Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
Looking at the stars does not imply that you interact with half the universe. You interact with those photons that are nearby. Even if those photons carry some information about the universe, you don't interact with it. Einstein locality limits the interaction to the spatial region covered by the spatial section of your backwards lightcone at the moment of your initial interaction. I also limited the system to information gained by interaction. So ensemble measurements are not applicable. And those rely on the measurement postulate too, even if you cannot single out individual events. – A.O.Tell Sep 24 '12 at 16:35
@A.O.Tell yes you don't interact with them in sense of field/forces, but any object in the universe "theoretically" affects how your wave function will actually collapse! because "probability waves" can travel at speed that exceeds speed of light. – TMS Sep 24 '12 at 18:04
TMS, that is first of all not related to my question above and secondly not true. The evolution of quantum states is generated by strictly local interactions. The state space contains nonlocal multi-particle states, but that does not change the locality of interaction. My question is very clearly formulated regarding the assumed system. Arnold's statement is not answering it at all, it's not even related to what I asked. – A.O.Tell Sep 24 '12 at 18:10
@A.O.Tell: The notion of nearby makes no sense for photons, as these cannot have a position. - In any case, we can only gather information by local interaction and still know an incredible amount about the whole universe. Your assumptions are therefore completely unrealistic. – Arnold Neumaier Sep 25 '12 at 7:19
@ArnoldNeumaier, you are missing the point entirely. The question is not what you learn about the past of the universe by observing photons, but what you learn about the current state sequence by interacting with your immediate environment. And even if photons don't have a position operator or spatial wavefunction, their interaction is still local and restricted to the past light cone. My assumptions are not unrealistic, they're in exact agreement. Where do I make the assumptions that say the observer cannot learn about his past by observing photons? – A.O.Tell Sep 25 '12 at 7:57

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.

share|cite|improve this answer
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. – user11547 Sep 24 '12 at 19:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.