# Dyson series

In scattering theory, a part of mathematical physics, the Dyson series, formulated by Freeman Dyson, is a perturbative series, and each term is represented by 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. 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 Template:Mvar, which we split into a free part Template:Mvar0 and an interacting part Template:Mvar, i.e. H = H0 + V.

We will work in the interaction picture here and assume units such that the reduced Planck constant Template:Mvar is 1.

In the interaction picture, the evolution operator Template:Mvar defined by the equation

$\Psi (t)=U(t,t_{0})\Psi (t_{0})$ is called the Dyson operator.

We have

$U(t,t)=I,$ $U(t,t_{0})=U(t,t_{1})U(t_{1},t_{0}),$ $U^{-1}(t,t_{0})=U(t_{0},t),$ and hence the Tomonaga–Schwinger equation,

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

$U(t,t_{0})=1-i\int _{t_{0}}^{t}{dt_{1}\ V(t_{1})U(t_{1},t_{0})}.$ ## Derivation of the Dyson series

This leads to the following Neumann series:

${\begin{array}{lcl}U(t,t_{0})&=&1-i\int _{t_{0}}^{t}{dt_{1}V(t_{1})}+(-i)^{2}\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t_{1}}{dt_{2}V(t_{1})V(t_{2})}}+\cdots \\&&{}+(-i)^{n}\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t_{1}}{dt_{2}\cdots \int _{t_{0}}^{t_{n-1}}{dt_{n}V(t_{1})V(t_{2})\cdots V(t_{n})}}}+\cdots .\end{array}}$ Here we have t1 > t2 > ..., > tn, so we can say that the fields are time-ordered, and it is useful to introduce an operator ${\mathcal {T}}$ called time-ordering operator, defining

$U_{n}(t,t_{0})=(-i)^{n}\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t_{1}}{dt_{2}\cdots \int _{t_{0}}^{t_{n-1}}{dt_{n}{\mathcal {T}}V(t_{1})V(t_{2})\cdots V(t_{n})}}}.$ We can now try to make this integration simpler. In fact, by the following example:

$S_{n}=\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t_{1}}{dt_{2}\cdots \int _{t_{0}}^{t_{n-1}}{dt_{n}K(t_{1},t_{2},\dots ,t_{n})}}}.$ Assume that K is symmetric in its arguments and define (look at integration limits):

$I_{n}=\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t}{dt_{2}\cdots \int _{t_{0}}^{t}{dt_{n}K(t_{1},t_{2},\dots ,t_{n})}}}.$ The region of integration can be broken in n! sub-regions defined by t1 > t2 > ... > tn, t2 > t1 > ... > tn, etc. Due to the symmetry of K, the integral in each of these sub-regions is the same and equal to $S_{n}$ by definition. So it is true that

$S_{n}={\frac {1}{n!}}I_{n}.$ Returning to our previous integral, it holds the identity

$U_{n}={\frac {(-i)^{n}}{n!}}\int _{t_{0}}^{t}{dt_{1}\int _{t_{0}}^{t}{dt_{2}\cdots \int _{t_{0}}^{t}{dt_{n}{\mathcal {T}}V(t_{1})V(t_{2})\cdots V(t_{n})}}}.$ Summing up all the terms, we obtain the Dyson series:

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

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

$|\Psi (t)\rangle =\sum _{n=0}^{\infty }{(-i)^{n} \over n!}\left(\prod _{k=1}^{n}\int _{t_{0}}^{t}dt_{k}\right){\mathcal {T}}\left\{\prod _{k=1}^{n}e^{iH_{0}t_{k}}Ve^{-iH_{0}t_{k}}\right\}|\Psi (t_{0})\rangle .$ Returning to the Schrödinger picture, for tf > ti,

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