Dyson Expansion as Formal Evolution

The Dyson expansion is the time-ordered series solution of the operator Schrödinger equation. It is the bridge between exact time-dependent evolution and perturbation theory: before any truncation, it is a formal rewriting of the evolution operator; after truncation, it becomes an approximation whose domain of validity must be checked.

The central problem is noncommutativity. If $H(t_1)$ and $H(t_2)$ do not commute, the evolution operator cannot usually be written as an ordinary exponential of $\int H(t')\,dt'$. The Dyson expansion keeps track of operator order.

Integral equation for the evolution operator

For $t\ge t_0$, the time-evolution operator satisfies

$$i\hbar \frac{\partial}{\partial t}U(t,t_0) = H(t)U(t,t_0), \qquad U(t_0,t_0)=I.$$

Integrating from $t_0$ to $t$ gives the equivalent integral equation

$$U(t,t_0) = I - \frac{i}{\hbar} \int_{t_0}^{t} dt_1\, H(t_1)U(t_1,t_0).$$

This equation is exact. The unknown $U$ appears on both sides, so it can be solved by iteration.

Iterative Solution

Substitute the same integral equation for $U(t_1,t_0)$:

$$U(t,t_0) = I - \frac{i}{\hbar} \int_{t_0}^{t} dt_1\,H(t_1) + \left(-\frac{i}{\hbar}\right)^2 \int_{t_0}^{t} dt_1 \int_{t_0}^{t_1} dt_2\, H(t_1)H(t_2) + \cdots .$$

The order of factors is fixed by the limits: $t_1$ is later than $t_2$, so $H(t_1)$ appears to the left of $H(t_2)$.

Continuing the iteration gives

$$U(t,t_0) = I + \sum_{n=1}^{\infty} \left(-\frac{i}{\hbar}\right)^n \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2 \cdots \int_{t_0}^{t_{n-1}}dt_n\, H(t_1)H(t_2)\cdots H(t_n).$$

This is the Dyson series.

Time-Ordered Exponential Notation

The same series is written compactly as

$$U(t,t_0) = \mathcal T \exp\left[ - \frac{i}{\hbar} \int_{t_0}^{t}H(t')\,dt' \right].$$

Equivalently,

$$U(t,t_0) = I + \sum_{n=1}^{\infty} \frac{1}{n!} \left(-\frac{i}{\hbar}\right)^n \times \int_{t_0}^{t}dt_1\cdots \int_{t_0}^{t}dt_n\, \mathcal T \left[ H(t_1)\cdots H(t_n) \right].$$

The factor $1/n!$ appears because the integrals run over the full $n$-dimensional time cube; the time-ordering operator divides it into ordered regions.

First and Second Order Terms

The first-order term is

$$U^{(1)}(t,t_0) = - \frac{i}{\hbar} \int_{t_0}^{t}dt_1\,H(t_1).$$

The second-order term is

$$U^{(2)}(t,t_0) = - \frac{1}{\hbar^2} \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2\, H(t_1)H(t_2).$$

The minus sign in $U^{(2)}$ comes from $(-i)^2=-1$. The operator product order is not optional. Replacing $H(t_1)H(t_2)$ with $H(t_2)H(t_1)$ changes the answer when the Hamiltonians do not commute.

Interaction-Picture Form

In perturbation theory one usually splits

$$H(t)=H_0+V(t),$$

and moves the exactly solvable $H_0$ evolution into the picture transformation. The interaction-picture state satisfies

$$i\hbar\frac{d}{dt}\lvert\psi_I(t)\rangle = V_I(t)\lvert\psi_I(t)\rangle.$$

The interaction-picture evolution operator therefore has its own Dyson series:

$$U_I(t,t_0) = \mathcal T \exp\left[ - \frac{i}{\hbar} \int_{t_0}^{t}V_I(t')\,dt' \right].$$

Detailed transition amplitudes begin when this series is truncated and matrix elements are taken. The approximation hierarchy is mapped in Time-Dependent Perturbation Theory and Transitions. Dyson Expansion for Transition Amplitudes develops ordered transition paths and intermediate-state sums, with the leading calculation continued in First-Order Transition Probability and the rate limit in Fermi’s Golden Rule.

Diagrammatic Preview

Each term in the Dyson expansion can be pictured as ordered insertions along a time line:

t0 ---- t_n ---- ... ---- t_2 ---- t_1 ---- t

The earliest insertion, on the right of the product, acts first on the ket; the latest insertion acts last. From evolution operators to time-ordered products explains how the same ordering enters Wick contractions and field-theory correlation functions.

Validity and Truncation

For a bounded, strongly measurable Hamiltonian with integrable operator norm on a finite interval, define the dimensionless integrated norm

$$M=\frac{1}{\hbar}\int_{t_0}^{t}\lVert H(s)\rVert\,ds.$$

The norm of the $n$th ordered term is at most $M^n/n!$. Consequently the remainder after order $N$ obeys the useful sufficient estimate

$$\lVert R_N\rVert \le e^M\frac{M^{N+1}}{(N+1)!}.$$

For the self-adjoint Hamiltonians considered here, unitarity improves this bound. Stop the Volterra substitution after $N+1$ integrations without expanding the remaining exact propagator. With $R_N=U-\sum_{n=0}^N U^{(n)}$ and $U^{(0)}=I$, the remainder is

$$\begin{aligned} R_N={}&\left(-\frac{i}{\hbar}\right)^{N+1} \int_{t_0}^t dt_1\int_{t_0}^{t_1}dt_2\cdots \int_{t_0}^{t_N}dt_{N+1}\\ &\qquad\times H(t_1)\cdots H(t_{N+1})U(t_{N+1},t_0). \end{aligned}$$

For $N=0$ this means one integral. Since the last factor has norm one, submultiplicativity and the ordered integration region give

$$\lVert R_N\rVert \le \frac{M^{N+1}}{(N+1)!}.$$

Thus this dimensionless bound controls the state-vector error for every normalized initial state. It can be useful without being small: if it is comparable to one, it supplies no small-error guarantee at that order. Neither estimate is automatically available for unbounded Hamiltonians; domains and existence theory then matter. The bounded-Hamiltonian propagator theorem supplies a sufficient construction and explains why backward evolution is needed to establish unitarity.

If a perturbation is written as $\lambda V_I(t)$, amplitudes have an expansion in powers of $\lambda$. A transition whose zeroth-order amplitude vanishes typically has probability beginning at order $\lambda^2$, not order $\lambda$. Squaring a truncated amplitude without consistent order bookkeeping can retain some higher-order terms while omitting others.

Unitarity is also order-by-order. Writing $U=I+U^{(1)}+U^{(2)}+\cdots$, self-adjoint evolution implies

$$U^{(1)}+U^{(1)\dagger}=0,$$

and

$$U^{(2)}+U^{(2)\dagger} +U^{(1)\dagger}U^{(1)}=0.$$

The Magnus Expansion instead expands the logarithm of $U$; finite Magnus truncations generated by anti-Hermitian exponents can preserve unitarity exactly, but have their own convergence conditions.

Under the integrable-norm hypothesis above, the factorial bound proves absolute operator-norm convergence. In general quantum mechanics, unbounded operators and domain questions require more care. Physics calculations usually use the Dyson series as a formal or perturbative expansion with assumptions stated by the model.

Truncating the series is not the same as exact time evolution. For a self-adjoint Hamiltonian, the full $U(t,t_0)$ is unitary. A finite truncation usually preserves unitarity only to the order being kept. A truncated expression can lie slightly outside the probability interval because it omits the terms that restore exact normalization. Compare the violation with the omitted-order estimate: a small discrepancy can be truncation error, while a large violation signals an uncontrolled approximation or an algebraic mistake.

Common Mistakes

• Treating the Dyson expansion as an approximation before it is truncated.
• Dropping time ordering for noncommuting Hamiltonians.
• Reversing the order of operator factors in the nested integrals.
• Forgetting the second-order sign from $(-i)^2=-1$.
• Using first-order transition formulas when the transition probability is no longer small.
• Treating diagrammatic intuition as a substitute for specifying the Hamiltonian, picture, and ordering convention.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• F. J. Dyson, “The Radiation Theories of Tomonaga, Schwinger, and Feynman,” Physical Review 75, 486–502 (1949), doi:10.1103/PhysRev.75.486.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• S. Weinberg, The Quantum Theory of Fields, Volume I, Cambridge University Press, 1995.

Exercises

Second-order term

Starting from the integral equation for $U(t,t_0)$, derive the second-order Dyson term.

Solution

The integral equation is

$$U(t,t_0) = I - \frac{i}{\hbar} \int_{t_0}^{t}dt_1\,H(t_1)U(t_1,t_0).$$

Now substitute the first iteration

$$U(t_1,t_0) = I - \frac{i}{\hbar} \int_{t_0}^{t_1}dt_2\,H(t_2) + O(H^2).$$

Then

$$U(t,t_0) = I - \frac{i}{\hbar} \int_{t_0}^{t}dt_1\,H(t_1) + \left(-\frac{i}{\hbar}\right)^2 \int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2\, H(t_1)H(t_2) + O(H^3).$$

Thus the second-order term is the ordered double integral.

Commuting Hamiltonians

Assume $[H(t_1),H(t_2)]=0$ for all times in the interval. Show that the Dyson series reduces to the ordinary exponential of $-\frac{i}{\hbar}\int H(t')\,dt'$.

Solution

If all Hamiltonians commute, then products at different times can be rearranged freely. The ordered integration region is one of $n!$ equivalent regions inside the full $n$-dimensional time cube:

$$\int_{t_0}^{t}dt_1 \int_{t_0}^{t_1}dt_2 \cdots \int_{t_0}^{t_{n-1}}dt_n\, H(t_1)\cdots H(t_n) = \frac{1}{n!} \left[ \int_{t_0}^{t}H(t')\,dt' \right]^n.$$

Substituting this into the Dyson series gives the ordinary exponential series.

First-order unitarity check

Let $K=\int_{t_0}^{t}H(t')\,dt'$ and assume $K=K^\dagger$. Show that $U_1=I-iK/\hbar$ is unitary up to first order in $K$. Then take $H=\hbar\Omega\sigma_x/2$ constant, $t_0=0$, and $\alpha=\Omega t/2$ with $\Omega>0$ and $t\ge0$. Find the operator-norm error of $U_1$ and compare it with the unitary remainder bound.

Solution

Compute

$$U_1^\dagger U_1 = \left(I+\frac{iK}{\hbar}\right) \left(I-\frac{iK}{\hbar}\right) = I+\frac{K^2}{\hbar^2}.$$

There is no first-order correction to $I$. The deviation begins at second order. This is why a first-order truncation preserves unitarity only to first order, not exactly.

For the two-level Hamiltonian, $U=\cos\alpha\,I-i\sin\alpha\,\sigma_x$ and $U_1=I-i\alpha\sigma_x$. In the $\sigma_x$ eigenbasis, the two eigenvalues of $U-U_1$ have the same magnitude, so

$$\lVert U-U_1\rVert =\sqrt{(1-\cos\alpha)^2+(\alpha-\sin\alpha)^2} \le\frac{\alpha^2}{2}.$$

The inequality follows from the remainder bound with $M=\alpha$ and $N=1$. At $\alpha=0.2$, the exact operator error is about $0.019978$, below the bound $0.02$. By contrast, the norm-squared excess $U_1^\dagger U_1-I=\alpha^2I$ is $0.04I$. An error in a propagator and its failure to conserve norm are different diagnostics.