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Unitary transformations

  1. Jun 14, 2010 #1
    Hi,

    I just have a quick question, I understand that [tex] (T\vec{u},T\vec{v})=(\vec{u},\vec{v})[/tex] defines a unitary transformation [tex] T [/tex], but how does one go from this relation to [tex] (T\vec{u},\vec{v})=(\vec{u},T^{-1}\vec{v}) [/tex] as the other way to define a unitary transformation?

    Thanks
     
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  3. Jun 14, 2010 #2

    Fredrik

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    You define the adjoint of T by [tex](T^\dagger\vec u,\vec v)=(\vec u,T\vec v)[/tex] (after proving that it always exists). Then you can define "unitary" by saying that T is unitary if [tex]T^{-1}=T^\dagger[/tex]. If T is unitary according to this alternative definition, we have

    [tex](T\vec u,T\vec v)=(T^\dagger T\vec u,\vec v)=(T^{-1}T\vec u,\vec v)=(\vec u,\vec v)[/tex]

    It's not much harder to prove that if T is unitary according to your definition, it's also unitary according to the alternative definition, but I'll let you do that one for yourself.
     
  4. Jun 14, 2010 #3

    Landau

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    It's impossible. You have to require that U is surjective. Of course, if U preserves the inner product (which is LAHLH's definition) then it is an isometry (actually, these are equivalent), hence injective. Surjectivity is needed to guarantee bijectiveness, else T^-1 does not make sense. In finite dimensions injective already implies surjective, but in general (infinite dimension) not.

    Consider E.g. the unilateral shift T on [itex]\ell(\mathbb{N})[/itex]which sends (x1,x2,x3,...) to (0,x1,x2,x3,...). It is an isometry, so it preserves the inner product. So T would be unitary according to the first definition. But T is not invertible (not surjective): its image does not contain sequences whose first coordinate is non-zero.
     
    Last edited: Jun 14, 2010
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