Thread: No-Interaction Theorem View Single Post
 P: 1,746 aspidistra, I think I understand the origin of your confusion. In order to clear things up I would suggest you to consider "instantaneous" observers, which exist and make their measurements only during very short time intervals. To make it more clear, let me consider two instantaneous observers. One of them is "John now". Another one is "John 10 minutes later". These two instantaneous observers are related to each other by a time translation transformation (t=10). Suppose that these observers look at the same physical system - for example a kettle on the stove. From the point of view of the observer "John now" the water in the kettle is cold. From the point of view of "John 10 minutes later" the water is boiling. To describe this situation mathematically we can consider water temperature, which is an observable pertinent to the observed system - the kettle. I will denote this observable T. The temperature measured by "John now" will be denoted T(0). The temperature measured by "John 10 minutes later" is T(10). The time evolution of this observable (i.e., the transition from T(0) to T(10)) can be described in the Hamiltonian formalism. The physical system (the kettle on the stove) is described by the Hamiltonian H, which is a generator of time translations. So, in order to connect observations of the two instantaneous observers we need to apply an exponent of the Hamiltonian to the observable. Then in your notation we can write (t=10) T(10) = e^{[H]t}T(0) = e^{[H] 10}T(0).....................(1) The chosen physical system (stove + kettle + water) is very complex. So, its Hamiltonian H cannot be written in a simple form, and it is very difficult to perform calculation (1) in any reasonable approximation. However, the same Hamiltonian rules of time evolution apply to all isolated physical systems both simple and complex. The good thing about using "instantaneous" observers is that the same approach that we've used above for time translations works for all other 9 types of transformations of the Poincare group. For example, consider a third observer "Bob now in another room". So, we can ask what are the results of measurements performed by this observer? This observer is displaced in space with respect to "John now". So, in order to answer the question we need to apply a space translation transformation to our observable - temperature. In order to do that we need to know the space translation operator (total momentum) P that is pertinent to our physical system. Then if the distance between John and Bob is 5 meters (x=5). Then the temperature measured by Bob can be found from formula T(x=5) = e^{[P]x}T(0) = e^{[P] 5}T(0).....................(5) If operator P is chosen correctly, then we should obtain the obvious result T(x=5) = T(0). Similarly, we can introduce the generator of rotations J and the generator of boosts K pertinent to the observed system (the kettle on the stove). The ten generators (H, P, J, K) must form a representation of the Lie algebra of the Poincare group. Their exponents e^{[H]t}, e^{[P]x}, e^{[J]a}, e^{[K]v} must form a representation of the Poincare group itself. Eugene.