Dyson Series and the Time-Ordered Evolution Operator

The interaction picture reduces the quantum problem to a single object: the evolution operator $\hat U_I(t,t_0)$ that propagates a state under the perturbation alone. This article solves for it perturbatively. Iterating the interaction-picture integral equation once, twice, and so on generates the Dyson series — an expansion of $\hat U_I$ in powers of the coupling, with the $n$ -th term an $n$ -fold nested time integral of $\hat V_I$ . It is the operator statement of time-dependent perturbation theory: every transition amplitude, every golden-rule rate, every scattering matrix element is a projection of one of these terms. Two facts make the series usable. First, the abstract time-ordered exponential $\hat U_I=\mathcal{T}\exp[-\tfrac{i}{\hbar}\int \hat V_I\,\mathrm{d}t']$ — the compact form and its $1/n!$ combinatorics — already belongs to the companion article on the time-ordered exponential, which we recall by name and do not re-derive. What this article owns is the perturbative realization: the explicit matrix-element amplitudes $c_f^{(n)}(t)$ that carry the Bohr-frequency phases $e^{i\omega_{fi}t}$ , and — the piece that is genuinely new here — the quantitative order-by-order unitarity of a truncated series. That is the owned result. A truncated Dyson sum is not unitary; the companion states this qualitatively. We make it sharp: the first-order truncation violates unitarity at order $V^2$ , the second-order truncation cancels that $V^2$ defect exactly and leaves a residual of order $V^3$ , and we compute the size of both on a concrete two-level drive. This is what tells you, in practice, how many orders you must keep for a target accuracy. The single-order-per-step iteration, the phase bookkeeping, and the unitarity ladder are the deliverables; the time-ordering symbol and its factorial are cited to their owner.

Key takeaways

Iterate the interaction-picture integral equation $|\psi_I(t)\rangle=|\psi_I(0)\rangle-\tfrac{i}{\hbar}\int_0^t\hat V_I(t')|\psi_I(t')\rangle\,\mathrm{d}t'$ : each pass adds one power of $\hat V_I$ and one nested time integral, generating $\hat U_I=\sum_n \hat U_I^{(n)}$ , the Dyson series.
The compact form is the time-ordered exponential $\hat U_I=\mathcal{T}\exp[-\tfrac{i}{\hbar}\int_0^t\hat V_I\,\mathrm{d}t']$ ; the $1/n!$ arises because the ordered simplex is $1/n!$ of the full $n$ -cube (owned by the time-ordered-exponential article, recalled here).
The matrix element of the first-order term is $c_f^{(1)}(t)=-\tfrac{i}{\hbar}\int_0^t V_{fi}(t')e^{i\omega_{fi}t'}\,\mathrm{d}t'$ , $\omega_{fi}=(E_f-E_i)/\hbar$ — the phase is the fingerprint of the unperturbed evolution.
Owned result: truncation breaks unitarity in a controlled ladder — first order violates $\hat U^\dagger\hat U=\mathbb{1}$ at $O(V^2)$ , second order cancels that and leaves $O(V^3)$ ; the defect sizes are computed on a two-level drive.
Time-ordering is essential exactly when $[\hat V_I(t_1),\hat V_I(t_2)]\neq0$ ; if the drive commutes with itself at all times, $\mathcal{T}$ drops and the ordinary exponential is exact.
Everything downstream — first-order transition probabilities, Fermi's golden rule, the QFT $S$ -matrix — is a projection of a Dyson term.
Key idea: the Dyson series is time-dependent perturbation theory written as one operator; its $n$ -th term is an $n$ -fold time-ordered integral of $\hat V_I$ , and truncating it costs unitarity at a computable order in the coupling.

The core insight

Feed the interaction-picture integral equation into itself. Each substitution promotes the accuracy by one power of $\hat V_I$ and adds one nested, time-ordered integral. The resulting sum for $\hat U_I(t,t_0)$ is the Dyson series; its compact form is a time-ordered exponential. Projecting the $n$ -th term between unperturbed states gives the $n$ -th-order transition amplitude with its $e^{i\omega_{fi}t}$ phases. The price of stopping at finite order is a loss of unitarity that is $O(V^{n+1})$ for the $n$ -th-order truncation — small, and computable.

Where this lives today

The interaction-picture Dyson series is the computational spine of every transition calculation in quantum mechanics. Its first term feeds first-order transition probabilities and Fermi's golden rule; its higher terms supply the corrections that matter for forbidden transitions and multiphoton processes. Carried into quantum field theory, the same series — with $\hat V_I$ the interaction Hamiltonian density — becomes the $S$ -matrix expansion whose terms are Feynman diagrams. In condensed matter its imaginary-time cousin generates the perturbative Green-function expansions of finite-temperature many-body theory. Modern quantum control keeps the unitarity lesson central: rather than truncate the raw series, one uses the Magnus expansion or split-operator exponentials that stay unitary at every order, precisely because a bare truncation loses probability at the computable rate this article makes explicit.

The physical picture

Think of the perturbation as a sequence of kicks. Between kicks the system rides its unperturbed phases $e^{-iE_n t/\hbar}$ ; each application of $\hat V_I$ is one kick that can redistribute amplitude among the unperturbed states. The Dyson series is the sum over all kick histories: zero kicks (the identity), one kick at some time $t_1$ , two kicks at ordered times $t_1>t_2$ , and so on. The $n$ -th term integrates over all ordered placements of $n$ kicks, weighted by $(-i/\hbar)^n$ and the matrix elements of $\hat V_I$ at those instants. Because the kicks are operators that generally do not commute, their order matters — a kick along $x$ then along $y$ is not the same rotation as $y$ then $x$ . That is why the integrals are time-ordered: the later time always sits to the left. The bookkeeping device that enforces this is the time-ordering symbol $\mathcal{T}$ , and the reason the compact exponential carries a $1/n!$ is purely geometric — the ordered region $t_1>t_2>\cdots>t_n$ is one of the $n!$ equal chambers of the full integration cube. Both facts are the property of the companion article on the time-ordered exponential; here they are the scaffolding, not the subject.

Figure

Kick histories: the Dyson series sums over all ordered placements of the perturbation

The second figure is the article's thesis. Truncate the series at some order and ask whether the resulting propagator still conserves probability. It does not — but the failure shrinks fast with order. Keeping only the first-order term leaves a unitarity defect of size $\sim V^2$ ; adding the second-order term cancels that piece exactly and drops the defect to $\sim V^3$ . The bars below are the actual measured defects for a concrete two-level drive: a clean geometric descent that tells you the price of stopping early.

Figure

Truncation defect ladder: the shrinking simplex fraction 1/n! and the O(V^{n+1}) unitarity cost

Building the series from the integral equation

The starting point is the interaction-picture integral equation, owned by The Interaction Picture and Time-Dependent Hamiltonians. It rewrites the interaction-picture Schrödinger equation $i\hbar\,\partial_t|\psi_I\rangle=\hat V_I(t)|\psi_I\rangle$ as an integral relation that already contains its own solution on the right-hand side, ready to be iterated.

|\psi_I(t)\rangle=|\psi_I(t_0)\rangle-\frac{i}{\hbar}\int_{t_0}^{t}\hat V_I(t')\,|\psi_I(t')\rangle\,\mathrm{d}t'.

where: $\hat V_I(t)=e^{i\hat H_0(t-t_0)/\hbar}\,\hat V(t)\,e^{-i\hat H_0(t-t_0)/\hbar}$ is the perturbation in the interaction picture, and $|\psi_I\rangle$ is the state stripped of its free evolution. The physical read: the state at $t$ is the initial state plus a correction accumulated by the perturbation acting all along the way — but the correction itself depends on the (unknown) state, which is exactly what iteration resolves. The same content, phrased for the propagator $\hat U_I(t,t_0)$ defined by $|\psi_I(t)\rangle=\hat U_I(t,t_0)|\psi_I(t_0)\rangle$ , reads $\hat U_I(t,t_0)=\mathbb{1}-\tfrac{i}{\hbar}\int_{t_0}^t \hat V_I(t')\,\hat U_I(t',t_0)\,\mathrm{d}t'$ . Iterating this operator equation is the whole game.

Derivation AIterating the integral equation into the Dyson series

We generate $\hat U_I$ order by order by substituting the integral equation into itself repeatedly. Each substitution replaces the $\hat U_I$ inside the integral by its own expansion, promoting the accuracy by one power of $\hat V_I$ . The goal is to exhibit the general $n$ -th term as an $n$ -fold nested time-ordered integral, so that every downstream amplitude is a projection of a known object.

Step 1. Zeroth order. Drop the integral entirely: $\hat U_I^{(0)}=\mathbb{1}$ . With no perturbation the interaction-picture state does not move — correct, since $\hat V_I=0$ leaves $i\hbar\partial_t|\psi_I\rangle=0$ .

Step 2. First order. Insert $\hat U_I\to\mathbb{1}$ under the integral: $\hat U_I^{(1)}=-\tfrac{i}{\hbar}\int_{t_0}^{t}\hat V_I(t_1)\,\mathrm{d}t_1$ . Why: to leading order the propagator inside the integral is just the identity, so the correction is a single time integral of one power of $\hat V_I$ .

Step 3. Second order. Insert the first-order result under the integral: $\hat U_I^{(2)}=\left(-\tfrac{i}{\hbar}\right)^2\int_{t_0}^{t}\mathrm{d}t_1\int_{t_0}^{t_1}\mathrm{d}t_2\,\hat V_I(t_1)\hat V_I(t_2)$ . Why: the inner integral runs only up to $t_1$ because the second kick must occur before the first — the propagator inside acted only up to time $t_1$ . This ordering $t_1>t_2$ is not a convention; it is forced by causality of the iteration.

Step 4. General order. Repeating gives $\hat U_I^{(n)}=\left(-\tfrac{i}{\hbar}\right)^n\int_{t_0}^{t}\mathrm{d}t_1\int_{t_0}^{t_1}\mathrm{d}t_2\cdots\int_{t_0}^{t_{n-1}}\mathrm{d}t_n\,\hat V_I(t_1)\hat V_I(t_2)\cdots\hat V_I(t_n)$ , with the times strictly ordered $t_1>t_2>\cdots>t_n$ and the operators written left-to-right in that same order. The full Dyson series is $\hat U_I=\sum_{n=0}^\infty \hat U_I^{(n)}$ .

Step 5. Consistency. Differentiate the sum with respect to $t$ : $\partial_t\hat U_I^{(n)}=-\tfrac{i}{\hbar}\hat V_I(t)\,\hat U_I^{(n-1)}$ , so $i\hbar\,\partial_t\hat U_I=\hat V_I(t)\hat U_I$ term by term. Why: this is exactly the interaction-picture evolution equation, so the series solves the dynamics to every order. Verified: verify_aq-qm-dyson-series-time-ordering.py reproduces $\hat U_I^{(1)}$ and $\hat U_I^{(2)}$ for a concrete two-level $\hat V_I(t)$ , and confirms the first-order composition $\hat U_I^{(1)}(0,2)=\hat U_I^{(1)}(0,1)+\hat U_I^{(1)}(1,2)$ to $3\times10^{-9}$ .

The time-ordered exponential

The nested-integral form is exact but clumsy. The compact rewriting uses the time-ordering symbol $\mathcal{T}$ , which by definition arranges any product of time-labelled operators with later times to the left. Its full theory — the abstract $\mathcal{T}$ -exponential $\hat U(t,t_0)=\mathcal{T}\exp[-\tfrac{i}{\hbar}\int_{t_0}^t\hat H(t')\,\mathrm{d}t']$ for a general time-dependent Hamiltonian, and the combinatorial proof that symmetrizing the ordered simplex over the full cube produces the $1/n!$ — is the property of the companion article Time-Ordered Exponential and the Dyson Series, which we recall by name.

\hat U_I(t,t_0)=\mathcal{T}\exp\!\left[-\frac{i}{\hbar}\int_{t_0}^{t}\hat V_I(t')\,\mathrm{d}t'\right]=\sum_{n=0}^\infty\frac{1}{n!}\left(-\frac{i}{\hbar}\right)^n\int_{t_0}^t\!\!\cdots\!\int_{t_0}^t \mathcal{T}\big[\hat V_I(t_1)\cdots\hat V_I(t_n)\big]\,\mathrm{d}t_1\cdots\mathrm{d}t_n.

where: the second form integrates over the full cube $[t_0,t]^n$ with $\mathcal{T}$ ordering the operators, and the $1/n!$ compensates for the $n!$ equal ordered chambers. The physical read: the ordered nested integral (no $1/n!$ ) and the symmetrized cube integral (with $1/n!$ ) are two ways of writing the same term; the companion article proves their equality. What we verify here is only that the fraction is right.

Go deeper · Derivation B · 1 minThe simplex fraction 1/n! — re-checked at the perturbative level

Derivation BThe simplex fraction 1/n! — re-checked at the perturbative level

The companion article proves the simplex-to-cube identity in the abstract. Because it is the load-bearing combinatorial fact of the whole series, we re-verify it concretely for the perturbative realization — the ordered nested integral equals $1/n!$ times the $n$ -th power of the single integral, for a genuinely time-dependent integrand.

Step 1. Take a concrete non-trivial scalar $f(t)=\sin t+t$ standing in for a matrix element of $\hat V_I$ , and let $I=\int_0^T f\,\mathrm{d}t$ .

Step 2. $n=2$ : compute $\int_0^T\mathrm{d}t_1\int_0^{t_1}\mathrm{d}t_2\,f(t_1)f(t_2)$ symbolically and compare to $\tfrac12 I^2$ . They are equal. Why: the region $t_1>t_2$ is exactly half of the square $[0,T]^2$ , and the integrand $f(t_1)f(t_2)$ is symmetric under $t_1\leftrightarrow t_2$ .

Step 3. $n=3$ : compute the triple ordered integral $t_1>t_2>t_3$ and compare to $\tfrac16 I^3$ . Equal. Why: the ordered simplex is one of the $3!=6$ equal chambers of the cube.

Step 4. Read off the SVG sequence: the ordered fraction of the $n$ -cube is $1/n!=1,\tfrac12,\tfrac16,\tfrac1{24},\dots$ for $n=1,2,3,4$ — the geometric collapse drawn in the second figure. Verified: verify_...py confirms the $n=2$ and $n=3$ simplex identities symbolically (exact zero difference) and the $1/n!$ sequence.

Transition amplitudes and the phase fingerprint

Projecting the Dyson series between unperturbed eigenstates turns operators into numbers — the transition amplitudes that time-dependent perturbation theory is built to compute. Writing $|\psi_I(t)\rangle=\sum_m c_m(t)|m\rangle$ with $c_m(0)=\delta_{mi}$ , the first-order coefficient is the matrix element of $\hat U_I^{(1)}$ .

c_f^{(1)}(t)=\langle f|\hat U_I^{(1)}|i\rangle=-\frac{i}{\hbar}\int_{0}^{t}V_{fi}(t')\,e^{i\omega_{fi}t'}\,\mathrm{d}t',\qquad \omega_{fi}=\frac{E_f-E_i}{\hbar}.

where: $V_{fi}(t)=\langle f|\hat V(t)|i\rangle$ is the Schrödinger-picture matrix element and $e^{i\omega_{fi}t}$ is the interaction-picture phase inherited from $\hat V_I=e^{i\hat H_0 t/\hbar}\hat V e^{-i\hat H_0 t/\hbar}$ . The physical read: the Bohr-frequency phase is the fingerprint of the unperturbed evolution — it decides which final states the perturbation can drive resonantly, and it is the object the first-order-transition-probability article squares to get the sinc-squared line shape. For a constant matrix element $V_{fi}$ switched on at $t=0$ , the integral is elementary: $c_f^{(1)}(t)=-\tfrac{V_{fi}}{\hbar\omega_{fi}}\big(e^{i\omega_{fi}t}-1\big)$ , whose modulus squared is $|c_f^{(1)}|^2=\tfrac{|V_{fi}|^2}{\hbar^2}\tfrac{\sin^2(\omega_{fi}t/2)}{(\omega_{fi}/2)^2}$ . This is the seed the companion first-order article develops in full; we display it here only to show that the perturbative amplitude falls straight out of the first Dyson term.

The owned result: unitarity of a truncated series

Go deeper · Derivation C · 1 minHow much unitarity a truncated Dyson series loses

Derivation CHow much unitarity a truncated Dyson series loses

The full series is unitary because $\hat H$ is Hermitian; any truncation is not. The companion article states this qualitatively. The owned contribution here is the quantitative ladder: we show that the first-order truncation violates $\hat U^\dagger\hat U=\mathbb{1}$ at order $V^2$ , that the second-order term cancels that $V^2$ piece exactly, leaving a residual of order $V^3$ , and we measure both defects on a concrete two-level drive.

Step 1. Note $\hat U_I^{(1)}=-\tfrac{i}{\hbar}\int\hat V_I\,\mathrm{d}t_1$ is anti-Hermitian, since $\hat V_I$ is Hermitian and the $-i$ is imaginary. Hence $(\hat U_I^{(1)})^\dagger+\hat U_I^{(1)}=0$ exactly — there is no linear-in- $V$ unitarity defect.

Step 2. First-order truncation $\hat U\approx\mathbb{1}+\hat U_I^{(1)}$ . Then $\hat U^\dagger\hat U-\mathbb{1}=(\hat U_I^{(1)})^\dagger+\hat U_I^{(1)}+(\hat U_I^{(1)})^\dagger\hat U_I^{(1)}=(\hat U_I^{(1)})^\dagger\hat U_I^{(1)}$ , which is $O(V^2)$ . Why: the linear term vanished in Step 1, so the leading defect is the quadratic self-product.

Step 3. Second-order truncation $\hat U\approx\mathbb{1}+\hat U_I^{(1)}+\hat U_I^{(2)}$ . The $O(V^2)$ part of $\hat U^\dagger\hat U-\mathbb{1}$ is now $(\hat U_I^{(1)})^\dagger\hat U_I^{(1)}+\hat U_I^{(2)}+(\hat U_I^{(2)})^\dagger$ . Why it cancels: $\hat U_I^{(2)}+(\hat U_I^{(2)})^\dagger$ is exactly the symmetrization that reconstructs $-(\hat U_I^{(1)})^\dagger\hat U_I^{(1)}$ — this is the operator content of the simplex-to-cube identity applied to $n=2$ . Numerically the $O(V^2)$ terms sum to $2\times10^{-17}$ , i.e. zero.

Step 4. Measure. For the Hermitian drive $\hat V_I(t)=V_0[\cos(\Omega t)\sigma_x+\sin(\Omega t)\sigma_y]$ with $V_0=0.05$ , $\Omega=1.3$ , over $t\in[0,2]$ : the first-order truncation defect is $\max|\hat U^\dagger\hat U-\mathbb{1}|\approx5.5\times10^{-3}$ , comparable to $V_0^2=2.5\times10^{-3}$ ; the second-order defect is $\approx1.7\times10^{-5}$ , comparable to $V_0^3=1.25\times10^{-4}$ . The defect drops by more than two orders of magnitude for one extra term.

Step 5. The rule this gives you: to keep $\hat U$ unitary to relative accuracy $\epsilon$ , retain terms through order $n$ where $V^{\,n+1}\lesssim\epsilon$ . That is the practical payoff of the ladder. Verified: verify_...py builds $\hat U_I^{(1)},\hat U_I^{(2)}$ numerically, confirms $\hat U_I^{(1)}$ is anti-Hermitian to $0$ , that the $O(V^2)$ unitarity terms cancel to $2\times10^{-17}$ , and that the first- and second-order defects scale as $V^2$ and $V^3$ respectively.

Go deeper · Derivation D · 1 minWhen time-ordering can be dropped

Derivation DWhen time-ordering can be dropped

Time-ordering is bookkeeping you would rather avoid, since it blocks you from simply exponentiating the integrated perturbation. It is dispensable in exactly one circumstance: when the drive commutes with itself at every pair of times. We identify that condition precisely, then check both sides — a drive that keeps a single operator direction (ordering unnecessary) against one that changes direction (ordering essential, and quantifiably so).

Step 1. If $[\hat V_I(t_1),\hat V_I(t_2)]=0$ for all $t_1,t_2$ , then $\mathcal{T}$ acts trivially and $\hat U_I=\exp[-\tfrac{i}{\hbar}\int\hat V_I\,\mathrm{d}t']$ — an ordinary exponential of the integrated operator.

Step 2. A drive with a single fixed operator structure, $\hat V_I(t)=g(t)\,\hat A$ , always commutes with itself: $[g(t_1)\hat A,g(t_2)\hat A]=g(t_1)g(t_2)[\hat A,\hat A]=0$ . Then time-ordering is unnecessary and the naive exponential is exact.

Step 3. A drive whose operator structure changes — say $\sigma_x$ during one interval and $\sigma_y$ during another — has $[\hat V_I(t_1),\hat V_I(t_2)]\propto[\sigma_x,\sigma_y]=2i\sigma_z\neq0$ . Here dropping $\mathcal{T}$ gives the wrong answer; the companion article's counterexample quantifies exactly how wrong (about twenty percentage points of population from swapping the order of two rotations). Verified: verify_...py confirms $[\hat V_I(t_1),\hat V_I(t_2)]\neq0$ for the $\sigma_x/\sigma_y$ drive ( $|[\cdot,\cdot]|=0.004$ ) and $=0$ for the single-operator drive (to $10^{-16}$ ).

Limiting cases

The Dyson series has clean regimes set by the coupling strength, the commutation structure of the drive, and the truncation order.

Case	Result	Meaning
Weak coupling, short time	first term dominates	$\hat U_I\approx\mathbb{1}+\hat U_I^{(1)}$ ; the physics is the single first-order amplitude $c_f^{(1)}$ — the regime of Fermi's golden rule.
Self-commuting drive	$\mathcal{T}$ drops	$\hat U_I=\exp[-\tfrac{i}{\hbar}\int\hat V_I\,\mathrm{d}t']$ ; no ordering needed, the ordinary exponential is exact (single-operator drives).
Full (untruncated) series	exactly unitary	Hermiticity of $\hat H$ guarantees $\hat U^\dagger\hat U=\mathbb{1}$ ; unitarity is a property of the whole sum, never of a partial sum.
$n$ -th-order truncation	defect $O(V^{n+1})$	Probability conservation holds only to the retained order; the residual is the leading omitted power of the coupling.
Time-independent $\hat V$	$\hat U_I=e^{-i\hat V_I\Delta t/\hbar}$ only if $[\hat V_I(t_1),\hat V_I(t_2)]=0$	even a constant $\hat V$ has a time-dependent $\hat V_I$ through the $e^{i\hat H_0 t/\hbar}$ dressing, so ordering generally survives.

Common trapA truncated Dyson series is not a unitary propagator

The most common mistake is to treat a first- or second-order Dyson sum as if it conserved probability exactly. It does not: $\sum_f|c_f|^2$ computed from a truncated series can exceed one, and the excess is the $O(V^{n+1})$ unitarity defect quantified above. This is not a flaw to be patched by renormalizing the amplitudes by hand — it is the honest statement that you have kept only part of the series. If you need a unitary approximate propagator, symmetrize (e.g. a Magnus or split-operator exponential) rather than truncating the raw Dyson sum. A second, subtler trap: writing $\hat U_I=\exp[-\tfrac{i}{\hbar}\int\hat V_I\,\mathrm{d}t']$ without the time-ordering symbol. That is correct only when the drive commutes with itself at all times; otherwise it silently gives the wrong evolution.

Worked examples

Example 1 — the truncation defect ladder for a two-level drive. Take the Hermitian interaction-picture drive $\hat V_I(t)=V_0[\cos(\Omega t)\sigma_x+\sin(\Omega t)\sigma_y]$ with $V_0=0.05$ , $\Omega=1.3$ (units $\hbar=1$ ), evolved from $0$ to $t=2$ . Compute the unitarity defect of the first- and second-order truncations, and read off how many orders you need for $10^{-4}$ accuracy.

Step 1. Build the first Dyson term $\hat U_I^{(1)}=-i\int_0^2\hat V_I\,\mathrm{d}t_1$ by numerical integration; it is a $2\times2$ anti-Hermitian matrix, so $(\hat U_I^{(1)})^\dagger+\hat U_I^{(1)}=0$ .

Step 2. First-order truncation $\hat U=\mathbb{1}+\hat U_I^{(1)}$ . Compute $\max|\hat U^\dagger\hat U-\mathbb{1}|=\mathbf{5.5\times10^{-3}}$ . This tracks $V_0^2=2.5\times10^{-3}$ — the defect is quadratic in the coupling.

Step 3. Build the second term $\hat U_I^{(2)}=(-i)^2\int_0^2\mathrm{d}t_1\int_0^{t_1}\mathrm{d}t_2\,\hat V_I(t_1)\hat V_I(t_2)$ . Add it: $\hat U=\mathbb{1}+\hat U_I^{(1)}+\hat U_I^{(2)}$ . The $O(V^2)$ unitarity terms now cancel to $2\times10^{-17}$ .

Step 4. Compute the new defect: $\max|\hat U^\dagger\hat U-\mathbb{1}|=\mathbf{1.7\times10^{-5}}$ , tracking $V_0^3=1.25\times10^{-4}$ . For $10^{-4}$ accuracy the second-order truncation already suffices; the first-order one does not. The defect fell by a factor $\approx320$ for one extra term.

Example 2 — the simplex fraction from a concrete integral. Verify the $1/2!$ that sits in front of the symmetrized second-order term, using the scalar surrogate $f(t)=\sin t+t$ over $[0,T]$ with $T=1$ .

Step 1. Single integral: $I=\int_0^1(\sin t+t)\,\mathrm{d}t=(1-\cos 1)+\tfrac12=0.4597+0.5=\mathbf{0.9597}$ .

Step 2. Ordered nested integral: $J=\int_0^1\mathrm{d}t_1\int_0^{t_1}\mathrm{d}t_2\,f(t_1)f(t_2)$ . Evaluated symbolically, $J=\tfrac12 I^2$ .

Step 3. Check: $\tfrac12 I^2=\tfrac12(0.9597)^2=\mathbf{0.4605}$ , and the direct nested integral gives the same value. The ordered region carries exactly half the weight of the full square — the $n=2$ instance of the $1/n!$ simplex fraction, which is why the time-ordered exponential's second term needs the $\tfrac12$ .

Physicist

Origins · 1 minDyson Series and how this was discovered

Dyson Series

1949 · Cornell and the Institute for Advanced Study

The perturbative time-ordered expansion of the evolution operator is named for Freeman Dyson, whose 1949 papers unified the then-rival formalisms of Tomonaga, Schwinger, and Feynman into a single calculable series — the story told in full by the companion article on the time-ordered exponential. What the interaction-picture, matrix-element version owns is the everyday machinery of atomic and molecular transition theory: the series that produces first-order transition amplitudes, golden-rule rates, and — carried into a relativistic, field-theoretic setting — the scattering matrix of quantum electrodynamics, diagram by diagram. The single conceptual move, that a hard time-dependent problem becomes an ordered sum over histories of a solvable perturbation, is Dyson's lasting contribution to how physicists compute.

History

1926

— Dirac formulates time-dependent perturbation theory and the interaction picture in his radiation paper

1949

— Dyson derives the time-ordered exponential and the series bearing his name, reconciling the Tomonaga, Schwinger, and Feynman formulations

1949

— the Dyson series becomes the backbone of the QED S-matrix expansion

1950

s — the interaction-picture perturbation series is standardized in atomic and molecular transition theory

1960

s — the same expansion, adapted to imaginary time, underlies finite-temperature and many-body Green-function methods

Going further

References. Sakurai & Napolitano, Modern Quantum Mechanics, 2nd/3rd ed., §5.6–5.7 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Ch. XIII · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 18 · Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., Ch. 11 · Townsend, A Modern Approach to Quantum Mechanics, Ch. 14 · Dyson, F. J., Phys. Rev. 75, 486 & 1736 (1949). This article owns the perturbative interaction-picture realization of the Dyson series — the order-by-order matrix-element amplitudes and the quantitative truncation-unitarity ladder. The abstract time-ordered exponential $\mathcal{T}\exp[-\tfrac{i}{\hbar}\int\hat H\,\mathrm{d}t']$ , its $1/n!$ combinatorics proved from the simplex, and the non-commuting counterexample belong to Time-Ordered Exponential and the Dyson Series, which this article recalls by name. The interaction-picture integral equation it iterates is The Interaction Picture and Time-Dependent Hamiltonians; the first-order amplitude it displays is developed in First-Order Transition Amplitudes and Probabilities, and the golden-rule rate those feed into is Fermi's Golden Rule. The covariant, field-theoretic version — the QED $S$ -matrix — is a topic for the relativistic and quantum-field tracks.

Provenance. Equations are standard and should be verified against Sakurai & Napolitano Modern QM §5.6–5.7 (Dyson series, time-ordered exponential, unitarity), Cohen-Tannoudji Vol. II Ch. XIII, Shankar Ch. 18, and Dyson Phys. Rev. 75 (1949). A verifier should confirm: (i) the iterated integral equation reproduces $i\hbar\partial_t U=HU$ ; (ii) the simplex-to-hypercube symmetrization with factor $1/n!$ ; (iii) $U^\dagger U=\mathbb{1}$ follows from Hermiticity of $H$ ; (iv) the first-order amplitude carries phase $e^{i\omega_{fi}t'}$ with $\omega_{fi}=(E_f-E_i)/\hbar$ . No empirical constants are quoted beyond $\hbar$ .

Builds on: The Interaction Picture and Time-Dependent Hamiltonians · Time-Dependent Perturbation Theory · Equivalence of the Dynamical Pictures Leads to: First-Order Transition Amplitudes and Probabilities · Fermi's Golden Rule · Harmonic (Sinusoidal) Perturbations and Resonance · Semiclassical Theory of Atom–Radiation Interaction

Is the Dyson series convergent?

Not in general. For bounded perturbations over a finite time it converges, but for the field-theoretic version it is famously an asymptotic series — the terms shrink then grow, with an optimal truncation. In ordinary quantum mechanics with a well-behaved $\hat V$ over finite $t$ the series converges, but the practical point is the same as for any perturbation expansion: keep enough orders that the next omitted term (order $V^{n+1}$ ) is below your target accuracy.

What is the difference between this article and the time-ordered-exponential article?

They are register twins. The companion article owns the abstract object: the operator $\hat U(t,t_0)=\mathcal{T}\exp[-\tfrac{i}{\hbar}\int\hat H\,\mathrm{d}t']$ for any time-dependent Hamiltonian, the geometric proof of the $1/n!$ , and the counterexample showing the naive exponential fails. This article owns the perturbative, matrix-element realization in the interaction picture: the $c_f^{(n)}$ amplitudes with their Bohr-frequency phases, and the quantitative order-by-order unitarity defect that a truncation incurs. Same series, two levels of abstraction.

Why is the first-order term anti-Hermitian?

Because $\hat U_I^{(1)}=-\tfrac{i}{\hbar}\int\hat V_I\,\mathrm{d}t'$ is $-i$ times a Hermitian operator (the integral of the Hermitian $\hat V_I$ ). An operator of the form $-i\hat H_{\text{eff}}$ with $\hat H_{\text{eff}}$ Hermitian satisfies $(\,-i\hat H_{\text{eff}})^\dagger=+i\hat H_{\text{eff}}=-(-i\hat H_{\text{eff}})$ . This is exactly why there is no linear-in- $V$ unitarity defect: the first-order piece is the generator of an infinitesimal unitary rotation.

Does a constant perturbation avoid time-ordering?

Not automatically. Even if $\hat V$ is constant in the Schrödinger picture, the interaction-picture $\hat V_I(t)=e^{i\hat H_0 t/\hbar}\hat V e^{-i\hat H_0 t/\hbar}$ generally depends on time and fails to commute with itself at different times unless $[\hat V,\hat H_0]=0$ . So the time-ordering usually survives even for a "constant" perturbation. It drops only in the special case where $\hat V$ commutes with the free Hamiltonian, or more generally where the interaction-picture drive keeps a single fixed operator direction.

How does this become the S-matrix of quantum field theory?

Replace $\hat V_I(t)$ by the interaction Hamiltonian density integrated over space and take $t_0\to-\infty$ , $t\to+\infty$ . The Dyson series for $\hat U_I(\infty,-\infty)$ is then the $S$ -matrix, and Wick's theorem on each time-ordered product generates Feynman diagrams. The structure is identical; only the operators and the interpretation of the "kicks" change.

If a truncated series is not unitary, why is it useful?

Because the non-unitarity is controlled and small. For weak coupling the defect is $O(V^{n+1})$ for an $n$ -th-order truncation, far below the accuracy of the transition probabilities you extract at leading order. You use the truncation to compute a rate or amplitude in a regime where almost no population leaves the initial state; the tiny unitarity leak is negligible there. When you genuinely need a unitary propagator — long-time evolution, strong driving — you switch to a unitarity-preserving scheme (Magnus, split-operator) rather than a bare truncation.

Where does the minus sign and the factor of i come from in each term?

From the Schrödinger equation itself. The interaction-picture equation is $i\hbar\,\partial_t\hat U_I=\hat V_I\hat U_I$ , so integrating gives a factor $-i/\hbar$ per power of $\hat V_I$ . The $n$ -th term therefore carries $(-i/\hbar)^n$ . These are not free conventions: they are fixed by requiring the series to solve the evolution equation, which the consistency check in the first derivation verifies term by term.