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Chapter 3 of volume 1 of Weinberg's QFT says the following regarding in and out states in scattering theory:

Implicit in the definition of the states is a choice of the inertial frame from which the observer views the system; different observers see equivalent state-vectors, but not the same state-vector. In particular, suppose that a standard observer $\mathcal{O}$ sets his or her clock so that $t=0$ is at some time during the collision process, while some other observer $\mathcal{O}'$ at rest with respect to the first uses a clock set so that $t'=0$ is at a time $t=\tau$; that is, the two observers' time coordinates are related by $t'=t-\tau$. Then if $\mathcal{O}$ sees the system to be in a state $\Psi$, $\mathcal{O}'$ will see the system in a state $U(1,-\tau)\Psi=\exp(-iH\tau)\Psi$. Thus the appearance of the state long before or long after the collision (in whatever basis is used by $\mathcal{O}$) is found by applying a time-translation operator $\exp(-iH\tau)$ with $\tau\to -\infty$ or $\tau\to+\infty$, respectively.

What's the difference between $\mathcal{O}$ seeing $\Psi$, waiting 5 seconds and still seeing $\Psi$, versus choosing $\tau=5$ seconds for the second observer $\mathcal{O}'$? From the above it looks like $\mathcal{O}'$ will see a different (equivalent) state, but isn't $\mathcal{O}'$ the same observer as $\mathcal{O}$, once 5 seconds have gone by? (And since the state doesn't change with time, they should also see $\Psi$). My qualm is mostly that it seems physically unreasonable for them to see different things.

Further, with regards to the last sentence in the quote: can't $\mathcal{O}$ by themselves know the appearance of the state long before/after the collision, simply because for them it will always be $\Psi$?

(Ps. This is in the Heisenberg picture.)

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I believe this passage is written in the Schrodinger picture, so $\Psi$ depends on time. Thus, in regards to your last point, the state is not always $\Psi$ at $t=0$, observer $\mathcal{O}$ needs to be able to evolve $\Psi$ to an earlier or later time to know the state at earlier or later times.

For the first part of the question:

Both $\mathcal{O}$ and $\mathcal{O}'$ will predict the same outcome of any physical experiment. But, the state according to $\mathcal{O}$ at $t=0$ is mathematically different than the state according to $\mathcal{O}'$ at $t'=\tau$. The difference is that the defined by $\mathcal{O}'$ has an extra phase factor $e^{-i H \tau}$ relative to the state defined by $\mathcal{O}$. Since the difference is an overall phase, physical predictions will be the same, but nevertheless the states are not literally equal to each other.

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  • $\begingroup$ No, in the paragraph right before this one the author explicitly notes that he's working in the Heisenberg picture. Sorry, I should have specified that in the question - I'll edit it now. And the first part of the question is a little imprecise - mostly I'm uncomfortable with the fact that another experimentalist, standing right next to me but with a clock that is a little bit out of phase with mine, will measure a physically equivalent, but different state (note that it's not just a phase factor if the state is not an energy eigenstate). $\endgroup$ Jul 12 '20 at 5:14
  • $\begingroup$ Ah ok, I think I see. In the Heisenberg picture, then $\Psi$ does not depend on time explicitly, but you need to choose a reference time to define the state $\Psi$. At the reference time, the Schrodinger and Heisenberg states will be equal. Usually in scattering processes one chooses to work with states in the asymptotically far past ("in states") and future ("out states"). I don't have Weinberg on hand, but perhaps he is pointing out this subtlety because he wants to carefully define a limiting procedure for constructing these asymptotic states. $\endgroup$
    – Andrew
    Jul 13 '20 at 7:11
  • $\begingroup$ For "physically equivalent, but different states" -- you are right I was too quick about saying the difference in states is just an overall phase. I think the point is that the reference times can be chosen differently when defining the state that defines the Heisenberg picture. Two different reference times will lead to two different states. However what really matters are the observable quantities. Then we just need to be careful what question we ask. If we want to know the average position of a particle at $t=0$, the two observers may disagree if their clocks are offset. (...) $\endgroup$
    – Andrew
    Jul 13 '20 at 7:15
  • $\begingroup$ (...) On the other hand, let's say there is a specific event (eg a flash of light) which the observers use to synchronize their measurements. Then the two observers will agree on the outcome. If the flash of light is at $t=0$, $\mathcal{O}'$ will need to add time evolution operators in the matrix element to compensate for the different time origin. However this is typically not the question we ask in scattering experiments, which are typically about relating in and out states, so these subtleties aren't important. If these comments are helpful I'll add it to the answer. $\endgroup$
    – Andrew
    Jul 13 '20 at 7:20

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