Exponentiate matrices

  • #1
Consider the operator Ô, choose a convenient base and obtain the representation of
[tex]
\ exp{(iÔ)}
[/tex]
[tex]
Ô =
\bigl(\begin{smallmatrix}
1 & \sqrt{3} \\
\sqrt{3} & -1
\end{smallmatrix}\bigr)
[/tex]

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 [tex] exp{({P_{}}^{-1}AP)} = {P_{}}^{-1}exp{(A)}P [/tex]

The proof is not so hard, we basically expand the exponential and note that: [tex] {({P_{}}^{-1}AP)}^m = P^{-1} A^mP[/tex]


So, finding the eigenvalues and eigenvectors of O, we get:
[tex] exp{(iÔ)} =
\begin{pmatrix}
-exp{(-2i)}-3exp{(2i)} & 3exp{(-2i)}-3exp{(2i)} \\
exp{(-2i)}-exp{(2i)} & -3exp{(-2i)}-exp{(2i)}
\end{pmatrix}
[/tex]

Now i'm asked to write [tex] |v> =
\begin{pmatrix}
1\\ 0

\end{pmatrix} [/tex]
on the eigenbase of [tex] exp{(iÔ)} [/tex]

Can you guys help me out? I don't really get what they ask me to do :/
 

Answers and Replies

  • #2
Strilanc
Science Advisor
607
224
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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