Physics Asked on March 12, 2021
In the book Scattering theory by John R. Taylor He has the following expression at page 138
$$
i lim _{epsilon downarrow 0} int_{0}^{infty} d tleftlanglemathbf{p}^{prime}right|V e^{ileft(E_{p}^{prime}+E_{p}+i epsilon-2 Hright) t}|mathbf{p}rangle=1 / 2 lim _{epsilon downarrow 0}leftlanglemathbf{p}^{prime}right|V Gleft(frac{E_{p^{prime}}+E_{p}}{2}+i epsilonright)|mathbf{p}rangle tag 1
$$
where $G$ is the green operator defined by
$G(z)=(z-H)^{-1}$
But for me we should have
$$
i lim _{epsilon downarrow 0} int_{0}^{infty} d tleftlanglemathbf{p}^{prime}right|V e^{ileft(E_{p}^{prime}+E_{p}+i epsilon-2 Hright) t}|mathbf{p}rangle=1 / 2 lim _{epsilon downarrow 0} leftlanglemathbf{p}^{prime}right|V Gleft(frac{E_{p^{prime}}+E_{p}+i epsilon}{2}right)|mathbf{p}rangle tag 2
$$
Since we have that
$$int dt e^{ileft(E_{p}^{prime}+E_{p}+i epsilon-2 Hright) t} =int dt e^{2ileft(frac{E_{p^{prime}}+E_{p}+i epsilon}{2}+ Hright) t} =frac{1}{2ileft(frac{E_{p^{prime}}+E_{p}+i epsilon}{2}+ Hright)}e^{2ileft(frac{E_{p^{prime}}+E_{p}+i epsilon}{2}+ Hright) t} tag 3$$
And from $(3)$ we have expression $(2)$.
I am not seeing how did he obtain expression $1$
Since $epsilon$ is a small parameter that is taken to zero at the end of the calculation, and no physical quantities depend on $epsilon$, it is conventional to "absorb" constant multiplicative factors into $epsilon$. In your example, you could define $epsilon'equiv epsilon/2$ and carry on with the calculation in terms of $epsilon'$ instead of $epsilon$. To save spending mental energy on a detail that is really irrelevant to a quite complex calculation, the convention is not to bother distinguishing $epsilon'$ from $epsilon$, and to refer to any $alpha epsilon$ (where $alpha$ is a constant, positive factor) as $epsilon$.
Correct answer by Andrew on March 12, 2021
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