Understanding Quantum Symmetry Breaking

In summary, the conversation discusses the concept of symmetry breaking in classical and quantum physics. It is stated that symmetry is broken when the associated charge of the symmetry, Q, does not equal 0 when acting on the vacuum state |0>. The definition of symmetry breaking is further explained as a subgroup H remaining invariant while the larger symmetry group G is broken. The conversation also explores the use of unitary and antiunitary operators in representing symmetry transformations and the role of the associated charge, Q, in maintaining symmetry. The question of whether symmetries represented by antihermitian operators also follow this pattern is raised.
  • #1
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?
Physics news on Phys.org
  • #2
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.

Suggested for: Understanding Quantum Symmetry Breaking