Callan-Symanzik equation for Effective Potential

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SUMMARY

The discussion focuses on the transition from the Callan-Symanzik equation for the Effective Action to the Effective Potential as outlined in Peskin & Schroeder's Quantum Field Theory (QFT) textbook. The user expresses confusion regarding the derivation of equations (13.24) and (13.25) and the relationship involving the classical field value, ##\phi_{cl}##, and the effective potential, ##V_{eff}(\phi_{cl})##. The derivation involves discretizing space and utilizing the functional derivative of the Effective Action, leading to a clearer understanding of the relationship between the Effective Action and the Effective Potential.

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  • Understanding of Quantum Field Theory (QFT) principles
  • Familiarity with the Callan-Symanzik equation
  • Knowledge of functional derivatives in field theory
  • Proficiency in mathematical manipulation of equations in QFT
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  • Study the derivation of the Callan-Symanzik equation in Peskin & Schroeder's QFT (specifically equations 13.24 and 13.25)
  • Learn about the Effective Action and its applications in Quantum Field Theory
  • Explore the concept of Effective Potential and its significance in particle physics
  • Review discretization techniques in field theory for better understanding of functional derivatives
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This discussion is beneficial for theoretical physicists, graduate students in quantum field theory, and researchers focusing on effective actions and potentials in particle physics.

thatboi
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Hey all,
I am looking equations (13.24),(13.25) in Peskin & Schroeder's QFT book and I am confused about how they change from the Callan-Symanzik equation for the Effective Action to the Effective Potential. I thought the relation for constant ##\phi_{cl}## was ##\Gamma[\phi_{cl}] = -(VT)\cdot V_{eff}(\phi_{cl})##, equation (11.50) in the book. But making such a substitution into (13.24), I do not understand how to get to (13.25).
Any advice would be appreciated, thanks.
 
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First imagine that the space is discretized, so that you can write
$$\int dx\, \phi(x)\frac{\delta}{\delta\phi(x)}\Gamma[\phi]=
\sum_{x=1,2,\cdots} \phi_x \frac{\partial}{\partial\phi_x} \Gamma(\phi_1,\phi_2,\ldots)$$
Then define the quantity
$$V(\phi)=\Gamma(\phi,\phi,\ldots)$$
Clearly
$$\frac{\partial V}{\partial\phi}=\sum_{x}\left(\frac{\partial \Gamma}{\partial\phi_x} \right)_{\phi_1=\phi_2=\cdots=\phi}$$
so
$$\phi\frac{\partial V}{\partial\phi}=
\sum_{x}\phi\left(\frac{\partial \Gamma}{\partial\phi_x} \right)_{\phi_1=\phi_2=\cdots=\phi} =
\left( \sum_{x} \phi_x\frac{\partial \Gamma}{\partial\phi_x} \right)_{\phi_1=\phi_2=\cdots=\phi}$$
Finally turn back to the continuous ##x##, so that the last formula can be written as
$$\left( \int dx\, \phi(x)\frac{\delta}{\delta\phi(x)}\Gamma[\phi]\right)_{\phi(x) =\phi, \; \forall x}
=\phi\frac{\partial V}{\partial\phi}$$
Now getting (13.25) should be obvious.
 
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