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Trotter Product Formula

The Trotter product formula explains how an exponential generated by a sum can be obtained as a limit of products of exponentials generated by the parts:

et(A+B)=lim⁡N→∞(etA/NetB/N)N.e^{t(A+B)} = \lim_{N\to\infty} \left( e^{tA/N}e^{tB/N} \right)^N.

In quantum mechanics, if

H=HA+HB,H=H_A+H_B,

then

e−iHt/ℏ=lim⁡N→∞(e−iHAt/(Nℏ)e−iHBt/(Nℏ))Ne^{-iHt/\hbar} = \lim_{N\to\infty} \left( e^{-iH_A t/(N\hbar)} e^{-iH_B t/(N\hbar)} \right)^N

under appropriate mathematical hypotheses. The formula is useful precisely when HAH_A and HBH_B are individually simpler than their sum, even when [HA,HB]≠0[H_A,H_B]\ne0.

If two Hamiltonian terms commute, then

e−i(HA+HB)t/ℏ=e−iHAt/ℏe−iHBt/ℏ.e^{-i(H_A+H_B)t/\hbar} = e^{-iH_A t/\hbar} e^{-iH_B t/\hbar}.

Most interesting decompositions do not commute: kinetic and potential energy, local terms in a many-body Hamiltonian, drift and control Hamiltonians, or terms assigned to different quantum gates. The naive factorization is then false at finite time.

The Trotter idea is that short-time errors can be made small, then controlled by taking many short steps.

Let

Δt=tN.\Delta t=\frac{t}{N}.

The first-order product approximation is

UN(t)=(e−iHAΔt/ℏe−iHBΔt/ℏ)N.U_N(t) = \left( e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/\hbar} \right)^N.

For bounded operators or finite matrices, UN(t)U_N(t) converges to the exact unitary:

lim⁡N→∞UN(t)=e−i(HA+HB)t/ℏ.\lim_{N\to\infty}U_N(t) = e^{-i(H_A+H_B)t/\hbar}.

For unbounded Hamiltonians, the statement requires care about domains and self-adjointness. In ordinary wave-mechanics applications, the formula is often used formally or under standard assumptions that make the relevant unitary groups well defined.

The Baker–Campbell–Hausdorff Formula gives the leading finite-step error. For one step,

e−iHAΔt/ℏe−iHBΔt/ℏ=exp⁡ ⁣[−iℏ(HA+HB)Δt−Δt22ℏ2[HA,HB]+O(Δt3)].\begin{aligned} & e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/\hbar} \\ &\quad = \exp\!\left[ -\frac{i}{\hbar}(H_A+H_B)\Delta t - \frac{\Delta t^2}{2\hbar^2}[H_A,H_B] + O(\Delta t^3) \right]. \end{aligned}

The leading correction is controlled by the commutator. If [HA,HB]=0[H_A,H_B]=0, the product is exact for every Δt\Delta t. If the commutator is large in the relevant norm or on the relevant states, many short steps may be needed.

The local one-step error is typically O(Δt2)O(\Delta t^2) for the first-order formula. Over N=t/ΔtN=t/\Delta t steps, the accumulated global error is typically O(tΔt)O(t\Delta t), with constants depending on commutators and higher nested commutators.

A common improvement is the symmetric, or Strang, splitting:

S2(Δt)=e−iHAΔt/(2ℏ)e−iHBΔt/ℏe−iHAΔt/(2ℏ).S_2(\Delta t) = e^{-iH_A\Delta t/(2\hbar)} e^{-iH_B\Delta t/\hbar} e^{-iH_A\Delta t/(2\hbar)}.

Then

e−i(HA+HB)t/ℏ≈[S2(Δt)]N.e^{-i(H_A+H_B)t/\hbar} \approx \left[S_2(\Delta t)\right]^N.

The symmetry under Δt↦−Δt\Delta t\mapsto-\Delta t cancels the leading even-order error terms in the exponent. For sufficiently regular bounded problems, the local error is O(Δt3)O(\Delta t^3) and the global error is O(tΔt2)O(t\Delta t^2).

The order of the half steps matters. The alternative

e−iHBΔt/(2ℏ)e−iHAΔt/ℏe−iHBΔt/(2ℏ)e^{-iH_B\Delta t/(2\hbar)} e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/(2\hbar)}

is another second-order splitting, but it is not the same finite-step approximation unless the terms commute.

For a Hamiltonian split into many parts,

H=∑ℓ=1LHℓ,H=\sum_{\ell=1}^L H_\ell,

a first-order step is

S1(Δt)=∏ℓ=1Le−iHℓΔt/ℏ,S_1(\Delta t) = \prod_{\ell=1}^L e^{-iH_\ell\Delta t/\hbar},

with a chosen operator order. The product order is part of the approximation.

A symmetric second-order step can be written as a forward sweep followed by a backward sweep:

S2(Δt)=e−iH1Δt/(2ℏ)e−iH2Δt/(2ℏ)⋯e−iHLΔt/ℏ×⋯e−iH2Δt/(2ℏ)e−iH1Δt/(2ℏ).\begin{aligned} S_2(\Delta t) &= e^{-iH_1\Delta t/(2\hbar)} e^{-iH_2\Delta t/(2\hbar)} \cdots e^{-iH_L\Delta t/\hbar} \\ &\quad \times \cdots e^{-iH_2\Delta t/(2\hbar)} e^{-iH_1\Delta t/(2\hbar)}. \end{aligned}

Different orderings can have the same formal order but different error constants.

Higher-order product formulas combine lower-order steps with carefully chosen fractional time steps. A standard recursive construction starts from a symmetric step S2k−2S_{2k-2} and defines

S2k(Δt)=S2k−2(pkΔt)2S2k−2((1−4pk)Δt)S2k−2(pkΔt)2,S_{2k}(\Delta t) = S_{2k-2}(p_k\Delta t)^2 S_{2k-2}((1-4p_k)\Delta t) S_{2k-2}(p_k\Delta t)^2,

where

pk=14−41/(2k−1).p_k= \frac{1}{4-4^{1/(2k-1)}}.

This cancels additional terms in the error expansion. The gain in asymptotic order comes with more exponentials per step, sometimes with negative or larger-than-one substeps. Practical performance depends on the Hamiltonian, commutators, hardware or numerical representation, and target accuracy.

For a particle with

H=T(p)+V(x),H=T(p)+V(x),

the second-order split-operator step is often written

e−iHΔt/ℏ≈e−iV(x)Δt/(2ℏ)e−iT(p)Δt/ℏe−iV(x)Δt/(2ℏ).e^{-iH\Delta t/\hbar} \approx e^{-iV(x)\Delta t/(2\hbar)} e^{-iT(p)\Delta t/\hbar} e^{-iV(x)\Delta t/(2\hbar)}.

This is powerful because V(x)V(x) is diagonal in position space and T(p)T(p) is diagonal in momentum space. A numerical implementation alternates between position and momentum representations using Fourier transforms.

This page records the operator identity and error logic. Detailed implementation issues such as grids, aliasing, absorbing boundaries, adaptive time steps, and benchmark tests belong to computational pages.

In Hamiltonian simulation, a many-term Hamiltonian may be decomposed as

H=∑ℓHℓ,H=\sum_{\ell}H_\ell,

where each e−iHℓΔt/ℏe^{-iH_\ell\Delta t/\hbar} is easier to implement than e−iHΔt/ℏe^{-iH\Delta t/\hbar}. Product formulas then turn continuous-time dynamics into a sequence of implementable unitary steps. The discrete-step viewpoint is developed in Quantum Maps and Discrete-Time Evolution.

The same mathematical idea appears in digital–analog simulation, lattice models, quantum chemistry algorithms, and control sequences. This page only states the dynamics-side principle; What Is Quantum Simulation? owns the target mapping, error ledger, resource accounting, and verification contract.

The product formula also explains why path integrals start from many short-time kernels. For

H=p22m+V(x),H=\frac{p^2}{2m}+V(x),

one inserts many small steps and uses a kinetic-potential splitting. In the continuum limit, the repeated integrations over intermediate positions lead to the path-integral expression. The path-integral page develops that construction without treating the product formula as a mere mnemonic.

  • Writing eA+B=eAeBe^{A+B}=e^Ae^B for noncommuting AA and BB.
  • Confusing the exact N→∞N\to\infty product formula with a finite-NN approximation.
  • Quoting an error order without checking commutator size, domains, or regularity.
  • Forgetting that product order matters at finite step size.
  • Assuming a higher-order formula is automatically faster in a real computation.
  • Applying a split-operator step on a grid without checking boundary conditions and Fourier conventions.
  • Treating Trotterization as the same thing as time ordering; both involve ordered products, but they solve different organizational problems.
  • H. F. Trotter, “On the Product of Semi-Groups of Operators,” Proceedings of the American Mathematical Society 10, 545-551, 1959.
  • M. Reed and B. Simon, Methods of Modern Mathematical Physics I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
  • G. Strang, “On the Construction and Comparison of Difference Schemes,” SIAM Journal on Numerical Analysis 5, 506-517, 1968.
  • M. Suzuki, “Generalized Trotter’s Formula and Systematic Approximants of Exponential Operators and Inner Derivations with Applications to Many-Body Problems,” Communications in Mathematical Physics 51, 183-190, 1976.
  • M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
  1. Show that the first-order product is exact at finite NN when [HA,HB]=0[H_A,H_B]=0.
Solution

If [HA,HB]=0[H_A,H_B]=0, then the exponentials commute and combine:

e−iHAΔt/ℏe−iHBΔt/ℏ=e−i(HA+HB)Δt/ℏ.e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/\hbar} = e^{-i(H_A+H_B)\Delta t/\hbar}.

Raising to the NNth power gives

(e−i(HA+HB)t/(Nℏ))N=e−i(HA+HB)t/ℏ.\left( e^{-i(H_A+H_B)t/(N\hbar)} \right)^N = e^{-i(H_A+H_B)t/\hbar}.
  1. Use the Baker–Campbell–Hausdorff expansion to identify the leading correction to one first-order step.
Solution

Set

A=−iℏHAΔt,B=−iℏHBΔt.A=-\frac{i}{\hbar}H_A\Delta t, \qquad B=-\frac{i}{\hbar}H_B\Delta t.

The BCH formula gives

eAeB=exp⁡(A+B+12[A,B]+O(Δt3)).e^Ae^B = \exp\left(A+B+\frac12[A,B]+O(\Delta t^3)\right).

Since

[A,B]=−Δt2ℏ2[HA,HB],[A,B] = -\frac{\Delta t^2}{\hbar^2}[H_A,H_B],

the exponent contains the correction

−Δt22ℏ2[HA,HB].-\frac{\Delta t^2}{2\hbar^2}[H_A,H_B].
  1. For H=T(p)+V(x)H=T(p)+V(x), explain why the split-operator step naturally alternates between position and momentum representations.
Solution

V(x)V(x) acts by multiplication in the position representation, so e−iV(x)Δt/(2ℏ)e^{-iV(x)\Delta t/(2\hbar)} is easy to apply to ψ(x)\psi(x). T(p)T(p) is diagonal in momentum representation, so e−iT(p)Δt/ℏe^{-iT(p)\Delta t/\hbar} is easy to apply to ψ~(p)\tilde\psi(p). A split step therefore applies the potential factor in position space, Fourier transforms to momentum space for the kinetic factor, then transforms back for the final potential factor.

  1. Why does a finite Trotter step preserve norm even though it is approximate?
Solution

If each HℓH_\ell is self-adjoint, then every factor e−iHℓΔt/ℏe^{-iH_\ell\Delta t/\hbar} is unitary. A product of unitary operators is unitary. Therefore the finite product preserves norm exactly, even though it approximates the unitary generated by the full sum only up to a finite-step error.