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The paper doesn't explain how their predictions would differ from those of non-collapse theories. Since the paper doesn't even discuss what would be predicted without collapse, it is difficult to see how it could rule out quantum theory without collapse. Quantum theory without collapse explains all of the predictions commonly attributed to quantum theory ...

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OK so why is it possible to use the tracing trick he uses on p9 eqn 1.16 if the b_i are not orthoganol basis sets?

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Thank you so much for responding. I'm afraid I did not quite follow your response. Perhaps if I explain by means of the workings below. If I trace rho S+E in the picture here using the quick method of the formula 1.16 on p9 then lines three and four below quickly give the correct result of eqn 1.76. However I should get exactly the same result using my ...

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Right, in general you're not going to see a straightforward equivalence there. We can use Dirac notation with $\hat P_b = |b\rangle\langle b|$ to see that $\langle \hat A \hat B \rangle = \sum_{a,b} a~b~\langle \psi | a \rangle~\langle a | b \rangle~\langle b | \psi \rangle$ and even inserting an identity matrix for $b$ (call it $b'$) gives: $$... 1 The problem is arising since you are trying to take the trace in a non-orthonormal basis including both |e_1\rangle and |e_2\rangle. If A is an operator, the trace of the matrix$$A_{mn}\equiv \langle e_m|A|e_n\rangle$$is not invariant under any basis |e_m\rangle. If this is the orthonormal eigenbasis, the trace is the sum of the eigenvalues. Any ... 0 You claim that there seem to be quantum jumps and ask why this does not refute non-collapse theories. To answer the question of whether there is evidence that rules out no-collapse theories, you first have to work out what would happen is no collapse took place. So suppose that you have an atom a in an excited state |e\rangle_a that has some half life ... 0 The no collapse people would say that if you write down the continuous evolution equations for the actual experimental setup in question that it predicts a perfectly continuous evolution of the wave vector into two parts that don't overlap. For instance a Stern-Gerlach device takes an incoming beam with a spin state and that beam branches into two. So the ... 2 A quantum system is described by a suitable *-algebra of observables. The quantum states are functionals of the observables, that when applied on observables yield the average value of it in the state. So, given an observable A, and a state \omega,$$\omega(A)$$is the evaluation of A, i.e. its average value on the state. Now the state (or ... 3 Having worked with elementary particles all my working life I can assure you that particles have a trajectory. Here is proof Another proof is the existence of accelerators which create the beams that we can scatter against targets, as in picture, or against each other and study the results statistically. That is how the standard model of particle ... 3 Most people would say that |\Psi(a)|^2 is the probability density that the particle would be found in some small neighborhood of a. Most people would not say it is the probability that it is in that location. This is because because that same kind of thinking (that it has a probability of having a property and that a measurement merely reveals the value ... 1 It seems your biggest issue is with the notion of measurement, so I'm going to give an answer that doesn't mention it at all. Suppose a compass needle is pointing in some direction. We say it is in state 'A' if it's pointing north/south, and in state 'B' if it's pointing west/east. So if it points north it's definitely A and definitely not B, but if it ... 1 The uncertainty principle is about when you pick a state. When you pick a state you might pick one with high uncertainty in one observable or high uncertainty in the other or one with high uncertainty for both. See http://physics.stackexchange.com/a/169757 But the uncertainty is 100% not about an experimental difficulty or a lack of knowledge of some ... 1 If you try to watch it then you just find out how watched electrons move. Or as dmckee reminds us, the quantum zeno effect might even stop them from having a chance to move. If you don't watch it then you don't see it so you don't know how they moved or whether they even just appear where you see them. You can try so called weak measurements to try a bit ... 0 So this should mean that you can know the exact position and momentum of a particle in a system, if you measure it. (1) A particle state with exact position and exact momentum doesn't exist. (2) A particle state with exact position or exact momentum doesn't exist. (3) One couldn't in principle measure exact position or exact momentum since our ... 5 It is correct that the uncertainty principle is not a statement about experimental precision as such. It is incorrect, however, to state that you can know position and momentum of a quantum system exactly, because it presupposes such a thing as "exact position" or "exact momentum" exists. It doesn't, and especially not simultaneously. Any two observables ... -1 I define the set of projectors P_{a} = |a\rangle\langle a|, what is the physical observable (like 'mopentum') that I need to measure to perform the operation? Even if you have some projections, even if they are each self adjoint, and even if they are mutually orthogonal and their range spans the whole space they still don't linearly combine into a ... 1 I think it is important to note that the wavefunction of quantum mechanics is not a field like the electric or magnetic field. It assigns complex numbers to configurations. So it is a function where the domain of the function is not space or even spacetime. So there is not a value of the field at a point in spacetime, so there isn't a thing to curve it. ... 1 Unfortunately, that question's answer depends on the theory of quantum gravity that you chose. But roughly here is the situation : If you take a classical spacetime and quantum matter fields (G_{\mu\nu} = \langle T_{\mu\nu} \rangle_\omega), also known as semiclassical gravity, then the stress energy tensor varies with the measurement. The standard ... 1 It sounds to me like you're asking about "Quantum Bayesianism", or "QBism" as it's called by David Mermin, one of its principal proponents, e.g., http://arxiv.org/abs/1311.5253 Or you can google those two terms in quotes for lots more references. Is that what you're talking about? And do their interpretations answer your questions? 0 What [is] a measurement [...] in the MWI? A Stern-Gerlach (SG) device has an incoming beam be split by the Schrödinger equation (applied to the actual experimental setup) into two beams, one going left and one going right. This is not the the measurement, this can be undone; and you can tell whether or not has been undone. Specifically, when a series ... 1 Let me try to use a programming analogy. A quantum mechanical object is like a class. It has certain attributes, which would be physical attributes, e.g. position, momentum. It also has certain methods, which are physical operations that can change or modify the object or modify the environment using the object. In physical terms these are the unitary ... 2 It is worse than lazy evaluation. Haskell I can treat a lazy value as if it were already there and manipulate it as such. In quantum mechanics you can't do that. What you have is something that tells you the relative frequency of getting lots of results for different interactions if you did one of the various interaction first. And you can't do them ... 0 You speak of the measurement problem, which as yet has no widely accepted resolution. But a pure quantum state after a measurement is in exactly the same kind of entity as it was before the measurement, namely a new pure quantum state. It's simply that the measurement forces (by whatever as yet less than fully understood mechanism that resolves the ... 0 In CHSH we can explain why quantum violate the inequality but it tells nothing about nonlocality : The operator S=AB-AB'+A'B'+A'B The measurement process in Copenhagen is used : the wavefunction collapses in an eigenstate of A remeasuring A will give the same result noted a The same for A' noted result a' Then a strange thing happens for the B side the ... 2 Consider a molecule of oxygen in a balloon. You know that at nonzero temperature all those molecules are bouncing around in all directions. Of course, the mass of air doesn't have any net motion in any direction. Indeed, the average velocity of each molecule is zero:$$\langle v \rangle = 0 \, . Of course, the average energy of any particular molecule is ...

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