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The question is simple: why do we seek to maximise $F_{p}(\gamma, \boldsymbol{\beta})=\langle\gamma, \boldsymbol{\beta}|H_{C}| \gamma, \boldsymbol{\beta}\rangle$? How does maximising this value correspond to finding the groundstate value of $H_{C}$ ? Why does this optimum value of $H_{C}$ correspond to a set of $ \gamma, \boldsymbol{\beta} $ that maximise our chances of getting the optimum solution?

We try to minimise the value of $\langle \gamma,\beta|H_{C}| \gamma, \beta\rangle$ as this gives us the groundstate, as pointed out in this answer.

The Fahri paper says we seek to maximise this value. Other sources try to minimise it: are these just equivalent attempts with a minus sign stuck in front? I know that when we encode the problem Hamiltonian, we do this in way that the groundstate corresponds to the optimum solution. Why then has it been found that the set of $ \gamma,\beta $ that give the best probability of success don't always correspond to the max value of $H_{c}$ ?

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The image is taken from https://arxiv.org/abs/1907.02359.

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Whether it is minimize or maximize, it is the same up to a minus sign indeed.

Why we optimize with this expression, is because the original theory tells you that, if you could have infinite depth ($p \rightarrow \infty$), this expression converges to the ground state. Think of it as a probability distribution over all possible bitstrings , that you drive with optimization, whose expectation value gets closer to the optimal value, while decreasing variance.

But in practice, nothing prevents you from optimizing a different objective (for instance maximizing the expectation over a quantile of measurements giving you the best values). Check this paper for instance. Then you will have different angles $\gamma, \beta$ for this objective.

In fact, optimizing $F_{p}(\gamma, \boldsymbol{\beta})$ is optimizing the average value, which in the process means you want a good performance in average, and you may in the process privilege bitstrings with good values rather than the optimal one(s) cause it will give you a better average output.

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  • $\begingroup$ Hi, thanks for this anwer, and sorry for my slow reponse. I thought that if $|\gamma, \beta \rangle$ was an eigenstate of $C$ then maximising $F_{p}(\gamma, \beta)$ corresponded to approximating the highest energy eigenstate of $H_{C}$ $\endgroup$
    – jolene
    Apr 21 at 12:08

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