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Broken Symmetries (Weinberg p215)

  1. Aug 16, 2008 #1
    A group G is proken to a subgroup H. Let [tex] t_{\alpha} [/tex] the generator of G and
    [tex]t_i[/tex] the generator of H. The t_i form a subalgebra. Take the x_a to be the other indipendent generator of G.
    Why any finite element of G may be expressed in the form [tex]g=exp[i\xi_ax_a]exp[i\theta_i t_i][/tex] even if [tex][t_i,x_a]\neq0[/tex]?
  2. jcsd
  3. Aug 16, 2008 #2


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    Because the http://en.wikipedia.org/wiki/Baker-Campbell-Hausdorff_formula" [Broken] says that
    [tex]\exp[i\xi_a x_a]\exp[i\theta_i t_i] = \exp[i\tilde\xi_a x_a + i\tilde\theta_i t_i][/tex]
    where the new parameters are complicated functions of the old ones.
    Last edited by a moderator: May 3, 2017
  4. Aug 17, 2008 #3


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    By definition of a group, you can always write the product of two group elements as a third group element. That's all there is to it.
  5. Aug 17, 2008 #4
    I don't understand... My problem is to express a generic element of the group [tex]g=exp[i\xi_ax_a+i\theta_i t_i][/tex] as the product of 2 element of the form
    [tex]g_1=exp[i\xi_ax_a] \ g_2=exp[i\theta_i t_i][/tex].

    Thank you
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