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A Unitary Matrix and Hermitian Matrix

  1. Dec 9, 2011 #1
    Its true that one can say a unitary matrix takes the form


    where [itex]H[/itex] is a Hermitian operator. Thats great, and it makes sense, but how can you compute the matrix form of [itex]H[/itex] if you know the form of the unitary matrix [itex]U[/itex]. For example, suppose you wanted to find [itex]H[/itex] given that the unitary matrix is one of the familiar rotation matrices (2 x 2) for simplicity. Let's say

    [itex]U=\left(\begin{array}{cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{array}\right)[/itex]

    What would the procedure be in finding the matrix form of [itex]H[/itex]? I suppose you could start by finding the eigensystem of the unitary matrix. Then, upon normalizing the eigenbasis of [itex]U[/itex], somehow you could find the matrix representation of [itex]H[/itex]. Any pointers or suggestions would be great.
  2. jcsd
  3. Dec 9, 2011 #2
    Can you do it for diagonal matrices??

    To extend it to nondiagonal matrices, notice that if [itex]D=e^{iH}[/itex], then

  4. Dec 10, 2011 #3
    So after some fiddling I find that the appropriate Hermitian matrix takes the form

    [itex]H=\left(\begin{array}{cc} 0 & i\theta \\ -i\theta & 0 \end{array}\right)[/itex]

    If this is indeed correct then I think I have what I need.
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