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Matrix equation in SL(2,C)

  1. May 21, 2009 #1
    Hello,

    Is the law for matrix multiplication in SL(2,C) the same as usual ? I try to solve the equation [tex]A_{k}k_{0}A_{k}^{\dagger}=k[/tex] where [tex]k_0[/tex] corresponds to the unit vector [tex]\{0,0,1\}[/tex] and [tex]k[/tex] is an arbitrary vector, i.e.:

    [tex]k0=
    \left( \begin{array}{cc}
    2 & 0 \\
    0 & 0 \\
    \end{array} \right)
    [/tex]

    [tex]k=
    \left( \begin{array}{cc}
    1+n_3 & n_- \\
    n_+ & 1-n_3 \\
    \end{array} \right)
    [/tex]

    If I try to solve for
    [tex]A_k=
    \left( \begin{array}{cc}
    a & b \\
    c & d \\
    \end{array} \right)
    [/tex]

    this gives (where [tex]a*[/tex] is the conjugate of [tex]a[/tex]):
    [tex]A_{k}k_{0}A_{k}^{\dagger}=
    \left( \begin{array}{cc}
    2aa* & 2ac* \\
    2ca* & 2cc* \\
    \end{array} \right)
    [/tex]

    So this gives conditions on [tex]\{a,c\}[/tex] but can [tex]\{b,c\}[/tex] be arbitrary ? How do I solve this equation and obtain the expression of [tex]A_k[/tex] involving only [tex]n_+, n_-[/tex] and [tex]n_3[/tex] ?

    Thanks a lot for your help!
     
  2. jcsd
  3. May 21, 2009 #2

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    I'm too lazy to think about the rest, but I multiplied the matrices together and got the same result you did, so that part seems to be OK.
     
  4. May 21, 2009 #3
    Well, thanks Fredrik but now I really would like to know how to do "the rest" !
     
  5. May 21, 2009 #4

    gabbagabbahey

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    Homework Helper
    Gold Member

    I think you need to know the effect of [itex]A_k[/itex] on at least two linearly independent vectors to completely determine the matrix.
     
  6. May 21, 2009 #5
    Thanks, actually I found an answer for [tex]A_{k}[/tex] without the details of the calculation. I tried to compute [tex]A_{k}k_{0}A_{k}^{\dagger}[/tex] but I don't get [tex]k[/tex] as would be expected.

    The proposed solution is:

    [tex]A_k=\frac{1}{\sqrt{2C(1+n_3)}}
    \left( \begin{array}{cc}
    C(1+n_3) & -n_- \\
    Cn_+ & 1+n_3 \\
    \end{array} \right)
    [/tex]
    i.e.:
    [tex]A_{k}=UB[/tex]
    where:
    [tex]U=\frac{1}{\sqrt{2(1+n_3)}}
    \left( \begin{array}{cc}
    (1+n_3) & -n_- \\
    n_+ & 1+n_3 \\
    \end{array} \right)
    [/tex]
    and
    [tex]B=
    \left( \begin{array}{cc}
    \sqrt{C} & 0 \\
    0 & \frac{1}{\sqrt{C}} \\
    \end{array} \right)
    [/tex]

    The [tex]C[/tex] that appears is actually a constant in [tex]k[/tex] that I ignored for simplification in my first posting:

    [tex]k=C
    \left( \begin{array}{cc}
    1+n_3 & n_- \\
    n_+ & 1-n_3 \\
    \end{array} \right)
    [/tex]

    Can somebody else try to see if this result is correct ?
     
  7. May 22, 2009 #6

    George Jones

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    Staff Emeritus
    Science Advisor
    Gold Member

    Without further restrictions, I don't think that there is a unique solution for [itex]A_k[/itex]. What happens when your [itex]A_k[/itex] is multiplied on the right by an element of the little group of [itex]k_0[/itex]?
     
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