Taylor series for $e^{i(H+\varepsilon A)}$

Question

$$e^{i(H+\varepsilon A)} = e^{iH}(I+i \varepsilon A + o(\varepsilon))$$

Is it correct? $H$ and $A$ are hermitian matrix and $[H,A]\neq0$. If the one I gave is not the exact solution how do I expand it in series?

Use the power-series representation of $e^X$ with $X = A + \epsilon B$ and use that $(A+\epsilon B)^n = A^n + \epsilon (A^{n-1}B + \ldots + B A^{n-1}) + \mathcal{o}(\epsilon)$ to get a series which will be something like $e^{A+\epsilon B} = e^{A}(1 + \epsilon B) + \epsilon (\ldots)$ (with the last term being compilated and containing all sorts of commutators). Will not be pretty. — Winther, Aug 06 '18 at 15:32
We can check that $$ e^{X+\epsilon Y} = e^X \left( 1 + \epsilon \int_{0}^{1} e^{-tX} Y e^{tX} , dt + \mathcal{O}(\epsilon^2) \right). $$ — Sangchul Lee, Aug 06 '18 at 15:41

score 1 · Accepted Answer · answered Aug 08 '18 at 15:30

Here is a simple trick to expand OP's matrix exponential. Write $X = iH$ and $Y = i\epsilon A$ for brevity. Next, introduce the following function

$$ f(t) = e^{-tX}e^{t(X+Y)}, \qquad a(t) = e^{-tX}Ye^{tX}. $$

Then it is easy to check that

$$ f'(t) = e^{-tX}Ye^{t(X+Y)} = a(t)f(t). $$

So it follows that $f$ is the ordered exponential of $a$. In particular,

$$ f(1) = 1 + \int_{0}^{1} a(t_1) \, dt_1 + \int_{0}^{1}\int_{0}^{t_1} a(t_1)a(t_2) \, dt_2 dt_1 + \cdots $$

From the estimate $\left\| \int_{0}^{1}\cdots\int_{0}^{t_{k-1}} a(t_1)\cdots a(t_k) \, dt_k \cdots dt_1 \right\| \leq \frac{e^{2k\|X\|}\|Y\|^k}{k!}$, we realize that the formula above correctly produces the expansion of $e^{X+Y}$ as perturbation of $e^X$. Plugging the substitution back, we obtain

$$ e^{i(H+\epsilon A)} = e^{iH} \left( 1 + i\epsilon \int_{0}^{1} e^{-itH}Ae^{itH} \, dt + \mathcal{O}(\epsilon^2) \right). $$

Taylor series for $e^{i(H+\varepsilon A)}$

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