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I Decomposition of Matrix

  1. Aug 15, 2016 #1
    Hi everyone.
    There is the ##2\times 2## matrix ##B##
    $$B=
    \left[
    \begin{array}{cc}
    B_{11} &B_{12} \\
    B_{21}&B_{22}
    \end{array}
    \right],~B_{ij}\in \mathbb{C}
    $$
    with property
    $$\vert B_{11}\vert^2 + \vert B_{12}\vert^2=1,$$
    $$\vert B_{21}\vert^2 + \vert B_{22}\vert^2=1,$$
    $$B_{11}B_{21}^{\ast}+B_{12}B_{22}^{\ast}=0.$$
    According to one of the texts, it is said that this matrix can be decomposed like
    $$B=e^{i\frac{\Lambda}{2}}
    \left[
    \begin{array}{cc}
    e^{i\frac{\Phi}{2}} & 0 \\
    0 & e^{-i\frac{\Phi}{2}}
    \end{array}
    \right]
    \left[
    \begin{array}{cc}
    \cos (\Theta/2) & \sin (\Theta/2) \\
    -\sin (\Theta/2) & \cos (\Theta/2)
    \end{array}
    \right]
    \left[
    \begin{array}{cc}
    e^{i\frac{\Psi}{2}} & 0 \\
    0 & e^{-i\frac{\Psi}{2}}
    \end{array}
    \right]
    $$
    $$\Lambda, \Phi, \Theta, \Psi \in \mathbb{R}$$.
    I don't know what kind of decomposition this is.
    Could someone tell me the name of this decomposition?

    Thanks.
     
  2. jcsd
  3. Aug 15, 2016 #2

    chiro

    User Avatar
    Science Advisor

    Hey Ken Gallock.

    The middle matrix is a rotation matrix (i.e. determinant equal to "positive" one and transpose is equal to "inverse") and the other two look like they are Hermitian (but you would have to double check everything at least once yourself to make sure).

    Look up matrix "decompositions" with rotations and Hermitian matrices and see if they fit the form for more information.
     
  4. Aug 15, 2016 #3

    fresh_42

    Staff: Mentor

  5. Aug 16, 2016 #4


    Thank you very much.
    I wasn't paying attention. ##B## is unitary matrix...

    Now, I have another question.
    Is this something to do with polar decomposition? When I first saw this decomposition, the first thing came up my mind was polar decomposition because I heard polar decomposition is similar to decomposition of complex number, that is
    $$z=\vert z \vert e^{i\theta}, (z\in \mathbb{C}).$$
     
  6. Aug 16, 2016 #5

    fresh_42

    Staff: Mentor

    Whether you write a complex number ##z## as ##z=a+ib## or as ##z=|z|e^{i\theta} , (\theta \in [0,2\pi))## doesn't matter. But you are right, the decomposition looks as if it is easier to calculate with polar coordinates. Only the factor in the middle, I guess, will have to be transformed by using ##\sin \varphi = \frac{1}{2i}(e^{i\varphi}-e^{-i\varphi})## and ##\cos \varphi = \frac{1}{2}(e^{i\varphi}+e^{-i\varphi})## then.
     
  7. Aug 16, 2016 #6

    chiro

    User Avatar
    Science Advisor

    You might want to look at every operator from right to left (compositions with these operators do their "work" from right to left since of how things are evaluated) and see the geometric intuition from that.

    For an interpretation of the outer linear operators I'd suggest a couple of things.

    The first is to "un-pack" the matrices (i.e. convert a 2x2 matrix into a 4x4 one if possible and use all real numbers) and the second is to see what a multiplication of two complex numbers do another complex number.

    Multiplication will always multiply the radii (the "r" term in the polar "decomposition") and add the angles so there is definitely a geometric intuition there (and the 4x4 un-packed definition should highlight the same dynamic).

    This means that according to what is happening, you are doing a set of multiplications (scaled ones) and rotations on two complex variables.

    That should give a bit more of the interpretation of what is going on.
     
  8. Aug 18, 2016 #7
    Thank you very much. Sorry for this late reply.
    Now I solved my problem.

    By the way, this decomposition can also be applied to quasi-unitary matrices. (but with ##\cos \rightarrow \cosh, \pm \sin \rightarrow \sinh##)
     
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