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About bases of complex (Hilbert) space

  1. Feb 16, 2009 #1

    KFC

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    Hi there,
    In 3-dimensional real linear space, the simplest bases can be taken as the canonical bases

    [tex]\hat{x} = \left(\begin{matrix}1 \\ 0 \\ 0\end{matrix}\right), \qquad \hat{y} = \left(\begin{matrix}0 \\ 1 \\0\end{matrix}\right), \qquad \hat{z} = \left(\begin{matrix}0 \\ 0 \\ 1\end{matrix}\right)[/tex]

    I wonder what's the simplest counterpart for 3-dimensional in complex (hilbert) space?
     
  2. jcsd
  3. Feb 16, 2009 #2
    the basis you gave for R^3 is also a basis for C^3
     
  4. Feb 16, 2009 #3

    KFC

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    Thanks. But can we construct a basis similar to those for R^3 but with the entries be complex? and how?
     
  5. Feb 17, 2009 #4

    Ben Niehoff

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    The exact same basis will do. Remember that your scalars are complex numbers.
     
  6. Feb 17, 2009 #5
    sure, if you replace your 1's with i's it's still a basis. all you have to do is pick 3 vectors so that no one of them is a linear combination of the other 2. if you want it to be an orthonormal basis (which is usually useful) then you need to also make sure that each vector is of unit length and orthogonal to the other 2.
     
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