Understanding Quantum Symmetry Breaking

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SUMMARY

This discussion focuses on the concept of quantum symmetry breaking, specifically the relationship between the vacuum state |0> and the associated charge Q of the broken symmetry. It establishes that symmetry is broken if and only if Q|0> is not equal to 0, indicating that the vacuum state is invariant under a subgroup H but not under the full symmetry group G. The participants clarify that any symmetry transformation can be represented by unitary or antiunitary operators, and the condition for symmetry manifestation is U|0> = |0>, leading to the conclusion that Q|0> must equal 0 for unbroken symmetry. The discussion also raises questions about the implications for antihermitian operators in symmetry transformations.

PREREQUISITES
  • Understanding of quantum mechanics and vacuum states
  • Familiarity with symmetry groups and their representations
  • Knowledge of unitary and antiunitary operators
  • Basic concepts of quantum field theory
NEXT STEPS
  • Study the implications of quantum symmetry breaking in quantum field theory
  • Learn about the role of unitary and antiunitary operators in quantum mechanics
  • Explore the mathematical formulation of symmetry groups and their subgroups
  • Investigate the properties of antihermitian operators in quantum transformations
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Physicists, quantum mechanics students, and researchers interested in quantum field theory and the mathematical foundations of symmetry breaking.

alphaone
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Hi,
I have done classical symmetry breaking and now want to understand the quantum one. I have seen the statement that the symmetry is broken if and only if Q|0> not 0. Where |0> is the vacuum and Q is the associated charge of the broken symmetry. Why does this imply symmetry breaking? The way I know it a symmetry group G is broken to a subgroup H if the theory is invariant under G whereas the vacuum is invariant only under H, meaning h|0>=|0> for any h element H and g|0> not equal |0> for any g not in H. Is this the right definition? And if so does this imply Q|0> not 0 in the case of a broken symmetry?
 
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I think I figured it out myself:
Any symmetry transformation can be represented by either a unitary operator U or an antiunitary operator A acting on the space of states, i.e. U|phi> is the transformed state(for the case of a unitary operator) now if I am not mistaken we can write U = exp(i*x*Q) (x being a parameter here and not spacetime position) where Q is the associated charge of the symmetry. Then for the symmetry to be mainfest we require U|0>=|0> which is equivalent to saying Q|0>=0 when expanding the exponential. A question I have about this is: Does this also hold for symetries being represented by antihermitian operators, i.e. if the states transform under the symmetry as A|phi> can I still write A=exp(i*x*Q) where Q is the conserved charge? In this case the charge would then be antihermitian instead of hermitian, I guess.
 

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