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Are Unitary Transformations Always Linear?

  1. Jul 10, 2010 #1
    Hello, I had a question regarding unitary transformations. The most common definition I see for unitary transformations is defined as a transformation between Hilbert spaces that preserves inner products. I was wondering if all unitary transformations between Hilbert spaces (according to this definition) are necessarily bounded linear transformations. (i.e. [tex]U( \alpha x + \beta y ) = \alpha U x + \beta U y [/tex] and [tex] U \in \mathcal{L} (H_1 , H_2) [/tex].) I have been trying to prove this to myself for the last hour but can't seem to show this for some reason.
     
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  3. Jul 10, 2010 #2
    Try showing that the quantity [tex]|U(x + y) - U(x) - U(y)|^2[/tex] is zero by writing it as an inner product, expanding, and finally using the preservation of the inner product by U.
     
  4. Jul 10, 2010 #3
    I think it follows from the linearity of the inner product. One might try to calculate [tex]<U( \alpha x + \beta y ) - \alpha U x + \beta U y,U( \alpha x + \beta y ) - \alpha U x + \beta U y> [/tex], using the axioms of the inner product. If it gives zero, you are home.
     
  5. Jul 10, 2010 #4
    Thanks. That's a cool way to show this, and then since it's an isometry it's bounded, great.
     
  6. Jul 10, 2010 #5

    Landau

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    Preserving inner product is equivalent to being an isometry, and this implies boundedness and linearity. However, unitary transformations are also (by definition) required to be surjective, or at least have dense range.
     
  7. Jul 10, 2010 #6
    How can you show that all surjective isometries between Hilbert spaces are linear?
     
  8. Jul 10, 2010 #7

    Landau

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    Surjectivity is not needed for linearity. Every isometry between inner product spaces is linear, as follows from showing that the quantity which element4 wrote equals zero.
     
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