Nielsen & Chuang Exercise 2.55: Prove that $exp left[ -frac{iH(t_2 - t_1)}{hbar} right]$ is unitary

Question

$newcommand{expterm}[0]{frac{-iH(t_2 - t_1)}{hbar}}
newcommand{exptermp}[0]{frac{iH(t_2 - t_1)}{hbar}}$Nielsen & Chuang (10th edition, page 82) states that $H$ is a fixed Hermitian operator known as the Hamiltonian. In exercise 2.54, we prove that if $A$ and $B$ are commuting Hermitian operators, then the following holds:
$$
exp(A)exp(B)=exp(A+B) tag{1}label{1}
$$
The goal is to prove
$$exp left[ expterm right] exp left[ exptermp right] = I.tag{2}label{2}$$
If $expterm$ is Hermitian, then we can plug $A=expterm$ and its Hermitian conjugate $B = exptermp$ into eqref{1} to prove eqref{2}. However, I don't see why $expterm$ is necessarily Hermitian. Take $H=I$ for example: $H$ is Hermitian but $expterm$ is not, so we can't use eqref{1}. Any thoughts?

glS · Accepted Answer

If $H$ is Hermitian, then $iH$ is not Hermitian, but rather skew-Hermitian: $(iH)^dagger = -i H^dagger =-iH$.
Still, the identity in (1) holds generally for commuting matrices, they don't have to be Hermitian:
$$e^{A+B} = sum_{k=0}^infty frac{(A+B)^k}{k!}=sum_{k=0}^infty frac{1}{k!} sum_{j=0}^k binom{k}{j} A^j B^{k-j} = sum_{k=0}^infty sum_{j=0}^k frac{A^j B^{k-j}}{j! (k-j)!}
 = sum_{n,m=0}^infty frac{A^n B^m}{n! m!} = e^A e^B,$$
where the commutativity was necessary to use Newton's formula in the second step, and in the penultimate step we changed the summation variables with $n=j, m=k-j$.
More generally, $e^A$ is unitary if $A$ is skew-Hermitian, as $$(e^A)^dagger e^A = e^{A^dagger} e^A = e^{-A}e^A=I,$$
and similarly for $e^{A}(e^A)^dagger=I$.
Vice-versa, for any unitary $U$ there is always a skew-Hermitian $A$ such that $U=e^A$, see this question on math.SE.
See also this similar question on math.SE.

Nielsen & Chuang Exercise 2.55: Prove that $exp left[ -frac{iH(t_2 - t_1)}{hbar} right]$ is unitary

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