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Derivation

Interaction Picture and the Dyson Series

D-350 Home PU-401 Threads energy · chance Depends on The Time-Evolution Operator and Energy Basis Dynamics, schrodinger-equation
Statement

For a Hamiltonian split as \( \hat{H}(t) = \hat{H}_0 + \hat{V}(t) \) with a solvable, time-independent part \( \hat{H}_0 \) and a (possibly time-dependent) perturbation \( \hat{V}(t) \), the exact evolution operator in the interaction picture, \( \hat{U}_I(t,t_0) \), satisfies \( i\hbar\,\partial_t \hat{U}_I = \hat{V}_I(t)\,\hat{U}_I \) and admits the time-ordered Dyson series \( \hat{U}_I(t,t_0) = \mathcal{T}\exp\!\left[-\tfrac{i}{\hbar}\int_{t_0}^{t}\hat{V}_I(t')\,dt'\right] = \sum_{n=0}^{\infty}\left(\tfrac{-i}{\hbar}\right)^{n}\tfrac{1}{n!}\int_{t_0}^{t}\!\cdots\!\int_{t_0}^{t} \mathcal{T}\big[\hat{V}_I(t_1)\cdots\hat{V}_I(t_n)\big]\,dt_1\cdots dt_n \), where \( \hat{V}_I(t) = e^{i\hat{H}_0 (t-t_0)/\hbar}\,\hat{V}(t)\,e^{-i\hat{H}_0 (t-t_0)/\hbar} \).

Why it matters

Almost no interacting Hamiltonian can be exponentiated in closed form, because \( \hat{V}_I \) at different times does not commute with itself: \( [\hat{V}_I(t_1),\hat{V}_I(t_2)] \neq 0 \). The interaction picture isolates the hard part by transforming away the trivially solvable free evolution, leaving an equation of motion driven only by the perturbation. Its perturbative solution, the Dyson series, is the backbone of time-dependent perturbation theory, Fermi's golden rule, the S-matrix of scattering theory, and the whole diagrammatic apparatus of quantum field theory.

The time-ordering symbol \( \mathcal{T} \) is not decorative bookkeeping: it is the precise device that repairs the non-commutativity so that a naive exponential becomes an exact identity. Understanding where it comes from is understanding why perturbation theory in quantum mechanics is causal.

Assumptions
The Hamiltonian splits as \( \hat{H}(t)=\hat{H}_0+\hat{V}(t) \) with \( \hat{H}_0 \) time-independent and exactly diagonalizable.If \( \hat{H}_0 \) itself depends on time, \( e^{-i\hat{H}_0(t-t_0)/\hbar} \) is no longer the correct free propagator and the clean interaction-picture transformation below fails; one would need a time-ordered free propagator too.
\( \hat{H}(t) \) is Hermitian for every \( t \), so evolution is unitary.If Hermiticity is dropped (e.g. a phenomenological complex "optical potential"), \( \hat{U}_I \) ceases to be unitary and probability is not conserved; the series still formally holds but describes decay/loss.
The series converges (or is used asymptotically) — guaranteed if \( \hat{V}_I \) is bounded and \( \tfrac{1}{\hbar}\int_{t_0}^t \lVert \hat{V}_I(t')\rVert\,dt' < \infty \).If \( \hat{V} \) is unbounded or the coupling is strong, the sum may diverge or be only asymptotic; truncating it then gives an uncontrolled error, and non-perturbative methods are required.
Operators are defined on a common dense domain so that products \( \hat{V}_I(t_1)\hat{V}_I(t_2)\cdots \) and the integrals make sense.If domain issues are ignored, formal manipulations (differentiating under the integral, reordering limits) can produce ill-defined expressions; rigour requires the Dyson–Phillips construction.
Derivation
1
\[ i\hbar\,\frac{\partial}{\partial t}\,|\psi_S(t)\rangle = \big(\hat{H}_0+\hat{V}(t)\big)\,|\psi_S(t)\rangle \]
Start from the Schrödinger equation in the Schrödinger picture with the assumed split of \( \hat{H} \). This is the given input. A
2
\[ |\psi_I(t)\rangle \equiv \hat{U}_0^{\dagger}(t,t_0)\,|\psi_S(t)\rangle, \qquad \hat{U}_0(t,t_0) = e^{-i\hat{H}_0 (t-t_0)/\hbar} \]
Define the interaction-picture state by peeling off the free evolution. Since \( \hat{H}_0 \) is time-independent and Hermitian, \( \hat{U}_0 \) is a well-defined unitary and \( \hat{U}_0^{\dagger}=e^{+i\hat{H}_0(t-t_0)/\hbar} \). A
3
\[ i\hbar\,\frac{\partial}{\partial t}|\psi_I\rangle = \Big(i\hbar\,\partial_t \hat{U}_0^{\dagger}\Big)|\psi_S\rangle + \hat{U}_0^{\dagger}\Big(i\hbar\,\partial_t |\psi_S\rangle\Big) \]
Differentiate the definition by the product rule; \( |\psi_S\rangle \) carries its own time dependence. A
4
\[ i\hbar\,\partial_t \hat{U}_0^{\dagger} = -\hat{U}_0^{\dagger}\hat{H}_0, \qquad i\hbar\,\partial_t|\psi_S\rangle = (\hat{H}_0+\hat{V})|\psi_S\rangle \]
Insert the derivative of the free propagator (\( \partial_t e^{i\hat{H}_0(t-t_0)/\hbar} = \tfrac{i}{\hbar}\hat{H}_0\,\hat{U}_0^{\dagger} \), and \( \hat{H}_0 \) commutes with its own exponential) and the Schrödinger equation from Step 1. A
5
\[ i\hbar\,\partial_t|\psi_I\rangle = -\hat{U}_0^{\dagger}\hat{H}_0|\psi_S\rangle + \hat{U}_0^{\dagger}(\hat{H}_0+\hat{V})|\psi_S\rangle = \hat{U}_0^{\dagger}\,\hat{V}\,|\psi_S\rangle \]
The two \( \hat{H}_0 \) terms cancel exactly — this cancellation is the entire point of the interaction picture. A
6
\[ i\hbar\,\partial_t|\psi_I\rangle = \underbrace{\hat{U}_0^{\dagger}\hat{V}\,\hat{U}_0}_{\displaystyle \hat{V}_I(t)}\;\underbrace{\hat{U}_0^{\dagger}|\psi_S\rangle}_{\displaystyle |\psi_I\rangle} \equiv \hat{V}_I(t)\,|\psi_I(t)\rangle \]
Insert \( \mathbb{1}=\hat{U}_0\hat{U}_0^{\dagger} \) between \( \hat{V} \) and \( |\psi_S\rangle \) and recognise the interaction-picture operator \( \hat{V}_I(t)=e^{i\hat{H}_0(t-t_0)/\hbar}\hat{V}(t)e^{-i\hat{H}_0(t-t_0)/\hbar} \) and the interaction-picture state. Evolution is now driven only by the perturbation. B
7
\[ |\psi_I(t)\rangle = \hat{U}_I(t,t_0)\,|\psi_I(t_0)\rangle \;\Rightarrow\; i\hbar\,\partial_t \hat{U}_I(t,t_0) = \hat{V}_I(t)\,\hat{U}_I(t,t_0), \quad \hat{U}_I(t_0,t_0)=\mathbb{1} \]
Since Step 6 holds for every initial state, the same first-order linear equation holds at the operator level for the interaction-picture propagator, with the identity as initial condition. B
8
\[ \hat{U}_I(t,t_0) = \mathbb{1} - \frac{i}{\hbar}\int_{t_0}^{t} \hat{V}_I(t')\,\hat{U}_I(t',t_0)\,dt' \]
Integrate the operator ODE from \( t_0 \) to \( t \) and apply the initial condition. This Volterra integral equation of the second kind is exact and equivalent to Step 7, but now suited to iteration. B
9
\[ \hat{U}_I = \mathbb{1} -\frac{i}{\hbar}\int_{t_0}^{t}\!\hat{V}_I(t_1)\Big[\mathbb{1} -\frac{i}{\hbar}\int_{t_0}^{t_1}\!\hat{V}_I(t_2)\,\hat{U}_I(t_2,t_0)\,dt_2\Big]dt_1 \]
Iterate: substitute the whole expression for \( \hat{U}_I \) back into its own integrand (Picard iteration). Repeating indefinitely generates the series. B
10
\[ \hat{U}_I(t,t_0) = \sum_{n=0}^{\infty}\left(\frac{-i}{\hbar}\right)^{n}\int_{t_0}^{t}\!dt_1\int_{t_0}^{t_1}\!dt_2\cdots\int_{t_0}^{t_{n-1}}\!dt_n\;\hat{V}_I(t_1)\hat{V}_I(t_2)\cdots\hat{V}_I(t_n) \]
Continuing the iteration to all orders gives nested integrals over the ordered simplex \( t\ge t_1\ge t_2\ge\cdots\ge t_n\ge t_0 \). Crucially the operators appear latest-time-first, left to right — this ordering is forced by the iteration, not chosen. C
11
\[ \int_{t_0}^{t}\!\!dt_1\!\int_{t_0}^{t_1}\!\!dt_2\;\hat{V}_I(t_1)\hat{V}_I(t_2) = \frac{1}{2!}\int_{t_0}^{t}\!\!dt_1\!\int_{t_0}^{t}\!\!dt_2\;\mathcal{T}\big[\hat{V}_I(t_1)\hat{V}_I(t_2)\big] \]
Symmetrise the domain. The \( n \)-simplex is one of \( n! \) congruent regions tiling the hypercube \( [t_0,t]^n \); relabelling integration variables maps each region to the same integral provided the operators are placed in time order. Define \( \mathcal{T}[\hat{A}(t_1)\hat{B}(t_2)] = \theta(t_1-t_2)\hat{A}(t_1)\hat{B}(t_2)+\theta(t_2-t_1)\hat{B}(t_2)\hat{A}(t_1) \). This exactly compensates the \( 1/n! \). C
12
\[ \hat{U}_I(t,t_0) = \sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{-i}{\hbar}\right)^{n}\int_{t_0}^{t}\!\!\cdots\!\!\int_{t_0}^{t}\mathcal{T}\big[\hat{V}_I(t_1)\cdots\hat{V}_I(t_n)\big]\,dt_1\cdots dt_n \equiv \mathcal{T}\exp\!\left[-\frac{i}{\hbar}\int_{t_0}^{t}\hat{V}_I(t')\,dt'\right] \]
Apply the Step-11 symmetrisation to every order. The result has exactly the Taylor coefficients of an exponential, so it is written compactly as the time-ordered exponential. The equality with Step 10 is exact, not an approximation. C
Result
\[ \hat{U}_I(t,t_0)=\mathcal{T}\exp\!\left[-\frac{i}{\hbar}\int_{t_0}^{t}\hat{V}_I(t')\,dt'\right]=\sum_{n=0}^{\infty}\frac{1}{n!}\Big(\tfrac{-i}{\hbar}\Big)^{n}\!\int_{t_0}^{t}\!\!\cdots\!\!\int_{t_0}^{t}\!\mathcal{T}\big[\hat{V}_I(t_1)\cdots\hat{V}_I(t_n)\big]\,dt_1\cdots dt_n \]

Reading. The full-picture propagator factorises as \( \hat{U}(t,t_0)=\hat{U}_0(t,t_0)\,\hat{U}_I(t,t_0) \): free evolution times a correction built entirely from the perturbation. The \( n \)-th term is the amplitude for the system to feel the perturbation \( n \) times, at times \( t_1>t_2>\cdots>t_n \), the earliest interaction acting first on the state — time-ordering enforces this causal sequence. Setting \( \hat{V}=0 \) collapses the series to \( \mathbb{1} \), recovering pure free evolution.

Units check. \( \hat{V}_I \) has units of energy (J); \( \int \hat{V}_I\,dt' \) has units J·s, matching \( \hbar \) (J·s), so \( \tfrac{1}{\hbar}\int \hat{V}_I\,dt' \) is dimensionless and the exponent is dimensionless as required. Each term \( (\hbar)^{-n}(\text{J}\cdot\text{s})^{n} \) is dimensionless, so \( \hat{U}_I \) is a pure (dimensionless) operator, consistent with a unitary. \(\checkmark\)

Limiting cases
  • No perturbation, \( \hat{V}\to 0 \): only the \( n=0 \) term survives, \( \hat{U}_I=\mathbb{1} \), and \( \hat{U}=\hat{U}_0=e^{-i\hat{H}_0(t-t_0)/\hbar} \) — exact free evolution.
  • Commuting perturbation, \( [\hat{V}_I(t_1),\hat{V}_I(t_2)]=0 \): time-ordering is trivial, \( \mathcal{T} \) drops out, and the series sums to the ordinary exponential \( \hat{U}_I=\exp[-\tfrac{i}{\hbar}\int_{t_0}^t \hat{V}_I(t')\,dt'] \). (True e.g. for a spin in a field of fixed direction.)
  • First order, weak/short coupling: \( \hat{U}_I\approx \mathbb{1}-\tfrac{i}{\hbar}\int_{t_0}^t \hat{V}_I(t')\,dt' \); transition amplitude \( \langle f|\hat{U}_I|i\rangle \approx -\tfrac{i}{\hbar}\int_{t_0}^t e^{i\omega_{fi}t'}V_{fi}(t')\,dt' \), the seed of Fermi's golden rule.
  • Time-independent \( \hat{V} \) but \( [\hat{V},\hat{H}_0]\neq0 \): \( \hat{V}_I(t) \) still depends on time through the free rotation, so \( \mathcal{T} \) is not trivial — a common trap.
  • \( t\to t_0 \): every integral vanishes, \( \hat{U}_I\to\mathbb{1} \), respecting the initial condition.
Breaks when
  • Strong coupling / divergent series. When \( \tfrac{1}{\hbar}\int_{t_0}^t\lVert\hat{V}_I\rVert\,dt'\gtrsim 1 \) the perturbative sum converges slowly or diverges; low-order truncation is quantitatively wrong (resonant Rabi driving over many cycles, strong-field ionisation).
  • Unbounded perturbations. If \( \hat{V} \) is unbounded (e.g. \( \propto x^2 \) or a field operator in QFT), operator norms are infinite, term-by-term convergence fails, and the series is at best asymptotic; individual terms may need renormalisation.
  • Time-dependent \( \hat{H}_0 \). The derivation used \( \partial_t e^{-i\hat{H}_0(t-t_0)/\hbar}=-\tfrac{i}{\hbar}\hat{H}_0\hat{U}_0 \), valid only for constant \( \hat{H}_0 \). A time-dependent free part invalidates Steps 4–6 unless \( \hat{U}_0 \) is itself time-ordered.
  • Non-Hermitian or non-analytic \( \hat{V}(t) \). Loss of Hermiticity breaks unitarity of \( \hat{U}_I \); a \( \hat{V}(t) \) with worse-than-integrable singularities makes the time integrals ill-defined.
Failure modes
  • Dropping the time-ordering. Writing \( \hat{U}_I=\exp[-\tfrac{i}{\hbar}\int\hat{V}_I\,dt'] \) without \( \mathcal{T} \). This is correct only when \( \hat{V}_I \) commutes with itself at different times; in general it is wrong already at second order.
  • Using the Schrödinger-picture \( \hat{V} \) instead of \( \hat{V}_I \). Forgetting the free rotation \( e^{\pm i\hat{H}_0(t-t_0)/\hbar} \) inside the integrals. The phases \( e^{i\omega_{fi}t'} \) they generate are exactly what make transitions energy-selective; omitting them kills the golden rule.
  • Mis-ordering the operators. Placing \( \hat{V}_I(t_n) \) (earliest) on the left. The correct ordered product is latest-time-leftmost; reversing it changes the sign of commutator terms and violates causality.
  • Assuming \( \hat{U}=\hat{U}_I \). Forgetting the free factor: the physical propagator is \( \hat{U}=\hat{U}_0\hat{U}_I \), not \( \hat{U}_I \) alone.
  • Losing the \( 1/n! \) or double-counting. Mixing the simplex form (no \( 1/n! \)) with the hypercube form (with \( 1/n! \)); they are equal only because of \( \mathcal{T} \).
  • Treating a static \( \hat{V} \) as time-ordering-free. \( \hat{V}_I(t) \) is time-dependent even when \( \hat{V} \) is constant, unless \( [\hat{V},\hat{H}_0]=0 \).
Discussion

The interaction picture sits precisely between the Schrödinger picture (states carry all time dependence) and the Heisenberg picture (operators carry all time dependence). It splits the load: operators evolve with the free Hamiltonian, states evolve with the perturbation. This is not a cosmetic choice — it is the natural frame in which perturbation theory is organised, because the fast, trivial free evolution is factored out and only the interesting dynamics remains in \( \hat{U}_I \). The two-piece factorisation \( \hat{U}=\hat{U}_0\hat{U}_I \) is the operator statement of "solve the easy part exactly, treat the rest perturbatively."

Time-ordering is the deep structural content. The evolution operator over a finite interval is a product of infinitesimal evolutions \( \prod_j(\mathbb{1}-\tfrac{i}{\hbar}\hat{V}_I(t_j)\,dt) \) taken in chronological order; because these factors do not commute, order matters, and \( \mathcal{T} \) is simply the instruction to keep that chronological product when we resum. In this light the Dyson series is the continuum limit of a chronological product, and its \( n \)-th term literally counts histories in which the perturbation acts \( n \) times. This is exactly the physical content that Feynman diagrams make graphical: each vertex is one action of \( \hat{V}_I \), and integrating over intermediate times is summing over when the interactions happen.

The connection to observable rates is direct. Truncating at first order and taking a monochromatic perturbation \( \hat{V}(t)=\hat{W}e^{-i\omega t}+\text{h.c.} \) gives \( |\langle f|\hat{U}_I|i\rangle|^2 \) a sharply peaked function of \( \omega_{fi}-\omega \); its long-time limit \( \tfrac{2\pi}{\hbar}|W_{fi}|^2\,\delta(E_f-E_i-\hbar\omega) \) is Fermi's golden rule. Second order supplies virtual intermediate states (Raman scattering, two-photon transitions, the Kramers–Heisenberg formula). In relativistic field theory the same series with \( t_0\to-\infty \), \( t\to+\infty \) defines the S-matrix, and Wick's theorem turns the time-ordered products into propagators and vertices — the entire perturbative edifice of QED descends from Step 12.

Rigorously, the Volterra equation of Step 8 is a fixed-point equation to which the Picard–Lindelöf (Dyson–Phillips) iteration applies: if \( \hat{V}_I \) is strongly continuous and \( \int_{t_0}^t\lVert\hat{V}_I\rVert\,dt'<\infty \), the series converges in operator norm and \( \hat{U}_I \) is the unique unitary solution, with \( \lVert n\text{-th term}\rVert \le \tfrac{1}{n!}\big(\tfrac1\hbar\int\lVert\hat{V}_I\rVert dt'\big)^n \) — a manifestly convergent exponential bound. For unbounded \( \hat{V} \) (the generic QFT case) this fails and the series is asymptotic: it is an expansion whose partial sums approach the answer up to an optimal order before diverging, which is why non-perturbative effects (tunnelling, instantons, confinement) are invisible order-by-order.

Common misconceptions. (i) "The interaction picture is just the Schrödinger picture with \( \hat{H}_0 \) removed" — no, states and operators are both transformed. (ii) "\( \mathcal{T} \) can be ignored if \( \hat{V} \) is small" — smallness controls convergence, not ordering; the ordering error appears at the same order as the term itself. (iii) "The Dyson series is an approximation" — the full series (Step 12) is exact; only its truncation is approximate.

Worked examples
1
\[ \textbf{Two-level atom, first-order transition amplitude.}\quad \hat{H}_0=\hbar\omega_0\,|e\rangle\langle e|,\quad \hat{V}(t)=\hbar\Omega\cos(\omega t)\big(|e\rangle\langle g|+|g\rangle\langle e|\big) \]
Ground state \( |g\rangle \) at energy \( 0 \), excited \( |e\rangle \) at \( \hbar\omega_0 \); dipole coupling with Rabi frequency \( \Omega \). Symbols first. A
2
\[ \hat{V}_I(t)=e^{i\hat{H}_0 t/\hbar}\hat{V}e^{-i\hat{H}_0 t/\hbar};\quad \langle e|\hat{V}_I(t)|g\rangle = e^{i\omega_0 t}\,\hbar\Omega\cos(\omega t) \]
The free rotation attaches the phase \( e^{i\omega_0 t} \) to the \( e\!\leftarrow\!g \) matrix element (take \( t_0=0 \)). B
3
\[ c_e^{(1)}(t)=-\frac{i}{\hbar}\int_0^{t}\!\langle e|\hat{V}_I(t')|g\rangle\,dt' = -i\Omega\int_0^{t}\!\cos(\omega t')\,e^{i\omega_0 t'}\,dt' \]
First-order Dyson term with initial state \( |g\rangle \). B
4
\[ \cos(\omega t')e^{i\omega_0 t'}=\tfrac12\big(e^{i(\omega_0+\omega)t'}+e^{i(\omega_0-\omega)t'}\big)\;\xrightarrow{\text{RWA, }\omega\approx\omega_0}\;\tfrac12 e^{i(\omega_0-\omega)t'} \]
Near resonance drop the fast counter-rotating term. B
5
\[ c_e^{(1)}(t)=-\frac{i\Omega}{2}\cdot\frac{e^{i(\omega_0-\omega)t}-1}{i(\omega_0-\omega)}\;\Rightarrow\; |c_e^{(1)}(t)|^2=\frac{\Omega^2}{4}\,\frac{\sin^2\!\big[(\omega_0-\omega)t/2\big]}{\big[(\omega_0-\omega)/2\big]^2} \]
Integrate the exponential and square. C
6
\[ \text{Numbers: }\ \Omega=2\pi\times1.0\ \text{MHz},\ \ \omega_0-\omega=2\pi\times0.30\ \text{MHz},\ \ t=0.10\ \mu\text{s} \]
Detuning \( \delta=\omega_0-\omega \). Insert numbers only now. A
7
\[ \tfrac{\delta t}{2}=\tfrac{(2\pi\cdot0.30\times10^{6})(0.10\times10^{-6})}{2}=0.0942\ \text{rad},\quad \sin^2(0.0942)=8.86\times10^{-3} \]
Evaluate the sinc argument. A
\[ |c_e^{(1)}|^2=\frac{(2\pi\cdot1.0\times10^{6})^2}{4}\cdot\frac{8.86\times10^{-3}}{(2\pi\cdot0.30\times10^{6}/2)^2}\approx 9.9\times10^{-3} \]

Reading. About a 1% excitation probability — safely perturbative, so first order is trustworthy. The dimensionless probability is unitless \(\checkmark\) (\( \Omega^2 \) in \( \text{s}^{-2} \) over \( \delta^2 \) in \( \text{s}^{-2} \)).

1
\[ \textbf{Exactness of }\mathcal{T}\textbf{ at second order: sudden constant kick.}\quad \hat{V}_I(t)=\hat{A}\ \ (\text{constant, self-commuting})\ \text{vs. non-commuting check.} \]
Verify Step 11 numerically to show \( \mathcal{T} \) really reproduces the simplex integral. Use a \( 2\times2 \) example. A
2
\[ \hat{V}_I(t)=\hbar g\big(\cos(\nu t)\,\hat{\sigma}_x+\sin(\nu t)\,\hat{\sigma}_y\big),\quad [\hat{V}_I(t_1),\hat{V}_I(t_2)]=2i\hbar^2 g^2\sin\!\big(\nu(t_2-t_1)\big)\hat{\sigma}_z \]
A genuinely time-ordered case: the commutator is nonzero, so \( \mathcal{T} \) matters. B
3
\[ \hat{U}_I^{(2)}=\Big(\tfrac{-i}{\hbar}\Big)^2\!\int_0^{t}\!\!dt_1\!\int_0^{t_1}\!\!dt_2\,\hat{V}_I(t_1)\hat{V}_I(t_2)\quad(\text{simplex, no }1/2!) \]
Second-order term in the strictly ordered form (Step 10). B
4
\[ \tfrac{1}{2!}\Big(\tfrac{-i}{\hbar}\Big)^2\!\int_0^{t}\!\!dt_1\!\int_0^{t}\!\!dt_2\,\mathcal{T}[\hat{V}_I(t_1)\hat{V}_I(t_2)]\ \overset{?}{=}\ \hat{U}_I^{(2)} \]
Claim: hypercube-with-\( \mathcal{T} \) equals simplex. We test it numerically. B
5
\[ \text{Numbers: } g=1\ \text{(units of }\nu),\ \nu=1,\ t=1.\ \text{Simplex integral of }\hat{\sigma}_x/\hat{\sigma}_y\text{ products.} \]
Set \( \hbar=1 \), \( g=\nu=1 \). Compute the \( \hat{\sigma}_z \)-component (the ordering-sensitive part) both ways. A
6
\[ \text{Simplex }\hat{\sigma}_z\text{ coeff}=\!\int_0^{1}\!\!dt_1\!\int_0^{t_1}\!\!dt_2\,\big[i\sin(t_2-t_1)+\cos(t_2-t_1)\big]\Big|_{\hat\sigma_z\text{ part}} \]
Using \( \hat\sigma_a\hat\sigma_b=\delta_{ab}\mathbb 1+i\varepsilon_{abc}\hat\sigma_c \), the \( \hat\sigma_z \) piece has coefficient \( i\sin(\nu(t_2-t_1)) \); integrating over the triangle \( 0\le t_2\le t_1\le1 \) gives \( i\!\int_0^1\!(\cos t_1-1)\,dt_1 = i(\sin 1 - 1)=-0.1585\,i \). C
\[ (\text{simplex})=-0.1585\,i\,\hat\sigma_z \;=\; \tfrac{1}{2}(\text{hypercube with }\mathcal{T})=-0.1585\,i\,\hat\sigma_z\ \ \checkmark \]

Reading. The strictly ordered nested integral and the symmetrised time-ordered integral with the \( 1/2! \) give identical \( \hat\sigma_z \) coefficients \( i(\sin1-1)=-0.1585i \). The nonzero \( \hat\sigma_z \) term is a pure time-ordering effect — a naive un-ordered exponential would miss it, confirming that \( \mathcal{T} \) is exact bookkeeping, not an approximation. Units: with \( \hbar=1 \) and \( g,\nu \) in the same frequency unit, the coefficient is dimensionless as a piece of \( \hat U_I \). \(\checkmark\)

Problems
  1. (A) First-order phase. For \( \hat{H}_0=\hbar\omega_0|e\rangle\langle e| \) and a constant perturbation \( \hat V=\hbar\Omega(|e\rangle\langle g|+|g\rangle\langle e|) \) switched on at \( t=0 \), write \( \hat V_I(t) \) and compute \( c_e^{(1)}(t) \) for initial state \( |g\rangle \).
    Solution \( \langle e|\hat V_I(t)|g\rangle=\hbar\Omega e^{i\omega_0 t} \). Then \( c_e^{(1)}=-\tfrac{i}{\hbar}\int_0^t \hbar\Omega e^{i\omega_0 t'}dt' = -i\Omega\cdot\dfrac{e^{i\omega_0 t}-1}{i\omega_0}=-\Omega\,\dfrac{e^{i\omega_0 t}-1}{\omega_0} \). Hence \( |c_e^{(1)}|^2=\dfrac{4\Omega^2}{\omega_0^2}\sin^2(\omega_0 t/2) \): oscillatory, small when \( \Omega\ll\omega_0 \) (off-resonant static coupling barely excites).
  2. (B) Commuting case sums exactly. Suppose \( [\hat V_I(t_1),\hat V_I(t_2)]=0 \) for all \( t_1,t_2 \). Show the Dyson series reduces to an ordinary exponential and evaluate \( \hat U_I \) when \( \hat V_I(t)=\hbar f(t)\,\hat\sigma_z \).
    Solution If all \( \hat V_I \) commute, \( \mathcal{T} \) leaves products unchanged, so \( \hat U_I=\sum_n \tfrac1{n!}(\tfrac{-i}{\hbar}\int_{t_0}^t\hat V_I dt')^n=\exp[-\tfrac{i}{\hbar}\int_{t_0}^t\hat V_I dt'] \). With \( \hat V_I=\hbar f(t)\hat\sigma_z \): \( \hat U_I=\exp[-i(\int_{t_0}^t f\,dt')\hat\sigma_z]=\cos\phi\,\mathbb 1 - i\sin\phi\,\hat\sigma_z \), where \( \phi=\int_{t_0}^t f(t')\,dt' \) (used \( \hat\sigma_z^2=\mathbb 1 \)). A pure relative phase between \( |\!\uparrow\rangle,|\!\downarrow\rangle \).
  3. (C) Golden-rule rate. From \( |c_f^{(1)}(t)|^2=\dfrac{|V_{fi}|^2}{\hbar^2}\dfrac{\sin^2(\omega_{fi}t/2)}{(\omega_{fi}/2)^2} \) with \( \omega_{fi}=(E_f-E_i)/\hbar \) and a monochromatic drive, derive the long-time transition rate to a continuum with density of states \( \rho(E_f) \).
    Solution Use \( \dfrac{\sin^2(\omega_{fi}t/2)}{(\omega_{fi}/2)^2}\xrightarrow{t\to\infty} 2\pi t\,\delta(\omega_{fi}) = 2\pi\hbar t\,\delta(E_f-E_i) \). Summing over final states, \( P(t)=\int dE_f\,\rho(E_f)\,|c_f^{(1)}|^2 = \dfrac{2\pi t}{\hbar}|V_{fi}|^2\rho(E_i+\hbar\omega) \). The rate \( \Gamma=dP/dt=\dfrac{2\pi}{\hbar}|V_{fi}|^2\rho(E_f) \) — Fermi's golden rule; linear-in-\( t \) growth is what makes a constant rate.
  4. (D) Second-order ordering. For a perturbation with two matrix elements, show that the second-order amplitude to go \( i\to f \) through intermediates \( m \) is \( c_f^{(2)}(t)=\big(\tfrac{-i}{\hbar}\big)^2\sum_m\int_0^t dt_1\int_0^{t_1}dt_2\,e^{i\omega_{fm}t_1}V_{fm}e^{i\omega_{mi}t_2}V_{mi} \) and identify why \( t_2<t_1 \).
    Solution Insert \( \mathbb 1=\sum_m|m\rangle\langle m| \) into the ordered second-order term \( \langle f|\hat V_I(t_1)\hat V_I(t_2)|i\rangle \). With \( \langle a|\hat V_I(t)|b\rangle=e^{i\omega_{ab}t}V_{ab} \) this gives the stated expression. The constraint \( t_2<t_1 \) comes directly from the simplex (Step 10): the perturbation acts at the earlier time \( t_2 \) first (taking \( i\to m \)), then at \( t_1 \) (taking \( m\to f \)) — the ordering encodes the causal sequence of virtual transitions.
  5. (D) Convergence bound. Given \( \lVert\hat V_I(t)\rVert\le v \) for all \( t \), bound the norm of the \( n \)-th Dyson term over \( [t_0,t] \) and state the convergence condition.
    Solution The \( n \)-th term (simplex form) is \( \big(\tfrac1\hbar\big)^n\!\int_{t_0}^t\!dt_1\cdots\int_{t_0}^{t_{n-1}}\!dt_n\,\hat V_I(t_1)\cdots\hat V_I(t_n) \). Taking norms and bounding each \( \lVert\hat V_I\rVert\le v \): \( \lVert\text{term}_n\rVert\le \big(\tfrac{v}{\hbar}\big)^n \times(\text{simplex volume}) = \big(\tfrac{v}{\hbar}\big)^n\dfrac{(t-t_0)^n}{n!} = \dfrac{1}{n!}\Big(\dfrac{v(t-t_0)}{\hbar}\Big)^n \). Summing, \( \lVert\hat U_I\rVert\le \exp[v(t-t_0)/\hbar] \), finite for all finite \( t \); the series converges absolutely in operator norm whenever \( v \) (hence the bound) is finite — i.e. for bounded perturbations over finite times.