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Math Help - exp(iH), where H is a matrix

  1. #1
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    exp(iH), where H is a matrix

    Is there any general way to express exp(iH), where H is a matrix, to another form? Of course, if H is trivial (one by one), there is, as well as if H = aI for a scalar a, or if H is a 2x2 rotation matrix, and other special cases. But otherwise? If it's any help, I'm only interested in unitary matrices.

    I suspect the answer is a simple no.
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  2. #2
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    Well, you can always use the Taylor's series form for the exponential:
    e^{iH}= I+ iH- \frac{1}{2}H^2- \frac{1}{6}iH^3+ \frac{1}{4!}H^4+ \cdot\cdot\cdot+ \frac{1}{n!}i^nH^n+ \cdot\cdot\cdot.

    In particular, if H is diagonalizable, that is, if H= PDP^{-1} where D is diagonal and P some invertible matrix, (D will have the eigenvalues of H on its diagonal, and P is the matrix with the eigenvectors of H as columns) then
    e^{ih}= Pe^{iD}P^{-1}= P(cos(D)+ i sin(D))P^{-1}
    where "cos(D)" is the diagonal matrix having cos(\lambda_i) on the diagona and "sin(D)" is the diagonal matrix having sin(\lambda_i) on the diagonal- \lambda_i being the eigenvalues of H.
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  3. #3
    A Plied Mathematician
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    Is H Hermitian?
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  4. #4
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    First, to Ackbeet: yes, it is Hermitian. Does this allow you to add anything? It would be most appreciated.

    To HallsofIvy: Silly me, I forgot to add a word "finite" before the word form. Yes, the Taylor Series expansion is the only way I can see to interpret any matrix that is not diagonizable, but I was hoping for a finite form. But the interpretation of sin(D) as sin(a) for every a in the diagonal is most enlightening: I did not know that would work. Thanks again.
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  5. #5
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    It works for a diagonal matrix. Again, you can see that by looking at the Taylor's series expansion of sine and cosine.
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  6. #6
    A Plied Mathematician
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    I believe that if H is Hermitian, it follows that e^{iH} is unitary. Look at the last property here. Note that iH is skew-Hermitian iff H is Hermitian.
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  7. #7
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    Very useful, both answers. Again, many thanks.
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  8. #8
    A Plied Mathematician
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    You're certainly very welcome for my contribution.
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