Dyson operator

Dyson series

Look first at S Matrix.

Def Dyson Operator

U (t, t 0) = U (t) U - 1 (t 0)

U (t 0, t 0) =

If $φ i$ is an interaction potential in the field operator $φ i$ , we can say that:

$i\dot U(t, t_0)=V(\phi_i)U(t, t_0)=V_iU(t, t_0)$

Thus:

$U(t,t_0)=1 - i \int_{t_0}^t{dt_1V_i(t_1)U(t_1,t_0)}$

This leads to the following Neumann's series (see John Von Neumann): $U(t,t_0)=1 - i \int_{t_0}^t{dt_1V_i(t_1)}+(-i)^2\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2V_i(t_1)V_i(t_2)}}+...+(-i)^n\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2...\int_{t_0}^{t_{n-1}}{dt_nV_i(t_1)V_i(t_2)...V_i(t_n)}}}$

If we assume that $t > t 1 > t 2 > ... > t n$ we can say that the fields are Time ordered, and so it is useful to introduce an operator called Time Ordering Operator. Defining:

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

We can now try to make this integration simpler. in fact, in the following example: $S_n=\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2...\int_{t_0}^{t_{n-1}}{dt_nK(t_1, t_2,...,t_n)}}}$

If K is symmetric in its arguments, we can define (look at integration limits):

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

And so it is true that:

$S_n=\frac{1}{n!}K_n$

Returning to our previous integral, it holds the identity:

$U_n=\frac{(-1)^n}{n!}\int_{t_0}^t{dt_1\int_{t_0}^t{dt_2...\int_{t_0}^t{dt_n\mathcal TV_i(t_1)V_i(t_2)...V_i(t_n)}}}$

Summing up all the terms we obtain the series (Dyson series):

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

Usually, it is a series in coupling constants, and each term is represented by Feynman diagrams. This series does not converge, but in quantum electrodynamics at the second order the difference from experimental data is in the order of $1 * 10 - 10$ . We got this result because the coupling constant of QED is much less than 1.

Categories: Quantum field theory

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