# S-matrix derived directly in terms of the interaction picture

Consider a quantum mechanical system with Hamiltonian

$$H=H_0+H_{\text{int}}.$$

Consider $H_0$ to be time-independent, so that its associated time-evolution operator is $U_0(t,t_0)=e^{-i(t-t_0)H_0}$.

Denote by $S(t,t_0)$ the time-evolution operator associated to $H$. What I understood from books like Peskin and Schwartz' is that in the case of QFT, the S-matrix ends up being defined by

$$\langle f |S|i\rangle = \lim_{t_\pm\to \pm\infty}\langle f | S(t_+,t_-)|i\rangle$$

where $|i\rangle,|f\rangle$ are eigenstates of $H_0$. In other words, $S$ is defined in terms of the time evolution operator of the full Hamiltonian.

It turns out in David Tong notes on QFT he does this with just the interaction picture time evolution operator. In other words, the operator $U(t,t_0)$ which evolves interaction picture states

$$|\psi(t)\rangle_I=U(t,t_0)|\psi(t_0)\rangle_I$$

He says

This means that we take the initial state $|i\rangle$ at $t\to -\infty$, and the final state $|f\rangle$ at $t\to +\infty$, to be eigenstates of the free Hamiltonian $H_0$. At some level, this sounds plausible: at $t\to-\infty$, the particles in a scattering process are far separated and don't feel the effects of each other. Furthermore, we intuitively expect these states to be eigenstates of the individual number operators $N$, which commute with $H_0$, but not $H_{\mathrm{int}}$. As the particles approach each other, they interact briefly, before departing again, each going on its own merry way. The amplitude to go from $|i\rangle$ to $|f\rangle$ is

$$\lim_{t_{\pm}\to\pm\infty}\langle f | U(t_+,t_-)|i\rangle = \langle f | S |i\rangle$$

where the unitary operator $S$ is known as the S-matrix.

his $U$ operator is the time evolution operator in the interaction picture, computed usin Dyson's formula in terms of $H_{\mathrm{int}}$ in the interaction picture.

What I want to understand is how to derive that the S-matrix can be computed as he does, using just the interaction picture time evolution $U$.

I don't seem to get it, because I have $S(t,t_0)=e^{iH_0(t-t_0)}U(t,t_0)$ thus

$$\langle f|S|i\rangle=\lim_{t_{\pm}\to \pm \infty}\langle f |S(t_+,t_-)|i\rangle=\lim_{t_{\pm}\to\pm \infty} \langle f | e^{iH_0(t_+-t_-)} U(t,t_0)|i\rangle$$

then using that $|f\rangle$ is eigenstate of $H_0$ I get

$$\langle f | S | i\rangle = \lim_{t_{\pm}\to \pm\infty} e^{i\omega_f (t_+-t_-)} \langle f | U(t_+,t_-)|i\rangle.$$

So the definitions aren't the same. There's that exponential in front. And worse, the exponential diverges.

What is going wrong here? How can I derive that these two approaches are the same?

## 2 Answers

The previous answer by user154997 already captures the main problem, let me just add some words to it.

The scattering matrix is defined by the following process: - Take a free state as your initial state in the infinite past. - Time evolve it to the infinite future. - Compute how likely it is to find your particle in a new free state (by taking the overlap.

You can choose to do this in any picture you like. In the interaction picture the time evolution operator involves only the interaction term of the Hamiltonian, in the Schrödinger picture it contains both the free and the interaction term. As a trade-off the states contain the free time evolution when you work in the interaction picture. This difference in the states is exactly what is missing in the comparison outlined by the OP.

Minor remark: I never understood why people talk about eigenstates when doing time-dependent scattering theory. It makes no sense to me.

What you call the "real" time evolution operator would correspond to the computation in the Schrödinger picture I guess. Then the interaction picture amounts to a redefinition of states and operators:

$$\langle f | U | i \rangle = \underbrace{\langle f | e^{-i H_0 t}}_{\langle f_I |}\ \underbrace{ e^{i H_0 t} U e^{-i H_0 t}}_{U_I}\ \underbrace{e^{i H_0 t} | i \rangle }_{| i_I \rangle}$$

where the subscript $I$ denotes the interaction picture. I hope I understood your question!