether we can use δ(1/4(q'+k')^2-m^2)=δ(k'^2+s-4m^2)δ(k'*q) directly, and the δ(k*q) is equivalent the δ(k'*q) because of the k' is integrated over total space?

No, you can't do that. You can't split one delta function in two. However, you already have two deltas:
$$
\delta\left(q^2+k'^2+2q\cdot k'-4m^2\right)\delta\left(q^2+k'^2-2q\cdot k'-4m^2\right).
$$
Now keep in mind that if you have some function [itex]f(x)[/itex] multiplied by a delta then [itex]\delta(x-x_0)f(x)=\delta(x-x_0)f(x_0)[/itex]. This is true also if the function is a delta itself.

Now, the second delta tells you that [itex]q^2+k'^2-4m^2=2q\cdot k'[/itex]. Using this is the first delta you obtain (I always omit the necessary Jacobian):
$$
\delta\left(q\cdot k'\right)\delta\left(q^2+k'^2-2q\cdot k'-4m^2\right)=\delta\left(q\cdot k'\right)\delta\left(q^2+k'^2-4m^2\right).
$$
Keep also in mind that [itex]q^2=s[/itex].
 
  • #5


As a scientist familiar with quantum chromodynamics (QCD), I can provide a response to the question regarding the integrals in the QCD textbook by Muta.

1. The delta functions in equation 2.3.153 can be dealt with by using the property that the integral of a delta function over its support is equal to 1. This means that we can rewrite the integral as the product of the integrand and the delta function evaluated at the integration variable, q'. This will result in equation 2.3.154, where the delta function has been replaced by its value at q'.

2. To calculate the integral from equation 2.3.154 to 2.3.156, we need to use the theta function which is defined as:

θ(x) = 1 for x > 0
θ(x) = 0 for x < 0

This function acts as a step function, ensuring that the integral is only evaluated for positive values of the integration variable, q'. This is necessary because the integrand in equation 2.3.154 is only valid for positive values of q'. The resulting integral can then be solved using standard integration techniques.

In summary, to deal with the delta functions in equation 2.3.153, we can use their property to rewrite the integral. And to calculate the integral from equation 2.3.154 to 2.3.156, we need to use the theta function to ensure that the integral is only evaluated for the appropriate values of q'. These techniques are essential in solving integrals in QCD and are commonly used by physicists in their research.
 

1. What is QCD and why is it important?

QCD stands for Quantum Chromodynamics and it is a theory that describes the strong nuclear force that binds quarks and gluons together to form protons, neutrons, and other hadrons. It is an important theory because it helps us understand the structure of matter and the interactions between particles.

2. What are integrals in QCD and how are they used?

In QCD, integrals are mathematical tools used to calculate the probability amplitudes of particle interactions. They are used to calculate various physical quantities such as cross sections and decay rates.

3. What is the role of muta in the study of integrals in QCD?

Muta is a physicist who wrote a textbook on QCD, which is widely used by students and researchers in the field. In this textbook, he provides a comprehensive understanding of integrals in QCD and their applications.

4. Can you explain the Feynman diagrams used in QCD integrals?

Feynman diagrams are graphical representations of mathematical expressions that describe the interactions between particles in QCD. They are used to simplify and visualize complex calculations and are an important tool in understanding QCD processes.

5. Are there any challenges associated with calculating QCD integrals?

Yes, there are several challenges in calculating QCD integrals, such as the need for high precision calculations, the complexity of the Feynman diagrams, and the need for advanced mathematical techniques. However, with advancements in technology and theoretical methods, these challenges can be overcome.

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