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:

$$e^{t(A+B)} = \lim_{N\to\infty} \left( e^{tA/N}e^{tB/N} \right)^N.$$

In quantum mechanics, if

$$H=H_A+H_B,$$

then

$$e^{-iHt/\hbar} = \lim_{N\to\infty} \left( e^{-iH_A t/(N\hbar)} e^{-iH_B t/(N\hbar)} \right)^N$$

under appropriate mathematical hypotheses. The formula is useful precisely when $H_A$ and $H_B$ are individually simpler than their sum, even when $[H_A,H_B]\ne0$.

Splitting a noncommuting Hamiltonian

If two Hamiltonian terms commute, then

$$e^{-i(H_A+H_B)t/\hbar} = e^{-iH_A t/\hbar} e^{-iH_B t/\hbar}.$$

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.

First-Order Formula

Let

$$\Delta t=\frac{t}{N}.$$

The first-order product approximation is

$$U_N(t) = \left( e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/\hbar} \right)^N.$$

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

$$\lim_{N\to\infty}U_N(t) = e^{-i(H_A+H_B)t/\hbar}.$$

For unbounded self-adjoint $H_A$ and $H_B$, a standard sufficient hypothesis is that $H_A+H_B$ is essentially self-adjoint on $D(H_A)\cap D(H_B)$; its self-adjoint closure defines $H$. The product then converges strongly on each state, not generally in operator norm. This existence theorem does not supply the finite-matrix error rates below without further regularity assumptions.

Local Error from BCH

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

$$e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/\hbar} = \exp\!\left[ -\frac{i}{\hbar}(H_A+H_B)\Delta t - \frac{\Delta t^2}{2\hbar^2}[H_A,H_B] + O(\Delta t^3) \right].$$

The leading correction is controlled by the commutator. If $[H_A,H_B]=0$, the product is exact for every $\Delta t$. 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 $O(\Delta t^2)$ for the first-order formula. Over $N=t/\Delta t$ steps, the accumulated global error is typically $O(t\Delta t)$, with constants depending on commutators and higher nested commutators.

Symmetric Second-Order Formula

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

$$S_2(\Delta t) = e^{-iH_A\Delta t/(2\hbar)} e^{-iH_B\Delta t/\hbar} e^{-iH_A\Delta t/(2\hbar)}.$$

Then

$$e^{-i(H_A+H_B)t/\hbar} \approx \left[S_2(\Delta t)\right]^N.$$

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

The order of the half steps matters. The alternative

$$e^{-iH_B\Delta t/(2\hbar)} e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/(2\hbar)}$$

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

More Than Two Terms

For a Hamiltonian split into many parts,

$$H=\sum_{\ell=1}^L H_\ell,$$

a first-order step is

$$S_1(\Delta t) = \prod_{\ell=1}^L e^{-iH_\ell\Delta t/\hbar},$$

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:

$$S_2(\Delta t) = e^{-iH_1\Delta t/(2\hbar)} e^{-iH_2\Delta t/(2\hbar)} \cdots e^{-iH_L\Delta t/\hbar} \times \cdots e^{-iH_2\Delta t/(2\hbar)} e^{-iH_1\Delta t/(2\hbar)}.$$

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

Higher-Order Suzuki–Trotter Preview

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

$$S_{2k}(\Delta t) = S_{2k-2}(p_k\Delta t)^2 S_{2k-2}((1-4p_k)\Delta t) S_{2k-2}(p_k\Delta t)^2,$$

for integers $k\ge2$, where

$$p_k= \frac{1}{4-4^{1/(2k-1)}}.$$

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.

Split-Operator Method

For a particle with

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

the second-order split-operator step is often written

$$e^{-iH\Delta t/\hbar} \approx e^{-iV(x)\Delta t/(2\hbar)} e^{-iT(p)\Delta t/\hbar} e^{-iV(x)\Delta t/(2\hbar)}.$$

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

An FFT implementation also imposes a finite periodic grid. Decrease the time step and refine the spatial grid separately; norm conservation alone cannot detect phase error or wraparound. The Trotter notebook compares step-size errors against an exact matrix solution.

Quantum Simulation Connection

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

$$H=\sum_{\ell}H_\ell,$$

where each $e^{-iH_\ell\Delta t/\hbar}$ is easier to implement than $e^{-iH\Delta t/\hbar}$. 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. Hamiltonian simulation studies the gate costs and error bounds of this construction.

Path-Integral Time Slicing

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

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

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. Propagators to path integrals develops the intermediate-position construction.

Common Mistakes

• Writing $e^{A+B}=e^Ae^B$ for noncommuting $A$ and $B$.
• Confusing the exact $N\to\infty$ product formula with a finite-$N$ 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.

References

• N. Hatano and M. Suzuki, “Finding exponential product formulas of higher orders”, in Quantum Annealing and Other Optimization Methods, Lecture Notes in Physics 679, 37–68, Springer, 2005.

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.

• 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.

• H. F. Trotter, “On the product of semi-groups of operators”, Proceedings of the American Mathematical Society 10, 545-551, 1959.

Exercises

1. Show that the first-order product is exact at finite $N$ when $[H_A,H_B]=0$.

Solution

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

$$e^{-iH_A\Delta t/\hbar} e^{-iH_B\Delta t/\hbar} = e^{-i(H_A+H_B)\Delta t/\hbar}.$$

Raising to the $N$th power gives

$$\left( e^{-i(H_A+H_B)t/(N\hbar)} \right)^N = e^{-i(H_A+H_B)t/\hbar}.$$

2. Use the Baker–Campbell–Hausdorff expansion to identify the leading correction to one first-order step.

Solution

Set

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

The BCH formula gives

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

Since

$$[A,B] = -\frac{\Delta t^2}{\hbar^2}[H_A,H_B],$$

the exponent contains the correction

$$-\frac{\Delta t^2}{2\hbar^2}[H_A,H_B].$$

3. For $H=T(p)+V(x)$, explain why the split-operator step naturally alternates between position and momentum representations.

Solution

$V(x)$ acts by multiplication in the position representation, so $e^{-iV(x)\Delta t/(2\hbar)}$ is easy to apply to $\psi(x)$. $T(p)$ is diagonal in momentum representation, so $e^{-iT(p)\Delta t/\hbar}$ is easy to apply to $\tilde\psi(p)$. 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.

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

Solution

If each $H_\ell$ is self-adjoint, then every factor $e^{-iH_\ell\Delta t/\hbar}$ 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.