12.03.03 · quantum / time-evolution

The interaction picture, Dyson series, and time-ordered exponentials

shipped3 tiersLean: nonepending prereqs

Anchor (Master): Weinberg — The Quantum Theory of Fields, Vol. 1 (Cambridge, 1995), §3.5

Intuition Beginner

You already know the Schrödinger and Heisenberg pictures from 12.03.01. In the Schrödinger picture the state moves and the operators sit still. In the Heisenberg picture the state is frozen and the operators move. The interaction picture — also called the Dirac picture — splits the difference. It is designed for problems where the Hamiltonian has two pieces: a "free" part $H_{0}$ that is easy to solve, and an "interaction" part $V$ that is small or complicated.

The interaction picture peels off the boring $H_{0}$ dynamics so you can focus on what $V$ does. The state evolves only under $V$ , and the operators evolve only under $H_{0}$ . If $V = 0$ , the interaction-picture state does not move at all — all the motion has been absorbed into the operators. When $V$ is nonzero but small, the residual motion of the state is slow and tractable, which is exactly the setup for perturbation theory.

Why not just work in the Schrödinger picture? You can, but the algebra is wasteful. The Schrödinger equation tracks the fast oscillations from $H_{0}$ alongside the slow corrections from $V$ . The interaction picture factors out the fast part analytically, leaving equations that contain only the physically interesting terms. This is the same principle as moving to a rotating frame in classical mechanics: you subtract the known rotation so you can see the perturbation distinctly.

The payoff is enormous. The interaction picture leads directly to the Dyson series — a systematic expansion of the time-evolution operator in powers of $V$ . At first order, the Dyson series gives the Born approximation in scattering and Fermi's golden rule for transition rates. At higher orders, it gives the full perturbation series of quantum field theory, organised into Feynman diagrams. Every scattering amplitude in the Standard Model is computed from the Dyson series applied to the quantum electrodynamics or QCD interaction.

Visual Beginner

Think of a spinning top. The top spins fast about its own axis — that is $H_{0}$ . A gentle breeze slowly pushes the top over — that is $V$ . In the Schrödinger picture you track both the fast spin and the slow tilt simultaneously. The interaction picture transforms into the top's rest frame: the spin disappears from view, and all you see is the slow tilt caused by the breeze. The Dyson series is the systematic description of that slow tilt, expanded in powers of the wind strength.

Worked example Beginner

A two-level atom has ground state $∣ g ⟩$ at energy $0$ and excited state $∣ e ⟩$ at energy $E_{0} = ℏ ω_{0}$ . The atom sits in a weak oscillating electric field that couples the two levels. The Hamiltonian is $H = H_{0} + V$ where $H_{0} = E_{0} ∣ e ⟩ ⟨ e ∣$ and $V = λ (∣ e ⟩ ⟨ g ∣ + ∣ g ⟩ ⟨ e ∣) cos ω t$ with $λ$ small.

Step 1: Transform to the interaction picture. The free evolution is $e^{- i H_{0} t /ℏ} ∣ g ⟩ = ∣ g ⟩$ and $e^{- i H_{0} t /ℏ} ∣ e ⟩ = e^{- i ω_{0} t} ∣ e ⟩$ . The interaction-picture coupling becomes $V_{I} (t) = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ}$ . A short calculation gives $V_{I} (t) = λ (e^{i ω_{0} t} ∣ e ⟩ ⟨ g ∣ + e^{- i ω_{0} t} ∣ g ⟩ ⟨ e ∣) cos ω t$ .

Step 2: Read off the frequencies. Using $cos ω t = \frac{1}{2} (e^{iω t} + e^{- iω t})$ , the coupling $∣ g ⟩ \to ∣ e ⟩$ carries the phase $e^{i ω_{0} t} e^{iω t} = e^{i (ω_{0} + ω) t}$ or $e^{i (ω_{0} - ω) t}$ . The near-resonant term — the one whose phase varies slowly when $ω \approx ω_{0}$ — is the piece proportional to $e^{i (ω_{0} - ω) t}$ . The interaction picture has isolated the slow-beating term automatically.

Step 3: First-order amplitude. If the atom starts in $∣ g ⟩$ , the first-order Dyson contribution gives the amplitude to be in $∣ e ⟩$ at time $t$ as $c_{e}^{(1)} (t) = - \frac{iλ}{2ℏ} \cdot \frac{e ^{i (ω_{0} - ω) t} - 1}{i ( ω _{0} - ω )}$ . For $ω = ω_{0}$ (exact resonance), the phase factor is constant and $∣ c_{e}^{(1)} ∣^{2} = λ^{2} t^{2} / (4 ℏ^{2})$ . The probability grows quadratically — the signature of resonant driving at short times.

The interaction picture stripped away the fast oscillation at frequency $ω_{0}$ and left only the slow detuning $(ω_{0} - ω)$ , making the resonance condition manifest.

Check your understanding Beginner

Exercise (medium, symbolic).

For a spin-1/2 particle in a magnetic field with $H_{0} = - \frac{ℏ ω}{2} σ_{z}$ and a small transverse perturbation $V = \frac{ℏΩ}{2} σ_{x}$ (constant), compute $V_{I} (t) = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ}$ . Show that $V_{I} (t) = \frac{ℏΩ}{2} (σ_{x} cos ω t - σ_{y} sin ω t)$ .

Hint

Use $e^{iω t σ_{z} /2} σ_{x} e^{- iω t σ_{z} /2} = σ_{x} cos ω t - σ_{y} sin ω t$ , which follows from the rotation formula for Pauli matrices.

Answer

The interaction-picture potential is $V_{I} (t) = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ} = \frac{ℏΩ}{2} e^{iω t σ_{z} /2} σ_{x} e^{- iω t σ_{z} /2}$ . Using the rotation identity for Pauli matrices: $e^{i θ σ_{z} /2} σ_{x} e^{- i θ σ_{z} /2} = σ_{x} cos θ - σ_{y} sin θ$ with $θ = ω t$ . Therefore $V_{I} (t) = \frac{ℏΩ}{2} (σ_{x} cos ω t - σ_{y} sin ω t)$ . The coupling rotates in the $x y$ -plane at the Larmor frequency $ω$ — the free precession has been absorbed into the operator.

Formal definition Intermediate+

Let $H$ be a Hilbert space and $H = H_{0} + V$ a Hamiltonian decomposition with $H_{0}$ self-adjoint. Define the interaction-picture transformation by

∣ ψ (t) ⟩_{I} = e^{i H_{0} t /ℏ} ∣ ψ (t) ⟩_{S}, A_{I} (t) = e^{i H_{0} t /ℏ} A_{S} e^{- i H_{0} t /ℏ} .

Substituting into the Schrödinger equation $i ℏ d ∣ ψ ⟩_{S} / d t = (H_{0} + V) ∣ ψ ⟩_{S}$ and cancelling the $H_{0}$ terms yields the interaction-picture equation of motion:

i ℏ \frac{d}{d t} ∣ ψ (t) ⟩_{I} = V_{I} (t) ∣ ψ (t) ⟩_{I}, V_{I} (t) = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ} .

Operators in the interaction picture evolve under $H_{0}$ alone, exactly as Heisenberg-picture operators would if the full Hamiltonian were $H_{0}$ :

\frac{d A _{I}}{d t} = \frac{i}{ℏ} [H_{0}, A_{I}] + (\frac{\partial A}{\partial t})_{I} .

The interaction picture interpolates between the Schrödinger picture (all dynamics in the state, $H_{0} = 0$ ) and the Heisenberg picture (all dynamics in the operators, $V = 0$ ). Setting $H_{0} = 0$ recovers the Schrödinger picture; setting $V = 0$ recovers the Heisenberg picture with respect to $H_{0}$ .

The Dyson time-evolution operator. The formal solution to the interaction-picture equation with initial condition $∣ ψ (t_{0}) ⟩_{I}$ is $∣ ψ (t) ⟩_{I} = U_{I} (t, t_{0}) ∣ ψ (t_{0}) ⟩_{I}$ , where $U_{I}$ satisfies

i ℏ \frac{d}{d t} U_{I} (t, t_{0}) = V_{I} (t) U_{I} (t, t_{0}), U_{I} (t_{0}, t_{0}) = 1.

Iterating the integrated form $U_{I} (t, t_{0}) = 1 - \frac{i}{ℏ} \int_{t_{0}}^{t} V_{I} (t_{1}) U_{I} (t_{1}, t_{0}) d t_{1}$ produces the Dyson series:

U_{I} (t, t_{0}) = T exp (- \frac{i}{ℏ} \int_{t_{0}}^{t} V_{I} (t^{'}) d t^{'}) = 1 + n = 1 \sum \infty (\frac{- i}{ℏ})^{n} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} \dots \int_{t_{0}}^{t_{n - 1}} d t_{n} V_{I} (t_{1}) V_{I} (t_{2}) \dots V_{I} (t_{n}) .

The symbol $T$ denotes time ordering: operators are arranged with later times to the left. The nested integration limits enforce $t_{1} > t_{2} > \dots > t_{n}$ , which is equivalent to integrating freely over $[t_{0}, t]^{n}$ and then applying $T$ . The equivalent symmetric form is

U_{I}^{(n)} (t, t_{0}) = \frac{1}{n ! ( i ℏ ) ^{n}} \int_{t_{0}}^{t} d t_{1} \dots \int_{t_{0}}^{t} d t_{n} T [V_{I} (t_{1}) \dots V_{I} (t_{n})] .

Connection to the full evolution operator. The full Schrödinger-picture evolution operator is recovered by sandwiching $U_{I}$ :

U (t, t_{0}) = e^{- i H_{0} (t - t_{0}) /ℏ} U_{I} (t, t_{0}) .

When $V = 0$ , $U_{I} = 1$ and $U = e^{- i H_{0} (t - t_{0}) /ℏ}$ — the free evolution. When $V$ is small, the Dyson series converges rapidly and truncating at order $V^{n}$ gives $n$ -th order perturbation theory.

Counterexamples to common slips

The interaction picture requires $H_{0}$ to be self-adjoint. If $H_{0}$ is merely symmetric (not self-adjoint), the transformation $e^{i H_{0} t /ℏ}$ may not be unitary and the interaction-picture state can fail to preserve its norm.
The Dyson series is not the ordinary exponential of $- \frac{i}{ℏ} \int V_{I} d t^{'}$ . The ordinary exponential $e^{- \frac{i}{ℏ} \int V_{I} d t^{'}}$ is correct only when $[V_{I} (t_{1}), V_{I} (t_{2})] = 0$ for all $t_{1}, t_{2}$ , which is generally false. The time-ordering symbol $T$ is essential and encodes the non-commutativity of $V_{I}$ at different times.
The interaction picture is not a new physical theory. It is a change of variables within the same unitary framework. Expectation values are identical to those computed in the Schrödinger or Heisenberg pictures.

Key theorem with proof Intermediate+

Theorem (Dyson series). Let $H_{0}$ be self-adjoint with domain $D (H_{0}) \subset H$ , and let $V$ be Hermitian and bounded with $∥ V ∥ \leq M < \infty$ . Define $V_{I} (t) = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ}$ . Then the series

U_{I} (t, t_{0}) = 1 + n = 1 \sum \infty (\frac{- i}{ℏ})^{n} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} \dots \int_{t_{0}}^{t_{n - 1}} d t_{n} V_{I} (t_{1}) \dots V_{I} (t_{n})

converges in operator norm for all $t, t_{0} \in R$ and defines a unitary operator satisfying $i ℏ \dot{U}_{I} = V_{I} U_{I}$ , $U_{I} (t_{0}, t_{0}) = 1$ , and $U_{I} (t, s) U_{I} (s, t_{0}) = U_{I} (t, t_{0})$ .

Proof.

Step 1: Norm bound on the $n$ -th term. Each $V_{I} (t_{j})$ is unitarily conjugated from $V$ , so $∥ V_{I} (t_{j}) ∥ = ∥ V ∥ \leq M$ . The volume of the nested integration domain ${(t_{1}, \dots, t_{n}) : t_{0} \leq t_{n} \leq \dots \leq t_{1} \leq t}$ is $(t - t_{0})^{n} / n!$ . Therefore

∥ U_{I}^{(n)} ∥ \leq \frac{1}{ℏ ^{n}} \cdot \frac{( t - t _{0} ) ^{n}}{n !} \cdot M^{n} = \frac{1}{n !} (\frac{M ( t - t _{0} )}{ℏ})^{n} .

Step 2: Convergence. The bound $\sum_{n} ∥ U_{I}^{(n)} ∥ \leq \sum_{n} \frac{1}{n !} (\frac{M ∣ t - t _{0} ∣}{ℏ})^{n} = e^{M ∣ t - t_{0} ∣/ℏ} < \infty$ . The series converges absolutely in operator norm, and the sum $U_{I} (t, t_{0})$ is a bounded operator with $∥ U_{I} ∥ \leq e^{M ∣ t - t_{0} ∣/ℏ}$ .

Step 3: Satisfies the differential equation. Differentiate the series term by term (justified by uniform convergence on bounded intervals). The $n$ -th term of $\dot{U}_{I}$ is

\frac{d}{d t} U_{I}^{(n)} = (\frac{- i}{ℏ})^{n} V_{I} (t) \int_{t_{0}}^{t} d t_{2} \dots \int_{t_{0}}^{t_{n - 1}} d t_{n} V_{I} (t_{2}) \dots V_{I} (t_{n}) = - \frac{i}{ℏ} V_{I} (t) U_{I}^{(n - 1)} .

Summing: $i ℏ \dot{U}_{I} = V_{I} \sum_{n = 1}^{\infty} U_{I}^{(n - 1)} = V_{I} U_{I}$ . The initial condition $U_{I} (t_{0}, t_{0}) = 1$ is immediate from the series (all integrals vanish at $t = t_{0}$ ).

Step 4: Unitarity. Define $\tilde{U}_{I} (t, t_{0}) = U_{I} (t_{0}, t)^{†}$ . One verifies that $\tilde{U}_{I}$ satisfies the same differential equation $i ℏ \dot{\tilde{U}}_{I} = V_{I} \tilde{U}_{I}$ with the same initial condition, so by uniqueness of solutions $\tilde{U}_{I} = U_{I}$ . Hence $U_{I} (t, t_{0})^{†} = U_{I} (t_{0}, t)$ and $U_{I}^{†} U_{I} = U_{I} (t_{0}, t) U_{I} (t, t_{0}) = U_{I} (t_{0}, t_{0}) = 1$ .

Step 5: Composition law. The operator $W (t, s) = U_{I} (t, s) U_{I} (s, t_{0})$ satisfies $i ℏ d W / d t = V_{I} (t) W$ with $W (s, s) = U_{I} (s, t_{0})$ . By uniqueness, $W (t, s) = U_{I} (t, t_{0})$ , establishing $U_{I} (t, t_{0}) = U_{I} (t, s) U_{I} (s, t_{0})$ . $□$

Corollary (First-order perturbation theory). If the system starts in eigenstate $∣ i ⟩$ of $H_{0}$ with energy $E_{i}$ , the first-order transition amplitude to eigenstate $∣ f ⟩$ ( $f \neq = i$ ) of $H_{0}$ with energy $E_{f}$ is

c_{f}^{(1)} (t) = - \frac{i}{ℏ} \int_{t_{0}}^{t} e^{i ω_{f i} t^{'}} ⟨ f ∣ V ∣ i ⟩ d t^{'}, ω_{f i} = \frac{E _{f} - E _{i}}{ℏ} .

This follows from $⟨ f ∣ U_{I}^{(1)} (t, t_{0}) ∣ i ⟩ = - \frac{i}{ℏ} \int_{t_{0}}^{t} ⟨ f ∣ V_{I} (t^{'}) ∣ i ⟩ d t^{'}$ and $V_{I} (t^{'}) ∣ i ⟩ = e^{i H_{0} t^{'} /ℏ} V e^{- i E_{i} t^{'} /ℏ} ∣ i ⟩$ , which evaluates to $e^{i ω_{f i} t^{'}} V_{f i} ∣ f ⟩$ plus off-diagonal terms.

Bridge. The Dyson series is the backbone of time-dependent perturbation theory 12.07.02, where the first-order term gives Fermi's golden rule. It reappears in scattering theory 12.08.01 as the S-matrix $S = U_{I} (+ \infty, - \infty)$ , and in the path integral 12.10.01 where the Trotter splitting provides an independent derivation. In QFT, the same series with $V$ replaced by the interaction Lagrangian density gives the perturbative expansion of the S-matrix in coupling constants — every Feynman diagram is a term in the Dyson series, and Wick's theorem reduces time-ordered products to contractions that become propagators.

Exercises Intermediate+

Exercise 2 (easy, multiple choice).

The time-ordering symbol $T$ in the Dyson series is necessary because:

A. $V_{I} (t_{1})$ and $V_{I} (t_{2})$ need not commute for $t_{1} \neq = t_{2}$ . B. The Hamiltonian is time-dependent. C. The series contains an infinite number of terms. D. $H_{0}$ and $V$ do not commute.

Hint

Consider what happens when you swap two interaction-picture potentials evaluated at different times.

Answer

A. The time-ordering is needed because $[V_{I} (t_{1}), V_{I} (t_{2})] \neq = 0$ in general. Without time ordering, the operator product $V_{I} (t_{1}) V_{I} (t_{2})$ depends on the order of the factors. Time ordering places later times to the left, giving an unambiguous result. Option D is a necessary condition for non-commutativity of $V_{I}$ at different times but is not the direct reason — the direct reason is the non-commutativity of $V_{I} (t_{1})$ and $V_{I} (t_{2})$ .

Exercise 3 (medium, symbolic).

Show that the time-ordering operator converts the nested integral into the symmetric form. Specifically, verify for $n = 2$ that

\int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} V_{I} (t_{1}) V_{I} (t_{2}) = \frac{1}{2} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t} d t_{2} T [V_{I} (t_{1}) V_{I} (t_{2})] .

Hint

Split the square $[t_{0}, t]^{2}$ into two triangles: $t_{1} > t_{2}$ and $t_{2} > t_{1}$ . On each triangle, $T$ orders the operators accordingly.

Answer

The square $[t_{0}, t]^{2}$ splits as ${(t_{1}, t_{2}) : t_{1} \geq t_{2}} \cup {(t_{1}, t_{2}) : t_{2} \geq t_{1}}$ (the diagonal has measure zero). On the first region, $T [V_{I} (t_{1}) V_{I} (t_{2})] = V_{I} (t_{1}) V_{I} (t_{2})$ ; on the second, $T [V_{I} (t_{1}) V_{I} (t_{2})] = V_{I} (t_{2}) V_{I} (t_{1})$ . Therefore:

\int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t} d t_{2} T [V_{I} (t_{1}) V_{I} (t_{2})] = \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} V_{I} (t_{1}) V_{I} (t_{2}) + \int_{t_{0}}^{t} d t_{2} \int_{t_{0}}^{t_{2}} d t_{1} V_{I} (t_{2}) V_{I} (t_{1}) .

Relabelling the second integral ( $t_{1} \leftrightarrow t_{2}$ ) shows it equals the first. Hence the right side is $2 \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} V_{I} (t_{1}) V_{I} (t_{2})$ , and dividing by 2 gives the stated identity. The general- $n$ case follows by the same partitioning of the hypercube $[t_{0}, t]^{n}$ into $n!$ simplices, each labelled by a permutation.

Exercise 4 (medium, symbolic).

A system has $H_{0} = ℏ ω_{0} a^{†} a$ (harmonic oscillator) and $V = λ (a + a^{†})$ (linear driving). Compute $V_{I} (t)$ and show that it oscillates at the natural frequency $ω_{0}$ .

Hint

Use the Heisenberg-picture results $a_{H} (t) = a e^{- i ω_{0} t}$ and $a_{H}^{†} (t) = a^{†} e^{i ω_{0} t}$ from 12.03.01.

Answer

The interaction picture evolves operators with $H_{0}$ alone, so $a_{I} (t) = e^{i H_{0} t /ℏ} a e^{- i H_{0} t /ℏ} = a e^{- i ω_{0} t}$ and $a_{I}^{†} (t) = a^{†} e^{i ω_{0} t}$ . Therefore:

V_{I} (t) = λ (a_{I} (t) + a_{I}^{†} (t)) = λ (a e^{- i ω_{0} t} + a^{†} e^{i ω_{0} t}) .

The driving term in the interaction picture oscillates at $ω_{0}$ — the free evolution has been stripped into the phase factors, leaving a harmonically modulated coupling. A resonant drive ( $ω = ω_{0}$ ) would appear as a slowly-varying (or constant) coupling in a co-rotating frame.

Exercise 5 (medium, symbolic).

Derive the interaction-picture equation of motion from scratch. Start with $∣ ψ (t) ⟩_{S} = e^{- i H_{0} t /ℏ} ∣ ψ (t) ⟩_{I}$ , differentiate, substitute into the Schrödinger equation $i ℏ d ∣ ψ ⟩_{S} / d t = (H_{0} + V) ∣ ψ ⟩_{S}$ , and show that the $H_{0}$ terms cancel to leave $i ℏ d ∣ ψ ⟩_{I} / d t = V_{I} (t) ∣ ψ ⟩_{I}$ .

Hint

Apply the product rule to $d / d t [e^{- i H_{0} t /ℏ} ∣ ψ ⟩_{I}]$ and use $i ℏ d (e^{- i H_{0} t /ℏ}) / d t = H_{0} e^{- i H_{0} t /ℏ}$ .

Answer

Differentiate: $i ℏ d ∣ ψ ⟩_{S} / d t = i ℏ (- \frac{i H _{0}}{ℏ}) e^{- i H_{0} t /ℏ} ∣ ψ ⟩_{I} + e^{- i H_{0} t /ℏ} i ℏ d ∣ ψ ⟩_{I} / d t = H_{0} e^{- i H_{0} t /ℏ} ∣ ψ ⟩_{I} + e^{- i H_{0} t /ℏ} i ℏ d ∣ ψ ⟩_{I} / d t$ .

Set equal to $(H_{0} + V) ∣ ψ ⟩_{S} = H_{0} e^{- i H_{0} t /ℏ} ∣ ψ ⟩_{I} + V e^{- i H_{0} t /ℏ} ∣ ψ ⟩_{I}$ .

The $H_{0}$ terms cancel on both sides. Multiply through by $e^{i H_{0} t /ℏ}$ :

i ℏ \frac{d}{d t} ∣ ψ ⟩_{I} = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ} ∣ ψ ⟩_{I} = V_{I} (t) ∣ ψ ⟩_{I} .

The cancellation of $H_{0}$ is the key algebraic step: it is why the interaction picture exists.

Exercise 6 (medium, symbolic).

Show that $U_{I} (t, t_{0})$ is unitary at first and second order by explicitly verifying $U_{I}^{†} U_{I} = 1 + O (V^{3})$ . Write $U_{I} = 1 + U^{(1)} + U^{(2)} + \dots$ and show that $U^{(1) †} + U^{(1)} = 0$ and $U^{(2) †} + U^{(1) †} U^{(1)} + U^{(2)} = 0$ .

Hint

$U^{(1)} = - \frac{i}{ℏ} \int_{t_{0}}^{t} V_{I} (t_{1}) d t_{1}$ . Its adjoint reverses the sign. For second order, split the unrestricted double integral by time ordering.

Answer

First order: $U^{(1)} = - \frac{i}{ℏ} \int_{t_{0}}^{t} V_{I} (t_{1}) d t_{1}$ , so $U^{(1) †} = + \frac{i}{ℏ} \int_{t_{0}}^{t} V_{I} (t_{1}) d t_{1} = - U^{(1)}$ (using $V_{I}^{†} = V_{I}$ ). Hence $U^{(1) †} + U^{(1)} = 0$ .

Second order: $U^{(2)} = (- \frac{i}{ℏ})^{2} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} V_{I} (t_{1}) V_{I} (t_{2})$ , so $U^{(2) †} = (- \frac{i}{ℏ})^{2} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} V_{I} (t_{2}) V_{I} (t_{1}) = - \frac{1}{ℏ ^{2}} \int_{t_{0}}^{t} d t_{2} \int_{t_{0}}^{t_{2}} d t_{1} V_{I} (t_{1}) V_{I} (t_{2})$ (relabelled). The product $U^{(1) †} U^{(1)} = - \frac{1}{ℏ ^{2}} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t} d t_{2} V_{I} (t_{1}) V_{I} (t_{2})$ . Split the square: $\int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t} d t_{2} = \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} + \int_{t_{0}}^{t} d t_{2} \int_{t_{0}}^{t_{2}} d t_{1}$ . Adding $U^{(2) †} + U^{(1) †} U^{(1)} + U^{(2)}$ cancels both triangular regions, leaving zero.

Exercise 7 (hard, symbolic).

Prove the composition law $U_{I} (t, t_{0}) = U_{I} (t, s) U_{I} (s, t_{0})$ by showing that both sides satisfy the same differential equation with the same initial condition. Why does uniqueness of solutions apply here?

Hint

Both $U_{I} (t, t_{0})$ and $U_{I} (t, s) U_{I} (s, t_{0})$ satisfy $i ℏ d W / d t = V_{I} (t) W$ as functions of $t$ . At $t = s$ both equal $U_{I} (s, t_{0})$ .

Answer

Fix $s$ and $t_{0}$ . Define $W (t) = U_{I} (t, s) U_{I} (s, t_{0})$ . Differentiate: $i ℏ d W / d t = i ℏ (d U_{I} (t, s) / d t) \cdot U_{I} (s, t_{0}) = V_{I} (t) U_{I} (t, s) U_{I} (s, t_{0}) = V_{I} (t) W (t)$ . So $W$ satisfies the same ODE as $U_{I} (t, t_{0})$ . At $t = s$ : $W (s) = U_{I} (s, s) U_{I} (s, t_{0}) = 1 \cdot U_{I} (s, t_{0}) = U_{I} (s, t_{0})$ . Meanwhile $U_{I} (s, t_{0})$ evaluated at $t = s$ gives $U_{I} (s, t_{0})$ — same initial value. Uniqueness of solutions to linear ODEs in Banach space (the standard result for $i ℏ \dot{X} = V_{I} (t) X$ with bounded $V_{I}$ ) guarantees $W (t) = U_{I} (t, t_{0})$ for all $t$ . The composition law is the group property of the time-evolution operator, inherited from the uniqueness theorem.

Exercise 8 (hard, symbolic).

For a constant perturbation $V$ (time-independent in the Schrödinger picture) acting on a system with $H_{0} = ℏ ω_{0} ∣ e ⟩ ⟨ e ∣$ and initial state $∣ g ⟩$ , compute the second-order Dyson amplitude $c_{e}^{(2)} (t)$ via the intermediate state sum. Show that when there is only one intermediate state ( $∣ e ⟩$ itself), the second-order contribution vanishes.

Hint

The second-order amplitude is $c_{f}^{(2)} = (- \frac{i}{ℏ})^{2} \sum_{m} \int_{0}^{t} d t_{2} \int_{0}^{t_{2}} d t_{1} V_{f m} (t_{2}) V_{mi} (t_{1})$ where the sums run over eigenstates of $H_{0}$ . If $i = g$ , $f = e$ , and the only candidate $m$ is $∣ g ⟩$ or $∣ e ⟩$ , check which matrix elements survive.

Answer

With only two states $∣ g ⟩$ and $∣ e ⟩$ , the second-order amplitude from $∣ g ⟩$ to $∣ e ⟩$ sums over $m \in {g, e}$ :

For $m = g$ : $V_{e g} (t_{2}) V_{g g} (t_{1})$ . The matrix element $V_{g g} = ⟨ g ∣ V ∣ g ⟩$ is a number (c-number, not an operator). $V_{e g} (t_{2}) = V_{e g} e^{i ω_{0} t_{2}}$ . The integral is $V_{e g} V_{g g} \int_{0}^{t} d t_{2} \int_{0}^{t_{2}} d t_{1} e^{i ω_{0} t_{2}}$ . This contributes but only shifts the first-order result by redefining the energy of $∣ g ⟩$ .

For $m = e$ : $V_{ee} (t_{2}) V_{e g} (t_{1}) = V_{ee} V_{e g} \int_{0}^{t} d t_{2} \int_{0}^{t_{2}} d t_{1} e^{i ω_{0} t_{1}}$ . The point is that for a generic $V$ coupling $∣ g ⟩ \leftrightarrow ∣ e ⟩$ , the diagonal elements $V_{g g}$ and $V_{ee}$ contribute energy shifts while the off-diagonal piece is captured at first order. The second-order correction does not vanish identically — it gives the energy-level shift (the AC Stark shift for oscillating fields). The claim in the problem statement should be refined: if $V$ is purely off-diagonal ( $V_{g g} = V_{ee} = 0$ ), then both second-order paths involve a vanishing matrix element, and $c_{e}^{(2)} = 0$ for this two-level system — the first non-vanishing correction is first order.

Exercise 9 (hard, symbolic).

Compute the first-order Dyson amplitude for a system with $H_{0} = ℏ ω_{0} ∣ e ⟩ ⟨ e ∣$ under a sinusoidal perturbation $V (t) = V_{0} cos ω t$ (in the Schrödinger picture) with $V_{0} = λ (∣ e ⟩ ⟨ g ∣ + ∣ g ⟩ ⟨ e ∣)$ . Starting from $∣ g ⟩$ , show that the amplitude $c_{e}^{(1)} (t)$ contains two resonances, one at $ω = ω_{0}$ and one at $ω = - ω_{0}$ , and identify which is the rotating-wave term and which is the counter-rotating term.

Hint

$V_{I} (t)$ contains $e^{i ω_{0} t} cos ω t = \frac{1}{2} (e^{i (ω_{0} + ω) t} + e^{i (ω_{0} - ω) t})$ in the $∣ g ⟩ \to ∣ e ⟩$ matrix element.

Answer

$V_{e g}^{(I)} (t) = λ e^{i ω_{0} t} cos ω t = \frac{λ}{2} (e^{i (ω_{0} + ω) t} + e^{i (ω_{0} - ω) t})$ . The first-order amplitude is:

c_{e}^{(1)} (t) = - \frac{iλ}{2ℏ} \int_{0}^{t} [e^{i (ω_{0} + ω) t^{'}} + e^{i (ω_{0} - ω) t^{'}}] d t^{'} = - \frac{λ}{2ℏ} [\frac{e ^{i (ω_{0} + ω) t} - 1}{ω _{0} + ω} + \frac{e ^{i (ω_{0} - ω) t} - 1}{ω _{0} - ω}] .

The second term is resonant at $ω = ω_{0}$ — this is the rotating-wave term. The first term is resonant at $ω = - ω_{0}$ , which is unphysical for positive $ω$ — this is the counter-rotating term. Near resonance ( $ω \approx ω_{0}$ ), the rotating-wave term has a large denominator that nearly vanishes, while the counter-rotating term oscillates at frequency $ω_{0} + ω \approx 2 ω_{0}$ and averages out. The rotating-wave approximation discards the counter-rotating term.

Exercise 10 (medium, symbolic).

The full Schrödinger-picture evolution operator is $U (t, t_{0}) = e^{- i H_{0} (t - t_{0}) /ℏ} U_{I} (t, t_{0})$ . Verify directly that $U (t, t_{0})$ satisfies $i ℏ d U / d t = (H_{0} + V) U$ by differentiating the product.

Hint

Apply the product rule. Use $i ℏ d (e^{- i H_{0} (t - t_{0}) /ℏ}) / d t = H_{0} e^{- i H_{0} (t - t_{0}) /ℏ}$ and $i ℏ d U_{I} / d t = V_{I} (t) U_{I}$ .

Answer

i ℏ \frac{d U}{d t} = i ℏ \frac{d}{d t} [e^{- i H_{0} (t - t_{0}) /ℏ} U_{I}] = H_{0} e^{- i H_{0} (t - t_{0}) /ℏ} U_{I} + e^{- i H_{0} (t - t_{0}) /ℏ} i ℏ \frac{d U _{I}}{d t} .

Substitute $i ℏ d U_{I} / d t = V_{I} (t) U_{I}$ :

= H_{0} e^{- i H_{0} (t - t_{0}) /ℏ} U_{I} + e^{- i H_{0} (t - t_{0}) /ℏ} V_{I} (t) U_{I} .

Since $V_{I} (t) = e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ}$ , the second term becomes $e^{- i H_{0} (t - t_{0}) /ℏ} e^{i H_{0} t /ℏ} V e^{- i H_{0} t /ℏ} U_{I} = e^{i H_{0} t_{0} /ℏ} V e^{- i H_{0} t /ℏ} U_{I}$ . For $t_{0} = 0$ : $V e^{- i H_{0} t /ℏ} U_{I} = V U (t, 0)$ . Hence $i ℏ d U / d t = H_{0} U + V U = (H_{0} + V) U$ as claimed.

Exercise 11 (hard, conceptual).

When does the interaction picture fail? Consider a Hamiltonian $H_{0} + V$ where $H_{0}$ has continuous spectrum and $V$ is unbounded (e.g., the Coulomb potential in scattering problems). Explain why the Dyson series may not converge in operator norm, and what replaces it in practice.

Hint

The norm bound $∥ U_{I}^{(n)} ∥ \leq (M ∣ t - t_{0} ∣/ℏ)^{n} / n!$ requires $∥ V ∥ < \infty$ . What happens when $V$ is unbounded?

Answer

The Dyson-series convergence proof (Step 1 of the theorem) requires $∥ V ∥ \leq M < \infty$ . For unbounded $V$ (Coulomb potential, the QED interaction, any local potential in 3D), the norm bound fails and the series is at best asymptotic. In practice, the Dyson series is still used as a formal expansion and individual terms are computed as time-ordered integrals; the series is then interpreted as a perturbative (asymptotic) expansion in the coupling constant. Convergence is replaced by Borel resummability in favourable cases, or by direct non-perturbative methods (variational, lattice, resummed Dyson–Schwinger equations). In QFT, Dyson's 1952 argument shows the perturbation series cannot converge for any nonzero coupling because the energy is non-analytic at $α = 0$ (negative $α$ would make the vacuum unstable). The series is asymptotic: truncating at optimal order gives an excellent approximation, but including more terms eventually diverges.

Advanced results Master

Time-ordered products and Wick's theorem

The $n$ -th order term of the Dyson series contains the time-ordered product $T [V_{I} (t_{1}) \dots V_{I} (t_{n})]$ . In quantum field theory, each $V_{I} (t)$ is an integral over space of the interaction Lagrangian density, and the time-ordered product becomes a product of field operators at different spacetime points. Wick's theorem reduces such products to sums of contractions — pairs of field operators replaced by the Feynman propagator:

T [ϕ (x) ϕ (y)] = ∙ ∙ ϕ (x) ϕ (y) ∙ ∙ + ⟨ 0∣ T [ϕ (x) ϕ (y)] ∣0 ⟩,

where the colons denote normal ordering and the vacuum expectation value $⟨ 0∣ T [ϕ (x) ϕ (y)] ∣0 ⟩ = D_{F} (x - y)$ is the Feynman propagator. Applied to the full time-ordered product, Wick's theorem generates all Feynman diagrams at order $n$ : each diagram corresponds to a specific pattern of contractions among the $n$ interaction vertices.

The identification between the Dyson series and Feynman diagrams was Dyson's great synthesis of 1949. He showed that Feynman's diagrammatic rules, Schwinger's operator formalism, and Tomonaga's super-many-time formalism were three representations of the same Dyson series — the time-ordered exponential of the interaction-picture Hamiltonian. The convergence of the three independent derivations on the same mathematical object was the decisive evidence that QED was a coherent theory.

The S-matrix and asymptotic limits

The S-matrix (scattering matrix) is defined as the Dyson time-evolution operator evaluated between the infinite past and the infinite future:

S = U_{I} (+ \infty, - \infty) = T exp (- \frac{i}{ℏ} \int_{- \infty}^{+ \infty} V_{I} (t) d t) .

The matrix elements $⟨ f ∣ S ∣ i ⟩$ give the transition amplitude from an initial state $∣ i ⟩$ (prepared in the far past) to a final state $∣ f ⟩$ (detected in the far future). The asymptotic limits require careful treatment: the free states $∣ i ⟩$ and $∣ f ⟩$ are eigenstates of $H_{0}$ , and the limits $t \to \pm \infty$ must be understood in the sense of strong limits on the Hilbert space, or equivalently through the adiabatic switching protocol discussed below.

The unitarity of $S$ ( $S^{†} S = 1$ ) encodes probability conservation in scattering: the sum of probabilities for all outgoing channels equals one. This is an important constraint and leads to the optical theorem, which relates the total cross-section to the imaginary part of the forward-scattering amplitude.

Adiabatic switching and the Gell-Mann–Low theorem

The Dyson series evaluated between $- \infty$ and $+ \infty$ requires a regularisation of the infinite-time integrals. The standard technique is adiabatic switching: replace $V$ by $V_{η} (t) = e^{- η ∣ t ∣} V$ with $η > 0$ small, compute the Dyson series for this modified interaction, and then take $η \to 0^{+}$ .

The physical interpretation is that the interaction is turned on infinitely slowly from $t = - \infty$ and turned off infinitely slowly as $t \to + \infty$ . The switching function $e^{- η ∣ t ∣}$ provides an infrared regularisation of the time integrals, converting oscillatory integrals into absolutely convergent ones via the Sokhotski–Plemelj identity:

\frac{1}{x + i η} = P.V. \frac{1}{x} - iπ δ (x) (η \to 0^{+}) .

The Gell-Mann–Low theorem (1951) provides the rigorous underpinning. It states that for a system with $H = H_{0} + λV$ and ground state $∣Ω ⟩$ of the full Hamiltonian, the ratio

\frac{⟨ 0∣ T [ ϕ _{I} ( x _{1} ) \dots ϕ _{I} ( x _{n} ) exp ( - \frac{i}{ℏ} λ \int V _{I} d t )] ∣0 ⟩}{⟨ 0∣ T [ exp ( - \frac{i}{ℏ} λ \int V _{I} d t )] ∣0 ⟩}

has a well-defined limit as $η \to 0^{+}$ order by order in $λ$ , and this limit equals the time-ordered Green's function $⟨ Ω∣ T [ϕ (x_{1}) \dots ϕ (x_{n})] ∣Ω ⟩$ computed in the interacting vacuum. The denominator cancels the disconnected vacuum bubbles; the numerator generates all connected diagrams. The theorem guarantees that the perturbative S-matrix, defined through the Dyson series with adiabatic switching, yields well-defined transition amplitudes between asymptotic free-particle states.

The adiabatic limit is subtle. For theories with massless particles (QED, QCD), the switching function can introduce spurious infrared divergences that cancel only after summing over soft final-state radiation (the Bloch–Nordsieck mechanism). For theories with a mass gap, the adiabatic limit is unproblematic and the Gell-Mann–Low theorem holds rigorously.

The Magnus expansion and unitary resummation

The Dyson series expands $U_{I}$ in powers of $V$ . An alternative, due to Magnus (1954), writes $U_{I} (t, t_{0}) = e^{Ω (t, t_{0})}$ where $Ω$ is anti-Hermitian, and expands $Ω$ in a series of nested commutators:

Ω = - \frac{i}{ℏ} \int_{t_{0}}^{t} V_{I} (t_{1}) d t_{1} - \frac{1}{2 ℏ ^{2}} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} [V_{I} (t_{1}), V_{I} (t_{2})] + \dots

The Magnus expansion has a crucial advantage: truncating at any finite order gives an exactly unitary operator ( $e^{Ω}$ with anti-Hermitian $Ω$ is unitary). The truncated Dyson series, by contrast, is only approximately unitary. This makes the Magnus expansion the preferred tool for numerical time propagation in quantum chemistry and lattice gauge theory, where preservation of probability at each time step is essential.

The Magnus expansion converges when $\frac{1}{ℏ} \int_{t_{0}}^{t} ∥ V_{I} (t^{'}) ∥ d t^{'} < π$ , a condition that limits the step size in numerical applications but is always satisfiable by taking small enough time steps. The higher-order terms involve nested commutators $[V_{I} (t_{1}), [V_{I} (t_{2}), V_{I} (t_{3})]]$ that encode the non-abelian structure of the time evolution — the Magnus expansion measures the degree to which the time-ordered exponential fails to be an ordinary exponential.

Connection to the path integral

The Dyson series has a dual representation as a path integral. The Trotter product formula splits the time-ordered exponential into a product of infinitesimal evolution operators:

U_{I} (t, t_{0}) = N \to \infty lim k = 1 \prod N (1 - \frac{i Δ t}{ℏ} V_{I} (t_{k})), Δ t = \frac{t - t _{0}}{N} .

Inserting complete sets of eigenstates between each factor converts the operator product into a multiple integral over field configurations — the path integral. Each Dyson term $U_{I}^{(n)}$ becomes a sum over paths with $n$ interaction vertices inserted along the trajectory. The time ordering is automatic: the product ordering of the Trotter factors ensures that later vertices act after earlier ones.

This connection is the foundation of perturbative quantum field theory. The Feynman rules are precisely the diagrammatic representation of the Dyson series in the path-integral formalism: each vertex is an insertion of $V_{I}$ , each propagator is a free-particle Green's function, and the sum over diagrams is the sum over all time-orderings of the vertices. The equivalence between the operator (Dyson) and functional (path-integral) formulations is the central bridge between canonical quantisation and path-integral quantisation.

Non-perturbative perspectives: resurgent trans-series

The Dyson series is an asymptotic expansion in the coupling constant $λ$ . For QED and QCD, the perturbation series diverges factorially: $∣ U_{I}^{(n)} ∣ \sim n!$ for large $n$ . This divergence is not a defect but a feature — it signals the presence of non-perturbative effects (instantons, renormalons, condensates) that are invisible at any finite order of perturbation theory.

The mathematical framework that unifies perturbative and non-perturbative contributions is the trans-series (also called a resurgent expansion): an expansion of the form

U_{I} = n = 0 \sum \infty a_{n} λ^{n} + e^{- S / λ} n = 0 \sum \infty b_{n} λ^{n} + e^{- 2 S / λ} n = 0 \sum \infty c_{n} λ^{n} + \dots,

where $S$ is the instanton action. The perturbative series ( $a_{n}$ terms) is the Dyson series; the exponentially suppressed terms represent tunnelling events (instantons) that the Dyson series cannot capture. Borel resummation reconstructs the exact $U_{I}$ from the asymptotic Dyson coefficients, provided the Borel transform has no singularities on the positive real axis. When it does (as in QED, due to the renormalon singularity), the ambiguity in the Borel sum is cancelled by the instanton terms — the trans-series resolves the ambiguity.

This is the modern understanding of the Dyson series: it is not a convergent expansion but the leading sector of a trans-series that encodes the complete physics. The practical implication is that perturbation theory, while formally divergent, gives extremely accurate predictions when truncated at the optimal order (typically $n \sim 1/ λ$ for QED with $α \approx 1/137$ ), and the residual error is exponentially small in $1/ λ$ .

Connections Master

To scattering theory 12.08.01. The S-matrix is the Dyson series evaluated between $t = - \infty$ and $t = + \infty$ . The Born approximation is the first-order Dyson term; the full perturbation series gives the complete S-matrix. The Lippmann–Schwinger equation is the position-space representation of the Dyson integral equation for the scattering state.

To time-dependent perturbation theory 12.07.02. Fermi's golden rule is the $n = 1$ term of the Dyson series, squared and summed over final states. The golden-rule rate is the long-time limit of $∣ c_{f}^{(1)} ∣^{2} / t$ . Higher-order Dyson terms give multi-photon processes, virtual intermediate states, and radiative corrections.

To the path integral 12.10.01. The Trotter discretisation of the time-ordered exponential is the lattice precursor of the Feynman path integral. The Dyson series in the operator formalism maps term by term onto the perturbative expansion in the path-integral formalism — this is the equivalence proved by Dyson in 1949.

To QFT 12.12.01 and renormalisation 12.16.06. In QED the interaction picture uses $H_{0} = H_{free}$ (the free Dirac + Maxwell Hamiltonian) and $V = \int d^{3} x e \overset{ˉ}{ψ} γ^{μ} A_{μ} ψ$ (the minimal coupling). The Dyson series with this $V$ generates all Feynman diagrams of QED. Ultraviolet divergences appear as divergent loop integrals in individual Dyson terms; renormalisation absorbs them into redefinitions of mass, charge, and field normalisation. The renormalised Dyson series is the perturbative S-matrix of the Standard Model.

To quantum information and control. The interaction picture is the standard framework for driven quantum systems: $H_{0}$ describes the qubit or oscillator, $V (t)$ describes the control pulses. The Dyson series computes the effect of imperfect pulses; the Magnus expansion provides unitary-averaging methods for robust control (NMR, quantum computing). The average Hamiltonian theory of Waugh and Haeberlen is the Magnus expansion applied to pulse-sequence design.

Historical notes Master

The interaction picture was introduced by Dirac in his 1927 paper on the quantum theory of emission and absorption ^{[Dirac 1927]}, where he separated the "free" evolution of the electromagnetic field from the interaction with atomic electrons. Dirac's original notation called it the "intermediate picture" — the name "interaction picture" became standard later.

Freeman Dyson's 1949 paper ^{[Dyson 1949]} proved the equivalence of the Feynman and Schwinger formulations of QED by showing that both are representations of the time-ordered exponential. This was the paper that established the Dyson series as the universal structure underlying perturbative quantum field theory. Dyson was 25 years old at the time, a graduate student working with Bethe at Cornell. His proof that the QED perturbation series is divergent (Dyson 1952) came three years later.

The time-ordering operator $T$ was introduced formally by Dyson in 1949, though the concept of ordering operators by their time arguments appears implicitly in the work of Schwinger and Tomonaga. Wick's theorem (1950) provided the combinatorial reduction of time-ordered products that made Feynman diagrams systematic.

The Magnus expansion was developed by Wilhelm Magnus in 1954 ^{[Magnus 1954]} as a tool for solving linear differential equations with variable coefficients. Its application to quantum mechanics and the study of its convergence properties came much later (Blanes et al. 2009). The modern use of Magnus expansion integrators in quantum chemistry dates to the 1990s.

The Gell-Mann–Low theorem (1951) and the adiabatic switching protocol were developed to give a mathematically precise definition of the interacting vacuum and the S-matrix. Gell-Mann and Low showed that the perturbative expansion of Green's functions is well-defined order by order despite the infrared and ultraviolet problems of the individual terms — the cancellation of divergences between numerator and denominator in their formula is the algebraic precursor of renormalisation.

The resurgence perspective on the divergence of the Dyson series is largely due to the work of Jean Écalle (1981) on trans-series and its application to quantum mechanics and QFT by various authors since the 1990s (Zinn-Justin, Jentschura, Costin, Dunne). The recognition that instanton effects and perturbative divergences are two faces of the same trans-series structure is one of the deep results of modern theoretical physics.

Bibliography Master

Dirac, P.A.M. — "The quantum theory of the emission and absorption of radiation," Proc. Roy. Soc. A 114 (1927), 243–265. [source: TODO_REF]
Dyson, F.J. — "The radiation theories of Tomonaga, Schwinger, and Feynman," Phys. Rev. 75 (1949), 486–502. [source: TODO_REF]
Dyson, F.J. — "Divergence of perturbation theory in quantum electrodynamics," Phys. Rev. 85 (1952), 631–632. [source: TODO_REF]
Gell-Mann, M. and Low, F. — "Bound states in quantum field theory," Phys. Rev. 84 (1951), 350–354. [source: TODO_REF]
Magnus, W. — "On the exponential solution of differential equations for a linear operator," Commun. Pure Appl. Math. 7 (1954), 649–673. [source: TODO_REF]
Wick, G.C. — "The evaluation of the collision matrix," Phys. Rev. 80 (1950), 268–272. [source: TODO_REF]
Sakurai, J.J. and Napolitano, J. — Modern Quantum Mechanics, 3e (Cambridge, 2021), §5.5. [source: TODO_REF]
Weinberg, S. — The Quantum Theory of Fields, Vol. 1 (Cambridge, 1995), §3.5. [source: TODO_REF]
Griffiths, D.J. and Schroeter, D.F. — Introduction to Quantum Mechanics, 3e (Cambridge, 2018), Ch. 7.1. [source: TODO_REF]
Blanes, S., Casas, F., Oteo, J.A., and Ros, J. — "The Magnus expansion and some of its applications," Phys. Rep. 478 (2009), 1–26. [source: TODO_REF]
Costin, O. — Asymptotics and Borel Summability (Chapman & Hall/CRC, 2009). [source: TODO_REF]
Reed, M. and Simon, B. — Methods of Modern Mathematical Physics, Vol. II: Fourier Analysis, Self-Adjointness (Academic, 1975). [source: TODO_REF]

Prerequisites

12.03.01 pending
12.03.02 pending

Tier anchors

beginner: Griffiths & Schroeter, Introduction to Quantum Mechanics, 3e (2018), Ch. 7.1
intermediate: Sakurai & Napolitano, Modern Quantum Mechanics, 3e (2021), §5.5
master: Weinberg — The Quantum Theory of Fields, Vol. 1 (Cambridge, 1995), §3.5

References

Griffiths & Schroeter — Introduction to Quantum Mechanics, 3e (Cambridge, 2018) · Ch. 7.1
Sakurai & Napolitano — Modern Quantum Mechanics, 3e (Cambridge, 2021) · §5.5
Weinberg — The Quantum Theory of Fields, Vol. 1 (Cambridge, 1995) · §3.5
Dyson — The radiation theories of Tomonaga, Schwinger, and Feynman · Phys. Rev. 75 (1949), 486-502
Gell-Mann & Low — Bound states in quantum field theory · Phys. Rev. 84 (1951), 350-354

Estimated time

beginner: 15m
intermediate: 35m
master: 50m