Prove $e^{i\alpha\hat{X}}=\cos(\alpha)I+i\sin(\alpha)\hat{X}$ if $\hat{X}^{2}=I$.

Question

If we have an operator $\hat{X}$ such that $\hat{X}^{2}=I$ (the identity), how do we prove that:

$$e^{i\alpha\hat{X}}=\cos(\alpha)I+i\sin(\alpha)\hat{X} \ ?$$

Answer 1

For simplicity, let me use $X$ instead of $\hat X$. \begin{equation} \begin{split} {\rm e} ^{{\rm i} \alpha X} &= \sum \limits _{n = 0} ^\infty \frac {{\rm i}^n \alpha^n X^n} {n!} \\&= \sum \limits _{n = 0 ; \ n \text{ even}} ^\infty \frac {{\rm i}^n \alpha^n X^n} {n!} + \sum \limits _{n = 0 ; \ n \text{ odd}} ^\infty \frac {{\rm i}^n \alpha^n X^n} {n!} \\&= \sum \limits _{k = 0} ^\infty \frac {{\rm i}^{2k} \alpha^{2k} X^{2k}} {(2k)!} + \sum \limits _{k = 0} ^\infty \frac {{\rm i}^{2k+1} \alpha^{2k+1} X^{2k+1}} {(2k+1)!} \\&= \sum \limits _{k = 0} ^\infty \frac {({\rm i}^2)^k \alpha^{2k} (X^2)^k} {(2k)!} + \sum \limits _{k = 0} ^\infty \frac {({\rm i}^2)^k {\rm i} \alpha^{2k+1} (X^2)^k X} {(2k+1)!} \\&= \sum \limits _{k = 0} ^\infty (-1)^k \frac {\alpha^{2k} I^k} {(2k)!} + {\rm i} X \sum \limits _{k = 0} ^\infty (-1)^k \frac {\alpha^{2k+1} I^k} {(2k+1)!} \\&= \sum \limits _{k = 0} ^\infty (-1)^k \frac {\alpha^{2k}} {(2k)!} I + {\rm i} X I \sum \limits _{k = 0} ^\infty (-1)^k \frac {\alpha^{2k+1}} {(2k+1)!} \\&= (\cos \alpha) I + {\rm i} (\sin \alpha) X , \end{split} \end{equation} where we have used the well-known series

$${\rm e}^z = \sum \limits _{n = 0} ^\infty \frac {z^n} {n!}, \quad \cos z = \sum \limits _{k = 0} ^\infty (-1)^k \frac {z^{2k}} {(2k)!}, \quad \sin z = \sum \limits _{k = 0} ^\infty (-1)^k \frac {z^{2k+1}} {(2k+1)!}$$

(in fact, these series are precisely the definitions of $\exp$, $\sin$ and $\cos$ in mathematical analysis).

Answer 2

Just remember that $e^{i \alpha X}$ is defined by its power series expansion. Recall for $e^y$ when $y$ is a real number: $$e^y = 1 + \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots \tag{1}$$ You may be familiar with De Moivre's Theorem where you replace $y$ by $iy$ in $(1)$ to get $$e^{iy} = \cos(y) + i \sin(y)\tag{De Moivre's Theorem}$$ To see this just look at the real and imaginary parts, using $(iy)^{2n} = (-1)^n y^{2n}$ and $(iy)^{2n+1} = i (-1)^n y^{2n+1}$ and remembering the power series expansions for $\cos(y)$ and $\sin(y)$. This also works if you replace $y$ in $(1)$ by $i\alpha X$, except now $X$ is a matrix. Just as before, expand the power series out this time using as well the property $X^{2n} = (X^2)^n = I^n = I$ and $X^{2n+1} = X\cdot X^{2n} = X\cdot I = X$.