What Are the Implications of Bounded 2HD Potential in Higgs Theory?

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SUMMARY

The discussion centers on the implications of bounded two Higgs doublets (2HD) potential in Higgs theory, emphasizing that any potential must be bounded from below to ensure the existence of a stable ground state. It is established that if the potential is unbounded, the theory lacks a stable vacuum, leading to instability as the field could escape the potential well. The example provided illustrates that for a potential V = a |φ|², it is bounded when a > 0 and unbounded when a < 0. The analysis extends to multiple fields, where each direction in the potential space must be considered independently to determine boundedness.

PREREQUISITES
  • Understanding of Higgs theory and the role of Higgs doublets
  • Familiarity with potential energy functions in quantum field theory
  • Knowledge of stability conditions in physical theories
  • Basic grasp of multivariable calculus for analyzing potentials in multiple dimensions
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  • Study the concept of boundedness in quantum field theories
  • Learn about the implications of potential energy stability in Higgs models
  • Explore textbooks on quantum field theory that cover potential analysis, such as "Quantum Field Theory" by Mark Srednicki
  • Investigate the mathematical techniques for analyzing multivariable potentials
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This discussion is beneficial for theoretical physicists, graduate students in particle physics, and researchers focusing on Higgs boson properties and stability conditions in quantum field theories.

Safinaz
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Hi all,

I found that any two Higgs doublets potential should be bounded from below - at ## V \to - \infty ##. I want to know why this bound is assumed or what does it mean ?

Also are there any textbooks to learn how to make this bound on any other general potential and so to constrain the potential's parameters ?

Best.
 
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If the potential is not bounded from below, your theory has no ground state.
 
The bounds are obtained just by looking at the potential...
For example V=a |\phi|^2 is bounded for a&gt;0 and unbound for a&lt;0. If you draw the potential you will see that.
In case you have more than one fields, I guess one has to look at each direction independently ... So the potential for two Higgses could be described by a sheet V(\phi_1, \phi_2) on the space of \phi_1, \phi_2, if it's bounded it means that it can't go to -\infty anywhere.

If there was no bound then even if you had a vacuum at some potential's minimum, this vacuum would not be stable. At some point the field would "escape" the well, and then start rolling down forever.
 

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