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If the eigenvalues of $A$ are $\ell /2^n$ with $0\le \ell <2^n$, exact quantum phase estimation on $U=e^{2\pi iA}$ uses $n$ controlled powers $U^{2^j}$. Their Hamiltonian simulation times reach $2\pi 2^{n-1}$. Counting each supplied controlled power as one oracle query is valid in that oracle model, but does not prove a polynomial time implementation from generic sparse-matrix access. In the ordinary HHL algorithm, one instead chooses finite precision appropriate to the output error and the condition number, or exploits additional structure giving efficient access to the needed powers.

The usual sufficient input promises for the HHL algorithm are efficient quantum state preparation of the nonzero vector $b$, efficient access to a sparse matrix $A$ with at most $\mathrm{poly}(n)$ nonzero entries per row, and a condition number $\kappa ={\lambda }_{\max }/{\lambda }_{\min }=\mathrm{poly}(n)$ after an efficiently known normalization. Both the positions and values of the nonzero matrix elements must be computable coherently in polynomial time. More generally, efficient Hamiltonian simulation can replace the sparsity promise. We assume $A$ is invertible; if zero eigenvalues occur, one must instead restrict $b$ to their orthogonal complement and specify the desired inverse on that support.

With $a={\lambda }_{\max }$ and a known bound $\kappa$, choose $c=a/\kappa \le {\lambda }_{\min }$. The HHL controlled reciprocal rotation then succeeds with probabilityThis is the required inverse-polynomial lower bound. In the usual approximate algorithm, inverse-polynomial requested error also gives polynomial runtime under these input promises. The original HHL paper states the dependence on sparsity, conditioning, and accuracy.

There is a qualification for a literally exact version. Representability of the eigenvalues in $n$ bits does not itself make exact quantum phase estimation efficient: using $e^{2\pi iA}$ requires controlled powers corresponding to evolution times as large as $2\pi 2^{n-1}$. Under the paper's idealization we may describe their exact action, but polynomial runtime for that exact circuit additionally requires efficient implementations of those controlled powers, or equivalent efficient exact spectral access. Assuming an operation executes exactly does not bound its cost. This distinction is recorded in cost of exact phase estimation on a dyadic spectrum.

Let the spectral decomposition of the Hermitian matrix beThe HHL algorithm proceeds as follows.

Efficiently prepare the normalized amplitude encoding $|b⟩$.
Apply quantum phase estimation to the simulated evolution $e^{iAt}$, producing an approximation of each eigenvalue: Add one ancilla and perform an eigenvalue-controlled rotationwhere $0<C\le {\min }_j|{\lambda }_j|$.
Uncompute the eigenvalue register. Conditional on measuring the ancilla as $1$, the system register isThe postselection probability can be increased with amplitude amplification.

The ingredients used here are efficient sparse Hamiltonian simulation of $e^{iAt}$ and quantum phase estimation, which converts an eigenphase of that evolution into a binary approximation of ${\lambda }_j$. For sparsity $s$, condition number $\kappa$, and error $ϵ$, the cost is polynomial in $s$, $\kappa$, $1/ϵ$, and $\log N$ under the stated access assumptions.

Finally estimate $⟨x|M|x⟩$ by repeated measurement of an efficient observable decomposition of $M$, or by a Hadamard test when $M$ is unitary. A general efficiently block-encoded Hermitian $M$ can similarly be measured through its block encoding. Repetition and a concentration inequality give additive sampling error $O(1/\sqrt{R})$ after $R$ independent preparations.

Each summand of $J_X$ acts on one qubit, while each summand $Z_i{Z}_j$ of $J_Z$ acts on two. Therefore $J=J_X+J_Z$ is a 2-local Hamiltonian. We haveSplit $t$ into $r$ steps of length $\delta =t/r$ and use the second-order product formulaFor one step, $\Lambda =O(\delta n^2)$ in the stated estimate, so the spectral-norm error is $O({\delta }^3{n}^6)$. The error bound for a product of unitary operators makes the total errorIt is therefore enough to choosewith $r$ also large enough that the small-step estimate applies.

All $Z_i{Z}_j$ terms commute. A factor $e^{-i\delta Z_i{Z}_j}$ uses a constant-size circuit of two controlled-NOT gates and one phase gate, up to a global phase, so one product-formula step costs $O(n^2)$ gates. The complete Hamiltonian simulation consequently has size

A two-qubit gate is a unitary operator on two qubit lines, extended by the identity on the others. It may be entangling, as a CNOT gate is, or a product of local operations. In a Hamiltonian simulation model allowing arbitrary two-qubit gates, the exponential of a Hermitian term supported on two lines counts as one such gate. Compiling it into a fixed discrete universal quantum gate set introduces a separate approximation cost.