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Measuring the expectation value of a hamiltonian is an essential step in some algorithms like QAOA.

I noticed that the procedure always starts with decomposing the hamiltonian into a sum of Pauli strings, then proceeding to building separate circuits to measure the expectation value, then adding them all.

It seems like it would be simpler to simply measure the state, compute the hamiltonian on a classical computer for the bit string measured, repeat and take the average.

Is there a reason why we like to measure the expectation value using the Pauli decomposition method instead? I'm guessing the answer would be it's more efficient. But I can't see why or how. It seems like we need to repeat many measurements for both procedures.

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  • $\begingroup$ What do you mean by “compute the Hamiltonian for the bitstring measured”? $\endgroup$
    – Lior
    Mar 14, 2022 at 6:41
  • $\begingroup$ @Lior so for example if I measure the state and I get "110", I can give this value to a classical computer that can compute H(6). If it's an Ising Hamiltonian like what we use for QAOA, it can be computed on a classical computer. $\endgroup$ Mar 14, 2022 at 7:00

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This paragraph from Peruzzo et al. can help.:

"Thus the evaluation of $\langle{H}\rangle$ reduces to the sum of a polynomial number of expectation values of simple Pauli operators for a quantum state $\mid\psi_i\rangle$, multiplied by some real constants. A quantum device can efficiently evaluate the expectation value of a tensor product of an arbitrary number of simple Pauli operators, therefore with an $n$-qubit state we can efficiently evaluate the expectation value of this $2^n \times 2^n$ Hamiltonian."

In other words, Pauli decomposition decomposes the Hamiltonian into states that are easily preparable on a quantum computer.

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  • $\begingroup$ this makes sense, so basically a polynomial number of gates is better than directly evaluating the hamiltonian $\endgroup$ Mar 16, 2022 at 5:31
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If I understand you correctly, then you suggest to calculate the expectation value of the Hamiltonian as $$\langle \hat H\rangle=\sum_{i=1}^{2^N}p_i \langle i|\hat H| i\rangle$$ where $N$ is the number of qubits and $p_i$ is the probability to measure state $|i\rangle$.

If you do this, you're neglecting off-diagonal elements. This can easiest be seen by considering the state $$|{\Psi}\rangle=\frac{1}{\sqrt 2}\left(|0\rangle+|1\rangle\right)$$ with expectation value $$\langle \Psi|\hat H| \Psi\rangle=\frac{1}{2}(\langle 1|\hat H| 1\rangle+\langle 0|\hat H| 0\rangle+\langle 0|\hat H| 1\rangle+\langle 1|\hat H| 0\rangle)$$

where off-diagonal elements of the Hamiltonian are nonzero in general.

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  • $\begingroup$ But this isn't the case for Ising Hamiltonians (what we use to encode cost functions for QAOA).They only have diagonal elements since the matrix is made up of only I's and Z's $\endgroup$ Mar 14, 2022 at 16:57

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