Commutators, Unitary Operators Let $[a, a^\dagger]=aa^\dagger-a^\dagger a = 1$ and $[b, b^\dagger]=bb^\dagger-b^\dagger b = 1$
Show that:
$e^{\theta (a^\dagger b - b^\dagger a)}$ is unitary, where $\theta$ is a constant.
 A: Let $k:=a^*b-b^*a$. Then $k^*=-k$. So, for every $\theta\in\mathbb{R}$, $\theta k$ and $(\theta k)^*=-\theta k$ commute, whence
$$
e^{\theta k}e^{(\theta k)^*}=e^{(\theta k)^*}e^{\theta k}=e^{\theta k+ (\theta k)^*}=e^{\theta k-\theta k}=e^0=1. 
$$
Since $(e^{\theta k})^*=e^{(\theta k)^*}$, the latter proves that $u=e^{\theta k}$ is unitary.
Note: the assumptions $[a,a^*]=[b,b^*]=1$ are useless. But we need $\theta$ to be real in this argument.
A: Your $\theta$ needs to be real for this to hold, and the conditions on the commutators are not needed. 
Indeed, one can check explicitly that $(a^*b-b^*a)(b^*a-a^*b)=(b^*a-a^*b)(a^*b-b^*a)$, and so the product of the exponentials will be the exponential of the sum. So
$$
\left(e^{\theta(a^*b-b^*a)}\right)^*e^{\theta(a^*b-b^*a)}=e^{\theta(b^*a-a^*b)}e^{\theta(a^*b-b^*a)}=e^{\theta(b^*a-a^*b+a^*b-b^*a)}=e^0=I.
$$
The computation to show that $e^{\theta(a^*b-b^*a)}\left(e^{\theta(a^*b-b^*a)}\right)^*=I$ is completely similar. 
What's happening here is that the exponent is skew-symmetric, i.e. it is an operator $y$ such that $y^*=-y$. This implies that $y$ is normal, i.e. $y^*y=-y^2=yy^*$, and so 
$$
(e^y)^*e^y=e^{-y}e^y=e^{-y+y}=e^0=I.
$$
