Dirac delta function definition in scattering theory

Question

I'm studying scattering theory from Sakurai's book. In the first pages he gets to the following expression:

$$\langle n|U_I(t, t_0)|i\rangle=\delta_{ni}-\frac{i}{\hbar}\langle n|V|i\rangle\int_{t_0}^t e^{i\omega_{ni}t'} dt',\tag{1.9}$$

where $U$ is the propagator in Dirac's interaction picture and $V$ is a potential operator.

So given that scattered states are only defined asymptotically we want to send $t \to \infty$ and $t_0 \to -\infty$, so I would say that the integral becomes immediately a Dirac's delta because that's just its integral representation! But he says: let's define a $T$ matrix such that:

$$\langle n|U_I(t, t_0)|i\rangle=\delta_{ni}-\frac{i}{\hbar}T_{ni}\int_{t_0}^t e^{i\omega_{ni}t'+\varepsilon t'} dt'.\tag{1.10}$$

And then keeps going. I don't get this! Why do we need this small parameter $\varepsilon$? He then says that it's going to be sent to zero and that it makes sure the integral does not diverge. I don't quite get this prescription. Can anyone help me understand this strategy?

Answer 1

I think the point is that if you took the limits to infinity without doing anything else, you'd be implicitly redefining the matrix element in order to make the equations consistent. So he just calls the redefined matrix element $T_{ni}$ instead of $V_{ni}$. Later he solves for $T_{ni}$ in terms of $V_{nj}$ (see the section "Solving for the T matrix").

Answer 2

If you go to these limit right away and get delta functions you might later face problems such as need to evaluate meaningless expressions, e.g. $\delta(x)^2$ or $\delta(0)$. This can often be avoided by adding so called regulators (in this case this role is played by $\epsilon$). These should be removed after all manipulations. If you get final results for physical quantities which are regular in the limit $\epsilon \to 0,$ you are happy.