On a quantum computer the trade differs from the classical one: each implicit step is itself a linear solve, so implicit stepping does not remove the quantum-linear-solve layer below — it invokes it repeatedly, or folds the whole trajectory into one larger block system and hands that over instead. Written out step by step it is one solve per time step, $T/h$ of them to reach time $T$, and each step's solve consumes the previous step's output as its right-hand side, so the chain is a quantum state passed forward and never a number read out. That is what makes the repetition expensive rather than merely long: a quantum linear solve succeeds only on a flagged branch, and the flags multiply down the chain, so the amplification bill compounds with the number of steps. Folding the trajectory into one banded system — which is what the all-at-once encodings do, and what the deliverable of this layer is — is how the published treatments spend that once instead of $T/h$ times. Still first order, so the precision dependence stays polynomial in $1/\varepsilon$. The nearest published quantum treatment, by Dong, Li and Xue, encodes diagonal Padé approximations of the matrix exponential into a large, block-sparse linear system solved via a quantum linear system algorithm; backward Euler is the subdiagonal $(0,1)$ approximant and is not among the schemes they analyse. No primary quantum source verified here gives an end-to-end complexity or conditioning bound for a pure backward-Euler encoding.

The generator $A(t)$, the interval $[0,T]$, an error tolerance $\varepsilon$, and a target algebraic form.

1. Linear ODE system → Linear system Ax = bsolve against I - hA at each step

   Each step solves $(I − hA)u_{k+1} = u_k + h b_{k+1}$, once per time step — $T/h$ of them to reach time $T$. approximation: First-order implicit stepping, so the precision dependence stays polynomial in $1/\varepsilon$.

   approximation

The discrete object, its truncation-error bound, and its conditioning bound.

Dong, Li and Xue (arXiv:2504.06948) set the off-diagonal Padé cases aside in one sentence of Section 3.1 — "we consider the case p = q = k in the rest of the paper" — and backward Euler is the (0,1) member. Everything they then measure is diagonal, so the paper's experiments are about schemes this record is not.

given  A, b, u_0, step size h, horizon T

for k = 0 … T/h − 1:
    # one quantum linear solve; being implicit does not remove the layer
    solve (I − hA) u_{k+1} = u_k + h b_{k+1}
    # u_{k+1} is a quantum state handed to the next step, never read out

return u_{T/h}

No verified source states an end-to-end cost for a pure backward-Euler encoding, and the field says so. The nearest treatment, Dong, Li and Xue, proves its complexity theorems for diagonal Padé approximants of order $k \ge 3$; backward Euler is the $(0,1)$ subdiagonal approximant, outside them. What the repetition itself charges — one flagged linear solve per step, $T/h$ of them — is in conditions.