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I'm learning about QAOA and I got curious about how they choose initial state. They somehow decided to choose initial state as equal superposition of all possible state and I wonder that there is any particular reason for that.

In 5.3 quantum circuit " We first implement 5 Hadamard H gates to generate the uniform superposition. "

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    $\begingroup$ regarding the last sentence: no, an equal superposition of the basis states for multiple qubits is not an entangled state of those qubits, it is the tensor product of equal superpositions on the individual qubits. I think you should remove that and only ask about the reasoning behind the initial state... $\endgroup$
    – M. Stern
    Commented Jul 26, 2021 at 8:32
  • $\begingroup$ Thank you for your comment, i edited it. I got too curious and mixed up my questions, sorry. Last sentence was : equal superposition of all possible state = maximally entangled state? $\endgroup$
    – hongildong1
    Commented Jul 27, 2021 at 1:44

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The reason is that state $H^{\otimes n} |0\rangle^ {\otimes n} = \frac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1}|i\rangle$, where $|i\rangle$ being a binary representation of decimal number $i$, is a ground state of Hamiltonian $$ \mathcal{H}_0 =\sum_{i=1}^{2^n} \sigma_i ^x, $$ where $\sigma_i ^x$ is $X$ gate applied on $i$th qubit whereas identity gate is applied on other qubits.

The Hamiltonian $H_0$ is used as an initial Hamiltonian for Ising model.

You can find more about this in my other answer.

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  • $\begingroup$ So did they just used ground state in the beginning? Shouldn't we supposed to find a ground state using variational method? I thought hamiltonian is fixed and we find ground state, but in your answer, it seems we already know the ground state and we are finding hamiltonian. Did i understand variational method wrong? $\endgroup$
    – hongildong1
    Commented Jul 28, 2021 at 5:56
  • $\begingroup$ @hongildong1: In fact, QAOA simulates a quantum annealer. The annealer is in a ground state all time. At the beginning in ground state of a known Hamiltonian, at the end in ground state of Hamiltonian whose ground state your are looking for. See the linked answer for more detailed explanation of quantum annealing. $\endgroup$
    – Martin Vesely
    Commented Jul 28, 2021 at 6:15
  • $\begingroup$ So QAOA uses adabatic theorem, but VQE or other variational algorithms don't use adiabatic theorem? In Variational quantum algorithms, is my understanding of variational method okay? $\endgroup$
    – hongildong1
    Commented Jul 28, 2021 at 6:31
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    $\begingroup$ @hongildong1: Yes, you are right. QAOA is actually simulation of a quantum annealer. VQE circuit is based on something another. However, the question was about QAOA :-) $\endgroup$
    – Martin Vesely
    Commented Jul 28, 2021 at 10:09
  • $\begingroup$ I'm so confusing... i heard that QAOA is sub-VQE. Your comment salvaged me. $\endgroup$
    – hongildong1
    Commented Jul 29, 2021 at 8:16
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A maximally entangled state is a state that has maximum mutual information of the random variables. I think you should first clear your concept about entanglement. http://www.cmi.ac.in/~neelraha/Resources/Internships/MayJuly2016/Maximally_Entangled_States.pdf

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  • $\begingroup$ I would suggest to post this as an comment. $\endgroup$
    – Martin Vesely
    Commented Jul 27, 2021 at 10:29
  • $\begingroup$ My answer was to the previous question. @hongildong1 if you have a new question do ask it separately next time. $\endgroup$
    – prathamesh bhole
    Commented Jul 27, 2021 at 13:47
  • $\begingroup$ I had a look at the original question and now I understand. No problem then. $\endgroup$
    – Martin Vesely
    Commented Jul 27, 2021 at 21:12
  • $\begingroup$ Sorry, my mistake. Your answer helped me a lot. Thank you $\endgroup$
    – hongildong1
    Commented Jul 28, 2021 at 6:06

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