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If $f(z)$ is a function with a Taylor series expansion $$f(z)=\sum _{ n=0 }^{ \infty }{c_n z^n },$$ then we define $$f(M)=\sum _{ n=0 }^{ \infty }{c_n M^n }.$$ First consider $M=\sigma_2=\sigma_y$. Using the definition for the function of an operator, show that $e^{i\sigma_y\theta}$ (here $\theta$ is any number) takes the form $$e^{i\sigma_y\theta}=A(\theta)1+B(\theta)\sigma_y$$ where $1$ is the identity operator, and $A(\theta)$ and $B(\theta)$ are functions that you are to determine. Repeat the same exercise, but now with $\sigma_x$ and $\sigma_z$.

Can anyone explain this question? Just a hint on how to solve it would be sufficient.

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As a hint: remember that $$\sigma_i^{2n}=\mathbb I$$ and that $$\sigma_i^{2n+1}=\sigma_i$$ so when you taylor expand $$e^{i\sigma_i\theta}$$ you will have the sum split in just two terms, one proportional to the identity and one to the sigma matrix.

EDIT:

an operator function of the form $$F=e^{\alpha M}$$ can be expanded as $$F=\sum \frac{(\alpha M)^n}{n!}$$ now if $\alpha=i \theta$ and $M=\sigma_i$ you will get$$F=\sigma_i\sum \frac{(i\theta)^{2n+1}}{(2n+1)!}+\mathbb I\sum\frac{(i\theta)^{2n}}{(2n)!}=A(\theta)\mathbb I+B(\theta)\sigma_i$$ where $$A(\theta)=\sum\frac{(i\theta)^{2n}}{(2n)!}\\B(\theta)=\sum \frac{(i\theta)^{2n+1}}{(2n+1)!}$$

as where the exponential comes from, i guess it's the rotation operator in the direction $i$ in the case of zero angular momentum.

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  • $\begingroup$ I don't understand where the $e^{i\sigma_y\theta}$ comes from. I get your other point but not the exponential one. $\endgroup$ – Paradox 101 Mar 2 '15 at 18:01
  • $\begingroup$ i've edited the answer $\endgroup$ – Fra Mar 2 '15 at 18:28
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    $\begingroup$ I think you could add explicit expressions to $A(\theta)$ and $B(\theta)$. For example $B(\theta) = \sum_{n=0}^\infty \frac{(i\theta)^{2n+1}}{(2n+1)!} = i\,\sum_{n=0}^\infty \frac{(-1)^n \,(\theta)^{2n+1}}{(2n+1)!} = i\,\sin(\theta) $ Similary $A(\theta) =\cos(\theta)$ $\endgroup$ – Héctor Mar 2 '15 at 21:00
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Try to compute $\sigma_2^2$ and $\sigma_2^3$ and see what you get. I think you can take it from here.

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  • $\begingroup$ I understand what I get by the squaring or raising $\sigma_y$ to some other powers, but i don't understand where the exponential comes from and what $A(\theta)$ and $B(\theta)$ might be $\endgroup$ – Paradox 101 Mar 2 '15 at 18:03

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