Hi guys, Quantum mechanics gives well-defined probabilistic predictions for the value we get when we measure position or momentum; one simply takes the absolute square of the wavefunction in either the x-basis or the p-basis. However, I am not so clear on how we would predict at what time we obtain the result, and whether the answer be definite or probabilistic, we should be able to answer this question. I have found and read this (relatively short) article: http://arxiv.org/abs/quant-ph/9802020 and would much appreciate it if you guys could read it and help answer a few questions I have. In the article, Rovelli defines the M operator that acts on the S-O system. He says that the question: when does O see a definite value of a dynamical variable? can be answered probabilistically by using a second observer O' and measuring the value of the observable corresponding to the M operator. If it's 1, the measurement has been made, and if it's 0, the measurement has not been made. This is a probabilistic answer to the question, because it involves measuring a quantum mechanical operator. I cannot comprehend this for the following reason: it seems as if we can define the M operator in many different ways. By defining M differently, we predict that with some other probability, the wavefunction of S collapses for the observer O under a different kind of projection operator. To put it really concisely: by suitably redefining the M operator, we predict a different type of wavefunction collapse, with a different probability in time. Furthermore, in real life, S and O are always weakly interacting, and I believe that the singular value decomposition theorem says that we can always find a set of basis vectors for the S-O system such that S and O are perfectly correlated. Even though the correlation might be for some operators corresponding to totally uninteresting observables, the correlation is still there. Then if we, again, suitably define the M operator, would we not predict (with certainty) that O sees S undergo a wavefunction collapse? Then would we not have wavefunction collapse all the time? True, the interaction between S and O is weak, and so the collapse means that the wavefunction of S with respect to O barely changes (I hope). However, one still wonders whether this would affect normal Schrodinger evolution. A really stupid sounding but serious question: if collapse happens all the time (continuously), then is there any "opportunity" for Schrodinger evolution to take place? As you guys can see, I'm pretty confused, so any help would be much appreciated. I'm really trying to understand this article. Thanks.