Common knowledge (logic)

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 (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:

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

Here we have t₁ > t₂ > ..., > t_n 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 > ... > t_n, t₂ > t₁ > ... > t_n, 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 t_f > t_i,

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

See also

• Magnus series

References

• Charles J. Joachain, Quantum collision theory, North-Holland Publishing, 1975, ISBN 0-444-86773-2 (Elsevier)