# Definition of the $S$-Matrix in Schwartz QFT-Book: Why is $\langle f, t_f | i, t_i \rangle$ in the Schroedinger picture, and not Heisenberg-picture?

On page 51, (equation 5.1), Mathew Schwartz introduces the $$S$$-matrix as

\begin{align} \langle f| S | i \rangle_{Heisenberg} = \langle f, \infty | i, -\infty \rangle_{Schrödinger} \end{align}

Were $$|i, t\rangle$$ is a Schrödinger-state at time $$t$$ (at least if I understood correctly).

So now my question is: Shouldn't it be exactly the other way around?

I think that it should be the other way around because of several things:

• If I set up a schrödinger state at time $$-\infty$$ and at $$\infty$$ then the time evolution of the system (which is the interesting part) doesn't have an effect on $$\langle f, \infty | i, -\infty \rangle$$ at all.

• Schwartz later, when prooving the LSZ-formula starting at equation 6.6, uses the fields and annihilation/creation operators in the Heisenberg picture. Yet he just calculates the $$S$$-matrix simply as the scalar product between the defined in- and out-states, as this question points out.

Can somebody clear things up here? Especially, if an answer writes something like "$$|i, t_i \rangle$$ is a state at time $$t$$", or "state $$|i, t_i \rangle$$ fixed at time $$t$$" it would be great if additionally one says what is meant by that, because that phrases meaning is very ambiguous.

This is a common confusion, and I don't think Schwartz's description is clear at all.

An in state $$|i, \text{in} \rangle$$ is a Heisenberg state, fixed at time $$t = 0$$, which would look like several widely separated, incoming particles if you evolved it like a Schrodinger state to $$t \to -\infty$$. (Explicitly, if $$U(t_f, t_i)$$ is the time evolution operator in Schrodinger picture, then $$|i, \text{in} \rangle = U(0, -\infty) |i \rangle$$. That is, it's the Schrodinger state evaluated at time $$t = 0$$.)

Similarly, an out state $$| f, \text{out} \rangle$$ is a Heisenberg state which has widely separated, outgoing particles as $$t \to \infty$$. Here, $$i$$ and $$f$$ stand for a specification like two particles, of momenta $$p^\mu$$ and $$q^\mu$$''.

We want to know the transition amplitudes $$\langle f, \text{out} | i, \text{in} \rangle$$. The $$S$$-matrix is defined by $$|i, \text{in} \rangle = S | i, \text{out} \rangle.$$ Upon using the unitarity property $$S^\dagger = S^{-1}$$, this implies $$\langle f, \text{out} | i, \text{in} \rangle = \langle f, \text{in} | S | i, \text{in} \rangle.$$ Again, all of these are Heisenberg states. But "in" and "out" states are a weird kind of Heisenberg state defined by how they would look if they were time evolved like Schrodinger states, even though they actually aren't, which is why books can be unclear on this point.

• Heisenberg state vectors do not depent on time at all. They may not be "widely separated" or labeled with time ans space variables. Schroedinger sates are time and space dependent, so theey depend explicitely on the interactions. In particular, the final state takes into account all the previous times when the interaction was significant. Jun 29, 2021 at 7:25
• @knzhou This may be a stupid question, but why do we do this whole "if it were time evolved like schrödinger states"-thing? Additionally, how do I relate your last line to the problematic line in schwartz in question? I shouldn't I simply because what he writes is wrong? Jun 29, 2021 at 7:54
• @knzhou since $i$ and $f$ in $|i\rangle$ stand for specifications (as you say), the actual states $|i\rangle$ and $|f\rangle$ also depend on the field operators $\Phi(\vec{x},t)$. At what time do we take these field operators (which are time dependent since we use the heisenberg picture) for that purpose? At time $t=0$, or at time $t=\pm \infty$? Jun 29, 2021 at 9:20
• I'll accept the answer, but I think the information you gave to me in another comment (especially, where the states in the schroedinger picture are evaluated at) are important. Could you edit it in? Jul 3, 2021 at 21:36
• @Quantumwhisp Sure thing! Jul 3, 2021 at 21:39