Recent content by PeterDonis

How many degrees of freedom does a symmetric second rank tensor have? More than 4.

Yes, in a general curved spacetime that is correct.

So if scenarios like the one Frauchiger and Renner describe are possible, the trust you describe is not justified, because you cannot trust that the mean, nasty things have not happened. You cannot trust anything you think you know. You cannot trust that you can draw inferences from whatever...

Yes, for that particular case, all of the things you say are correct.

For that particular case, yes, you can define the "distance" between the worldlines so that it is constant, and so that the numerical value of the calculated two-way speed of light is ##c##.

That the round-trip light travel time between them is constant, as measured by an observer following each worldline.

Careful. The measurement involves only one clock, so the round-trip travel time of the light is invariant. But converting that to a speed requires knowing the distance that the light traveled, and depending on how you interpret "distance", that can depend on your choice of coordinates. Also...

But what is invariant is the light cones, not the numerical value of ##c##. One can always find a coordinate transformation that changes the coordinate speed of light. But one cannot find a coordinate transformation that changes the light cones.

But someone else could then come in and do a unitary operation on the friend that would undo the change that Wigner induced in the friend by doing this. And the friend would have no way of knowing that this had not happened. Or, for that matter, someone else could do a unitary operation on...

Your "bump" is being added to the Hamiltonian, not the wave function.

It seems like you actually might be thinking of it as a classical harmonic oscillator. They're not the same. Ok. This doesn't make sense if it's a quantum oscillator. The state is a function of position (the wave function), not a position. Perhaps you mean that the wave function, if it starts...

The bolded statement in the quote above should have been a huge red flag to you. The short answer is, no, the "explanation" you reference does not support the claim you are making. The fact that it seems to say the same thing as your claim is an illusion, a combination of poorly chosen phrasing...

The Heisenberg principle does not say this. Your arguments are based on false premises.

No, it's just describing what the basic math of QM does and does not tell us. Not "to interact", to be measured. "Measurement" is a narrower category than "interaction". There is also no need to assume that it is not. The basic math of QM tells us nothing either way. Nor is it the reason why...

If the particle's state is a wave packet in position space, then our knowledge of its position is only probabilistic: we can't say the particle has this position or that position, all we can give is the probability of it having position ##x##, as a function of ##x##. That doesn't mean the...