$\newcommand{\expterm}[0]{\frac{-iH(t_2 - t_1)}{\hbar}} \newcommand{\exptermp}[0]{\frac{iH(t_2 - t_1)}{\hbar}}$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?

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    $\begingroup$ 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 $\endgroup$
    – C. Kang
    Commented Sep 3, 2020 at 19:09
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    $\begingroup$ @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? $\endgroup$
    – Attila Kun
    Commented Sep 3, 2020 at 20:05

1 Answer 1


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.


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