# How to derive equation 6.1.26 in Sakurai's quantum mechanics? [closed]

I am trying to get to equation 6.1.26 in Sakurai's quantum mechanics and I am getting confused with one of the steps in the derivation. Similar questions have been asked about this question on this site but they don't cover this issue. It starts in the Solving for the $$T$$ matrix section.

$$\langle n|U_I(t,-\infty)|i\rangle = \delta_{ni} + \frac{1}{\hbar}T_{ni}\frac{e^{\left(i\omega_{ni}+\epsilon\right)t}}{-\omega_{ni}+i\epsilon} \tag{6.1.23}$$ $$\langle n|U_I(t,-\infty)|i\rangle = \delta_{ni} - \frac{i}{\hbar}\sum_mV_{nm}\int_{-\infty}^t e^{i\omega_{nm}t'}\langle m|U_I(t',-\infty)|i\rangle dt'\tag{6.1.24}$$

Insert the first equation into the second:

$$\langle n|U_I(t,-\infty)|i\rangle = \delta_{ni} - \frac{i}{\hbar} V_{ni}\int_{-\infty}^te^{i\omega_{ni}t'} dt' - \frac{i}{\hbar^2}\sum_m V_{nm}T_{mi}\int_{-\infty}^t \frac{e^{i\omega_{nm}t'}e^{\left(i\omega_{mi}+\epsilon\right)t'}}{-\omega_{mi}+i\epsilon} dt'$$

This is where I get confused. In the text it says that the second term of equation 3 is the same as the second term of equation 1 with $$T$$ replaced with $$V$$. I don't see how the epsilon is introduced into the second term of this equation. My best guess is that its linked through the $$T$$ matrix definition.

$$\langle n|U_I(t,-\infty)|i\rangle = \delta_{ni} - \frac{i}{\hbar}V_{ni}\int_{-\infty}^te^{i\omega_{ni}t'}dt'\tag{6.1.9}$$ $$\langle n|U_I(t,-\infty)|i\rangle = \delta_{ni} - \frac{i}{\hbar}T_{ni}\int_{-\infty}^te^{i\omega_{ni}t'+\epsilon t'}dt'\tag{6.1.10}$$ Equating the second terms of these equations would make the integral solvable but this runs into problems.

Insert the first equation into the second and compare with equation 6.1.23:

$$\langle n|U_I(t,-\infty)|i\rangle = \delta_{ni} - \frac{i}{\hbar}T_{ni}\int_{-\infty}^te^{i\omega_{ni}t'+\epsilon t'}dt' - \frac{i}{\hbar^2}\sum_m V_{nm}T_{mi}\int_{-\infty}^t \frac{e^{\left(i\omega_{ni}+\epsilon\right)t'}}{-\omega_{mi}+i\epsilon} dt'$$ $$\langle n|U_I(t,-\infty)|i\rangle = \delta_{ni} + \frac{1}{\hbar}T_{ni}\frac{e^{\left(i\omega_{ni}+\epsilon\right)t}}{-\omega_{ni}+i\epsilon} \tag{6.1.23}$$

Comparing terms: $$\frac{1}{\hbar}T_{ni}\frac{e^{\left(i\omega_{ni}+\epsilon\right)t}}{-\omega_{ni}+i\epsilon} = - \frac{i}{\hbar}T_{ni}\frac{e^{\left(i\omega_{ni}+\epsilon\right)t}}{i\omega_{ni}+\epsilon} - \frac{i}{\hbar^2}\sum_m V_{nm}T_{mi}\frac{e^{\left(i\omega_{ni}+\epsilon\right)t}}{\left(-\omega_{mi}+i\epsilon\right)\left(i\omega_{ni}+\epsilon\right)}$$ The first terms obviously cancel so this does not yield equation 6.1.26 as desired.

$$\sum_m V_{nm}T_{mi}\frac{e^{\left(i\omega_{ni}+\epsilon\right)t}}{\left(-\omega_{mi}+i\epsilon\right)\left(i\omega_{ni}+\epsilon\right)} = 0$$

Not equal to: $$T_{ni} = V_{ni} + \frac{1}{\hbar}\sum_m \frac{V_{nm}T_{mi}}{\left(-\omega_{mi}+i\epsilon\right)}$$

I am using Sakurai 2nd edition and the relevant pages are 386-390. Any help that can be provided would be appreciated and feel free to ask any questions for clarification.

• Disappointed that this has not been answered and has even been closed as I am having the exact same problem. It seems to be very much on topic given that it touches on the derivation of a fundamentally important equation in physics Commented Nov 7, 2022 at 9:00