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Proove that U is unitary (Cayley transform)

  1. May 5, 2013 #1
    Function [itex]f(t)=\frac{t-i}{t+i}[/itex] for [itex]t\in \mathbb{R}[/itex] maps real ax into complex circle. Show that for any hermitian operator [itex]H[/itex] operator [itex]U:=(H-iI)(H+iI)^{-1}[/itex] is unitary (where [itex]H+iI[/itex] is reversible)

    If I understand correctly [itex]U[/itex] is unitary when [itex]U=U^{T}[/itex] right?

    So I tried to show that [itex]U[/itex] is unitary like this (hopefully it is ok):
    [itex]U=(H-iI)(H+iI)^{-1}[/itex] because H hermitian than [itex]H^{*}=H[/itex]
    [itex]U=(H^{*}-iI)(H^{*}+iI)^{-1}[/itex] but Identity does not change if I conjugate it and transpose it
    [itex]U=(H^{*}-iI^{*})(H^{*}+iI^{*})^{-1}=((H-iI)^{-1}(H+iI))^{*}[/itex]
    [itex]U=((H+iI)^{-1}(H-iI))^{T}[/itex] so [itex]U=U^{T}[/itex]
    Right or wrong?

    Than I have to calculate the inverse function [itex]f^{-1}=(\zeta ))i\frac{1+\zeta }{1-\zeta } [/itex] and show that [itex]f^{-1}(U)[/itex] is hermitian operator for any unitary operator [itex]U[/itex], where [itex]I-U[/itex] is reversible operator.

    And here I have no idea how to start... :/
     
  2. jcsd
  3. May 5, 2013 #2
    Ok first part is clearly wrong.

    For unitary operator this has to be true: [itex]UU^{*}=I or U=(U^{*})^{-1}[/itex]. After realizing that I had no problem proving that [itex]U[/itex] is unitary.

    But I am still having problems with second part: If [itex]f^{-1}[/itex] is hermitian, than [itex]f^{-1}(U)=(f^{-1}(U))^{*}[/itex]
    [itex]f^{-1}(U)=i\frac{I+U}{I-U}=i(I+U)(I-U^{-1})[/itex] but [itex]U[/itex] is unitary, so [itex]UU^{*}=I[/itex]
    [itex]f^{-1}(U)=i(I+(U^{*})^{-1})(I-U^{*})=i(I+U^{-1})^{*}(I-U)^{*}[/itex]

    now I don't know how to get rid of [itex]U^{-1}[/itex] and that - that comes from [itex]i^{*}[/itex] in order to prove that [itex]f^{-1}(U)=(f^{-1}(U))^{*}[/itex]
     
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