Interaction Picture and Dyson Expansion

The previous pages built free relativistic fields as collections of harmonic oscillators. That construction is powerful because the free theory is exactly solvable: one knows its energy eigenstates, its mode expansion, and eventually its propagators. But the physics we actually want is not free. Interactions scatter particles, shift masses, mix states, and create correlations between field operators at different spacetime points.

The first problem of perturbative QFT is therefore not to draw diagrams. It is to write the exact time evolution in a form that separates the part we can solve from the part we will expand in. The interaction picture does precisely that. The free Hamiltonian $H_0$ is used to evolve operators, while the interaction $V$ evolves states. The result is the Dyson expansion, a time-ordered exponential that is the operator ancestor of Feynman diagrams.

For a complementary treatment that emphasizes ordered integration domains, switching, regulators, and the limits imposed by Haag’s theorem, see The Interaction Picture and Dyson Series.

Splitting the Hamiltonian

How literally to read the interaction picture

For a finite-dimensional quantum system, the interaction picture is an exact change of picture. In continuum QFT, the same formulas are best understood as regulated perturbative statements: one introduces cutoffs, finite volume, or adiabatic switching and expands in the interaction. This is the level at which Feynman perturbation theory is used in these notes. The formalism is extremely effective, but it should not be mistaken for a proof that the exact interacting Hilbert space is unitarily identical to the free Fock space.

Hamiltonian split and time-ordering choice

We write the Hamiltonian as

$$H=H_0+V,$$

where $H_0$ is exactly solvable and $V$ is treated perturbatively. To keep formulas uncluttered, many equations take the initial time to be $t_0=0$. The generalization to an arbitrary $t_0$ is written explicitly where it matters. If the Schrödinger-picture interaction has explicit time dependence, it should be included before conjugating by $H_0$. The time-ordering symbol is denoted by $\mathcal T$ and places later times to the left.

In the Schrödinger picture, states evolve and operators are time-independent unless they have explicit time dependence:

$$i\frac{d}{dt}|\psi_S(t)\rangle=H|\psi_S(t)\rangle, \qquad H=H_0+V.$$

If $V=0$, this equation is solved by

$$|\psi_S(t)\rangle=e^{-iH_0(t-t_0)}|\psi_S(t_0)\rangle.$$

The idea of the interaction picture is to remove this known free evolution from the state:

$$|\psi_I(t)\rangle=e^{iH_0(t-t_0)}|\psi_S(t)\rangle.$$

At the same time, every Schrödinger-picture operator $A_S$ is converted into a free-evolving interaction-picture operator

$$A_I(t)=e^{iH_0(t-t_0)}A_S e^{-iH_0(t-t_0)}.$$

Thus the interaction picture is halfway between the Schrödinger and Heisenberg pictures. In the Schrödinger picture, all dynamics is carried by the state. In the Heisenberg picture, all dynamics is carried by operators. In the interaction picture, the exactly solvable free dynamics is carried by operators, while the remaining interaction dynamics is carried by the state.

The interaction picture splits the dynamics. Operators evolve with $H_0$, while states evolve with the interaction Hamiltonian $V_I(t)$. This is the natural framework for perturbation theory because the free theory supplies known oscillators, states, and propagators.

Different pictures are not different theories. They are different ways of distributing the same unitary time evolution between states and operators. The interaction picture is useful because it makes the perturbative small parameter appear in the equation of motion.

The interaction-picture equation

Differentiate the interaction-picture state:

$$\frac{d}{dt}|\psi_I(t)\rangle = iH_0e^{iH_0(t-t_0)}|\psi_S(t)\rangle +e^{iH_0(t-t_0)}\frac{d}{dt}|\psi_S(t)\rangle.$$

Using the Schrödinger equation,

$$\frac{d}{dt}|\psi_S(t)\rangle=-i(H_0+V)|\psi_S(t)\rangle,$$

we obtain

$$\begin{aligned} \frac{d}{dt}|\psi_I(t)\rangle &= iH_0|\psi_I(t)\rangle -i e^{iH_0(t-t_0)}(H_0+V)|\psi_S(t)\rangle \\ &= -i e^{iH_0(t-t_0)}V e^{-iH_0(t-t_0)}|\psi_I(t)\rangle. \end{aligned}$$

Therefore

$$i\frac{d}{dt}|\psi_I(t)\rangle=V_I(t)|\psi_I(t)\rangle,$$

where

$$V_I(t)=e^{iH_0(t-t_0)}V e^{-iH_0(t-t_0)}.$$

This equation looks like the Schrödinger equation, but with two important differences. First, the Hamiltonian on the right is only the interaction, not the full Hamiltonian. Second, even if $V$ has no explicit time dependence in the Schrödinger picture, $V_I(t)$ usually has time dependence because it is conjugated by $e^{iH_0t}$.

For example, if $|n\rangle$ are eigenstates of $H_0$ with energies $E_n$, then matrix elements of the interaction-picture interaction are

$$\langle m|V_I(t)|n\rangle =e^{i(E_m-E_n)(t-t_0)}\langle m|V|n\rangle.$$

The phases $e^{i(E_m-E_n)t}$ are the free oscillations. Perturbation theory is built by integrating these oscillations against interaction matrix elements.

The evolution operator

Define $U_I(t,t_0)$ by

$$|\psi_I(t)\rangle=U_I(t,t_0)|\psi_I(t_0)\rangle.$$

Then $U_I$ obeys

$$i\frac{\partial}{\partial t}U_I(t,t_0)=V_I(t)U_I(t,t_0), \qquad U_I(t_0,t_0)=1.$$

It also satisfies the composition rule

$$U_I(t_2,t_1)U_I(t_1,t_0)=U_I(t_2,t_0),$$

and, since $V_I(t)$ is Hermitian when $V$ is Hermitian, it is unitary:

$$U_I(t,t_0)^{\dagger}U_I(t,t_0)=1.$$

The relation between full Schrödinger-picture evolution and interaction-picture evolution is

$$e^{-iH(t-t_0)}=e^{-iH_0(t-t_0)}U_I(t,t_0),$$

or equivalently

$$U_I(t,t_0)=e^{iH_0(t-t_0)}e^{-iH(t-t_0)}$$

when $H$ is time independent. This is the exact identity being expanded in powers of $V$.

The equation for $U_I$ can be written as an integral equation:

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

This form is the starting point of Dyson’s expansion. Iterating once gives

$$U_I(t,t_0)=1-i\int_{t_0}^{t}dt_1\,V_I(t_1) +(-i)^2\int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\,V_I(t_1)V_I(t_2)+\cdots.$$

After $n$ iterations,

$$U_I(t,t_0)=\sum_{n=0}^{\infty}(-i)^n \int_{t_0}^{t}dt_1\int_{t_0}^{t_1}dt_2\cdots \int_{t_0}^{t_{n-1}}dt_n\, V_I(t_1)V_I(t_2)\cdots V_I(t_n).$$

The nested limits enforce

$$t_1>t_2>\cdots>t_n.$$

This ordering is not cosmetic. Operators at different times need not commute:

$$[V_I(t_1),V_I(t_2)]\neq 0.$$

So the product $V_I(t_1)V_I(t_2)$ is genuinely different from $V_I(t_2)V_I(t_1)$.

Time ordering

For two bosonic operators, time ordering means

$$\mathcal T\{A(t_1)B(t_2)\} =\theta(t_1-t_2)A(t_1)B(t_2) +\theta(t_2-t_1)B(t_2)A(t_1),$$

where $\theta(t)$ is the step function. For three operators, $\mathcal T$ sums over all six possible time orderings. For example, the term for $t_2>t_1>t_3$ is

$$B(t_2)A(t_1)C(t_3).$$

In general, $\mathcal T$ places the operator with the latest time at the far left, the next latest time next, and so on. For fermionic fields, each exchange of fermionic operators introduces a minus sign; the bosonic definition above is enough for the present page, but the sign matters later for Dirac fields.

The nested second-order integral can be rewritten over the full square $[t_0,t]\times[t_0,t]$:

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

The factor $1/2$ compensates for the two equivalent triangular regions. At order $n$, the same reasoning gives a factor $1/n!$ because the $n$-dimensional cube is divided into $n!$ ordered simplices.

The second-order term first appears as an integral over the ordered region $t_1>t_2$. Time ordering lets us integrate over the full square and divide by $2!$. At order $n$, the same geometry replaces one ordered simplex by the full $n$-cube divided by $n!$.

Therefore the Dyson series becomes

$$U_I(t,t_0)=\sum_{n=0}^{\infty}\frac{(-i)^n}{n!} \int_{t_0}^{t}dt_1\cdots\int_{t_0}^{t}dt_n\, \mathcal T\{V_I(t_1)\cdots V_I(t_n)\}.$$

This is abbreviated as the time-ordered exponential

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

The notation means the series above. It is not the ordinary exponential unless all the $V_I(t)$ commute with one another.

Combining this with the full evolution operator gives

$$\boxed{ e^{-iH(t-t_0)} =e^{-iH_0(t-t_0)} \mathcal T\exp\left[-i\int_{t_0}^{t}dt'\,V_I(t')\right]. }$$

This is one of the central identities of perturbative quantum field theory.

Ordered insertions as perturbation theory

The expansion has a simple picture. The system evolves freely except at a sequence of interaction insertions. If the insertions occur at times $t_1,t_2,\ldots,t_n$, the latest insertion is written on the left because operators act on states from right to left.

A term in the Dyson expansion is a free history interrupted by interaction insertions. The ordered region $t_1>t_2>t_3$ contributes the product $V_I(t_1)V_I(t_2)V_I(t_3)$. The symbol $\mathcal T$ performs the corresponding ordering automatically for all regions of integration.

This is already the skeleton of Feynman perturbation theory. In a field theory, $V_I(t)$ is typically an integral of an interaction Hamiltonian density,

$$V_I(t)=H_{\mathrm{int},I}(t)=\int d^3x\,\mathcal H_{\mathrm{int},I}(t,\mathbf x).$$

Then the Dyson expansion contains spacetime integrals of products of free fields. Wick’s theorem, introduced later, will turn those time-ordered products of free fields into propagators and diagrams.

A common example is scalar $\phi^4$ theory. When there are no derivative interactions, one often has

$$\mathcal H_{\mathrm{int},I}(x)=-\mathcal L_{\mathrm{int},I}(x),$$

so if

$$\mathcal L_{\mathrm{int}}=-\frac{\lambda}{4!}\phi^4,$$

then

$$V_I(t)=\int d^3x\,\frac{\lambda}{4!}\phi_I(t,\mathbf x)^4.$$

The first-order term in $U_I$ is therefore

$$-i\int dt\,V_I(t) =-i\frac{\lambda}{4!}\int d^4x\,\phi_I(x)^4.$$

This is the operator origin of the familiar vertex factor in perturbation theory. The factor of $-i$ comes from time evolution; the spacetime integral comes from summing over where the interaction occurs; the $1/4!$ is the symmetry normalization of the interaction.

Why the ordinary exponential is wrong

It is tempting to write

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

This is generally wrong. To see why, expand the ordinary exponential to second order:

$$1-i\int dt_1\,V_I(t_1) -\frac12\int dt_1dt_2\,V_I(t_1)V_I(t_2)+\cdots.$$

The second-order term has the same operator order $V_I(t_1)V_I(t_2)$ throughout the square. But the correct expression must use $V_I(t_1)V_I(t_2)$ in the region $t_1>t_2$ and $V_I(t_2)V_I(t_1)$ in the region $t_2>t_1$. The two are equal only if

$$[V_I(t_1),V_I(t_2)]=0$$

for all times. Quantum fields do not generally satisfy this condition.

This failure is the same noncommutativity that makes the Baker–Campbell–Hausdorff formula nontrivial. The interaction picture gives a useful version of that formula. For fixed operators $A$ and $B$, define

$$B_I(s)=e^{-sA}B e^{sA}.$$

Then

$$e^{A+B}=e^A\,\mathcal T\exp\left[\int_0^1ds\,B_I(s)\right],$$

where the parameter $s$ is ordered just like time. Expanding to first order in $B$ gives Duhamel’s formula,

$$e^{A+B}=e^A+e^A\int_0^1ds\,e^{-sA}B e^{sA}+O(B^2).$$

This is the same mechanism as

$$e^{-i(H_0+V)t}=e^{-iH_0t}\, \mathcal T\exp\left[-i\int_0^t dt'\,V_I(t')\right].$$

The free Hamiltonian is treated exactly, and the interaction is conjugated into the frame that rotates with the free motion.

Transition amplitudes and energy conservation

The simplest physical use of the Dyson expansion is the first-order transition amplitude. Suppose $|i\rangle$ and $|f\rangle$ are eigenstates of $H_0$ with energies $E_i$ and $E_f$. To first order,

$$\langle f|U_I(t,t_0)|i\rangle =\delta_{fi}-i\int_{t_0}^{t}dt_1\,\langle f|V_I(t_1)|i\rangle+O(V^2).$$

Using

$$\langle f|V_I(t_1)|i\rangle =e^{i(E_f-E_i)(t_1-t_0)}V_{fi},$$

we get

$$\langle f|U_I(t,t_0)|i\rangle^{(1)} =-iV_{fi}\int_{t_0}^{t}dt_1\,e^{i(E_f-E_i)(t_1-t_0)}.$$

For a long time interval $T=t-t_0$, the integral is sharply peaked when $E_f=E_i$:

$$\int_0^Tdt\,e^{i\Delta E t} =e^{i\Delta E T/2}\frac{2\sin(\Delta E T/2)}{\Delta E}.$$

The one-sided integral from $0$ to $T$ becomes increasingly narrow as $T$ grows, but it does not by itself converge pointwise to $2\pi\delta(\Delta E)$. In the $S$-matrix one instead uses an integral over the whole time axis, understood with symmetric limits or adiabatic switching. Distributionally,

$$\lim_{\epsilon\to0^+} \int_{-\infty}^{\infty}dt\,e^{i\Delta E t}e^{-\epsilon|t|} =2\pi\delta(\Delta E).$$

This is the first appearance of a central theme: conservation laws in scattering amplitudes arise from integrating interaction insertions over spacetime. In actual scattering calculations one also uses wave packets, so the delta distribution is paired with smooth external-state profiles rather than treated as an ordinary square-integrable function. These qualifications prepare the later passage from amplitudes to rates and cross sections.

Toward correlation functions

The Dyson expansion also explains why time-ordered products dominate perturbative QFT. If $A_I(t)$ and $B_I(t')$ are free-evolving operators, then an interacting matrix element can be expanded schematically as

$$\langle \text{out}|\mathcal T\{A_I(t)B_I(t')e^{-i\int d\tau\,V_I(\tau)}\}|\text{in}\rangle.$$

The exponential supplies all possible interaction insertions. The time-ordering symbol places those insertions and the observed operators into a single chronological product. With adiabatic switching and the asymptotic limits understood, the Gell-Mann–Low formula for vacuum expectation values contains the normalized ratio

$$\frac{\langle0|\mathcal T\{A_I B_I e^{-i\int V_I}\}|0\rangle} {\langle0|\mathcal T e^{-i\int V_I}|0\rangle},$$

where $|0\rangle$ is the free vacuum. The limiting prescription is essential: without it, the displayed ratio is only schematic and need not relate the free and interacting vacua. Its denominator removes vacuum bubbles disconnected from the observed operators. This cancellation will become explicit when generating functionals are introduced. In the next pages, the numerator structure becomes the time-ordered Green function

$$G(t,t')=\langle 0|\mathcal T\{\phi(t)\phi(t')\}|0\rangle,$$

and then Wick’s theorem turns the free-theory time-ordered products into contractions. That is the bridge from the operator identity derived here to Feynman diagrams.

Summary

The interaction picture is the correct perturbative frame for QFT. It uses the free Hamiltonian $H_0$ to define the simple oscillatory time dependence of fields and uses the interaction $V$ to evolve states. In regulated finite systems this split is exact; in continuum QFT it is the controlled formal starting point for perturbative expansions. The approximation enters when the interaction-picture evolution operator is expanded in powers of $V$ and truncated.

The central result is Dyson’s formula,

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

The time-ordering symbol is essential because interaction-picture operators at different times need not commute. Its expansion converts nested ordered integrals into full time integrals with a factor $1/n!$. In field theory, those ordered products become the raw material for Wick contractions, propagators, vertices, and eventually scattering amplitudes.

Physically, perturbation theory describes free propagation interrupted by interaction events at all possible times. The Dyson expansion sums over the number, location, and chronological ordering of those events.

Common pitfalls

1. Dropping time ordering. The formula $U_I=\exp[-i\int V_I]$ is valid only when $[V_I(t),V_I(t')]=0$ for all times. In QFT this is not generally true.

2. Confusing $V$ with $V_I(t)$. The interaction-picture interaction is not usually time-independent. It is $V_I(t)=e^{iH_0t}Ve^{-iH_0t}$, so it carries free-theory phases.

3. Putting the earliest operator on the left. Time ordering places the latest time on the left. This agrees with the fact that operators act on states from right to left.

4. Forgetting fermion signs. For bosonic operators, time ordering just reorders the operators. For fermionic fields, each exchange of fermionic operators contributes a minus sign.

5. Assuming $\mathcal H_{\mathrm{int}}=-\mathcal L_{\mathrm{int}}$ always. This relation is common for non-derivative interactions, but derivative interactions and constrained systems require more care.

Exercises

Exercise 1: deriving the interaction-picture equation

Let

$$|\psi_I(t)\rangle=e^{iH_0(t-t_0)}|\psi_S(t)\rangle, \qquad H=H_0+V.$$

Derive the interaction-picture Schrödinger equation

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

Solution

Differentiate:

$$\frac{d}{dt}|\psi_I(t)\rangle =iH_0e^{iH_0(t-t_0)}|\psi_S(t)\rangle +e^{iH_0(t-t_0)}\frac{d}{dt}|\psi_S(t)\rangle.$$

The Schrödinger-picture state obeys

$$i\frac{d}{dt}|\psi_S(t)\rangle=(H_0+V)|\psi_S(t)\rangle,$$

so

$$\frac{d}{dt}|\psi_S(t)\rangle=-i(H_0+V)|\psi_S(t)\rangle.$$

Substitute this into the derivative of $|\psi_I(t)\rangle$:

$$\begin{aligned} \frac{d}{dt}|\psi_I(t)\rangle &=iH_0|\psi_I(t)\rangle -i e^{iH_0(t-t_0)}(H_0+V)|\psi_S(t)\rangle \\ &=-i e^{iH_0(t-t_0)}V e^{-iH_0(t-t_0)}|\psi_I(t)\rangle. \end{aligned}$$

Therefore

$$i\frac{d}{dt}|\psi_I(t)\rangle=V_I(t)|\psi_I(t)\rangle,$$

with

$$V_I(t)=e^{iH_0(t-t_0)}V e^{-iH_0(t-t_0)}.$$

Exercise 2: from ordered triangles to time ordering

Show that

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

Solution

The full square of integration is the union of two triangular regions:

$$t_1>t_2, \qquad t_2>t_1.$$

The diagonal $t_1=t_2$ has measure zero and does not affect the integral. By definition,

$$\mathcal T\{V_I(t_1)V_I(t_2)\} =\theta(t_1-t_2)V_I(t_1)V_I(t_2) +\theta(t_2-t_1)V_I(t_2)V_I(t_1).$$

Thus

$$\begin{aligned} &\int_{t_0}^{t}dt_1\int_{t_0}^{t}dt_2\, \mathcal T\{V_I(t_1)V_I(t_2)\} \\ &=\int_{t_1>t_2}dt_1dt_2\,V_I(t_1)V_I(t_2) +\int_{t_2>t_1}dt_1dt_2\,V_I(t_2)V_I(t_1). \end{aligned}$$

In the second integral, exchange the dummy labels $t_1\leftrightarrow t_2$. It becomes another copy of the first integral. Therefore the full square integral is twice the ordered triangular integral, giving the desired factor $1/2$.

Exercise 3: first-order transition amplitude

Let $H_0|n\rangle=E_n|n\rangle$. Compute the first-order transition amplitude from $|i\rangle$ to $|f\rangle$ over the time interval $0<t<T$.

Solution

At first order,

$$U_I(T,0)=1-i\int_0^Tdt\,V_I(t)+O(V^2).$$

For $f\neq i$,

$$\langle f|U_I(T,0)|i\rangle^{(1)} =-i\int_0^Tdt\,\langle f|V_I(t)|i\rangle.$$

Since

$$V_I(t)=e^{iH_0t}Ve^{-iH_0t},$$

we have

$$\langle f|V_I(t)|i\rangle =e^{i(E_f-E_i)t}V_{fi}.$$

Therefore

$$\langle f|U_I(T,0)|i\rangle^{(1)} =-iV_{fi}\int_0^Tdt\,e^{i(E_f-E_i)t}.$$

If $\Delta E=E_f-E_i$, then

$$\int_0^Tdt\,e^{i\Delta E t} =e^{i\Delta E T/2}\frac{2\sin(\Delta E T/2)}{\Delta E}.$$

For large $T$, the magnitude is sharply peaked near $\Delta E=0$. The one-sided finite-time integral itself should not be identified with a delta function. In scattering theory the energy-conserving delta distribution comes from the whole time axis with an adiabatic regulator:

$$\lim_{\epsilon\to0^+} \int_{-\infty}^{\infty}dt\,e^{i\Delta E t}e^{-\epsilon|t|} =2\pi\delta(\Delta E).$$

Exercise 4: when time ordering becomes unnecessary

Assume $[V_I(t),V_I(t')]=0$ for all $t,t'$. Show that the Dyson expansion reduces to the ordinary exponential.

Solution

If all the $V_I(t)$ commute, time ordering no longer changes any product:

$$\mathcal T\{V_I(t_1)\cdots V_I(t_n)\}=V_I(t_1)\cdots V_I(t_n).$$

The Dyson series becomes

$$U_I(t,t_0)=\sum_{n=0}^{\infty}\frac{(-i)^n}{n!} \left(\int_{t_0}^{t}dt'\,V_I(t')\right)^n.$$

This is exactly the Taylor series of the ordinary exponential:

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

Thus time ordering is needed precisely because the interaction-picture interactions generally fail to commute at different times.

References and further reading

• Sidney Coleman, Lectures on Quantum Field Theory, Chapter 7, especially the interaction picture and Dyson’s formula.
• Mark Srednicki, Quantum Field Theory, Sections 6–9, for the connection between time evolution, perturbation theory, and path integrals.
• Steven Weinberg, The Quantum Theory of Fields, Volume I, Sections 3.5 and 6.1, for perturbation theory, the Dyson series, and Feynman rules.
• A. Zee, Quantum Field Theory in a Nutshell, Chapters I.7–I.9, for a compact physical introduction to Feynman diagrams and perturbing the vacuum.