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

$$\newcommand{\expterm}{\frac{-iH(t_2 - t_1)}{\hbar}} \newcommand{\exptermp}{\frac{iH(t_2 - t_1)}{\hbar}}$$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?

• Am I missing something here: why isn't $\frac{-i H(t_2 - t_1)}{h}$ hermitian when $H$ hermitian? You are just multiplying $H$ by a constant Sep 3 '20 at 19:09
• @C.Kang Let $H \equiv I$ and $K\equiv \frac{-i H(t_2 - t_1)}{h}$. Then $K^\dagger = \left( \frac{-i I(t_2 - t_1)}{h} \right)^\dagger = \frac{i I(t_2 - t_1)}{h}$. So $K^\dagger=-K$ but we need $K^\dagger=K$ for Hermiticity, right? Sep 3 '20 at 20:05

If $$H$$ is Hermitian, then $$iH$$ is not Hermitian, but rather skew-Hermitian: $$(iH)^\dagger = -i H^\dagger =-iH$$.
$$e^{A+B} = \sum_{k=0}^\infty \frac{(A+B)^k}{k!}=\sum_{k=0}^\infty \frac{1}{k!} \sum_{j=0}^k \binom{k}{j} A^j B^{k-j} = \sum_{k=0}^\infty \sum_{j=0}^k \frac{A^j B^{k-j}}{j! (k-j)!} \\ = \sum_{n,m=0}^\infty \frac{A^n B^m}{n! m!} = e^A e^B,$$ 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.