Dyson Expansion for Transition Amplitudes

The Dyson expansion organizes interaction-driven evolution by the number of times the interaction Hamiltonian acts. After taking matrix elements, each term becomes a transition amplitude: first order connects the initial and final channels directly, second order sums over intermediate channels and ordered interaction times, and higher orders generate longer chains.

Required background. Interaction Picture for Perturbation Theory supplies the transformed interaction; Dyson Expansion as Formal Evolution derives ordered evolution; Completeness and Orthonormal Bases supplies the intermediate-state sum.

Expanding the Interaction-Picture Propagator

For

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

the interaction-picture propagator satisfies

$$i\hbar \frac{\partial U_I(t,t_0)}{\partial t} = \lambda V_I(t)U_I(t,t_0),$$

with

$$U_I(t_0,t_0)=I.$$

Its integral equation is

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

Iterating produces

$$U_I(t,t_0) ={} I+\lambda U_I^{(1)}(t,t_0) +\lambda^2U_I^{(2)}(t,t_0) +\cdots.$$

The superscript counts interaction insertions, not powers already hidden in the definition of $V$. If a physical interaction itself contains several coupling constants, their powers must be restored explicitly when interpreting the result.

First-Order Term

The first operator coefficient is

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

For initial and final channels chosen from an orthonormal discrete basis, define

$$\mathcal A_{fi}(t,t_0) = \langle f\rvert U_I(t,t_0)\lvert i\rangle.$$

Then

$$\mathcal A_{fi} = \delta_{fi} +\lambda\mathcal A_{fi}^{(1)} +\lambda^2\mathcal A_{fi}^{(2)} +\cdots,$$

with

$$\mathcal A_{fi}^{(1)} = -\frac{i}{\hbar} \int_{t_0}^{t}dt_1\, \langle f\rvert V_I(t_1)\lvert i\rangle.$$

For a time-independent reference Hamiltonian,

$$\langle f\rvert V_I(t_1)\lvert i\rangle = e^{i\omega_{fi}(t_1-t_0)}V_{fi}(t_1).$$

The first-order result is therefore a direct matrix element filtered by its accumulated phase. Its detailed finite-time resonance structure is derived in First-Order Transition Probability.

Second-Order Term

The next iteration gives

$$U_I^{(2)}(t,t_0) ={} \left(-\frac{i}{\hbar}\right)^2 \times \int_{t_0}^{t}dt_2 \int_{t_0}^{t_2}dt_1\, \times V_I(t_2)V_I(t_1).$$

The nested limits impose

$$t_0\le t_1\le t_2\le t.$$

The rightmost operator acts first. The second-order amplitude is

$$\mathcal A_{fi}^{(2)} ={} \left(-\frac{i}{\hbar}\right)^2 \times \int_{t_0}^{t}dt_2 \int_{t_0}^{t_2}dt_1\, \times \langle f\rvert V_I(t_2)V_I(t_1) \lvert i\rangle.$$

Insert the identity

$$I=\sum_n\lvert n\rangle\langle n\rvert$$

between the two interactions:

$$\mathcal A_{fi}^{(2)} ={} \left(-\frac{i}{\hbar}\right)^2 \sum_n \times \int_{t_0}^{t}dt_2 \int_{t_0}^{t_2}dt_1\, \times \langle f\rvert V_I(t_2)\lvert n\rangle \times \langle n\rvert V_I(t_1)\lvert i\rangle.$$

Every complete set gives the same exact answer. A basis adapted to $H_0$, symmetry, or the spectrum usually makes the phases and selection rules transparent.

For time-independent $H_0$,

$$\mathcal A_{fi}^{(2)} ={} \left(-\frac{i}{\hbar}\right)^2 \sum_n \times \int_{t_0}^{t}dt_2 \int_{t_0}^{t_2}dt_1\, \times e^{i\omega_{fn}(t_2-t_0)} V_{fn}(t_2) \times e^{i\omega_{ni}(t_1-t_0)} V_{ni}(t_1).$$

This expression separates three filters:

1. $V_{ni}$ must connect the initial channel to an intermediate channel;
2. $V_{fn}$ must connect that channel to the final channel;
3. the two time integrals must add coherently rather than cancel by phase oscillation.

At second order, the interaction at $t_1$ acts before the interaction at $t_2$. Inserting completeness turns the operator product into a sum over intermediate states $\lvert n\rangle$. The ordered triangle has half the area of the full time square; the time-ordering operator provides an equivalent full-square representation.

Time-Ordering Notation

The second-order operator can also be written

$$U_I^{(2)}(t,t_0) ={} \frac12 \left(-\frac{i}{\hbar}\right)^2 \times \int_{t_0}^{t}dt_2 \int_{t_0}^{t}dt_1\, \times \mathcal T \left\{ V_I(t_2)V_I(t_1) \right\}.$$

Hamiltonian insertions use ordinary time ordering, including on a fermionic Hilbert space; exchanging the two interaction Hamiltonians introduces no fermionic minus sign. Define $\mathcal O_{21}=\mathcal T\{V_I(t_2)V_I(t_1)\}$. Then

$$\mathcal O_{21} = \begin{cases} V_I(t_2)V_I(t_1),&t_2\gt t_1, \\ V_I(t_1)V_I(t_2),&t_1\gt t_2. \end{cases}$$

The factor $1/2!$ compensates for integrating both labeled orderings over the full square. The convention at $t_1=t_2$ does not affect ordinary integrals unless singular equal-time terms are present.

The same construction extends to $r$ insertions, with $1/r!$ for the full time cube. The nested form is convenient for low-order amplitudes; the all-order operator formula is derived in Dyson Expansion as Formal Evolution.

A Sequential-Pulse Example

Consider three channels $\lvert i\rangle$, $\lvert n\rangle$, and $\lvert f\rangle$. Let the interaction-picture Hamiltonian contain

$$V_I(t) = g_1(t)\lvert n\rangle\langle i\rvert +g_2(t)\lvert f\rangle\langle n\rvert +\text{h.c.}$$

Assume $g_1$ acts in an early time window and $g_2$ in a later, nonoverlapping window. There is no direct $i\to f$ matrix element, so

$$\mathcal A_{fi}^{(1)}=0.$$

At second order, the only relevant ordered path is

$$\lvert i\rangle \xrightarrow{\,g_1(t_1)\,} \lvert n\rangle \xrightarrow{\,g_2(t_2)\,} \lvert f\rangle.$$

Its amplitude is

$$\mathcal A_{fi}^{(2)} ={} \left(-\frac{i}{\hbar}\right)^2 \times \int dt_2\,g_2(t_2) \int^{t_2}dt_1\,g_1(t_1).$$

Because every $g_1$ time precedes every $g_2$ time, the ordered integral factorizes:

$$\mathcal A_{fi}^{(2)} ={} \left(-\frac{i}{\hbar}\right)^2 \times \left(\int dt_2\,g_2(t_2)\right) \times \left(\int dt_1\,g_1(t_1)\right).$$

If the pulse order is reversed and the Hamiltonian contains no other path, this particular second-order amplitude vanishes: the operator that requires population in $\lvert n\rangle$ acts before $\lvert n\rangle$ has been created. Time ordering is therefore physical causal bookkeeping, not decorative notation.

Virtual Intermediate States

An intermediate state in the completeness sum need not be populated as an asymptotic or measured outcome. In an off-resonant two-step process, its contribution can remain transient and enter through an energy denominator after the time integrals are evaluated.

For a constant interaction switched on over $0\le t\le T$,

$$\mathcal A_{fi}^{(2)}(T) ={} \left(-\frac{i}{\hbar}\right)^2 \sum_nV_{fn}V_{ni} \times \int_0^Tdt_2\,e^{i\omega_{fn}t_2} \times \int_0^{t_2}dt_1\,e^{i\omega_{ni}t_1}.$$

For $\omega_{ni}\ne0$, the inner integral is

$$\int_0^{t_2}dt_1\,e^{i\omega_{ni}t_1} = \frac{e^{i\omega_{ni}t_2}-1}{i\omega_{ni}}.$$

The factor $1/\omega_{ni}$ is the finite-time ancestor of an intermediate-state energy denominator. Precise infinite-time denominators and their $i0$ prescriptions depend on switching, boundary conditions, and whether one is computing bound-state response or scattering. They should not be inserted by memory before the time protocol is specified.

If an intermediate detuning becomes small, the denominator is a warning rather than an enhancement formula to trust blindly. The near-resonant state may need to be included in a coupled model space and evolved nonperturbatively.

Selection Rules by Order

Suppose a symmetry makes

$$\langle f\rvert V_I(t)\lvert i\rangle=0.$$

Then the first-order amplitude vanishes. A second-order amplitude may still exist if there are intermediate states for which

$$\langle n\rvert V_I(t_1)\lvert i\rangle\ne0$$

and

$$\langle f\rvert V_I(t_2)\lvert n\rangle\ne0.$$

For example, if $H_0$ commutes with parity and $V(t)$ is odd at every time, one insertion changes parity. Two insertions have even overall parity, so states of the same parity can be connected at second order through opposite-parity intermediate states.

If $f\ne i$ and $\mathcal A_{fi}^{(1)}=0$, then

$$\mathcal A_{fi} = \lambda^2\mathcal A_{fi}^{(2)} +O(\lambda^3),$$

and the leading probability is

$$P_{i\to f} = \lambda^4 \lvert\mathcal A_{fi}^{(2)}\rvert^2 +O(\lambda^5).$$

Calling such a probability “second order” is ambiguous. It is second order in the amplitude and fourth order in the coupling at the probability level. State which convention is being used.

Probability Through the Next Order

For a previously empty final channel,

$$\mathcal A_{fi} = \lambda\mathcal A_{fi}^{(1)} +\lambda^2\mathcal A_{fi}^{(2)} +O(\lambda^3).$$

Squaring gives

$$P_{i\to f} ={} \lambda^2 \lvert\mathcal A_{fi}^{(1)}\rvert^2 +2\lambda^3 \operatorname{Re} \left[ \mathcal A_{fi}^{(1)*} \mathcal A_{fi}^{(2)} \right] +O(\lambda^4).$$

The second-order amplitude is needed for the cubic correction to a generic transition probability. To know the probability completely through $O(\lambda^4)$, one also needs the third-order amplitude because it interferes with the first-order term. The fourth-order contribution alone is:

$$\Delta P_{i\to f}^{(4)} ={} \lambda^4 \left\lvert\mathcal A_{fi}^{(2)}\right\rvert^2 +2\lambda^4 \operatorname{Re} \left[ \mathcal A_{fi}^{(1)*} \mathcal A_{fi}^{(3)} \right].$$

Computing “the probability to second order” therefore requires a declared counting convention.

Unitarity Order by Order

The exact $U_I$ is unitary. Insert

$$U_I=I+\lambda U_I^{(1)}+\lambda^2U_I^{(2)}+\cdots$$

into $U_I^\dagger U_I=I$. At first order,

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

At second order,

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

Taking an initial-state matrix element and inserting completeness gives

$$2\operatorname{Re}\mathcal A_{ii}^{(2)} + \sum_f \left\lvert\mathcal A_{fi}^{(1)}\right\rvert^2 = 0.$$

The sum here includes $f=i$: its diagonal term describes a first-order phase, not probability transferred to a different state. Combining that term with the second-order forward amplitude gives the survival probability below. This relation is the finite-dimensional form of the unitarity constraint behind the optical theorem.

The survival probability becomes

$$P_{i\to i} ={}1 -\lambda^2 \sum_{f\ne i} \left\lvert\mathcal A_{fi}^{(1)}\right\rvert^2 +O(\lambda^3).$$

Thus probability loss from the initial channel appears at the same order as probability gained by the other channels. Keeping only the first-order survival amplitude misses this normalization mechanism.

Convergence and Truncation

For a bounded interaction on a finite interval, define

$$\eta(t) = \frac{\lvert\lambda\rvert}{\hbar} \int_{t_0}^{t}dt'\, \lVert V_I(t')\rVert.$$

The volume of the ordered $r$-simplex gives the bound

$$\left\lVert \lambda^rU_I^{(r)} \right\rVert \le \frac{\eta(t)^r}{r!}.$$

Consequently, the operator-norm remainder after order $N$ obeys a conservative estimate of the form

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

This estimate establishes convergence for bounded integrable interactions, but it ignores oscillatory cancellation and is often much less sharp than channel-specific phase estimates. For unbounded operators, domain and relative-boundedness questions require additional analysis; the formal Dyson series is not automatically an operator-norm theorem.

Even when the series converges for every finite time, a low-order truncation can become inaccurate when $\eta$ or a resonant channel amplitude is not small. Convergence of the infinite series and usefulness of the first few terms are different questions.

Finite-order unitarity

A truncated Dyson polynomial is not exactly unitary, although it satisfies unitarity identities through its retained order when used consistently. If exact finite-order unitarity is operationally important, a Magnus Expansion exponentiates anti-Hermitian generators and may be a better organization. Magnus has its own convergence and commutator costs; it is not universally superior.

Diagrammatic Intuition and the QFT Bridge

The inserted completeness relation suggests a diagrammatic reading:

• each interaction insertion is a vertex labeled by its time and operator;
• each segment between vertices carries an intermediate state and its phase;
• one sums over all allowed intermediate states;
• one integrates over all ordered vertex times;
• symmetry can remove entire paths by setting a vertex matrix element to zero.

This is useful bookkeeping, but it is not yet a Feynman diagram expansion. In quantum field theory, fields contain creation and annihilation operators, Wick’s theorem reorganizes time-ordered products into contractions, propagators encode field correlations, and diagram symmetry factors must be tracked.

The exact bridge begins at From Evolution Operators to Time-Ordered Products and QFT Bridge: S-Matrix. The reusable idea is already present here: amplitudes are ordered sums over interaction histories, and probabilities require coherent interference among those histories before squaring.

Exercises

Show the two second-order forms agree

Prove that the nested second-order integral equals the full-square time-ordered expression with a factor $1/2!$.

Solution

Split the square $[t_0,t]^2$ into the two triangles $t_2\ge t_1$ and $t_1\ge t_2$. Then

$$\frac12 \int_{t_0}^{t}dt_2 \int_{t_0}^{t}dt_1\, \mathcal T\{V_I(t_2)V_I(t_1)\} ={} \frac12 \int_{t_2\ge t_1}dt_2dt_1\, V_I(t_2)V_I(t_1) + \frac12 \int_{t_1\ge t_2}dt_2dt_1\, V_I(t_1)V_I(t_2).$$

Relabel $t_1\leftrightarrow t_2$ in the second triangle. It becomes identical to the first, so the factor $1/2$ cancels the two equal contributions:

$$\int_{t_0}^{t}dt_2 \int_{t_0}^{t_2}dt_1\, V_I(t_2)V_I(t_1).$$

Insert the intermediate states

Starting from $\langle f\rvert V_I(t_2)V_I(t_1)\lvert i\rangle$, derive the intermediate-state sum and state what changes if part of the spectrum is continuous.

Solution

Insert a complete discrete basis:

$$I=\sum_n\lvert n\rangle\langle n\rvert.$$

Then

$$\langle f\rvert V_I(t_2)V_I(t_1)\lvert i\rangle = \sum_n {}\langle f\rvert V_I(t_2)\lvert n\rangle \times \langle n\rvert V_I(t_1)\lvert i\rangle.$$

For a mixed discrete-continuum spectrum, completeness has the schematic form

$$I = \sum_{n\in\mathrm{disc}} \lvert n\rangle\langle n\rvert + \int dE\, \lvert E\rangle\langle E\rvert,$$

with normalization-dependent measures and channel labels included as needed. The intermediate sum then contains both a sum and an integral.

Reverse the sequential pulses

In the three-channel example, suppose the $g_2$ pulse is entirely earlier than the $g_1$ pulse. Show why the displayed path gives zero second-order amplitude.

Solution

The ordered amplitude contains

$$\int dt_2\,g_2(t_2) \int^{t_2}dt_1\,g_1(t_1).$$

Whenever $g_2(t_2)$ is nonzero, every time at which $g_1$ is nonzero lies later than $t_2$. The inner domain requires $t_1\le t_2$, so it contains no support of $g_1$ and vanishes.

For this three-state chain, even the exact reversed sequence gives no final $f$ population: the first pulse annihilates the initial $i$ state and the second couples only $i$ to $n$. Additional couplings can change that conclusion.

Determine the leading probability order

Assume $H_0$ preserves parity and $V(t)$ is odd at every time. For distinct orthogonal even-parity initial and final states, what are the leading possible amplitude and probability orders?

Solution

One odd operator changes parity, so

$$\mathcal A_{fi}^{(1)}=0$$

for even $i$ and even $f$. Two odd insertions have even overall parity and may connect through an odd intermediate state:

$$\mathcal A_{fi} = \lambda^2\mathcal A_{fi}^{(2)} +O(\lambda^4).$$

The leading probability is therefore

$$P_{i\to f} = \lambda^4 \left\lvert\mathcal A_{fi}^{(2)}\right\rvert^2 +O(\lambda^6),$$

unless another symmetry also removes the second-order paths. All odd insertion orders vanish between these two even states, which accounts for the stronger remainder orders.

Derive the unitarity identity

Expand $U_I^\dagger U_I=I$ through second order and derive the relation between the forward amplitude and the sum of first-order transition probabilities.

Solution

Write

$$U_I=I+\lambda U_1+\lambda^2U_2+O(\lambda^3).$$

Multiplication gives

$$U_I^\dagger U_I ={}I +\lambda(U_1+U_1^\dagger) +\lambda^2(U_2+U_2^\dagger) +\lambda^2U_1^\dagger U_1 +O(\lambda^3).$$

The second-order coefficient must vanish. Taking its expectation value in $\lvert i\rangle$ gives

$$2\operatorname{Re}\langle i\rvert U_2\lvert i\rangle + \langle i\rvert U_1^\dagger U_1\lvert i\rangle = 0.$$

Insert completeness into the last term:

$$\langle i\rvert U_1^\dagger U_1\lvert i\rangle = \sum_f \left\lvert\langle f\rvert U_1\lvert i\rangle\right\rvert^2.$$

Identifying the amplitude coefficients yields

$$2\operatorname{Re}\mathcal A_{ii}^{(2)} + \sum_f\left\lvert\mathcal A_{fi}^{(1)}\right\rvert^2 = 0.$$

Bound the ordered integral

Assume $\lVert V_I(t)\rVert\le V_{\max}$ for an interval of duration $T$. Bound the norm of the $r$th Dyson contribution including its coupling factor.

Solution

Submultiplicativity gives

$$\left\lVert V_I(t_r)\cdots V_I(t_1) \right\rVert \le V_{\max}^r.$$

The ordered integration region has volume

$$\frac{T^r}{r!}.$$

Therefore

$$\left\lVert \lambda^rU_I^{(r)} \right\rVert \le \frac{1}{r!} \left( \frac{\lvert\lambda\rvert V_{\max}T}{\hbar} \right)^r.$$

This bound is sufficient but often pessimistic because it discards phase cancellation and detailed matrix-element structure.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 2, Wiley, 1977.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• F. J. Dyson, “The radiation theories of Tomonaga, Schwinger, and Feynman,” Physical Review 75, 486–502, 1949.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• S. Weinberg, The Quantum Theory of Fields, Vol. 1, Cambridge University Press, 1995.