1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Symmetry Breaking

  1. Mar 6, 2007 #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?
  2. jcsd
  3. Mar 6, 2007 #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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Symmetry Breaking
  1. Symmetry breaking (Replies: 11)