Error Scaling of Product Formulas

How large is the error you make when you replace true evolution by a product formula? The answer controls everything practical: how many Trotter steps you need, and therefore the depth and runtime of your circuit. This lesson turns the abstract big- $O$ bounds into concrete numbers.

The first-order error term

Split $H = A + B$ and compare exact evolution to the first-order step. Expanding both sides to second order in $t$ using the Baker–Campbell–Hausdorff identity,

e^{-iHt} = I - iHt - \tfrac{1}{2}H^2 t^2 + O(t^3),

e^{-iAt}e^{-iBt} = I - i(A+B)t - \tfrac{1}{2}\bigl(A^2 + 2AB + B^2\bigr)t^2 + O(t^3).

The linear terms match. The quadratic terms differ only by the commutator: using $H^2 = A^2 + AB + BA + B^2$ ,

e^{-iHt} - e^{-iAt}e^{-iBt} = \tfrac{1}{2}\,[A, B]\,t^2 + O(t^3).

If $A$ and $B$ commute the formula is exact; all Trotter error comes from non-commutativity. Taking operator norms gives the standard single-step bound

\bigl\lVert e^{-iHt} - e^{-iAt}e^{-iBt}\bigr\rVert \le \frac{t^2}{2}\,\bigl\lVert [A, B]\bigr\rVert + O(t^3).

Worked estimate for $H = X + Z$

Take the toy single-qubit Hamiltonian from the previous lesson, $A = X$ , $B = Z$ . The commutator is

[X, Z] = XZ - ZX = -2iY,

(recall $XZ = -iY$ and $ZX = +iY$ ). A Pauli operator scaled by a constant has spectral norm equal to that constant's magnitude, so $\lVert [X, Z]\rVert = \lVert -2iY\rVert = 2$ . The single-step error bound at $t = 0.1$ is therefore

\frac{t^2}{2}\,\lVert [X, Z]\rVert = \frac{0.1^2}{2}\cdot 2 = 0.01.

Splitting into $r$ steps

To evolve for a long time $t$ accurately, divide it into $r$ steps of length $t/r$ . Each step costs $\tfrac{(t/r)^2}{2}\lVert[A,B]\rVert$ , and the $r$ errors add up:

\text{total error} \le r \cdot \frac{(t/r)^2}{2}\,\lVert[A,B]\rVert = \frac{t^2}{2r}\,\lVert[A,B]\rVert.

So total error scales as $O(1/r)$ — to reach accuracy $\varepsilon$ you need $r = O\!\left(\frac{t^2 \lVert[A,B]\rVert}{2\varepsilon}\right)$ steps. This linear-in- $1/\varepsilon$ cost is the headline weakness of first-order Trotter.

Higher-order improvement

The symmetric 2nd-order step has per-step error $O((t/r)^3)$ , so the total error is $O(t^3/r^2)$ and $r = O(1/\sqrt{\varepsilon})$ . More generally, the order- $2k$ Suzuki formula gives $r = O\!\left((1/\varepsilon)^{1/2k}\right)$ — each extra order pushes the exponent of $1/\varepsilon$ down, which is why high-precision simulations favour 4th order and beyond.

Try it

Compute the leading first-order Trotter error bound for one step of $H = X + Z$ at $t = 0.1$ , using $\lVert [X, Z]\rVert = 2$ . Return the number; the grader checks it equals $0.01$ .

Run your code to see the quantum state.