Renormalisation group equation

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In summary, the renormalization group equations for the n-point Green's function in a four-dimensional massless field theory can be written as a function of the coupling g defined at mass scale ##\mu## or as a function of the beta parameter and n. However, in the equation ##\frac{\partial \tilde \Gamma}{\partial Z_{\psi}} = n (Z_{\psi})^{n-1}\tilde \Gamma##, the missing factor of ##\gamma## leads to incorrect results.
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CAF123
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Homework Statement


The renormalization group equations for the n-point Green’s function ##\Gamma(n) = \langle \psi_{x_1} \dots \psi_{x_n}\rangle ## in a four-dimensional massless field theory is $$\mu \frac{d}{d \mu} \tilde{\Gamma}(n) (g) = 0$$ where the coupling g is defined at mass scale ##\mu##.

Show that this is equivalent to $$ (\beta \frac{\partial}{\partial g} + n )\tilde{\Gamma}(n) = 0, $$ where ##\beta(g) = \mu \frac{d g}{d \mu}. ##The field ##\psi## has mass dimension one and the Green’s function is a homogeneous function of degree n in the field.

Homework Equations


[/B]
function of homogenous degree n is one in which the exponents of each term all add up to n.

Renormalisation of fields

The Attempt at a Solution


In renormalisation, ##\psi \rightarrow Z_{\psi} \psi## and given that the Green's function is a homogenous function of degree n, in the renormalised Green's function, we now have a factor of ##(Z_{\psi})^n## in each term. So, $$\frac{d}{d \mu} \tilde \Gamma = \frac{\partial \tilde \Gamma}{\partial \mu} + \frac{\partial \tilde \Gamma}{\partial Z_{\psi}} \frac{\partial Z_{\psi}}{\partial \mu}$$ I would say that $$\frac{\partial \tilde \Gamma}{\partial Z_{\psi}} = n (Z_{\psi})^{n-1}\tilde \Gamma$$ but this does not seem to give me correct result.

Did I assume something incorrect? Thanks!
 
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CAF123 said:

Homework Statement


The renormalization group equations for the n-point Green’s function ##\Gamma(n) = \langle \psi_{x_1} \dots \psi_{x_n}\rangle ## in a four-dimensional massless field theory is $$\mu \frac{d}{d \mu} \tilde{\Gamma}(n) (g) = 0$$ where the coupling g is defined at mass scale ##\mu##.

Show that this is equivalent to $$ (\beta \frac{\partial}{\partial g} + n )\tilde{\Gamma}(n) = 0, $$

You are missing a ##\gamma## in that equation, next to the factor of n , did you realize this?
 

What is the Renormalisation Group Equation?

The Renormalisation Group Equation (RGE) is a mathematical tool used in theoretical physics and statistical mechanics to study the behavior of physical systems at different length scales. It describes how certain parameters, such as coupling constants, change as a function of energy scale.

What is the purpose of the Renormalisation Group Equation?

The RGE allows us to study the behavior of physical systems at different length scales, which is important because many physical phenomena, such as phase transitions, are scale-dependent. It also helps us understand the underlying structure of theories and identify the relevant degrees of freedom.

How is the Renormalisation Group Equation derived?

The RGE is derived using a combination of statistical mechanics and quantum field theory. It involves calculating the change in a system's partition function under a change in energy scale, and then using this to derive an equation that relates the behavior of the system at different length scales.

What is the significance of the Renormalisation Group Equation in physics?

The RGE has many applications in physics, including in the study of phase transitions, critical phenomena, and the behavior of quantum field theories. It has also been used to develop models for the behavior of complex systems, such as in condensed matter physics and high-energy physics.

What are the limitations of the Renormalisation Group Equation?

The RGE is a powerful tool, but it has some limitations. It typically only applies to systems close to a critical point or phase transition, and may not accurately describe behavior at extreme energy scales. It also relies on certain assumptions, such as the existence of a fixed point, which may not always hold true.

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