Casimir Effect and Time-Energy form of the Uncertainty Principle

[Alan Grayson]

This issue is addressed primarily to Clark who applies the Time-Energy form of the Uncertainty Principle to establish the Casmir Effect using Quantum EM theory. Please define the operator T used in the Time-Energy form of the Uncertainty Principle. TY, AG

[Cosmin Visan]

Obey the religion!

[Alan Grayson]

On Saturday, May 3, 2025 at 1:17:39 PM UTC-6 Cosmin Visan wrote:

Obey the religion!

You're an incorrigible asshole who can't learn. I explained what science is and isn't, but you keep insisting on your BS. AG 

[Alan Grayson]

On Saturday, May 3, 2025 at 11:00:13 AM UTC-6 Alan Grayson wrote:

This issue is addressed primarily to Clark who applies the Time-Energy form of the Uncertainty Principle to establish the Casmir Effect using Quantum EM theory. Please define the operator T used in the Time-Energy form of the Uncertainty Principle. TY, AG

Clark; as you must know, unless you can justify using the Time-Energy form of the Uncertainty Principle in justifying the Casmir Effect, your claim is without merit. Time is not an operator, so what does the T mean in this form of the Uncertainty Principle? I anxiously await your reply in this matter. AG 

[Cosmin Visan]

Science clearly isnt "AI! AI! AI!". That is just religious exaltation.

[Alan Grayson]

On Sunday, May 4, 2025 at 12:54:58 AM UTC-6 Cosmin Visan wrote:

Science clearly isnt "AI! AI! AI!". That is just religious exaltation.

Cosmin; Why do you insist on being a prick who refuses to learn? If everything is conscious, or is a manifestation of consciousness, it's certainly a legitimate discussion whether AI is conscious, particularly if it passes the Turing Test (which it does).  Bringing religion into this discussion is misleading and obfuscating. Scientists make measurements but I think there's huge disagreement about what underlies these measurements. Does anyone really believe that the wave properties of particles -- which are idealizations of concentrations of "energy" -- imply for certain that some underlying "substance" is responsible for these phenomena? You make huge assumptions about what science and religions are, but they reflect nothing more than your ignorance of both points of view. I believe, based on your past behavior, that you will continue your ignorant ranting, without any comprehension how stupid you make yourself appear to those reading your nonsense. AG

JC; I await and would appreciate your response to whether Time is an operator, and can be used in the Time-Energy form of the Uncertainty Principle. Your answer to this question critically influences your claim that the Casimir Effect can be explained by applying Quantum EM theory. Also, why do you apparently accept the physical existence of virtual particles even though they contradict Conservation of Energy. Are you aware of this fact? TY, AG

[Cosmin Visan]

That's like saying "if I paint Niagara falls it will make me wet", because in my monkey brain "what I see is what I get".

[Cosmin Visan]

Also, AI passes Turing Test for retarded people, like you for example. It doesnt passes that test for me. Also, that test is like saying that a rock passes turing test for a worm (i.e. the worm believes the rock is alive), so the rock must be alive at least at the level of worm.

Retards gonna retard. What can you do ?

On Sunday, 4 May 2025 at 23:37:31 UTC+3 Alan Grayson wrote:

[John Clark]

On Sun, May 4, 2025 at 4:37 PM Alan Grayson <agrays...@gmail.com> wrote:

> JC; I await and would appreciate your response to whether Time is an operator, and can be used in the Time-Energy form of the Uncertainty Principle.

 

Time is not an operator because it is not an observable in the way that position, momentum and energy are, instead it's a parameter, the stage against which things happen. We can say that a particle has a position, a momentum and an energy but a particle doesn't have a time. Mathematically that means that in quantum mechanics  position, momentum and energy are all Hermitian operators but time is not. However it remains true that Δt × ΔE ≥ ℏ/2, and of course Δx × Δp ≥ ℏ/2

 > Your answer to this question critically influences your claim that the Casimir Effect can be explained by applying Quantum EM theory. 

It's not my claim, it's the claim of Hendrik Casimir, he used Quantum Electrodynamics to predict it in 1948, although it wasn't until 1997 that it was experimentally proven that Casimir's prediction was correct.  

 

 > Also, why do you apparently accept the physical existence of virtual particles 

 

I believe virtual particles exist for a number of reasons, the most important being experiments insist that they do. If you assume that virtual particles are real you are then able to make the most precise theoretical prediction not just in physics but in all of science, one part in 10 trillion. The theory of Quantum Electrodynamics, which is about virtual particles, predicted that the electron's magnetic moment should be approximately 0.001,159,652,181,643, the most accurate experimental measurement says it's approximately 0.001,159,652,180.

> even though they contradict Conservation of Energy. 

 In quantum mechanics you can borrow mass/energy from nowhere but you have to pay it back, and the more mass/energy you borrow the shorter amount of time you have before you have to pay it back; that's why you will never be able to directly observe a virtual particle, however you can observe effects caused by those virtual particles.  

 > Are you aware of this fact?

Of course I'm aware of it! In Quantum Mechanics the law of conservation of energy had to be modified. And it's not just Quantum Mechanics, in General Relativity the law of conservation of energy is true locally but not at a cosmic scale.  

John K Clark    See what's on my new list at  Extropolis

kfa

 

[Brent Meeker]

On 5/5/2025 5:36 AM, John Clark wrote:

Time is not an operator because it is not an observable in the way that position, momentum and energy are, instead it's a parameter, the stage against which things happen. We can say that a particle has a position, a momentum and an energy but a particle doesn't have a time. Mathematically that means that in quantum mechanics  position, momentum and energy are all Hermitian operators but time is not. However it remains true that Δt × ΔE ≥ ℏ/2, and of course Δx × Δp ≥ ℏ/2

It remains true if you correctly interpret Δt as the time for the expected value of the energy of the variable E to change by a standard deviation.  In other words by using whatever energy E refers to as a clock.  That's why it remains true even though there is no t operator in the sense of measuring a universal time.

Brent

[Alan Grayson]

On Monday, May 5, 2025 at 6:37:07 AM UTC-6 John Clark wrote:

On Sun, May 4, 2025 at 4:37 PM Alan Grayson <agrays...@gmail.com> wrote:

> JC; I await and would appreciate your response to whether Time is an operator, and can be used in the Time-Energy form of the Uncertainty Principle.

 

Time is not an operator because it is not an observable in the way that position, momentum and energy are, instead it's a parameter, the stage against which things happen. We can say that a particle has a position, a momentum and an energy but a particle doesn't have a time. Mathematically that means that in quantum mechanics  position, momentum and energy are all Hermitian operators but time is not. However it remains true that Δt × ΔE ≥ ℏ/2, and of course Δx × Δp ≥ ℏ/2

Your first equation is nonsense since Δt represents the statistical uncertainty of an operator, but t isn't an operator. Unless you can find a way around this problem, your proof, using Quantum EM is fatally flawed. Moreover, it has recently been shown that the Casimir Effect can be calculated independently of vacuum energy and virtual particles.

From Wiki: Relativistic van der Waals force[edit]

Alternatively, a 2005 paper by Robert Jaffe of MIT states that "Casimir effects can be formulated and Casimir forces can be computed without reference to zero-point energies. They are relativistic, quantum forces between charges and currents. The Casimir force (per unit area) between parallel plates vanishes as alpha, the fine structure constant, goes to zero, and the standard result, which appears to be independent of alpha, corresponds to the alpha approaching infinity limit", and that "The Casimir force is simply the (relativistic, retarded) van der Waals force between the metal plates."^[19] Casimir and Polder's original paper used this method to derive the Casimir–Polder force. In 1978, Schwinger, DeRadd, and Milton published a similar derivation for the Casimir effect between two parallel plates.^[24] More recently, Nikolic proved from first principles of quantum electrodynamics that the Casimir force does not originate from the vacuum energy of the electromagnetic field,^[25] and explained in simple terms why the fundamental microscopic origin of Casimir force lies in van der Waals forces.^[26]

AG

 > Your answer to this question critically influences your claim that the Casimir Effect can be explained by applying Quantum EM theory. 

It's not my claim, it's the claim of Hendrik Casimir, he used Quantum Electrodynamics to predict it in 1948, although it wasn't until 1997 that it was experimentally proven that Casimir's prediction was correct.  

As I read Wiki, Casimir didn't use QED to make his prediction. AG

 

 > Also, why do you apparently accept the physical existence of virtual particles 

 

I believe virtual particles exist for a number of reasons, the most important being experiments insist that they do. If you assume that virtual particles are real you are then able to make the most precise theoretical prediction not just in physics but in all of science, one part in 10 trillion. The theory of Quantum Electrodynamics, which is about virtual particles, predicted that the electron's magnetic moment should be approximately 0.001,159,652,181,643, the most accurate experimental measurement says it's approximately 0.001,159,652,180.

> even though they contradict Conservation of Energy. 

 In quantum mechanics you can borrow mass/energy from nowhere but you have to pay it back, and the more mass/energy you borrow the shorter amount of time you have before you have to pay it back; that's why you will never be able to directly observe a virtual particle, however you can observe effects caused by those virtual particles.  

I don't think this borrowing concept comes from the postulates of Quantum EM, but is an ad hoc add-on. AG 

[Brent Meeker]

On 5/5/2025 1:28 PM, John K Clark wrote:

 

I believe virtual particles exist for a number of reasons, the most important being experiments insist that they do. If you assume that virtual particles are real you are then able to make the most precise theoretical prediction not just in physics but in all of science, one part in 10 trillion. The theory of Quantum Electrodynamics, which is about virtual particles, predicted that the electron's magnetic moment should be approximately 0.001,159,652,181,643, the most accurate experimental measurement says it's approximately 0.001,159,652,180.

Virtual particles are just part of infinite series expansions of Green's functions in Feynman diagrams QED.  Your assertion is like saying some term in the Fourier expansion of a square wave exists.

Brent

[Alan Grayson]

If we're measuring the Casimir force between plates, how can that be a clock? TY, AG 

[Cosmin Visan]

You're just debating religions. Nothing of what you say is actually real.

[John Clark]

On Mon, May 5, 2025 at 4:28 PM Alan Grayson <agrays...@gmail.com> wrote:

 

>> Time is not an operator because it is not an observable in the way that position, momentum and energy are, instead it's a parameter, the stage against which things happen. We can say that a particle has a position, a momentum and an energy but a particle doesn't have a time. Mathematically that means that in quantum mechanics  position, momentum and energy are all Hermitian operators but time is not. However it remains true that Δt × ΔE ≥ ℏ/2, and of course Δx × Δp ≥ ℏ/2

> Your first equation is nonsense

No it is not. It expresses a limitation on how quickly a quantum system can evolve between distinguishable states.

>> It's not my claim, it's the claim of Hendrik Casimir, he used Quantum Electrodynamics to predict it in 1948, although it wasn't until 1997 that it was experimentally proven that Casimir's prediction was correct.  

As I read Wiki, Casimir didn't use QED to make his prediction. AG

Hendrik Casimir didn't mention the terms "zero point energy" or "virtual particle" in his original 1948 paper because back then quantum field theory wasn't mature and the terms hadn't been invented, but Casimir certainly used quantum electrodynamics. Your original claim was that Casimir didn't even need to use quantum mechanics and that classical physics was sufficient, and that is nonsense. You use R. L. Jaffe's 2005 paper to back up your claim:

The Casimir Effect and the Quantum Vacuum

However clear as day right there in the abstract Jaffe says Casimir effects "are Relativistic, Quantum forces between charges and currents". 

> it has recently been shown that the Casimir Effect can be calculated independently of vacuum energy and virtual particles.

In physics, as in mathematics, things can almost always be derived in more than just one way; and that is especially true if Quantum Mechanics is involved. 

John K Clark    See what's on my new list at  Extropolis

#$&

[John Clark]

On Mon, May 5, 2025 at 7:23 PM Brent Meeker <meeke...@gmail.com> wrote:

> Virtual particles are just part of infinite series expansions of Green's functions in Feynman diagrams QED.  Your assertion is like saying some term in the Fourier expansion of a square wave exists.

Yes but.... some terms in the Fourier expansion of a square wave DO exist. As I said in my previous post, there is almost always more than one way to derive something in physics or mathematics. 

John K Clark    See what's on my new list at  Extropolis

5r!

[Brent Meeker]

On 5/6/2025 4:57 AM, John Clark wrote:

On Mon, May 5, 2025 at 7:23 PM Brent Meeker <meeke...@gmail.com> wrote:

> Virtual particles are just part of infinite series expansions of Green's functions in Feynman diagrams QED.  Your assertion is like saying some term in the Fourier expansion of a square wave exists.

Yes but.... some terms in the Fourier expansion of a square wave DO exist. As I said in my previous post, there is almost always more than one way to derive something in physics or mathematics.

But those terms don't exist in the wavelet expansion of a square wave or many other decompositions of the square wave.  So they only "exist" for certain interpretations of "exist".

Brent

John K Clark    See what's on my new list at  Extropolis

5r!

On 5/5/2025 1:28 PM, John K Clark wrote:

 

I believe virtual particles exist for a number of reasons, the most important being experiments insist that they do. If you assume that virtual particles are real you are then able to make the most precise theoretical prediction not just in physics but in all of science, one part in 10 trillion. The theory of Quantum Electrodynamics, which is about virtual particles, predicted that the electron's magnetic moment should be approximately 0.001,159,652,181,643, the most accurate experimental measurement says it's approximately 0.001,159,652,180.

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[Alan Grayson]

On Tuesday, May 6, 2025 at 5:30:51 PM UTC-6 Brent Meeker wrote:

On 5/6/2025 4:57 AM, John Clark wrote:

On Mon, May 5, 2025 at 7:23 PM Brent Meeker <meeke...@gmail.com> wrote:

> Virtual particles are just part of infinite series expansions of Green's functions in Feynman diagrams QED.  Your assertion is like saying some term in the Fourier expansion of a square wave exists.

Yes but.... some terms in the Fourier expansion of a square wave DO exist. As I said in my previous post, there is almost always more than one way to derive something in physics or mathematics.

But those terms don't exist in the wavelet expansion of a square wave or many other decompositions of the square wave.  So they only "exist" for certain interpretations of "exist".

Brent

Perfect! I think you nailed it! AG 

[Alan Grayson]

On Tuesday, May 6, 2025 at 5:30:51 PM UTC-6 Brent Meeker wrote:

On 5/6/2025 4:57 AM, John Clark wrote:

On Mon, May 5, 2025 at 7:23 PM Brent Meeker <meeke...@gmail.com> wrote:

> Virtual particles are just part of infinite series expansions of Green's functions in Feynman diagrams QED.  Your assertion is like saying some term in the Fourier expansion of a square wave exists.

Yes but.... some terms in the Fourier expansion of a square wave DO exist. As I said in my previous post, there is almost always more than one way to derive something in physics or mathematics.

But those terms don't exist in the wavelet expansion of a square wave or many other decompositions of the square wave.  So they only "exist" for certain interpretations of "exist".

Brent

Maybe someone can explain this; if, say, the momentum operator always returns an eigenvalue of the momentum of the system being measured, then when used in the UP, how can there be an uncertainty in momentum to give a statistical variance? TY, AG

[Brent Meeker]

On 5/6/2025 7:47 PM, Alan Grayson wrote:

Maybe someone can explain this; if, say, the momentum operator always returns an eigenvalue of the momentum of the system being measured, then when used in the UP, how can there be an uncertainty in momentum to give a statistical variance? TY, AG

You misunderstand what the HUP refers to.   There is often confusion between preparing a system in state, which is limited by the uncertainty principle, and making a destructive measurement on the system which can be more precise (but not more accurate) that the uncertainty principle.  In the literature the former, a preparation, is referred to as an ideal measurement…but then the “ideal” gets dropped and people assume that it applies to any measurement.  Then there’s a confusion between precision of single measurements and the scatter of measurements of the same system state. 

Heisenberg’s uncertainty  principle is commonly misinterpreted as saying you cannot make a precise (i.e. to arbitrarily many decimal places) measurement of both the x-axis momentum and the position along the x-axis at the same time or on the same particle.  This is untrue. It comes from confusing the concept of preparing particles in a state and measuring the state of a particle.  Heisenberg actually contributed to this confusion with his microscope thought experiment

The theory only says you cannot prepare a particle so that it has precise values of both momentum and position.  The distinction is that you can measure both x and p and get precise values, but when you repeat the process with exactly the same preparation of the particle the measurement will yield different values.  So even though you measure precise values there is no sense in which you can say the particle had those values independent of the measurement.  Your measurement has been precise, but not accurate.  And if you repeat the experiment many times, the scatter in the measured values will satisfy the uncertainty principle.  See Ballantine, “Quantum Mechanics, A Modern Development” pp 225–227 for more complete exposition.

To illustrate, you can prepare particles so that their position has only a small uncertainty and when you measure their position and momentum you will get a scatter plot like the blue points below.



 Each point is a precise measurement of both momentum, p, and position, q.  But because the scatter in position is small the scatter in momentum will be big - and the uncertainty principle will the satisfied.  The green points illustrate the complementary case in which the particles have a small scatter in momentum, but a big scatter in position.  The red points illustrate an intermediate case, a preparation in which the momentum and position scatters are similar.

It might also mention that in practice, i.e. in the laboratory and in colliders like the LHC, the error scatter due to instrument uncertainties is usually much bigger than that due to the theoretical limit of the Heisenberg uncertainty,  So both position and momentum are measured and the uncertainty in their value arises from instrument limitations, not from QM

Brent

[Cosmin Visan]

Talking theory on a forum is the most wastest of time ever. And the reason is trivial: you have no idea if it works in practice. Professional science is not meant as a revealer of truth, it is only meant as an empirical enterprise. It's validity goes only as far as experiments go. So if you are really obsessed about these things, go and do experiments. You will see instantly how your whole debate collapses instantly, because in practice F=ma will give you random results. For example, the mass that you will measure will be 1.525123541325312, the acceleration will be 52.45662436. And when you will multiply them, you will get 5345.2351235125132, which will contradict the theory by 0.0001%. The question is: what are you going to do ? Are you going to repeat the experiment at infinity in order to obtain the perfect result ? Or are you going to change at infinity the theory in order to match the result ? As such, you will realize that F=ma is just an empirical approximation. It says nothing about reality. As such, any theoretical debate on a forum is absurd. The point of science is to be a practical activity to produce technology. If you want theoretical debates, talk about consciousness, because only that is what exists and you have empirical access to it at all times, so any theoretical debates on a forum can be instantly compared to the empirical experience of consciousness and decide on the spot if they are good or not.

[John Clark]

On Tue, May 6, 2025 at 7:30 PM Brent Meeker <meeke...@gmail.com> wrote:

> So they only "exist" for certain interpretations of "exist"

Yes but you could say exactly the same thing about just about ANYTHING. For example, if the Many Worlds idea is correct, and I think it probably (but not certainly) is, then in the quantum realm the entire idea of a "particle", virtual or otherwise, is just an imperfect analogy. But if our monkey brains are going to understand what's going on to the greatest extent they are capable of then analogies are important, even if they're imperfect.     

John K Clark    See what's on my new list at  Extropolis

uhv

[Cosmin Visan]

You people are so confused. There is only 1 meaning of "to exist", namely "to have quality"/"to have form". As such, only qualia exist.

[Brent Meeker]

On 5/7/2025 3:28 AM, John Clark wrote:

On Tue, May 6, 2025 at 7:30 PM Brent Meeker <meeke...@gmail.com> wrote:

> So they only "exist" for certain interpretations of "exist"

Yes but you could say exactly the same thing about just about ANYTHING. For example, if the Many Worlds idea is correct,

Do you think "correct" and "exists" mean the same thing?  Have you still not looked into Jacob Barandes formulation of QM? 

Brent

and I think it probably (but not certainly) is, then in the quantum realm the entire idea of a "particle", virtual or otherwise, is just an imperfect analogy. But if our monkey brains are going to understand what's going on to the greatest extent they are capable of then analogies are important, even if they're imperfect.     

John K Clark    See what's on my new list at  Extropolis

uhv

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[spudb...@aol.com]

You have to have a foundational theory on how things are? Theories require some evidence, yes? So where does you hypothesis come from? 

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[Alan Grayson]

On Wednesday, May 7, 2025 at 3:06:18 PM UTC-6 spudb...@aol.com wrote:

You have to have a foundational theory on how things are? Theories require some evidence, yes? So where does you hypothesis come from? 

No. As Feynman asserts, theories are mostly guesses about how nature behave, which are verified, or not, by experiments. AG 

[Alan Grayson]

So the scatter does not arise from QM, and yet, as I recall, that when QM is developed axiomatically, the UP follows. I'm not sure I get it. AG 

Brent

[Brent Meeker]

Where did I say that??  It arises because in QM you cannot prepare a particle to have zero scatter in both position and the conjugate momentum.  But that doesn't mean you can't measure them both precisely.  The scatter is an ensemble property.

Brent

and yet, as I recall, that when QM is developed axiomatically, the UP follows. I'm not sure I get it. AG 

Brent

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[Alan Grayson]

On Wednesday, May 7, 2025 at 6:58:20 PM UTC-6 Brent Meeker wrote:

On 5/7/2025 5:46 PM, Alan Grayson wrote:

On Tuesday, May 6, 2025 at 9:53:17 PM UTC-6 Brent Meeker wrote:

On 5/6/2025 7:47 PM, Alan Grayson wrote:

Maybe someone can explain this; if, say, the momentum operator always returns an eigenvalue of the momentum of the system being measured, then when used in the UP, how can there be an uncertainty in momentum to give a statistical variance? TY, AG

You misunderstand what the HUP refers to.   There is often confusion between preparing a system in state, which is limited by the uncertainty principle, and making a destructive measurement on the system which can be more precise (but not more accurate) that the uncertainty principle.  In the literature the former, a preparation, is referred to as an ideal measurement…but then the “ideal” gets dropped and people assume that it applies to any measurement.  Then there’s a confusion between precision of single measurements and the scatter of measurements of the same system state. 

Heisenberg’s uncertainty  principle is commonly misinterpreted as saying you cannot make a precise (i.e. to arbitrarily many decimal places) measurement of both the x-axis momentum and the position along the x-axis at the same time or on the same particle.  This is untrue. It comes from confusing the concept of preparing particles in a state and measuring the state of a particle.  Heisenberg actually contributed to this confusion with his microscope thought experiment

The theory only says you cannot prepare a particle so that it has precise values of both momentum and position.  The distinction is that you can measure both x and p and get precise values, but when you repeat the process with exactly the same preparation of the particle the measurement will yield different values.  So even though you measure precise values there is no sense in which you can say the particle had those values independent of the measurement.  Your measurement has been precise, but not accurate.  And if you repeat the experiment many times, the scatter in the measured values will satisfy the uncertainty principle.  See Ballantine, “Quantum Mechanics, A Modern Development” pp 225–227 for more complete exposition.

To illustrate, you can prepare particles so that their position has only a small uncertainty and when you measure their position and momentum you will get a scatter plot like the blue points below.



 Each point is a precise measurement of both momentum, p, and position, q.  But because the scatter in position is small the scatter in momentum will be big - and the uncertainty principle will the satisfied.  The green points illustrate the complementary case in which the particles have a small scatter in momentum, but a big scatter in position.  The red points illustrate an intermediate case, a preparation in which the momentum and position scatters are similar.

It might also mention that in practice, i.e. in the laboratory and in colliders like the LHC, the error scatter due to instrument uncertainties is usually much bigger than that due to the theoretical limit of the Heisenberg uncertainty,  So both position and momentum are measured and the uncertainty in their value arises from instrument limitations, not from QM

So the scatter does not arise from QM,

Where did I say that?? 

You wrote, "So both position and momentum are measured and the uncertainty in their value arises from instrument limitations, not from QM"  AG

 

It arises because in QM you cannot prepare a particle to have zero scatter in both position and the conjugate momentum.  But that doesn't mean you can't measure them both precisely.  The scatter is an ensemble property.

Do you mean, the scatter of each observable is caused by the unavoidable variations in how the observables are prepared, and then, by applying QM, one gets the UP? So what has this to do with "instrument limitations"? AG 

[John Clark]

On Wed, May 7, 2025 at 8:58 PM Brent Meeker <meeke...@gmail.com> wrote:

> that doesn't mean you can't measure them both precisely

But thanks to Heisenberg's uncertainty principle you can't predict what that measurement will be, and even if you repeat conditions exactly you will not get the same measurements for momentum and position (or energy and time) if you perform the experiment again. 

John K Clark    See what's on my new list at  Extropolis

ujq

[Cosmin Visan]

@John. If you focus your attention you can force the wavefunction to collapse as per your attention. Dean Radin showed this.

https://www.youtube.com/watch?v=nRSBaq3vAeY

[spudb...@aol.com]

I am being verbal here. In other words how the Universe ends may be by fire, ice, or rip. Evidence? Does the math for one's hypothesis have any evidence? Obeservation? Mass, Dark matter, etc. I am content with letting astronomers and physicists do the cosmological work, and put forth ideas. 

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[John Clark]

On Wed, May 7, 2025 at 4:04 PM Brent Meeker <meeke...@gmail.com> wrote:

  > Have you still not looked into Jacob Barandes formulation of QM? 

Barandes says when a particle hits a screen it shows up add a specific spot, and I agree with that, but then Barandes decrees the detection of that spot is the only thing that is "real"; and if you detect a particle at position X and a bit later you detect the same particle at position Y then those are the only positions that are "real", so you're not allowed to think about where the particle was or what it was doing between X and Y. It seems to be all he's done is redefined the word "real", and he's coming perilously close to "Shut Up And Calculate".  

Barandes says that at the quantum level processes are fundamentally random. Well maybe, they certainly seem random to us but I'm not so sure they're fundamentally random, if Many Worlds is correct then they are not. But then, in what seems to me to be a contradiction of what he just said, he claims that quantum processes are non-Markovian, behavior doesn't just depend on the current situation, particles have a memory of past events. Barandes doesn't even attempt to explain where or how that information is encoded.  

Barandes's idea does not produce any new predictions, he says his goal is to make the quantum world more intuitive and less philosophically perplexing. In that Barandes has certainly failed, at least in my case. 

 John K Clark    See what's on my new list at  Extropolis

xrx

[Brent Meeker]

On 5/7/2025 9:03 PM, Alan Grayson wrote:

On Wednesday, May 7, 2025 at 6:58:20 PM UTC-6 Brent Meeker wrote:

On 5/7/2025 5:46 PM, Alan Grayson wrote:

On Tuesday, May 6, 2025 at 9:53:17 PM UTC-6 Brent Meeker wrote:

On 5/6/2025 7:47 PM, Alan Grayson wrote:

Maybe someone can explain this; if, say, the momentum operator always returns an eigenvalue of the momentum of the system being measured, then when used in the UP, how can there be an uncertainty in momentum to give a statistical variance? TY, AG

You misunderstand what the HUP refers to.   There is often confusion between preparing a system in state, which is limited by the uncertainty principle, and making a destructive measurement on the system which can be more precise (but not more accurate) that the uncertainty principle.  In the literature the former, a preparation, is referred to as an ideal measurement…but then the “ideal” gets dropped and people assume that it applies to any measurement.  Then there’s a confusion between precision of single measurements and the scatter of measurements of the same system state. 

Heisenberg’s uncertainty  principle is commonly misinterpreted as saying you cannot make a precise (i.e. to arbitrarily many decimal places) measurement of both the x-axis momentum and the position along the x-axis at the same time or on the same particle.  This is untrue. It comes from confusing the concept of preparing particles in a state and measuring the state of a particle.  Heisenberg actually contributed to this confusion with his microscope thought experiment

The theory only says you cannot prepare a particle so that it has precise values of both momentum and position.  The distinction is that you can measure both x and p and get precise values, but when you repeat the process with exactly the same preparation of the particle the measurement will yield different values.  So even though you measure precise values there is no sense in which you can say the particle had those values independent of the measurement.  Your measurement has been precise, but not accurate.  And if you repeat the experiment many times, the scatter in the measured values will satisfy the uncertainty principle.  See Ballantine, “Quantum Mechanics, A Modern Development” pp 225–227 for more complete exposition.

To illustrate, you can prepare particles so that their position has only a small uncertainty and when you measure their position and momentum you will get a scatter plot like the blue points below.



 Each point is a precise measurement of both momentum, p, and position, q.  But because the scatter in position is small the scatter in momentum will be big - and the uncertainty principle will the satisfied.  The green points illustrate the complementary case in which the particles have a small scatter in momentum, but a big scatter in position.  The red points illustrate an intermediate case, a preparation in which the momentum and position scatters are similar.

It might also mention that in practice, i.e. in the laboratory and in colliders like the LHC, the error scatter due to instrument uncertainties is usually much bigger than that due to the theoretical limit of the Heisenberg uncertainty,  So both position and momentum are measured and the uncertainty in their value arises from instrument limitations, not from QM

So the scatter does not arise from QM,

Where did I say that?? 

You wrote, "So both position and momentum are measured and the uncertainty in their value arises from instrument limitations, not from QM"  AG

You left out the preceding sentence that conditional the "So...".

 

It arises because in QM you cannot prepare a particle to have zero scatter in both position and the conjugate momentum.  But that doesn't mean you can't measure them both precisely.  The scatter is an ensemble property.

Do you mean, the scatter of each observable is caused by the unavoidable variations in how the observables are prepared, and then, by applying QM, one gets the UP?

Yes.  The measurements the UP applies to are ideal measurements that leave the system in the measured state, i.e. prepare it.

So what has this to do with "instrument limitations"? AG

It doesn't.  That's why instrument limitations are noted separately as swamping the UP in most applications.

Brent

 
Brent

and yet, as I recall, that when QM is developed axiomatically, the UP follows. I'm not sure I get it. AG 

Brent

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[Brent Meeker]

On 5/8/2025 3:58 AM, John Clark wrote:

On Wed, May 7, 2025 at 8:58 PM Brent Meeker <meeke...@gmail.com> wrote:

> that doesn't mean you can't measure them both precisely

But thanks to Heisenberg's uncertainty principle you can't predict what that measurement will be, and even if you repeat conditions exactly you will not get the same measurements for momentum and position (or energy and time) if you perform the experiment again.

Yes, exactly.  But my point is that the fact tha you can measure both precisely for the same particle is commonly denied.  The problem in is that the HUP applies to an ideal measurement, one that leaves the system having the measured value, in other words a preparation.

Brent

John K Clark    See what's on my new list at  Extropolis

ujq

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[Alan Grayson]

On Thursday, May 8, 2025 at 5:01:00 PM UTC-6 Brent Meeker wrote:

On 5/7/2025 9:03 PM, Alan Grayson wrote:

On Wednesday, May 7, 2025 at 6:58:20 PM UTC-6 Brent Meeker wrote:

On 5/7/2025 5:46 PM, Alan Grayson wrote:

On Tuesday, May 6, 2025 at 9:53:17 PM UTC-6 Brent Meeker wrote:

On 5/6/2025 7:47 PM, Alan Grayson wrote:

Maybe someone can explain this; if, say, the momentum operator always returns an eigenvalue of the momentum of the system being measured, then when used in the UP, how can there be an uncertainty in momentum to give a statistical variance? TY, AG

You misunderstand what the HUP refers to.   There is often confusion between preparing a system in state, which is limited by the uncertainty principle, and making a destructive measurement on the system which can be more precise (but not more accurate) that the uncertainty principle.  In the literature the former, a preparation, is referred to as an ideal measurement…but then the “ideal” gets dropped and people assume that it applies to any measurement.  Then there’s a confusion between precision of single measurements and the scatter of measurements of the same system state. 

Heisenberg’s uncertainty  principle is commonly misinterpreted as saying you cannot make a precise (i.e. to arbitrarily many decimal places) measurement of both the x-axis momentum and the position along the x-axis at the same time or on the same particle.  This is untrue. It comes from confusing the concept of preparing particles in a state and measuring the state of a particle.  Heisenberg actually contributed to this confusion with his microscope thought experiment

The theory only says you cannot prepare a particle so that it has precise values of both momentum and position.  The distinction is that you can measure both x and p and get precise values, but when you repeat the process with exactly the same preparation of the particle the measurement will yield different values.  So even though you measure precise values there is no sense in which you can say the particle had those values independent of the measurement.  Your measurement has been precise, but not accurate.  And if you repeat the experiment many times, the scatter in the measured values will satisfy the uncertainty principle.  See Ballantine, “Quantum Mechanics, A Modern Development” pp 225–227 for more complete exposition.

To illustrate, you can prepare particles so that their position has only a small uncertainty and when you measure their position and momentum you will get a scatter plot like the blue points below.



 Each point is a precise measurement of both momentum, p, and position, q.  But because the scatter in position is small the scatter in momentum will be big - and the uncertainty principle will the satisfied.  The green points illustrate the complementary case in which the particles have a small scatter in momentum, but a big scatter in position.  The red points illustrate an intermediate case, a preparation in which the momentum and position scatters are similar.

It might also mention that in practice, i.e. in the laboratory and in colliders like the LHC, the error scatter due to instrument uncertainties is usually much bigger than that due to the theoretical limit of the Heisenberg uncertainty,  So both position and momentum are measured and the uncertainty in their value arises from instrument limitations, not from QM

So the scatter does not arise from QM,

Where did I say that?? 

You wrote, "So both position and momentum are measured and the uncertainty in their value arises from instrument limitations, not from QM"  AG

You left out the preceding sentence that conditional the "So...".

 

It arises because in QM you cannot prepare a particle to have zero scatter in both position and the conjugate momentum.  But that doesn't mean you can't measure them both precisely.  The scatter is an ensemble property.

Do you mean, the scatter of each observable is caused by the unavoidable variations in how the observables are prepared, and then, by applying QM, one gets the UP?

Yes.  The measurements the UP applies to are ideal measurements that leave the system in the measured state, i.e. prepare it.

So what has this to do with "instrument limitations"? AG

It doesn't.  That's why instrument limitations are noted separately as swamping the UP in most applications.

Brent

I can see that the measurement spreads due to instrument limitations are usually immensely larger than the much smaller spreads accounted for by the UP, but what causes these much smaller spreads? Is this a quantum effect? AG 

[Brent Meeker]

On 5/9/2025 7:08 PM, Alan Grayson wrote:

I can see that the measurement spreads due to instrument limitations are usually immensely larger than the much smaller spreads accounted for by the UP, but what causes these much smaller spreads? Is this a quantum effect? AG 

Yes.  Quantum evolution is unitary, i.e. the state vector just rotates in a complex Hilbert space so that probability is preserved.  Consequently the infinitesimal time translation operator is U=1+e6/6t or in common notation 1-i(e/h)H where H=ih6/6t and h is just conversion factor because we measure energy in different units than inverse time. It's not mathematics, but an empirical fact that h is a universal constant.

Brent 

[Alan Grayson]

If one wants to prepare a system in some momentum state to be measured, doesn't this imply a pre-measurement measurement, and the observable to be measured remains in that state on subsequent measurements? If so, how can the unitary operator, which just changes the state of the system's wf, create the quantum spread? TY, AG 

[Brent Meeker]

On 5/12/2025 1:58 PM, Alan Grayson wrote:

On Friday, May 9, 2025 at 10:40:42 PM UTC-6 Brent Meeker wrote:

On 5/9/2025 7:08 PM, Alan Grayson wrote:

I can see that the measurement spreads due to instrument limitations are usually immensely larger than the much smaller spreads accounted for by the UP, but what causes these much smaller spreads? Is this a quantum effect? AG 

Yes.  Quantum evolution is unitary, i.e. the state vector just rotates in a complex Hilbert space so that probability is preserved.  Consequently the infinitesimal time translation operator is U=1+e6/6t or in common notation 1-i(e/h)H where H=ih6/6t and h is just conversion factor because we measure energy in different units than inverse time. It's not mathematics, but an empirical fact that h is a universal constant.

Brent 

If one wants to prepare a system in some momentum state to be measured, doesn't this imply a pre-measurement measurement,

Right, given that it's an ideal measurement.  Most measurements don't leave the system in the eigenstate that is the measurement result.  An ideal measurement is one that leaves the system in the state that the measurement yielded.

and the observable to be measured remains in that state on subsequent measurements?

Only if they're ideal measurements of that same variable or of other variables that commute with it.

If so, how can the unitary operator, which just changes the state of the system's wf, create the quantum spread?

You don't need a change in the wf to "create the quantum spread".  Having prepared in an eigenstate of A just measure some other variable B that doesn't commute with A.  In general A will be a superposition of other variables, say A=xC+yD; that's just a change of coordinates.  But the system is not in an eigenstate of C or D.

Brent

TY, AG 

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[Alan Grayson]

On Monday, May 12, 2025 at 4:15:52 PM UTC-6 Brent Meeker wrote:

On 5/12/2025 1:58 PM, Alan Grayson wrote:

On Friday, May 9, 2025 at 10:40:42 PM UTC-6 Brent Meeker wrote:

On 5/9/2025 7:08 PM, Alan Grayson wrote:

I can see that the measurement spreads due to instrument limitations are usually immensely larger than the much smaller spreads accounted for by the UP, but what causes these much smaller spreads? Is this a quantum effect? AG 

Yes.  Quantum evolution is unitary, i.e. the state vector just rotates in a complex Hilbert space so that probability is preserved.  Consequently the infinitesimal time translation operator is U=1+e6/6t or in common notation 1-i(e/h)H where H=ih6/6t and h is just conversion factor because we measure energy in different units than inverse time. It's not mathematics, but an empirical fact that h is a universal constant.

Brent 

If one wants to prepare a system in some momentum state to be measured, doesn't this imply a pre-measurement measurement,

Right, given that it's an ideal measurement.  Most measurements don't leave the system in the eigenstate that is the measurement result.  An ideal measurement is one that leaves the system in the state that the measurement yielded.

and the observable to be measured remains in that state on subsequent measurements?

Only if they're ideal measurements of that same variable or of other variables that commute with it.

If so, how can the unitary operator, which just changes the state of the system's wf, create the quantum spread?

You don't need a change in the wf to "create the quantum spread".  Having prepared in an eigenstate of A just measure some other variable B that doesn't commute with A.  In general A will be a superposition of other variables, say A=xC+yD; that's just a change of coordinates.  But the system is not in an eigenstate of C or D.

Brent

Sorry, I really don't get it. Not at all! If we want to prepare a particle with some momentum p, why would we measure it with some non-commuting operator, and why would this, if done repeatedly, result in a spread of momentum? And what has this to do with a unitary operator which advances time? TY, AG 

[Alan Grayson]

Is the spread in momentum caused by an imprecision in preparing a particle in some particular momentum? Generally speaking, how is that done? TY, AG 

[Brent Meeker]

The HUP doesn't limit how precisely you can prepare a particle's momentum.  The HUP just says that the more precisely the momentum is determined the less precisely defined will be the conjugate position.  Think of a searchlight.  How sharply you can focus the beam isn't limited in principle, it just depends on how big you make the reflector.  So the lateral spread of the beam is "caused" by the finite size, if you want to call that an "imprecision". 

I don't think there is any "generally speaking way it's done".  Generally experiments do not attempt to reach the HUP limit.  For example in the LHC  they're aiming for the highest possible energy.  The accelerating frequency causes the groups of protons to have a certain length and hence a certain scatter in energy, which is small but not necessarily at the HUP limit.

Brent

[Alan Grayson]

On Sunday, May 18, 2025 at 4:16:26 PM UTC-6 Brent Meeker wrote:

On 5/18/2025 10:02 AM, Alan Grayson wrote:

On Tuesday, May 13, 2025 at 4:54:55 AM UTC-6 Alan Grayson wrote:

On Monday, May 12, 2025 at 4:15:52 PM UTC-6 Brent Meeker wrote:

On 5/12/2025 1:58 PM, Alan Grayson wrote:

On Friday, May 9, 2025 at 10:40:42 PM UTC-6 Brent Meeker wrote:

On 5/9/2025 7:08 PM, Alan Grayson wrote:

I can see that the measurement spreads due to instrument limitations are usually immensely larger than the much smaller spreads accounted for by the UP, but what causes these much smaller spreads? Is this a quantum effect? AG 

Yes.  Quantum evolution is unitary, i.e. the state vector just rotates in a complex Hilbert space so that probability is preserved.  Consequently the infinitesimal time translation operator is U=1+e6/6t or in common notation 1-i(e/h)H where H=ih6/6t and h is just conversion factor because we measure energy in different units than inverse time. It's not mathematics, but an empirical fact that h is a universal constant.

Brent 

If one wants to prepare a system in some momentum state to be measured, doesn't this imply a pre-measurement measurement,

Right, given that it's an ideal measurement.  Most measurements don't leave the system in the eigenstate that is the measurement result.  An ideal measurement is one that leaves the system in the state that the measurement yielded.

and the observable to be measured remains in that state on subsequent measurements?

Only if they're ideal measurements of that same variable or of other variables that commute with it.

If so, how can the unitary operator, which just changes the state of the system's wf, create the quantum spread?

You don't need a change in the wf to "create the quantum spread".  Having prepared in an eigenstate of A just measure some other variable B that doesn't commute with A.  In general A will be a superposition of other variables, say A=xC+yD; that's just a change of coordinates.  But the system is not in an eigenstate of C or D.

Brent

Sorry, I really don't get it. Not at all! If we want to prepare a particle with some momentum p, why would we measure it with some non-commuting operator, and why would this, if done repeatedly, result in a spread of momentum? And what has this to do with a unitary operator which advances time? TY, AG 

Is the spread in momentum caused by an imprecision in preparing a particle in some particular momentum? Generally speaking, how is that done? TY, AG

The HUP doesn't limit how precisely you can prepare a particle's momentum.  The HUP just says that the more precisely the momentum is determined the less precisely defined will be the conjugate position. 

I know. What I don't know is the cause of the spread. AG

[Brent Meeker]

On 5/18/2025 9:58 PM, Alan Grayson wrote:

On Sunday, May 18, 2025 at 4:16:26 PM UTC-6 Brent Meeker wrote:

On 5/18/2025 10:02 AM, Alan Grayson wrote:

On Tuesday, May 13, 2025 at 4:54:55 AM UTC-6 Alan Grayson wrote:

On Monday, May 12, 2025 at 4:15:52 PM UTC-6 Brent Meeker wrote:

On 5/12/2025 1:58 PM, Alan Grayson wrote:

On Friday, May 9, 2025 at 10:40:42 PM UTC-6 Brent Meeker wrote:

On 5/9/2025 7:08 PM, Alan Grayson wrote:

I can see that the measurement spreads due to instrument limitations are usually immensely larger than the much smaller spreads accounted for by the UP, but what causes these much smaller spreads? Is this a quantum effect? AG 

Yes.  Quantum evolution is unitary, i.e. the state vector just rotates in a complex Hilbert space so that probability is preserved.  Consequently the infinitesimal time translation operator is U=1+e6/6t or in common notation 1-i(e/h)H where H=ih6/6t and h is just conversion factor because we measure energy in different units than inverse time. It's not mathematics, but an empirical fact that h is a universal constant.

Brent 

If one wants to prepare a system in some momentum state to be measured, doesn't this imply a pre-measurement measurement,

Right, given that it's an ideal measurement.  Most measurements don't leave the system in the eigenstate that is the measurement result.  An ideal measurement is one that leaves the system in the state that the measurement yielded.

and the observable to be measured remains in that state on subsequent measurements?

Only if they're ideal measurements of that same variable or of other variables that commute with it.

If so, how can the unitary operator, which just changes the state of the system's wf, create the quantum spread?

You don't need a change in the wf to "create the quantum spread".  Having prepared in an eigenstate of A just measure some other variable B that doesn't commute with A.  In general A will be a superposition of other variables, say A=xC+yD; that's just a change of coordinates.  But the system is not in an eigenstate of C or D.

Brent

Sorry, I really don't get it. Not at all! If we want to prepare a particle with some momentum p, why would we measure it with some non-commuting operator, and why would this, if done repeatedly, result in a spread of momentum? And what has this to do with a unitary operator which advances time? TY, AG 

Is the spread in momentum caused by an imprecision in preparing a particle in some particular momentum? Generally speaking, how is that done? TY, AG

The HUP doesn't limit how precisely you can prepare a particle's momentum.  The HUP just says that the more precisely the momentum is determined the less precisely defined will be the conjugate position. 

I know. What I don't know is the cause of the spread. AG

See attached.

Brent

[Alan Grayson]

Your attachment shows how to establish the HUP, not why there is a spread in momentum. Classically, energy and momentum are related by a simple formula. So if one wants to prepare a system in some specific momentum, one needs to control the energy of the particle. Presumably, this can never be done precisely; hence we get the spread. Is this not a sufficient explanation for the spread? AG

[Brent Meeker]

As far as the HUP is concerned the cause of spread in momentum is that the spread in conjugate position must be finite, and vice versa.  Of course as I pointed out there are usually much bigger causes just in classical sources of instrument uncertainties.

Brent

[Alan Grayson]

Are all the momenta in the spread, eigenvalues of the momentum operator? AG

[Brent Meeker]

Yes.  But they have different probabilities of being found when measured.

Brent

[Alan Grayson]

But if one always gets a spread, how can any particular momentum in the spread be measured? AG 

[Alan Grayson]

Oh, I see. The spread is an ensemble view, but on a single trial one gets a single momenta to measure. AG  

[Brent Meeker]

You can't choose which value you get measuring a random variable.  You just measure momentum and you get a certain value.  Then you repeat the experiment and you get a different value.  You repeat this a thousand times and you can plot the distribution function of momenta and measure the spread.

Brent

[Cosmin Visan]

Why don't you guys go and actually perform the experiments that you hallucinate about here ? Let's see how you will manage when the results obtained diverge by 30% from your delirium here on the forum.

[Alan Grayson]

Your attachment shows how to establish the HUP, not why there is a spread in momentum. Classically, energy and momentum are related by a simple formula. So if one wants to prepare a system in some specific momentum, one needs to control the energy of the particle. Presumably, this can never be done precisely; hence we get the spread. Is this not a sufficient explanation for the spread? AG

As far as the HUP is concerned the cause of spread in momentum is that the spread in conjugate position must be finite, and vice versa. 

Are all the momenta in the spread, eigenvalues of the momentum operator? AG

Yes.  But they have different probabilities of being found when measured.

Brent

But if one always gets a spread, how can any particular momentum in the spread be measured? AG

You can't choose which value you get measuring a random variable.  You just measure momentum and you get a certain value.  Then you repeat the experiment and you get a different value.  You repeat this a thousand times and you can plot the distribution function of momenta and measure the spread.

Brent

Presumably, if it's momentum that's being measured, and one always measures eigenvalues, why is the spread larger on "imprecise" measuring devices, as opposed to undefined "ideal" measurements? And what is an ideal measurement? AG 

[Brent Meeker]

An ideal measurement is one that leaves the system in the eigenstate corresponding to the measured eigenvalue.  It's effectively a preparation.  So it excludes destructive measurement, like hitting photographic film.  I was assuming ideal measurements.  Of course in real measurements the instrument noise may be bigger than the interval between eignvalues and so introduces additional spread.

Brent

[Alan Grayson]

But regardless of the increased spread, won't the noise still result in eigenvalues of the momentum operator? AG 

[Brent Meeker]

In general when noise is small it can be treated as additive so when you measure you get some eigenvalue+noise, not the true value.  Of course if the system is in a single definite eigenstate, not a superposition of many eigenstates, you can repeat the measurement many times and the noise term will average to zero.

Brent

[Cosmin Visan]

So easy, you just measure, you just observe, bla-bla. Go do the measurements yourselves! See how easy they are! =))))))))))))))))))))))

[Alan Grayson]

On Wednesday, May 21, 2025 at 1:04:35 AM UTC-6 Cosmin Visan wrote:

So easy, you just measure, you just observe, bla-bla. Go do the measurements yourselves! See how easy they are! =))))))))))))))))))))))

You keep demonstrating schmuck-consciousness and apparently have zero consciousness of what you're doing. Moreover, in your quest for juvenile attention, you don't even know the content of the questions. So, in conclusion, you're a juvenile fool who should STFU. AG 

[Cosmin Visan]

@Alan. So when do you test your hallucinations ?