Recent content by normvcr

Are you saying ##\ket{x}## is not a physical state, because no measuring device can be so accurate? That makes sense. I like the example of the Hilbert space of band-limited functions, whose spectral domain is contained in the interval ##[-1/2, 1/2]##. An orthonormal basis is $$sinc(t-N), N =...

Agreed, if you are referring to the action taking place in the separable Hilbert space ## L^2(R) ##. In this case, ##P_a\Psi (x) = \Psi(x-a)##, and things are continuous. The problem, of course, is that the x-operator is defined as ##Q\Psi (x) = x\Psi(x)##, whose eigen states are not in the...

I did read the RHS section 1.4, but did not think that RHS was adopted by the text as the place to model QP -- only that this is one of the directions people are thinking about. For example, the postulates of section 2 assume Hermitian operators i.e. Hilbert spaces. Unitary operators are...

I agree, and that is precisely my point: The group action is not continuous. Is this not a problem? Certainly conceptually, but also technically, as it is the continuity of the group action that guarantees the group generators have a common, invariant, dense domain in the Hilbert space.

Am reading a book (Ballentine, "Quantum Mechanics: A modern development) which I have found very helpful. Am now puzzled by section 3.4, where the position operator satisfies Q|x> = x |x> (I have simplified from 3 dims to 1 dim). Here, x is any real number. There are, thus, uncountably many...

The paper was turned down by the journal, for reasons that I accept -- "this is an interesting mathematics article, but does not contain sufficient philosophical/conceptual insights to be publishable in ...". This is quite reasonable, given the nature of the journal. I raised two such insights...

Clever argument. The fly in the ointment is that since P(T|S)=P(S|T), then we always will have P(T)=P(S), even when not at equilibrium. The analysis needs to take place in a somewhat more complex context: P(T|S) manifests itself when a measurement is done for T, in initial state S, and vice...

This got me to thinking. There is some principal that a system should be able to evolve to maximum entropy. Perhaps, this can form the core of an argument that the transition probabilities are symmetric. This might also not pan out at all ...

Again, thanks for the detailed response. I appreciate the time it takes to do this. Here, I am commenting on the response to my question: Why is state space a closed subset of the ambient vector space? I do not understand what is meant by, "Well actually it isn't". For example, the Bloch...

Thanks for the detailed responses to my two questions. Here, I comment on the response to the one question Why is the probability of state transitions symmetric? i.e. P( S transitions to T) = P( T transitions to S ) I read through the discussions on POVM, Gleason's Th, QFT and What is a...

Thanks for all the interesting comments, and clarification of what can be discussed in this forum. My paper references, of course, Hardy's paper, another paper published in Studies in History and Philosophy of Modern Physics, and three books. I will take V50's suggestion and see if the journal...

The state spaces satisfy axioms very similar to the ones that Hardy proposes e,g, level 2 state spaces sit in R^4 and level 3 state spaces sit in R^9. If you postulate connectivity of state space, you get QP, as Hardy demonstrates. If you do not postulate connectivity, it turns out the level...

The paper is a contribution, but not "cutting edge", so I am somewhat in a quandry of where to submit the paper.

Lucien Hardy's Quantum Theory From Five Reasonable Axioms has deepened my understanding of QP foundations, and motivated me to write a paper. The essence of my paper is that "connectedness" of state space (or the acting Lie group), need not be assumed, but can be deduced. Before linking to the...

Time symmetry is an interesting perspective. The difficulty I have with this, though, is that the two directions of state transition require two different experiments. BTW, it is directly probability, as the p's add up to 1, owing to the states having trace 1.