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Help with math operators

  1. Sep 6, 2006 #1


    where H is an operator.

    1. If

    [tex]H= \left(\begin{array}{cc}a & b\\c & d\end{array}\right)[/tex]

    in its matrix representation. Then what is U in its matrix representation.
    Im confused, is it

    [tex]U= \left(\begin{array}{cc}e^{iH(1,1)} & e^{iH(1,2)} \\e^{iH(2,1)} & e^{iH(2,2)} \end{array}\right)[/tex]

    where H(i,j) is the elements in H's matrix?

    2. The hermitian adjoint of U is


    (+ represents hermitian adjoint...couldnt find the correct symbol)

    2b. In matrix form the hermitian adjoint is the complex conjugate transposed?

    2.c In operator form ? lets say H=i*f(x) then


    Last edited: Sep 6, 2006
  2. jcsd
  3. Sep 6, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    Well, first of all, you need to review the chapter where you learnt about the exponential of a matrix.

    Second, the "\dagger" produces [itex] H^{\dagger} [/itex].

    Third, yes to the question preceeding the last. And no the last question.

    Operator analysis requires many things: specifications whether linear or not, boundedness, continuity, domains, ranges,... It's not that simple to find the adjoint, if it exists at all for a given operator.

  4. Sep 7, 2006 #3
    Thanks for your answer.

    Now i have read a little about functions of matrices.

    [tex]f(M) = \sum_{n} a_nM^n [/tex]

    Assume M is hermitian so it can be diagonlized.

    D=U\dagger MU

    Then we have

    [tex]f(M) = \sum_{n} a_nUD^n U\dagger
    = \sum_{n} a_nU\left(\begin{array}{cc}\lambda^n_1 & 0\\0 &\lambda^n_n\end{array}\right)U\dagger
    = U\left(\begin{array}{cc}\sum_{n} a_n \lambda^n_1 & 0\\0 & \sum_{n} a_n\lambda^n_n\end{array}\right)U\dagger


    where the 2*2 matrix really is a n*n diagonal matrix with D's eigenvalues in the diagonal. Wasnt sure how to do that in latex.
    Is the calculations above correct? If so all you need to do to get f(M) is to find the matrix U that diagonalize M and compute f(D) which is easy according to the last step in the calculations above.
    Last edited: Sep 7, 2006
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