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In scattering theory, the Dyson series, formulated by British-born American physicist 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 (the reduced Planck constant) satisfies ħ = 1.

The Dyson operator

We suppose we have a Hamiltonian H which we split into a "free" part H0 and an "interacting" part V i.e. H = H0 + V. We will work in the interaction picture here and assume units such that the reduced Planck constant \hbar is 1.

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

$$\Psi(t)=U(t,t_0) \Psi(t_0) \$$

is called 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 then (Tomonaga-Schwinger equation)

$$i{d \over dt} U(t,t_0)\Psi(t_0) = V(t) U(t,t_0)\Psi(t_0).$$

Thus:

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

Derivation of the Dyson series

This leads to the following Neumann series:

$$\begin{array}{lcl} U(t,t_0) & = & 1 - i \int_{t_0}^{t}{dt_1V(t_1)}+(-i)^2\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2V(t_1)V(t_2)}}+\cdots \\ & &{} + (-i)^n\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2 \cdots \int_{t_0}^{t_{n-1}}{dt_nV(t_1)V(t_2) \cdots V(t_n)}}} +\cdots. \end{array}$$

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:

$$U_n(t,t_0)=(-i)^n\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2\cdots\int_{t_0}^{t_{n-1}}{dt_n\mathcal TV(t_1)V(t_2)\cdots V(t_n)}}}.$$

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

$$S_n=\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2\cdots\int_{t_0}^{t_{n-1}}{dt_nK(t_1, t_2,\dots,t_n)}}}.$$

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

$$I_n=\int_{t_0}^t{dt_1\int_{t_0}^{t}{dt_2\cdots\int_{t_0}^t{dt_nK(t_1, t_2,\dots,t_n)}}}.$$

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 Sn by definition. So it is true that:

$$S_n=\frac{1}{n!}I_n.$$

Returning to our previous integral, it holds the identity:

$$U_n=\frac{(-i)^n}{n!}\int_{t_0}^t{dt_1\int_{t_0}^t{dt_2\cdots\int_{t_0}^t{dt_n\mathcal TV(t_1)V(t_2)\cdots V(t_n)}}}.$$

Summing up all the terms we obtain the Dyson series:

$$U(t,t_0)=\sum_{n=0}^\infty U_n(t,t_0)=\mathcal Te^{-i\int_{t_0}^t{d\tau V(\tau)}}.$$

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,

$$\langle\psi_f;t_f|\psi_i;t_i\rangle=\sum_{n=0}^\infty (-i)^n \underbrace{\int dt_1 \cdots dt_n}_{t_f\,\ge\, t_1\,\ge\, \cdots\, \ge\, t_n\,\ge\, t_i}\, \langle\psi_f;t_f|e^{-iH_0(t_f-t_1)}Ve^{-iH_0(t_1-t_2)}\cdots Ve^{-iH_0(t_n-t_i)}|\psi_i;t_i\rangle.$$