I am reading the book Abrikosov methods of quantum field theory in statistical physics. I got stuck at some derivation of the field operators for the free particle. The reasoning in the book is as follows. First particle field operators are defined as
$ \Psi(\xi)=\sum_{i}\varphi_{i}(\xi)a_{i} $
$ \Psi^{+}(\xi)=\sum_{i}\varphi^{*}_{i}(\xi)a^{+}_{i} $
where $\varphi_{i}(\xi)$ are wavefunctions of a particle in state i. $\xi$ is a combination of spin and position coordinates. $a_{i}$ and $a^{+}_{i}$ are the ladder operators. The field operators satisfy the same commutation/anticommutation relations as the ladder operators
$ [\Psi(\xi),\Psi^{+}(\xi^{'})]=\delta(\xi-\xi^{'}) $ and $[\Psi(\xi),\Psi(\xi^{'})]=0$ and $[\Psi^{+}(\xi),\Psi^{+}(\xi^{'})]=0$.
Then a free particle wavefunction (without spin) is introdcued
$ \frac{1}{\sqrt{V}}e^{i\mathbf{p}\mathbf{r}} $
where $\mathbf{p}$ is the momentum and $\mathbf{r}$ is the position.
The field operator $\Psi(\xi)$ in the Schroedinger picture for the free spinless particle is
$ \Psi(\mathbf{r})=\sum_{\mathbf{p}}e^{i\mathbf{pr}}a_{\mathbf{p}} $
The Hamiltonian is $ H = \sum_{\mathbf{p}}n_{\mathbf{p}}\epsilon_{0}(\mathbf{p}) $
where $n_{\mathbf{p}}=a^{+}(\mathbf{p})a(\mathbf{p})$ is the number of particle operator and $\epsilon_{0}(\mathbf{p})$ is the energy of a particle with momentum $\mathbf{p}$.
Now the field operator in the Heisenberg picture is introduced as $a(t)=e^{iHt}ae^{-iHt}$.
This is done for the free particle field operator as
$ \tilde{\Psi}(\mathbf{r},t)=\frac{1}{\sqrt{V}} \sum_{\mathbf{p}}e^{i\sum_{\mathbf{p}^{'}}\epsilon_{0}(\mathbf{p}^{'})n_{\mathbf{p}^{'}}t} a_{\mathbf{p}}e^{-i\sum_{\mathbf{p}^{''}}\epsilon_{0}(\mathbf{p}^{''})n_{\mathbf{p}^{''}}t} e^{i\mathbf{pr}}=\frac{1}{\sqrt{V}}\sum_{\mathbf{p}}a_{\mathbf{p}}e^{i(\mathbf{pr}-\epsilon_{0}(\mathbf{p})t)} $
This is exactly where I got stuck. Because I don't understand the mathematical identity or what "trick" was used to obtain the result after the second equal sign in the last equation.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Use Heisenberg Equation of Motion: \begin{align} \ic\,\totald{a_{\bf p}\pars{t}}{t} & = \bracks{a_{\bf p}\pars{t},H} = \expo{\ic Ht}\ \underbrace{\bracks{a_{\bf p},H}} _{\ds{\epsilon_{\bf p}\,a_{\bf p}}}\expo{-\ic Ht} = \epsilon_{\bf p}\,a_{\bf p}\pars{t} \\[5mm] \implies & a_{\bf p}\pars{t} = a_{\bf p}\expo{-\ic\epsilon_{\bf p}\,t} \end{align}