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Why? exp(ixA)=cos(x)I+isin(x)A if A*A=I

  1. Jun 16, 2007 #1
    why? exp(ixA)=cos(x)I+isin(x)A if A*A=I
  2. jcsd
  3. Jun 16, 2007 #2

    Gib Z

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    Your notation is not the friendliest..is I the same as i?

    Anyway [tex]e^{ixA}=\cos (xA) + i \sin (xA)[/tex] So i don't think it does...
  4. Jun 16, 2007 #3
    I think in the post I is the identity matrix, and A some matrix.
    The way you can see the identity you gave is true is if you expand the operator [itex]e^{ixA}[/itex] as a power series in x,

    e^{ixA} = \sum_{k=0}^{\infty}\frac{(ixA)^{k}}{k!}

    For all the even terms in the above summation [itex]A^{2k}=I[/itex], while for all the odd terms [itex]A^{2k+1}=A[/itex]. Resum the remaining series and you end up with the expression you quoted.
  5. Jun 16, 2007 #4

    George Jones

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    I assume [itex]A[/itex] is a linear operator or a matrix, that the space of operators has a norm, and there are convergence properties that justify my rearrangements (below) of infinites series. In physics, we typically pretend that everything is okay and proceed formally.

    From [itex]A^2 = 1[/itex], [itex]A^n = 1[/itex] if [itex]n[/itex] is even, and [itex]A^n = A[/itex] if [itex]n[/itex] is odd.

    \exp{ixA} &= 1 + ixA + \frac{1}{2!}\left(ixA\right)^2 + \frac{1}{3!}\left(ixA\right)^3 ...\\
    &= \left(1 - \frac{1}{2!} x^2A^2 + ...\right) + \left(ixA - \frac{1}{3!}ix^3A^3 + ...\right)\\
    &= \left(1 - \frac{1}{2!}x^2 + ... \right) + i\left(x - \frac{1}{3!}x^3 + ... \right)A
    Last edited: Jun 16, 2007
  6. Jun 16, 2007 #5
    thank you very much!!!

    thank you very much!!!
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