Let us assume that we have some Hamiltonian and we know its spectrum $$H_0 \psi_n = E_n \psi_n .$$ We define the Green's function in as $$ G(x,y,E) =\sum_m \frac{\psi_m^*(x)\psi_m(y)}{E-E_m}, $$ and the reduced Green's function as $$G'_n(x,y,E_n) =\sum_{m\neq n} \frac{\psi_m^*(x)\psi_m(y)}{E_n-E_m}.$$ Now let us assume that we have some perturbation $\delta H$. In the relativistic perturbation theory the second order correction to the energy can be calculated with the help of the reduced Green's function: $$\Delta E_n^{(2)}=\int d^3x d^3 y \psi_n(x) \delta H(x) G'_n(x,y,E_n)\delta H(y)\psi_n(y).$$ However in the QFT, when I calculate any Feynman diagram I am forced to use the full Green's function. For example, when we calculate the Lamb shift, we use the full Green's function (see S. Weinberg, The Quantum Theory of Fields, Vol. 1 eq. 14.3.26 and 14.3.27). My question is: what is the relation between the perturbation theory in the relativistic quantum mechanics and the diagrammatic approach of QFT? Why in the QFT I need the full Green's function, while in the quantum mechanics I need only the reduced Green's function?
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