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QM - linear algebra

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm given the Taylor expansion of [itex]e^{i\Theta M}[/itex]
    Then the question says "Compute the exponential for the given matrix M and verify that the resulting matrix is unitary.

    3. The attempt at a solution

    I really just don't know what they're asking...
  2. jcsd
  3. Sep 23, 2007 #2
    What's the given matrix? And do you know what unitary means?
  4. Sep 23, 2007 #3
    Given matrix [[3/4, 1/4][1/4, 3/4]] and yeah I know what unitary means.
  5. Sep 23, 2007 #4


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    Cscott, is that a hermitian matrix ?
  6. Sep 23, 2007 #5
    Yes it is.

    I just really don't know what they're asking me to do.
  7. Sep 23, 2007 #6
    So if you have the expression for exponentiating a matrix, and the matrix to exponentiate, it seems to me like you can do it. What's the problem?
  8. Sep 23, 2007 #7
    They give me the Taylor expansion for that exponential (I was just too lazy to latex it)... wouldn't it be stupid to evaluate a random amount of terms?

    [tex]e^{i\theta M} = \Sigma\frac{\left(i\theta\right)^n}{n!}M^n[/tex] from n=0 to infinity
  9. Sep 23, 2007 #8
    I suppose having the diagonalized version of the matrix would help for raising it to a power.

    Is it correct that the diagonalized version would be [[2, 0][0, 1]]?
  10. Sep 23, 2007 #9
    If you put M into the form UDU* where U is a unitary matrix, D is diagonal, and U* is the inverse (or transpose) of U, does that help?

    And no, that diagonalised form is not quite right.
    Last edited: Sep 23, 2007
  11. Sep 23, 2007 #10
    Mmm... how about D= [[1, 0][0, 1/2]] ?

    Is there a maple command for doing this?
  12. Sep 23, 2007 #11
    I don't know about Maple. But it's pretty easy to do by hand -- it's not like a 2x2 matrix is that big...
  13. Sep 23, 2007 #12
    Yeah I know but I just wanted a quick way to verify it.
  14. Sep 23, 2007 #13
    In the diagonal form, the elements are the eigenvalues. To find the U matrix I mentioned above, you'll need the eigenvectors anyway. So just act M on the eigenvectors and you should end up with the respective eigenvalues. That should be check enough, and quick.
  15. Sep 23, 2007 #14
    I'm pretty sure the answer is D= [[1, 0][0, 1/2]] but if I don't know how high n should go how can I get a matrix out of the Taylor expansion they gave me?
  16. Sep 23, 2007 #15
    n is for all n: 0 to infinity. D^n for any diagonal matrix should be easy to evaluate. U*U = I for unitary matrices. And remember that matrix multiplication is distributive: AB+AC=A(B+C). Does that help?
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