A Unitary Matrix and Hermitian Matrix

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SUMMARY

A unitary matrix U can be expressed as U=e^{iH}, where H is a Hermitian operator. To compute the matrix form of H from a known unitary matrix, such as the rotation matrix U=\left(\begin{array}{cc} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{array}\right), one must first determine the eigensystem of U. After normalizing the eigenbasis, the Hermitian matrix H can be derived, resulting in H=\left(\begin{array}{cc} 0 & i\theta \\ -i\theta & 0 \end{array}\right) for the given example. This method can also be extended to non-diagonal matrices.

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Demon117
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Its true that one can say a unitary matrix takes the form

[itex]U=e^{iH}[/itex]

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.
 
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Can you do it for diagonal matrices??

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

[tex]ADA^{-1}=e^{iAHA^{-1}}[/tex]
 
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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