# Initial state for QAOA

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. "

• 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... Jul 26, 2021 at 8:32
• 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? Jul 27, 2021 at 1:44

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.