I try to understand precisely the link between AQC and QAOA, through the Trotter-Suzuki formula. A similar question is Derivation of QAOA from AQC, but I was asked by moderators to post my question independently. So here it is.
In quantum adiabatic computing, the Hamiltonian is as follows : \begin{equation}\tag{1} \mathcal{H} (t)= s(t)\mathcal{H}_{c}+ (1-s(t))\mathcal{H}_{m} \end{equation} with $s(0)=0$ and $s(T)=1$. The duration time $T$ is supposed to guarantee that the final state $|\psi\rangle(T)$ is $\varepsilon > 0$ close to the ground state of $\mathcal{H}_{c}$, the smaller $\varepsilon$, the greater $T$.
The Schrödinger equation gives : \begin{equation} U(t)= e^{-\frac{i}{\hbar}\int_{0}^{T}\mathcal{H}(u)\,du}= e^{-\frac{i}{\hbar}\biggl(\mathcal{H}_{c}\int_{0}^{T}s(u)\,du+\mathcal{H}_{m}\int_{0}^{T}[1-s(u)]\,du\biggr)} = e^{-\frac{i}{\hbar}\biggl(\mathcal{H}_{c}S(T)+\mathcal{H}_{m}[T-S(T)]\biggr)} \end{equation}
Applying the Trotter-Suzuki formula would give : \begin{equation}\tag{2} U(T)\approx \biggl[e^{-\frac{i}{\hbar}\mathcal{H}_{m} \frac{T-S(T)}{n}}e^{-\frac{i}{\hbar}\mathcal{H}_{c}\frac{S(T)}{n}}\biggr]^{n} \end{equation}
for $n\gg 1$ sufficiently high.
This formula is great as it lets one approach $U(T)$ by successively applying Hamiltonian $\mathcal{H}_c$ or $\mathcal{H}_m$ (the mixer) with some coefficients.
On the other hand, the idea usually developed to draw a link between AQC and QAOA is : \begin{equation}\tag{3} U(T)= U(t_n, t_{n-1})U(t_{n-1}, t_{n-2})\dots U(t_{2}, t_{1})U(t_{1}, t_{0}) \end{equation} with $t_0 = 0$ and $t_n = T$.
This formula derives from the exact formula : \begin{equation} U(t)= e^{-\frac{i}{\hbar}\int_{0}^{T}\mathcal{H}(u)\,du}= e^{-\frac{i}{\hbar}\biggl[\int_{0}^{t_1}\mathcal{H}(u)\,du+\dots+\int_{t_{n-1}}^{t_n}\mathcal{H}(u)\,du\biggr]} =e^{-\frac{i}{\hbar}\int_{0}^{t_1}\mathcal{H}(u)\,du}\dots e^{-\frac{i}{\hbar}\int_{t_{n-1}}^{t_n}\mathcal{H}(u)\,du} \end{equation}
For each time interval : \begin{equation} U(t_{i+1},t_{i}) = e^{-\frac{i}{\hbar}\int_{t_{i}}^{t_{i+1}}\mathcal{H}(u)\,du}= e^{-\frac{i}{\hbar}(\mathcal{H}_{c}\gamma_{i}+\mathcal{H}_{m}\beta_{i})} \end{equation}
Given that : \begin{equation} \begin{split} \gamma_{i}&= \int_{t_{i}}^{t_{i+1}}s(u)\,du\simeq s(t_{i})\,dt_{i}\simeq s(t_{i}) \frac{T}{n}\\ \beta_{i}&= \int_{t_{i}}^{t_{i+1}}[1-s(u)]\,du\simeq [1 -s(t_{i})]\,dt_{i}\simeq [1 -s(t_{i})]\frac{T}{n} \end{split} \end{equation}
The time segment can be approached again by Trotter Suzuki formula : \begin{equation} U(t_{i+1},t_{i}) = e^{-\frac{i}{\hbar}\int_{t_{i}}^{t_{i+1}}\mathcal{H}(u)\,du}= e^{-\frac{i}{\hbar}(\mathcal{H}_{c}\gamma_{i}+\mathcal{H}_{m}\beta_{i})} \simeq e^{-\frac{i}{\hbar}\mathcal{H}_{c}\gamma_{i}} e^{-\frac{i}{\hbar}\mathcal{H}_{m}\beta_{i}} \end{equation} where we used Eq. (2) but with $n=1$ (not the same "n" as in $t_n$) as $\gamma_i$ and $\beta_i$ are already "small" enough.
One could also have used a second order approximation :
\begin{equation} U(t_{i+1},t_{i}) \simeq e^{-\frac{i}{\hbar}\mathcal{H}_{c}\frac{\gamma_{i}}{2}} e^{-\frac{i}{\hbar}\mathcal{H}_{m}\beta_{i}} e^{-\frac{i}{\hbar}\mathcal{H}_{c}\frac{\gamma_{i}}{2}} \end{equation}
Then, this draws a direct link between Eq. 1 (AQC) and Eq. 2 (QAOA) : \begin{equation} \begin{split} |\psi\rangle (T)&= U_{m}(\beta_{p}) U_{c}(\gamma_{p})U_{m}(\beta_{p-1}) U_{c}(\gamma_{p-1})\dots U_{m}(\beta_{1}) U_{c}(\gamma_{1})|\psi (0)\rangle\\ &= e^{-i\beta_{p}H_{m}} e^{-i\gamma_{p}H_{c}}e^{-i\beta_{p-1}H_{m}} e^{-i\gamma_{p-1}H_{c}}\dots e^{-i\beta_{1}H_{m}} e^{-i\gamma_{1}H_{c}}|\psi (0)\rangle \end{split} \end{equation}
where $|\psi (0)\rangle$ is the ground state of $\mathcal{H}_m$.
Is my understanding correct ?
