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In the book Scattering theory by John R. Taylor He has the following expression at page 138

$$ i \lim _{\epsilon \downarrow 0} \int_{0}^{\infty} d t\left\langle\mathbf{p}^{\prime}\right|V e^{i\left(E_{p}^{\prime}+E_{p}+i \epsilon-2 H\right) t}|\mathbf{p}\rangle=1 / 2 \lim _{\epsilon \downarrow 0}\left\langle\mathbf{p}^{\prime}\right|V G\left(\frac{E_{p^{\prime}}+E_{p}}{2}+i \epsilon\right)|\mathbf{p}\rangle \tag 1 $$

where $G$ is the green operator defined by $G(z)=(z-H)^{-1}$

But for me we should have $$ i \lim _{\epsilon \downarrow 0} \int_{0}^{\infty} d t\left\langle\mathbf{p}^{\prime}\right|V e^{i\left(E_{p}^{\prime}+E_{p}+i \epsilon-2 H\right) t}|\mathbf{p}\rangle=1 / 2 \lim _{\epsilon \downarrow 0} \left\langle\mathbf{p}^{\prime}\right|V G\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}\right)|\mathbf{p}\rangle \tag 2 $$ Since we have that

$$\int dt e^{i\left(E_{p}^{\prime}+E_{p}+i \epsilon-2 H\right) t} =\int dt e^{2i\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}+ H\right) t} =\frac{1}{2i\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}+ H\right)}e^{2i\left(\frac{E_{p^{\prime}}+E_{p}+i \epsilon}{2}+ H\right) t} \tag 3$$ And from $(3)$ we have expression $(2)$. I am not seeing how did he obtain expression $1$

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1 Answer 1

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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 $\epsilon$.

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