It means that no actual experimental apparatus can report a result that is from a continuum of values. Observations must necessarily draw from a countable set of possible results.
In other words, intermediate (and initial) states may be represented as a continuous set, but the observable outcomes of a quantum event must come from a countable set of states.
I thought this chain of replies was in response to
Perhaps you were referring to something else, in which case we’re simply talking past each other and not disagreeing.
Did you see my post above? For a free particle, the range of possible energies or momenta is continuous. Of course, you can always only measure with finite accuracy (I mean, where would you write down an infinite accuracy measurement?), but the theoretically possible values form a continuum, in contrast to, say, harmonic oscillator expectation values. Besides, one could use the same argument to claim that there is only a finite amount of possible outcomes, since you will only ever make finitely many measurements.
What he said. Talk all you want about ‘many worlds’, but keep in mind, you’re not talking physics, you’re talking 3 AM dorm room pass-the-bong-dude speculations.
There’s no right answer.
Even with perfect accuracy, an actual apparatus can only report one of a countable set of values. Any “ruler” you use will necessarily be composed of discrete parts–atoms or counts of cycles, etc.
It’s a convenient approximation that a continuum of values can be measured, but the quantum nature of the real world prevents that. If the instrument is quantized in sufficiently small intervals, the approximation is a good one and we can ignore the quantum nature of our instrument. But don’t forget it’s only an approximation, in general we must include the quantum instrument in the experiment.
Despite the popularity of the Copenhagen interpretation, there are in fact no classical measuring devices. There are only quantum devices, although they may closely approximate classical ones.
A quantum event is an interaction between “particles” governed by the physics of quantum mechanics. As opposed to non-quantum physics, like gravitation and classical electromagnetics.
I am stating that the outcomes of quantum interactions are quantized. Giving an example like the X operator (measuring the position of a state) does not address that statement because that state need not have come from a quantum interaction.
This isn’t right, unless you assume space-time to be discrete. Generally, a free particle simply doesn’t have a discrete expectation value distribution, for instance. Its energy (momentum, position…) distribution thus gives a ruler that isn’t composed of discrete parts.
But this is only tangentially related to your original assertion, which was that an electron’s momentum could only assume one of countably many values. In fact, according to the formalism of quantum mechanics, a free electron’s momentum can take a continuum of values, whether you can measure that or not.
Pleonast, You seem to be meandering – none of the above defends the point I thought you were defending:
All quantum events are discrete questions, by definition. The election will have one of a finite number of states–not a continuum of possibilities.
The electron can have one of an infinite number of states, regardless of whether the measurement apparatus is quantized. You can have a ruler than measures in millimeter intervals, and you can have a ruler than measures in millimeters*1.0001 intervals, and so on. There is no physical boundary condition that requires all rulers to constrain the possible set of observed states to be finite.
I still don’t know what you mean by “Giving an example like the X operator (measuring the position of a state) does not address that statement because that state need not have come from a quantum interaction.” The state need not have come from a quantum interaction? Where did it come from then? What are you talking about?
It is an observable and it does take on conitnious values. I don’t know what you mean by representing a quantum interaction.
The only cavaet is that, for an operators representing quantum mechanical observables with continious eigenvalue spectra, wavefunctions representing exactly defined values for observables are not square-integrable, whereas quantum mechanics is usually taken as demanding that the wavefunction is square-integrable.
This is usually avoided by simply saying that where an observable has a continious set of values it cannot take an exact value, but this doesn’t make it’s values discrete.
I’m not assuming space-time is discrete. I don’t see how a free particle can be used as a ruler.
A free electron does not have definite energy until it is measured. Until that measurement, the energy (i.e., the distribution of values that the energy could be measured) is simply a bookkeeper’s note. For example, it’s possible for a particle to have negative energy in intermediate steps of an experiment. But the observed energy will never be negative because our instrument cannot measure that. Likewise, although a particle could have any of a continuum of values of energy, it’s observed energy will always be from a countable set of values, because that’s what our instrument will be able to measure.
When you make a measurement, you are using a particular ruler. I think the crux of our issue is that you want to separate the state of the particle from the instrument used to measure it. That is not possible, in general.
I’m at a loss to explain quantum interaction any better.
You seem to be operating under the assumption that the instrument itself is in a collapsed state. Generally speaking, both the measuring apparatus and the particle are continuous in some variables. Correlations between the apparatus and the particle may be such that possible measured values corresponding to each collapsed observer state take on discrete spectra, but those discrete spectra are from a dense set specified by the continuous probability distribution corresponding to the wave function.
tl;dr – orthodox QM: continuous wave functions are continuous; continuous probability distributions are continuous
An eigenvalue of an operator (representing a quantum mechanical observable) corresponds to the result of a measurement. The previous sentence has the status of a postulate in quantum mechanics.
Some operators representing quantum mechanical observables have continious eigenvalue spectra. This is simply because the possible results of measurements on that observable are assumed to be continious. The postion operator X of a free particle is the most obvious example of such an operator.
Countabilty is a bit of a red herring, but if you did try to restrict the allowed values of a measurement to a countable set of values I think you’d find the restriction is cumbersome and artificial.
Can you describe a human-readable instrument that, when measuring the position (or momentum or energy) of a particle, returns a value from a continuous set?
Completely irrelevant to what we were talking about, which was the set of electron states after scattering, not the set of electron states after scattering “for a given scattering angle of measurement”.
As far as I can tell, Pleonast must believe that rather than a continuous wave function existing independent of observer, that there are numerous pairs of {wf, observer } states, where each wf is non-continuous. In other words, if you approach a wave function with ruler A, it will get scared and be non-continuous in such a way as to correspond to the possible values ruler A can measure. And if you approach the wave function with another ruler B, it will get morph into another non-continuous distribution in such a way as to correspond to the possible values ruler B can measure. This is, of course, an absurd way of viewing things, and certainly not internally consistent. The wave function is continuous. Sample it as you please. The phase space is continuous.
This is not about the technicaltieis measuring instruments, can you provide me with any objections against such an instrument that wouldn’t equally apply to classicla physics?
In quantum physics we assume the possible results of such measuremnts are continious just as we assume so in classical physics.
Yes, the wavefunction is continuous. The set of observable values is not.
The wavefunction is for bookkeeping and exists only in the sense that measurements of it can be made. Measurements, because of the quantized nature of the universe, can only return results selected from a countable set of values.
Assuming measurements are continuous is simply an approximation. It is often a good approximation, but it does not represent the general case.
It’s not an approximation, it would be easier if we only had to deal with operators with discrete spectra, but unfortunately as I said any such restriction would artificial and cumbersome.