Dyson series

Suppose that 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 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,

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

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:

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

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

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

Returning to the Schrödinger picture, for t_f > t_i,

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

• Magnus series
• Picard iteration

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