In scattering theory, a part of mathematical physics, the Dyson series, formulated by Freeman Dyson, is a perturbative expansion of the time evolution operator in the interaction picture. Each term can be represented by a sum of Feynman diagrams.

This series diverges asymptotically, but in quantum electrodynamics (QED) at the second order the difference from experimental data is in the order of 10−10. This close agreement holds because the coupling constant (also known as the fine structure constant) of QED is much less than 1.[clarification needed]

Notice that in this article Planck units are used, so that ħ = 1 (where ħ is the reduced Planck constant).

## The Dyson operator

Suppose that we have a Hamiltonian H, which we split into a free part H = H0 and an interacting part VS(t), i.e. H = H0 + VS(t).

We will work in the interaction picture here, that is,

${\displaystyle V_{I}(t)=\mathrm {e} ^{\mathrm {i} H_{0}(t-t_{0})}V_{S}(t)\mathrm {e} ^{-\mathrm {i} H_{0}(t-t_{0})},}$

where ${\displaystyle V_{S}(t)}$ is the possibly time-dependent interacting part of the Schrödinger picture. To avoid subscripts, ${\displaystyle V(t)}$ stands for ${\displaystyle V_{\text{I))(t)}$ in what follows. We choose units such that the reduced Planck constant ħ is 1.

In the interaction picture, the evolution operator U defined by the equation

${\displaystyle \Psi (t)=U(t,t_{0})\Psi (t_{0})}$

is called the Dyson operator.

We have

${\displaystyle U(t,t)=I,}$
${\displaystyle U(t,t_{0})=U(t,t_{1})U(t_{1},t_{0}),}$
${\displaystyle U^{-1}(t,t_{0})=U(t_{0},t),}$

and hence the Tomonaga–Schwinger equation,

${\displaystyle i{\frac {d}{dt))U(t,t_{0})\Psi (t_{0})=V(t)U(t,t_{0})\Psi (t_{0}).}$

Consequently,

${\displaystyle U(t,t_{0})=1-i\int _{t_{0))^{t}{dt_{1}\ V(t_{1})U(t_{1},t_{0})}.}$

## Derivation of the Dyson series

This leads to the following Neumann series:

{\displaystyle {\begin{aligned}U(t,t_{0})={}&1-i\int _{t_{0))^{t}dt_{1}V(t_{1})+(-i)^{2}\int _{t_{0))^{t}dt_{1}\int _{t_{0))^{t_{1))\,dt_{2}V(t_{1})V(t_{2})+\cdots \\&{}+(-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(t_{1})V(t_{2})\cdots V(t_{n})+\cdots .\end{aligned))}

Here we have ${\displaystyle t_{1}>t_{2}>\cdots >t_{n))$, so we can say that the fields are time-ordered, and it is useful to introduce an operator ${\displaystyle {\mathcal {T))}$ called time-ordering operator, defining

${\displaystyle U_{n}(t,t_{0})=(-i)^{n}\int _{t_{0))^{t}dt_{1}\int _{t_{0))^{t_{1))dt_{2}\cdots \int _{t_{0))^{t_{n-1))dt_{n}\,{\mathcal {T))V(t_{1})V(t_{2})\cdots V(t_{n}).}$

We can now try to make this integration simpler. In fact, by the following example:

${\displaystyle S_{n}=\int _{t_{0))^{t}dt_{1}\int _{t_{0))^{t_{1))dt_{2}\cdots \int _{t_{0))^{t_{n-1))dt_{n}\,K(t_{1},t_{2},\dots ,t_{n}).}$

Assume that K is symmetric in its arguments and define (look at integration limits):

${\displaystyle I_{n}=\int _{t_{0))^{t}dt_{1}\int _{t_{0))^{t}dt_{2}\cdots \int _{t_{0))^{t}dt_{n}K(t_{1},t_{2},\dots ,t_{n}).}$

The region of integration can be broken in ${\displaystyle n!}$ sub-regions defined by ${\displaystyle t_{1}>t_{2}>\cdots >t_{n))$, ${\displaystyle t_{2}>t_{1}>\cdots >t_{n))$, etc. Due to the symmetry of K, the integral in each of these sub-regions is the same and equal to ${\displaystyle S_{n))$ by definition. So it is true that

${\displaystyle S_{n}={\frac {1}{n!))I_{n}.}$

Returning to our previous integral, the following identity holds

${\displaystyle U_{n}={\frac {(-i)^{n)){n!))\int _{t_{0))^{t}dt_{1}\int _{t_{0))^{t}dt_{2}\cdots \int _{t_{0))^{t}dt_{n}\,{\mathcal {T))V(t_{1})V(t_{2})\cdots V(t_{n}).}$

Summing up all the terms, we obtain Dyson's theorem for the Dyson series:[clarification needed]

${\displaystyle U(t,t_{0})=\sum _{n=0}^{\infty }U_{n}(t,t_{0})={\mathcal {T))e^{-i\int _{t_{0))^{t}{d\tau V(\tau ))).}$

## Wavefunctions

Then, going back to the wavefunction for t > t0,

${\displaystyle |\Psi (t)\rangle =\sum _{n=0}^{\infty }{(-i)^{n} \over n!}\left(\prod _{k=1}^{n}\int _{t_{0))^{t}dt_{k}\right){\mathcal {T))\left\{\prod _{k=1}^{n}e^{iH_{0}t_{k))V(t_{k})e^{-iH_{0}t_{k))\right\}|\Psi (t_{0})\rangle .}$

Returning to the Schrödinger picture, for tf > ti,

${\displaystyle \langle \psi _{\rm {f));t_{\rm {f))\mid \psi _{\rm {i));t_{\rm {i))\rangle =\sum _{n=0}^{\infty }(-i)^{n}\underbrace {\int dt_{1}\cdots dt_{n)) _{t_{\rm {f))\,\geq \,t_{1}\,\geq \,\cdots \,\geq \,t_{n}\,\geq \,t_{\rm {i))}\,\langle \psi _{\rm {f));t_{\rm {f))\mid e^{-iH_{0}(t_{\rm {f))-t_{1})}V_{S}(t_{1})e^{-iH_{0}(t_{1}-t_{2})}\cdots V_{S}(t_{n})e^{-iH_{0}(t_{n}-t_{\rm {i)))}\mid \psi _{\rm {i));t_{\rm {i))\rangle .}$