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I disagree with most of that. I disagree with the overall drift
and with many of the details.
Quantum mechanics is right, so far as we know. However, a greatLet us see it.
deal of what is /said/ about quantum mechanics is wrong.
(Citing from my previous statement)Actually, quantization is known already in CM, e.g., frequency
quantization of a finite string or elastic rod.
That seems kinda inconsistent with the previous warning aboutThere are no inconsistencies here. The frequency f of a vibrating string in the sufficiently high frequency range may be practically continuous observable in the sense that the ratio (delta f) / f = (f(n+1)-f(n)) / f(n) <<1 at n >>1. A measuring device designed for this region of spectrum may be totally insensitive to see delta f , so f here is continuous for all practical purposes. And yet it does not conflict with the fact that all actual set of harmonics is quantized.
arguments based on classical mechanics.
In fact if you look at the oscillations of a string, rod, organ pipe,This fact implicitly assumes dissipation, which was not assumed in the initial discussion. Dissipation is (almost!) always present, but it can be negligible in many situations. The line width is there in all Spectroscopy, and yet we distinguish between discrete and continuous spectra. The same is true about atomic energy spectra, which can be explained only by QM. But what is really important here, is the fact that even with the broad lines, the corresponding radiated field is quantized, with the only distinction that each respective photon is a non-monochromatic wave packet. So while we can, strictly speaking, dispute energy quantization in the narrow sense (as a discrete set) by saying that observed values are merely spectral maxima, it does not eliminate the reality of quanta. As JD himself admits below, the corresponding Hilbert space is is not restricted to the basis of pure energy eigenstates. And as of angular momentum, even mathematical purists cannot deny today its strict quantization as a discrete set.
et cetera, you find that the resonances are not particularly sharp.
There is a nontrivial linewidth.
In accordance with Floquet's theorem,
if you drive the thing at frequency f, it will respond at frequency f,
whether or not that's one of the resonant frequencies.
The same is true of atoms (whether or not you consider this analogous
to classical resonances). I've never seen an atom emit a photon
"instantly". I've never seen an atomic transition with zero linewidth.
There is not "overwhelming" evidence of this; indeed there is no
evidence at all.
An experiment optimized to look for "instant" or near-"instant" transitionsFalse. Time-energy relationship has not the same meaning as momentum-coordinate relationship, because time (even in SR!) is not the same as space. In (Delta E)(delta t) in the Heisenberg relationship delta t ) has the meaning of the average lifetime of the system for the given transition, not the transition time.
would be incompatible with assigning a definite energy to the photons.
Here "incompatible" has a precise technical meaning, as explained by
Heisenberg.
One of the most fundamental principles of quantum mechanics says thatTrue. But, as Landau used to say, "I can measure the energy and look at the watch." So one can know the energy at any moment. And for a stationary system, the energy can be known exactly at any moment.
it doesn't suffice to /talk/ about the energy. If you want to know
the energy, you have to measure it.
The same goes for other observablesTrue. So, the angular momentum is quantized!
such as the spin components Sx, Sy, and Sz.
*IF* you measure Sx, the result is quantized.
*IF* you measure Sy, the result is quantized.
*IF* you measure Sz, the result is quantized.
However, you can't measure all three of them at the same time. You can'tTrue.
measure any one of them without greatly disturbing the others. If you
measure Sx, there are some things you can say about Sy and Sz, but quite
a few things that you cannot say.
You ought no even imagine that theyFalse. If I have an electron with its spin up, I know that it is an equally-weighted superposition of eigenstates |Sx+> and |Sx->, or |Sy+> and |Sy->. The exact value of the respective net spin component in such superposition is indeterminate, but In no way does it undermine quantization of the corresponding eigenvalues.
are quantized.
We see this All The Time in atomic physics. An atomic electron hasFalse. An atomic electron with "a perfectly well defined position" is no longer atomic, since a perfectly defined position involves infinite energy due to QM indeterminacy which JD himself so vividly describes.
a perfectly well defined position.
You can measure the position ifTrue, but, using JD's own expression,"That seems kinda inconsistent with the previous warning..."
you want. However such a measurement is incompatible with measuring
the energy, incompatible with the spectroscopic N,l,m quantum numbers.
If you measure position you forfeit the option of measuring the energy
and vice versa.