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Is the unitary operator unique?

  1. Sep 14, 2011 #1
    So there is a theorem at the beginning of section 1.5 in Sakurai that states the following:

    Given two sets of base kets, both satisfying orthonormality and completeness. there exists a unitary operator [itex]U[/itex] such that

    [itex]|b^{(1)}> = U|a^{(1)}>,|b^{(2)}> = U|a^{(2)}>,...,|b^{(n)}> = U|a^{(n)}> [/itex]

    By a unitary operator we mean an operator fulfilling the conditions

    [itex]U^{t}U=1[/itex]

    as well as

    [itex]UU^{t}=1[/itex]

    So this is not difficult to prove. But my real question is can we prove that [itex]U[/itex] is unique or is that just not the case and why?
     
  2. jcsd
  3. Sep 14, 2011 #2

    Physics Monkey

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    The elements of U are [itex] U_{nm} = \langle a_n | U | a_m \rangle = \langle a_n | b_m \rangle, [/itex] so if you fix the basis sets then you fix U.
     
  4. Sep 14, 2011 #3
    That is what I thought, but I wanted a second opinion. Thank you.
     
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