Ignoring SSB in massless phi^4 theory

  • Thread starter Thread starter geoduck
  • Start date Start date
  • Tags Tags
    Massless Theory
geoduck
Messages
257
Reaction score
2
If a theory is spontaneously broken, you usually make calculations by expanding the field about the ground-state expectation value \nu, so that you define a new field \rho such that \phi=\nu+\rho such that \langle \rho \rangle =0.

What if you just calculate perturbation theory with \phi instead of the new field \rho?

It seems that massless \phi^4 is spontaneously broken at the 1-loop level. So do you have to do perturbation theory with the field \rho instead of \phi?
 
Physics news on Phys.org
It is possible to do perturbation theory with the field \phi, however it is not necessarily the most accurate or efficient way to approach the problem. Expanding the field about the ground-state expectation value \nu by defining a new field \rho allows you to find the exact ground-state energy of the system as well as calculate the mass of the particle created due to the spontaneous breaking. This method is often thought of as the more effective way to approach the problem.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
Back
Top