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Commuting operators

  1. Sep 21, 2014 #1
    Hi. Say a, A(a) and b are well behaving functions. Then say [a,b] = 0, i.e. a and b commute.

    Why will this automatically mean that A=A(a) will commute with b? Can somebody give me an intuitive explanation, or link me to some proof?
  2. jcsd
  3. Sep 21, 2014 #2


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    I think one way is taylor expanding A(a). Then [itex] [A(a),b]=c_0 [1,b]+c_1 [a,b]+c_2[a^2,b]+c_3[a^3,b]+...=0 [/itex].
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