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Rotation in hilbert space?

  1. Feb 5, 2009 #1

    KFC

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    In real linear space, we can use the rotation matrix in terms of Euler angle to rotate any vector in that space. I know in hilbert space, the corresponding rotation matrix is so-called unitary operator. I wonder how do I construct such matrix to rotate a complex vector in hilbert space? Can I use the real matrix (for real linear space) to rotate the real and imaginary part separately?
     
  2. jcsd
  3. Feb 5, 2009 #2
    Are you mostly interested in the two dimensional complex vector space [itex]\mathbb{C}^2[/itex], or in general Hilbert spaces?
     
  4. Feb 5, 2009 #3

    KFC

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    I am interesting in 3 dimensional complex vector space. But as starting, 2D complex vector space will do.
     
  5. Feb 5, 2009 #4
    ok, anyway, you cannot rotate real and imaginary parts separately. I don't have any ready formulas available now, but you can try to solve (with some small [itex]n[/itex]) what kind of matrices [itex]U\in\mathbb{C}^{n\times n}[/itex] satisfy the condition

    [tex]
    \sum_{l=1}^n U^*_{lk}U_{lm} = \delta_{kl}
    [/tex]

    which is equivalent with [itex]U^{-1}=U^{\dagger}[/itex].
     
  6. Feb 5, 2009 #5
    Suppose [itex]\boldsymbol{z}\in\mathbb{C}^n[/itex] some (vertical) vector. Then [itex]\boldsymbol{z}=\boldsymbol{x} + i\boldsymbol{y}[/itex] with some [itex]\boldsymbol{x},\boldsymbol{y}\in\mathbb{R}^n[/itex]. If [itex]C\in\mathbb{C}^{n\times n}[/itex] is some complex matrix, you can write it in form [itex]C=A+iB[/itex], where [itex]A,B\in\mathbb{R}^{n\times n}[/itex] are real matrices. Then

    [tex]
    C\boldsymbol{z} = (A+iB)(\boldsymbol{x}+i\boldsymbol{y}) = (A\boldsymbol{x} - B\boldsymbol{y}) + i(B\boldsymbol{x} + A\boldsymbol{y}),
    [/tex]

    so you can reduce linear mappings [itex]\mathbb{C}^n\to\mathbb{C}^n[/itex] into linear mappings [itex]\mathbb{R}^n\to\mathbb{R}^n[/itex] like this. It is easy to see that in general acting with unitary matrices on complex vectors will not be the same as acting on the real and imaginary parts with the usual real rotations.
     
    Last edited: Feb 5, 2009
  7. Feb 5, 2009 #6

    StatusX

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    Homework Helper

    Think of it like this: a unitary matrix is to a complex vector space as an orthogonal matrix is to a real vector space (and, if you ever come across it, as a symplectic matrix is to a quaternionic vector space). Most of your intuition for ordinary rotations can be applied to unitary matrices acting on complex vectors. For example, it's always possible to apply a unitary transformation to a vector to get it in the form (a,0,0,...,0), for some real a>0. To see why, first try to see why the corresponding thing is true for rotations of real vectors, and see if you can adapt the argument to the complex case.

    Incidentally, a matrix that is both unitary and real is just an orthogonal matrix. Since it's real, it doesn't mix up the real and imaginary parts of a vector, so if you write the vector as [itex]\vec u + i \vec v[/itex], then a real unitary matrix R takes this to [itex] (R\vec u) + i( R \vec v) [/itex], where [itex]\vec u[/itex] and [itex]\vec v[/itex] are ordinary real vectors and R is an ordinary rotation. But these aren't the most general unitary transformations, there are others whose matrix elements aren't all real. For example, the best you could do with a real unitary matrix is rotate your vector into the form (a,b,0,0,...,0) (can you see why this is true? - think about rotating u and v simultaneously with R), so clearly the complex unitary matrices are important.
     
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