Recent content by tom.stoer

I agree, sorry for being imprecise (using rays the problem of equivalent vectors differing by a phase vanishes)

Agreed. I can reformulate my statement that two states are observably identical if there is no observable for which we expect different measurement results for the two states ##|a\rangle## and ##|b\rangle##. Theoretically we can chose the projectors ##E_\lambda## to one-dim. eigenspaces of...

So let's discuss observables. Two states are observably different if there is at least one observable represented by a self-adjoint operator ##\mathcal{O}## for which we expect different measurement results for the two states ##|a\rangle## and ##|b\rangle##.

yes, of course, that's what I wrote

You are absolutely right. Measurement as described in orthodox quantum mechanics with collapse violates unitarity. But Hawking radiation based on QFT plus classical GR violates unitarity entirely on the mathematical level w/o any additional consideration of the measurement problem. So whereas...

That is a valid position - but not the only one. Platonists think different. FAPP is of course a valid position, but does not comply with "always" or "universal". Why not take Everett's view as fundamental hypothesis, work on it and try to falsify it - either theoretically or experimentally?

I cannot say what "they" mean. For me there are basically two options. 1) If you look at the fundamental axioms of quantum mechanics in many cases "measurement" is mentioned but not defined, or it is defined w.r.t to some external observer. IMHO all these scenarios are dead ends when talking...

This is basically the unitary time evolution according to the Schrödinger equation. The cosine of the angle between two state vectors ##|a\rangle## and ##|b\rangle## is given by the scalar product $$\cos\phi_{ab} = \langle a|b\rangle$$ The Schrödinger equation says that a state vector...

The ground state ##|1s\rangle## and the excited state ##|2p\rangle## are orthogonal, i.e. ##\langle 1s|2p\rangle = 0##. In non-rel. qm they stay different for any finite time. But this scenario is a little bit obscure from a fundamental perspective b/c the quantum system is not isolated, one...

I fully agree. Starting with uniform non-zero charge ##\rho## yields a class of non-trivial solutions for ##E \neq 0##. Starting with maximum symmetry = homogeneity and isotropy yields ##\rho = 0## and ##E = 0##.

Of course one could follow a different approach by requiring maximal symmetry instead of specifying boundary conditions. [We do that in general relativity quite frequently. As an example: we do not require asymptotic flatness but homogenity plus isotropy, resulting in FRW cosmologies; the main...

The symmetry is encoded in ##\rho## and in the boundary conditions for ##E##. ##\rho## alone does not specify the symmetry completely. Look at the 1-dim toy model. They system and its symmetry can be specified by 1) ##\partial_x E = \rho## 2) ##E(0) = 0## This breaks translational invariance...

It depends on your starting point. I said "Given ..." so I started with non-zero charge density which was the original question; this rules out maximal symmetry. If you start with maximal symmetry then this rules out non-zero charge. Obviously not - only in terms of the charge, not in terms of...

##E=0## does not solve ##\nabla E = \rho = \text{const} \neq 0## The symmetry of the problem is not completely specified w/o boundary condition. Given ##\rho = \text{const} \neq 0## there are two choices: 1) a boundary condition compatible with ##\rho = \text{const} \neq 0##; then the allowed...

String theory certainly yes, LQG is not clear as far as I can see (they have results like a graviton propagator, but the classical limit is by no means fully understood).