# There are two definitions of S operator (or S matrix) in quantum field theory. Are they equivalent?

I read several textbooks of QFT and found that there are two kinds of definition of $S$ operator (or S matrix).

• First kind:

Define $\hat{S}$ is map from out space to in space $$\hat{S}\left|\beta,\text{out}\right\rangle：=\left|\beta,\text{in}\right\rangle,$$ so that $$S_{\beta\alpha}:= \left \langle \beta,\text{out} | \alpha,\text{in}\right\rangle= \left \langle \beta,\text{out} \middle |\hat{S}\middle | \alpha,\text{out}\right\rangle= \left \langle \beta,\text{in}\middle |\hat{S}\middle | \alpha,\text{in}\right\rangle.$$ I understand that all these vectors are defined in the Heisenberg picture.

• Second definition: $$S_{\beta\alpha}:={}_I \left \langle \beta \middle |\hat{S}\middle | \alpha \right\rangle_I$$ where subscript $_I$ means vector are in interacting picture. In this definition, then, $$\hat{S}=U_I(+\infty,-\infty),$$ where $U_I(+\infty,-\infty)$ is the evolution operator in interacting picture.

Are these two definitions equivalent? I am confused about it.

Remark: I konw that the matrix element $S_{\beta\alpha}$ is the same in these two pictures, what I want to ask is whether the operator $\hat{S}$ is same in these two definitions. Thanks!

## 1 Answer

I don't completely understand the two sets of statements you've written, but I think I understand the essence of your question. Maybe this helps:

The S-matrix (operator) is a transfer function from in states to out states.

1. If your states are not evolving (Heisenberg/Interaction picture) then you need to insert an evolution operator between the states.

2. If your states are in the Schrodinger picture and they're evolving with time, then $| out, t = \infty \rangle = (\textrm{evolution operator}) \; | out , t = 0 \rangle$

So the definition (convention) for the S-matrix depends on your convention for defining the Hilbert space at late times (whether it is the same Hilbert space as initial times, or if it is the time-evolved Hilbert space). This is equivalent to whether you're in the Schrodinger or the Heisenberg picture. Physically, I hope it's now clear why both descriptions/conventions are the same object.