# knowledge of an internal observer

I would like to discuss the consequences of the concept of an internal observer in quantum theory. If we assume that we have a universe that evolves unitarily at a global scale and an observer is defined as a subsystem of that universe, what can this observer learn about the universe?

Some assumptions are that the evolution of the universe is generated by interaction. Interactions are local and the observer gathers information by interacting with his environment. But most importantly, there is no measurement postulate that the observer could use to perform a quantum measurement.

I would argue that everything the observer learns about the universe is contained in the state history of the subsystem he has interacted with. If one assumes that he can derive a law for the evolution of this state history, then he can only reconstruct the state history up to isomorphisms. Meaning, if he finds plausible state history psi(t) which evolves unitarily like $\psi(t) = U(t,t_0)[\psi(t_0)]$, then any $\phi(t)$ for which a bijection $f$ exists so that $\phi(t) = f(\psi(t))$ and $\psi(t) = f^{-1}(\phi(t))$ and which evolves with the same law $\phi(t) = U(t,t_0)[\phi(t_0)]$ is an equivalent description of the state history of the system he is part of. (Note that psi is not necessarily a ket. I would like to stay agnostic of the actual representation for now)

A very simple example for such a bijection would be the multiplication with a nonzero complex number. Another would be a spacetime symmetry transformation.

Do you think this argumentation is correct so far? I will continue the argument to the point where it actually gets more interesting, but I would like to see some confirmation so that we can later discuss the consequences.

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With one gaze to the sky in a clear night the subsystem the observer interacts with is already half of the universe.

We are all observers embedded into a huge quantum system called the universe. But we have little trouble to gather enough information about this quantum system to write many books on it.

Most of our measurements don't make use of Born's measurement postulate, as most of the time we are only measuring mean properties of quantum observables. This gives us enough information on the state of the quantum system to survive in it.

But when we are presumptuous enough to try to know very detailed microscopic properties of the state of this quantum system, we are forced to restrict ourselved to tiny subsystems external to us. Then Born's rule might apply - but only if we perform a perfect von-Neumann experiment.

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I am going to butcher notation, but to answer this we must first consider that the Unitary operator $U(t,t_0)$ can be represented as a Dyson series:

$$U_n(t,t_0) = (-i)^n\int_{t_0}^{t}dt'\int_{t_0}^{t'}dt''\dots\int_{t_0}^{t^{n-1}}dt^{(n)}T\{{H_I(t')H_I(t'')\dots H_I(t^{(n)})}\}$$

Where $H_I$ is the interaction Hamiltonian and $T$ is the time ordering operator.

In order to find a solution for state evolution, we need to solve:

$$i\dfrac{\partial}{\partial t}[\psi(t)]=H_I[\psi(t)]$$

We can approach this in terms of the scattering matrix S:

$$[\psi^*(t)]S[\psi(t_0)] = S_{fi}$$

Where the probability of transitioning from some initial state to some final state is given as:

$$\textbf{P}_{fi} = S_{fi}S^*_{fi}$$ and:

$$S_{fi} = \lim_{t_0\to - \infty}_{t\to +\infty} [\psi^*(t)]U(t,t_0)[\psi(t_0)]$$

and the S matrix then is:

$$S = \sum_{n=0}^{\infty}(-i)^n\int_{-\infty}^{t}dt'\int_{-\infty}^{t'}dt''\dots\int_{-\infty}^{t^{n-1}}dt^{n}{H_I(t_1)H_I(t_2)\dots H_I(t_n)}$$

What is important is that in this example we can see the scattering matrix as representing a type of square root of a probability matrix controlling the probability of state to state transitions. The point being is that this is not a history of observed states, so while the bijection you propose may be possible, it does not mean that the seemingly identical functions would automatically give you the same observed history. It only suggests that the probabilities governing transitions would be the same.

Note: This discussion borrowed heavily from David McMahon's discussion in chapter 7 of Quantum Field Theory Demystified under Perturbation Theory.

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 Thank you for your answer, however it does not really relate to what I intended to ask. Maybe I have to reformulate my question. The point was that an observer who only gathers information by interaction cannot distinguish certain state representations. Your distinction is based on probabilities, i.e. the measurement postulate. – A.O.Tell Sep 24 '12 at 19:02 McMahon also has a nice quote, "any transitions in the state are due to interactions" and this is because if $H_I$ goes to zero, then states are constant. I am not sure you can sensibly talk about interactions without talking about observation of outcomes. I may be misinterpreting your meaning, but when I see measurement postulate I see wave function collapse, which is an entirely different discussion, since the wave function never collapses it seems making that connection is immaterial. – Hal Swyers Sep 24 '12 at 19:15