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Atomic emissions are expected to be very exact quanta, based on the conditions. They are not subject to the odds of dice or cards.
In my opinion statistics is not consistent with quantum theory
I would guess
Quote from: puppypower on 31/10/2021 18:51:44Atomic emissions are expected to be very exact quanta, based on the conditions. They are not subject to the odds of dice or cards.Yes they are.It's just that you don't understand it.https://en.wikipedia.org/wiki/Spectral_line#Broadening_due_to_local_effectsQuote from: puppypower on 31/10/2021 18:51:44In my opinion statistics is not consistent with quantum theoryIn my observation, you understand neither.Quote from: puppypower on 31/10/2021 18:51:44I would guess Stop guessing: learn.
You still have not answered the topic question?
Stop guessing: learn.
A quantum universe is composed of distinct quanta with quantum gaps between. In that respect probability is either 1.0 or 0.0. These are two discontinuous places in statistics. This is why I was wondering if anyone can explain how statistical theory is consistent with quantum theory?
Quote from: puppypower on 31/10/2021 18:51:44A quantum universe is composed of distinct quanta with quantum gaps between. In that respect probability is either 1.0 or 0.0. These are two discontinuous places in statistics. This is why I was wondering if anyone can explain how statistical theory is consistent with quantum theory?Statistics is more complicated than just rolling dice, this could be the source of your confusion. Try looking at this wiki article https://en.wikipedia.org/wiki/Quantum_mechanics. Notice how probability crops up time and time again in the calculations? That is because statistics is an integral part of QM. It doesn't matter if you 'like' statistics or not, it is still part of QM.
There are a lot of smart people on these forums, but nobody can or will answer the question; how is statistics consistent with quantum theory.
Quote from: puppypower on 01/11/2021 11:01:33You still have not answered the topic question?If you ask me how the radiator of a car drives the wheels all I can say is that it doesn't.I can't say how it does it.I can't answer a question which makes no sense.The answer to "How are statistics consistent with a quantum universe?" is "Nicely, thanks."The problem isn't with how stats and QM mesh, but with your understanding of QM.I already gave you the best available answer to that.Quote from: Bored chemist on 31/10/2021 19:38:23Stop guessing: learn.
However, if statistical theory is not consistent with quantum theory, then this approach is not correct, even if long accepted tradition. This is why I ask, how is statistics consistent with quantum theory
where the quantum universe appears to exist.
Quote from: puppypower on 02/11/2021 14:33:46where the quantum universe appears to exist. It only appears that way to you.
You still have not answered the question; how is statistics consistent with quantum theory?
Your evasion tells us you cannot answer this question.
A quantum universe is not based on continuous functions
Stalling and waiting for me to move on
Probability was designed for continuous functions
You still have not answered the question; how is statistics consistent with quantum theory? Your evasion tells us you cannot answer this question.
This is not because you are not intelligent and well read,
Quote from: puppypower on 04/11/2021 11:06:20You still have not answered the question; how is statistics consistent with quantum theory? Your evasion tells us you cannot answer this question.No, you seem to be ignoring the answers.I assume when you say statistics you mean the following:This is from wiki on QM. https://en.wikipedia.org/wiki/Quantum_mechanicsApplying the Born rule to these amplitudes gives a probability density function for the position that the electron will be found to have when an experiment is performed to measure it. This is the best the theory can do; it cannot say for certain where the electron will be found. The Schrödinger equation relates the collection of probability amplitudes that pertain to one moment of time to the collection of probability amplitudes that pertain to another.After the measurement, if result was obtained, the quantum state is postulated to collapse to , in the non-degenerate case, or to , in the general case. The probabilistic nature of quantum mechanics thus stems from the act of measurement.Quote from: puppypower on 04/11/2021 11:06:20This is not because you are not intelligent and well read, You however do appear to be having this very problem! Why don't you read up on QM starting with the cited article and educate yourself.
The difference is always the same quanta of energy.
It's just that you don't understand it.https://en.wikipedia.org/wiki/Spectral_line#Broadening_due_to_local_effects
Quote from: puppypower on 05/11/2021 18:48:03The difference is always the same quanta of energy.THIS IS STILL WRONG.PLEASE READ WHAT YOU HAVE BEEN SHOWN.Quote from: Bored chemist on 31/10/2021 19:38:23It's just that you don't understand it.https://en.wikipedia.org/wiki/Spectral_line#Broadening_due_to_local_effects
this approach is not conceptually consistent.
Quote from: puppypower on 06/11/2021 13:22:49 this approach is not conceptually consistent. Because you have the wrong concepts.The energy released when an electron drops from an excited state to the ground state is not always exactly the same.This is not a measurement effect.We can measure the energy sufficiently accurately that we know that the variation is intrinsic to the transition.Not only that but we can predict it. All other things being equal, the energy uncertainty is proportional to the cube of the energy released.Please stop trying to ignore the uncertainty principle.
The probabilistic nature of quantum mechanics thus stems from the act of measurement.
Quote from: puppypower on 07/11/2021 22:52:10The probabilistic nature of quantum mechanics thus stems from the act of measurement.No, it does not.See if this helpshttps://www.thenakedscientists.com/forum/index.php?topic=83459.msg659621#msg659621
What advantage does this create?