Why? exp(ixA)=cos(x)I+isin(x)A if A*A=I

[dream_chaser]

why? exp(ixA)=cos(x)I+isin(x)A if A*A=I

 

[Gib Z]

Your notation is not the friendliest..is I the same as i?

Anyway $$e^{ixA}=\cos (xA) + i \sin (xA)$$ So i don't think it does...

 

[jpr0]

I think in the post I is the identity matrix, and A some matrix.
The way you can see the identity you gave is true is if you expand the operator $e^{ixA}$ as a power series in x,

$$e^{ixA} = \sum_{k=0}^{\infty}\frac{(ixA)^{k}}{k!}$$

For all the even terms in the above summation $A^{2k}=I$, while for all the odd terms $A^{2k+1}=A$. Resum the remaining series and you end up with the expression you quoted.

 

[George Jones]

I assume $A$ is a linear operator or a matrix, that the space of operators has a norm, and there are convergence properties that justify my rearrangements (below) of infinites series. In physics, we typically pretend that everything is okay and proceed formally.

From $A^2 = 1$, $A^n = 1$ if $n$ is even, and $A^n = A$ if $n$ is odd.

$$\begin{equation*} \begin{split} \exp{ixA} &= 1 + ixA + \frac{1}{2!}\left(ixA\right)^2 + \frac{1}{3!}\left(ixA\right)^3 ...\\ &= \left(1 - \frac{1}{2!} x^2A^2 + ...\right) + \left(ixA - \frac{1}{3!}ix^3A^3 + ...\right)\\ &= \left(1 - \frac{1}{2!}x^2 + ... \right) + i\left(x - \frac{1}{3!}x^3 + ... \right)A \end{split} \end{equation*}$$

 

[dream_chaser]

thank you very much!

thank you very much!