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Finding the matrix A such that, exp(sA) is in SU(2)

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Finding the matrix A such that, exp(sA) is in SU(2)

    2. Relevant equations

    My attempt is in trying to solve

    [tex]\left(e^{sA}\right)^{t} B \left(e^{sA}\right) = B [/tex]

    for A, where A is some 2x2 (complex?) matrix.

    and B is the matrix representing the group of SU(2) matrices. Trouble is I'm not sure what B is, but have been using the matrix of the general form of SU(2)

    B = [ [tex]\alpha, -\beta*; \beta, \alpha*[/tex]], where * denotes the conjugate

    3. The attempt at a solution

    [tex]\left(e^{sA}\right)^{t} B \left(e^{sA}\right) = B [/tex]

    simplifying to the solving of

    [tex]A^{t}[/tex]B + B A = 0

    which I'm not really having much success at doing. Anyone who knows more about this than me will see that I not really sure what I'm up to. Some help would be appreciated.

    Thanks, in anticipation..
     
  2. jcsd
  3. Oct 21, 2009 #2

    Dick

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    I don't see what B has to do with anything. You just want S=exp(sA) to be is SU(2), right? That means S is orthogonal, i.e. S^+=S^(-1) (where ^+ is hermitian conjugate) and det(S)=1. For the first one, S^(-1)=exp(-sA) and S^+=exp(sA^+) and compare the expansions of the exponentials (I'm assuming you mean s to be real). For the second condition use det(exp(A))=exp(trace(A)).
     
  4. Oct 21, 2009 #3
    Thanks, Dick - much appreciated.

    I got went down the track of using the approach with B as it was used on another example that I had available to me. Your approach seems like common sense ... now.

    Am I right that det(exp(sA))=exp(trace(A)) so the s does not figure in that calc?

    Therefore, I require, exp(trace(A)) = 1, hence A is traceless?

    Also, as an aside, am I correct in thinking that if s was not real, say s=i, then I can show that exp(iA) is unitary if A is Hermitian, so any traceless Hermitian matrix would be in SU(2)?

    Thanks again - I'll carry on working through what you've given me..
     
  5. Oct 21, 2009 #4

    Dick

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    Well, no, det(exp(sA))=exp(trace(sA)), you can't just leave the s out on one part. But if A is traceless, then sA is traceless. You reach the same conclusion. And, yes, if s=i then A must be Hermitian. Can you use the essentially the same argument to show if s is real, then A must be antihermitian? I.e. A^(+)=(-A)?
     
  6. Oct 21, 2009 #5
    Actually, I've just done the expansions of S^(-1)=exp(-sA) and S^+=exp(sA^+), and comparing these brings me to exactly that conclusion, that A^(+)=(-A), ie. A must be antihermitian.

    Adding the requirement, det(exp(sA))=1, would leave me concluding that A must be any traceless, antihermitian matrix. Does that sound on track?
     
  7. Oct 21, 2009 #6

    Dick

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    I believe you.
     
  8. Oct 21, 2009 #7
    Got it! Brilliant ... thanks again for your time.
     
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