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Expression in John R. Taylor Scattering theory book

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 $1$

One Answer

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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