Exponentiating Matrices: Representation of \exp{(iÔ)}

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The discussion focuses on the representation of the matrix exponential exp(iÔ) for the operator Ô given by a specific 2x2 matrix. It highlights the method of diagonalizing the matrix when possible and using the lemma that relates the exponential of a transformed matrix to the original matrix. The eigenvalues and eigenvectors of Ô are computed to facilitate the exponentiation process. The final representation of exp(iÔ) is provided, but there is confusion regarding how to express a specific vector |v> in the eigenbasis of exp(iÔ). The thread seeks clarification on performing the eigendecomposition and reconstructing the matrix with the exponentiated eigenvalues.
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Consider the operator Ô, choose a convenient base and obtain the representation of
<br /> \ exp{(iÔ)}<br />
<br /> Ô =<br /> \bigl(\begin{smallmatrix}<br /> 1 &amp; \sqrt{3} \\<br /> \sqrt{3} &amp; -1<br /> \end{smallmatrix}\bigr)<br />

Attempt at solution:
So, i read on Cohen-Tannjoudji's Q.M. book that if the matrix is diagonal you can just exponentiate the diagonal elements and that's pretty much it. However, when the matrix is not diagonal, you have to use something like the following:
Lemma: A, P complex matrices, n by n, let's suppose P is invertible, then exp{({P_{}}^{-1}AP)} = {P_{}}^{-1}exp{(A)}P

The proof is not so hard, we basically expand the exponential and note that: {({P_{}}^{-1}AP)}^m = P^{-1} A^mPSo, finding the eigenvalues and eigenvectors of O, we get:
exp{(iÔ)} = <br /> \begin{pmatrix}<br /> -exp{(-2i)}-3exp{(2i)} &amp; 3exp{(-2i)}-3exp{(2i)} \\<br /> exp{(-2i)}-exp{(2i)} &amp; -3exp{(-2i)}-exp{(2i)}<br /> \end{pmatrix}<br />

Now I'm asked to write |v&gt; = <br /> \begin{pmatrix}<br /> 1\\ 0<br /> <br /> \end{pmatrix}
on the eigenbase of exp{(iÔ)}

Can you guys help me out? I don't really get what they ask me to do :/
 
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Compute the eigendecomposition of the matrix. Exponentiate the eigenvalues. Put the matrix back together with the new eigenvalues.

##M = \sum_\lambda \lambda \;|\lambda\rangle \langle \lambda|##

##F(M) = \sum_\lambda F(\lambda) \; |\lambda\rangle \langle \lambda|##

Where the ##\lambda##s in the sum are the eigenvalues/vectors.
 
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