# Link between AQC and QAOA

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 ?