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Unitary transformation of pure states to other pure states

  1. May 30, 2014 #1
    Is it true that there always exist a unitary matrix that can take a state vector of an arbitrary pure state to another arbitrary pure state ? (of course assuming same hilbert space). If true, how do we prove it ? it look like it is true via geometrical arguments but i have not been able to convince myself completely. I am sure in the case of 1 qubit system; what about multiqubit system ? Thanks.
     
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  3. May 30, 2014 #2

    Simon Bridge

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    Conversely, can you find an example where the proposition is not true?

    You would start your proof by defining your terms, then showing what a transformation between states would look like without assuming that the transformation is unitary. Enjoy.
     
  4. May 30, 2014 #3

    atyy

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    Not exactly what you are asking, but related, I think. Probably the article on reachability is the closes to what you are asking about.

    http://arxiv.org/abs/quant-ph/0010031
    Complete controllability of quantum systems
    S. G. Schirmer, H. Fu, A. I. Solomon

    http://arxiv.org/abs/quant-ph/0102017
    Complete controllability of finite-level quantum systems
    H. Fu, S. G. Schirmer, A. I. Solomon

    http://arxiv.org/abs/quant-ph/0110171
    Criteria for reachability of quantum states
    S. G. Schirmer, A. I. Solomon, J. V. Leahy
     
  5. May 31, 2014 #4

    rubi

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    Let's say you have ##\psi, \phi \in \mathcal H## with ##\lVert\psi\rVert = \lVert\phi\rVert = 1## and you want ##U\psi = \phi##. Let ##(v_\alpha)_\alpha## be an basis for ##\mathcal H##. Apply Gram-Schmidt to ##(\psi,(v_\alpha)_\alpha)## to get an ONB ##(x_\alpha)_\alpha## with ##x_0=\psi## and repeat the procedure to get an ONB ##(y_\alpha)_\alpha## with ##y_0 = \phi##. Now define ##U\sum_\alpha c_\alpha x_\alpha = \sum_\alpha c_\alpha y_\alpha##. This is clearly a surjective (for ##\sum_\alpha c_\alpha y_\alpha## in ##\mathcal H##, choose ##\sum_\alpha c_\alpha x_\alpha##) linear operator with ##U\psi = \phi## and it is also unitary because ##\left<U\sum_\alpha c_\alpha x_\alpha,U\sum_\beta d_\beta x_\beta\right> = \left<\sum_\alpha c_\alpha y_\alpha,\sum_\beta d_\beta y_\beta\right> = \sum_\alpha c_\alpha d_\alpha = \left<\sum_\alpha c_\alpha x_\alpha,\sum_\beta d_\beta x_\beta\right>## by orthogonality.
     
  6. May 31, 2014 #5

    Simon Bridge

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    ... etc. But m~ray has to do it in order for it to be properly convincing.
     
  7. May 31, 2014 #6

    rubi

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    Well, of course it would have been better if he had done it himself, but he said that he was unable to do it. I think my proof sketch is pretty clear, so it should be convincing. (If not, don't hesitate to ask, m~ray.)
     
  8. May 31, 2014 #7

    dextercioby

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    An isometric operator defined on all vectors in a Hilbert space is called a unitary operator. An isometric operator has unit (strong) norm and reverse. Hence if a,b arbitrary vectors of unit norm and U is an endomorphism of H taking a into b, then ||U|| = max ||Ua|| = max ||b|| = 1. , forall ||a|| = 1. If U is unit norm, it's isometric. If it's isometric and maximal on H, it's unitary.
     
  9. Jun 1, 2014 #8

    tom.stoer

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    Two pure states define a 2-dim. subspace

    [tex]V_{a,a^\prime} = \text{span}\{|a\rangle, |a^\prime\rangle\}[/tex]

    of the full Hilbert space ##H##. So for the existence of a unitary operator ##U## with

    [tex]U\,|a\rangle = |a^\prime\rangle[/tex]

    it's sufficient that ##U## acts as a rotation operator on ##V_{a,a^\prime}## and as the identity on the orthogonal complement ##V_\perp## of ##V_{a,a^\prime}##.

    [tex]U = u_{V_{a,a^\prime}} \otimes \text{id}_{V_\perp}[/tex]

    From what I said above it should become clear that it's sufficient to proof this for the 2-dim. case, i.e. for ##u_{V_{a,a^\prime}}##. And this is just linear algebra
     
    Last edited: Jun 1, 2014
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