Classical verses Quantum EM theory

[Alan Grayson]

Presumably, if Maxwell's equations are solved in empty space, in the absence of charged particles, the E and B fields will be identically zero. If the same problem is solved in Quantum EM theory, won't we also get the same classical result, but then won't the HUP fail to be satisfied? Something seems awry here. Enlightening replies would be appreciated. AG

[Cosmin Visan]

All kinds of religions. What would you expect ?

[John Clark]

On Mon, Apr 28, 2025 at 9:21 PM Alan Grayson <agrays...@gmail.com> wrote:

> Presumably, if Maxwell's equations are solved in empty space, in the absence of charged particles, the E and B fields will be identically zero.

Yes.

 

> If the same problem is solved in Quantum EM theory, won't we also get the same classical result,

No. If the electric, magnetic, gravitational, or any other type of field was precisely zero then you would know exactly what the energy density was in any arbitrarily short amount of time, zero. And quantum mechanics says that is impossible.  So the amount of energy must be greater than zero for short amounts of time, the shorter the time the greater the energy. This is why there are virtual particles, they can't be detected directly but they produce effects that can be detected, the simplest and most dramatic is the Casimir Effect.

 

> but then won't the HUP fail to be satisfied?

I don't know about that but HUP fails to satisfy me because IHA. 

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

hqc

[Alan Grayson]

Never heard of the HUP, the Heisenberg Uncertainty Principle? How can it be applied in quantum EM field theory other than on an ad-hoc basis. That is, can the HUP be derived from the princples of the theory, or is it just ad hoc? As for the Casimir Effect, it can be predicted classically, so it proves nothing about virtual particles. AG

[Cosmin Visan]

Everything is ad-hoc. Just guesses until one works.

[John Clark]

On Tue, Apr 29, 2025 at 9:58 AM Alan Grayson <agrays...@gmail.com> wrote:

>>> but then won't the HUP fail to be satisfied?

>> I don't know about that but HUP fails to satisfy me because IHA. 

> Never heard of the HUP, the Heisenberg Uncertainty Principle?

Yes of course I've heard of  the Heisenberg Uncertainty Principle, but I've NEVER heard it referred to as "HUP".

 

 > As for the Casimir Effect, it can be predicted classically,

BULLSHIT!! Classical physics doesn't have zero-point energy or vacuum fluctuations! Schrodinger's equation insists those things must exist while classical physics insists they do not. Numerous experiments, including observations of the Casimir effect, prove that Schrodinger's equation is correct and classical physics is wrong. 

> so it proves nothing about virtual particles.

 BULLSHIT!

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

7ss

[Cosmin Visan]

AI!

[Brent Meeker]

On 4/29/2025 3:46 AM, John Clark wrote:

On Mon, Apr 28, 2025 at 9:21 PM Alan Grayson <agrays...@gmail.com> wrote:

> Presumably, if Maxwell's equations are solved in empty space, in the absence of charged particles, the E and B fields will be identically zero.

Yes.

 

> If the same problem is solved in Quantum EM theory, won't we also get the same classical result,

No. If the electric, magnetic, gravitational, or any other type of field was precisely zero then you would know exactly what the energy density was in any arbitrarily short amount of time, zero. And quantum mechanics says that is impossible. 

No, QM says that if you measure the field there is an uncertainty relation between the energy and duration measured values.   Measure also refers to interactions that imply values.

Brent

So the amount of energy must be greater than zero for short amounts of time, the shorter the time the greater the energy. This is why there are virtual particles, they can't be detected directly but they produce effects that can be detected, the simplest and most dramatic is the Casimir Effect.

 

> but then won't the HUP fail to be satisfied?

I don't know about that but HUP fails to satisfy me because IHA. 

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

hqc

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[Brent Meeker]

If you make some rather abstract assumptions about the physics, then you can derive the HUP. You
assume that the state of a physical system is described by a normalized vector ψ in a complex
valued vector space of high dimension. This vector is a function of the physical variables, q , of the
system, ψ(q). So the values of physical variables are given by projections of this state vector onto
the axes of the vector space. Then we consider unitary transformations, ψ′=Uψ of the state vector.
Unitary means the norm isn’t changed, so it’s a rotation, U=exp( θi ). In two dimensions θ is just an
angle, but in this higher dimension space it needs to be an operator, i.e. a matrix. Let it be an
infinitesimal transformation θ=−ϵG. Then the infinitesimal unitary transformation is given
by U=1− ϵi G where G is called the generator of the transformation.

ψ′=ψ−iϵGψ

Then we compare this to the form of a change of basis, q′μ=qμ−ϵμ

ψ′(q‘)=ψ(q−ϵ)=ψ(q)−ϵ∂ψ(q)/∂q

It is seen that the generator is

G=-i∂/∂q

If q is position then G is the generator of position, i.e. momentum. Then to make this look more familiar we define p as:

p≡ℏG=-iℏ ∂/∂q

Note that ℏ is just an arbitrary constant we introduce because we want to measure p in units other
than the inverse of q . Then it follows that 

[p,q]ψ=-iℏψ.

So Heisenberg’s uncertainty relation comes out of the mathematics AFTER you’ve put some
assumptions about how physical systems are represented in a high dimensional vector space by a
normalized vector that just rotates. The HUP is just the statement that the infinitesimal rotation
corresponding to a change in a variable q fails to commute with the conjugate variable p , we call
“momentum” along q , which is not surprising since rotations don’t commute in general.

Brent

[John Clark]

On Tue, Apr 29, 2025 at 4:53 PM Brent Meeker <meeke...@gmail.com> wrote:

 >> If the electric, magnetic, gravitational, or any other type of field was precisely zero then you would know exactly what the energy density was in any arbitrarily short amount of time, zero. And quantum mechanics says that is impossible. 

> No, QM says that if you measure the field there is an uncertainty relation between the energy and duration measured values.   Measure also refers to interactions that imply values.

If you place two macroscopic conductive plates close to each other the Casimir Effect will cause the two plates to attract each other; this occurs regardless of if you make any measurements or not. It happens because there are fewer virtual particles between the two plates than there are outside the plates. And virtual particles exist because it's impossible for the energy in the electromagnetic field to be exactly zero for any arbitrary length of time; and the shorter the time the greater the deviation from zero it's likely to be.   

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

hjd

[Brent Meeker]

That's why the qualification about measure like interactions.  The two conductive plates exclude longer wavelengths.  I don't recall that the effect depended on duration.  There are a lot of different forces between plates near one another that have been referred to as Casimir effects.  Here's a review article,

 https://courses.physics.ucsd.edu/2014/Fall/physics215a/project/Casimir-Review.pdf

Brent

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

hjd

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[Alan Grayson]

The Casimir Effect can be predicted classically, as Van der Waals forces. BTW, HUP is used frequently. I'm suprised you haven't noticed it. As for virtual particles, they're "off shell", meaning they violate conservation of energy. You're out on a limb to place any faith in their existence. AG

https://en.wikipedia.org/wiki/Casimir_effect

"The treatment of boundary conditions in these calculations is controversial. In fact, "Casimir's original goal was to compute the van der Waals force between polarizable molecules" of the conductive plates. Thus it can be interpreted without any reference to the zero-point energy (vacuum energy) of quantum fields." 7ss

[Cosmin Visan]

Obey the Bible!

[John Clark]

On Tue, Apr 29, 2025 at 8:09 PM Brent Meeker <meeke...@gmail.com> wrote:

>> If you place two macroscopic conductive plates close to each other the Casimir Effect will cause the two plates to attract each other; this occurs regardless of if you make any measurements or not. It happens because there are fewer virtual particles between the two plates than there are outside the plates. And virtual particles exist because it's impossible for the energy in the electromagnetic field to be exactly zero for any arbitrary length of time; and the shorter the time the greater the deviation from zero it's likely to be.  

>That's why the qualification about measure like interactions.  The two conductive plates exclude longer wavelengths. 

Yes.

 > I don't recall that the effect depended on duration. 

Heisenberg's uncertainty principle is not just about the relationship between momentum and position, it also insists there is a similar relationship between energy and time; the shorter amount of time the greater the random variation from a zero value there is. And without that there wouldn't be any wavelengths (or virtual particles) inside or outside those plates and the Casimir Effect would not exist. 

 

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

6tc

 

 

[John Clark]

On Wed, Apr 30, 2025 at 2:38 AM Alan Grayson <agrays...@gmail.com> wrote:

> The Casimir Effect can be predicted classically, as Van der Waals forces. BTW,

No. They are different phenomena, for example the Casimir Effect is proportional to 1/d^4 where d is distance, but for the Van der Waals force it's proportional to 1/d^6.  More importantly, the Van der Waals force exists because of quantum mechanical fluctuations in the distribution of electrons in atoms or molecules. The Casimir Effect requires Quantum Electrodynamics for a full description, while Van der Waals forces can be described using simpler quantum mechanical perturbation theory. 

It's true that Van der Waals first introduced his theory in the 1870s and that was decades before quantum mechanics, so his explanation had to be based just on classical physics.  Van der Waals amended the ideal gas law by injecting  not one but two new mathematical constants into it, but he gave no explanation as to why they had the particular values that they had, he obtained them by empirically fitting experimental data. The full explanation wouldn't come until 1930 using quantum mechanical perturbation theory. The Casimir Effect needed something more advanced, full Quantum Electrodynamics, so it wasn't derived until 1948. But neither could have been predicted by classical theory alone.

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

3b0

[Brent Meeker]

On 4/30/2025 4:29 AM, John Clark wrote:

On Tue, Apr 29, 2025 at 8:09 PM Brent Meeker <meeke...@gmail.com> wrote:

>> If you place two macroscopic conductive plates close to each other the Casimir Effect will cause the two plates to attract each other; this occurs regardless of if you make any measurements or not. It happens because there are fewer virtual particles between the two plates than there are outside the plates. And virtual particles exist because it's impossible for the energy in the electromagnetic field to be exactly zero for any arbitrary length of time; and the shorter the time the greater the deviation from zero it's likely to be.  

>That's why the qualification about measure like interactions.  The two conductive plates exclude longer wavelengths. 

Yes.

 > I don't recall that the effect depended on duration. 

Heisenberg's uncertainty principle is not just about the relationship between momentum and position, it also insists there is a similar relationship between energy and time; the shorter amount of time the greater the random variation from a zero value there is.

In quantum mechanics energy and the time per unit change of a variable are conjugate variables. So they satisfy an Heisenberg uncertainty relation, often written $\Delta E \Delta t \geq \hbar$ . This is sloppy though and not quite right. What is right is given any operator $A$ and the Hamiltonian $H$ defining the time evolution of $A$, then $\Delta A \Delta H \geq \frac{1}{2} \hbar [d<A>/dt]$ . In this case I don't see what is the time per unit change in the expected value of the energy density between the plates?  The plates are assumed stationary.

Brent

And without that there wouldn't be any wavelengths (or virtual particles) inside or outside those plates and the Casimir Effect would not exist. 

 

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

6tc

 

 

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[Alan Grayson]

On Wednesday, April 30, 2025 at 2:41:43 PM UTC-6 Brent Meeker wrote:

On 4/30/2025 4:29 AM, John Clark wrote:

On Tue, Apr 29, 2025 at 8:09 PM Brent Meeker <meeke...@gmail.com> wrote:

>> If you place two macroscopic conductive plates close to each other the Casimir Effect will cause the two plates to attract each other; this occurs regardless of if you make any measurements or not. It happens because there are fewer virtual particles between the two plates than there are outside the plates. And virtual particles exist because it's impossible for the energy in the electromagnetic field to be exactly zero for any arbitrary length of time; and the shorter the time the greater the deviation from zero it's likely to be.  

>That's why the qualification about measure like interactions.  The two conductive plates exclude longer wavelengths. 

Yes.

 > I don't recall that the effect depended on duration. 

Heisenberg's uncertainty principle is not just about the relationship between momentum and position, it also insists there is a similar relationship between energy and time; the shorter amount of time the greater the random variation from a zero value there is.

In quantum mechanics energy and the time per unit change of a variable are conjugate variables. So they satisfy an Heisenberg uncertainty relation, often written $\Delta E \Delta t \geq \hbar$ . This is sloppy though and not quite right. What is right is given any operator $A$ and the Hamiltonian $H$ defining the time evolution of $A$, then $\Delta A \Delta H \geq \frac{1}{2} \hbar [d<A>/dt]$ . In this case I don't see what is the time per unit change in the expected value of the energy density between the plates?  The plates are assumed stationary.

Brent

In the time-energy form of the HUP, what is the role of time as an operator? What does time per unit change of a variable mean? Which variable is referenced? About virtual particles; aren't they elements of a perturbation expansion and thus not to be considered real since those terms violate conservation of energy? TY, AG

[Brent Meeker]

That's why I include the equations (although I see they didn't get converted to display).  It can be any variable whose change is encoded by the Hamiltonian, A and H respectively in the equation.  It doesn't have anything to do with how you might solve the equations; which is where perturbation expansions and virtual particles enter.

Brent

And without that there wouldn't be any wavelengths (or virtual particles) inside or outside those plates and the Casimir Effect would not exist. 

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[Cosmin Visan]

I think Allah is greater than God!

[Alan Grayson]

On Wednesday, April 30, 2025 at 11:06:07 PM UTC-6 Brent Meeker wrote:

On 4/30/2025 5:54 PM, Alan Grayson wrote:

On Wednesday, April 30, 2025 at 2:41:43 PM UTC-6 Brent Meeker wrote:

On 4/30/2025 4:29 AM, John Clark wrote:

On Tue, Apr 29, 2025 at 8:09 PM Brent Meeker <meeke...@gmail.com> wrote:

>> If you place two macroscopic conductive plates close to each other the Casimir Effect will cause the two plates to attract each other; this occurs regardless of if you make any measurements or not. It happens because there are fewer virtual particles between the two plates than there are outside the plates. And virtual particles exist because it's impossible for the energy in the electromagnetic field to be exactly zero for any arbitrary length of time; and the shorter the time the greater the deviation from zero it's likely to be.  

>That's why the qualification about measure like interactions.  The two conductive plates exclude longer wavelengths. 

Yes.

 > I don't recall that the effect depended on duration. 

Heisenberg's uncertainty principle is not just about the relationship between momentum and position, it also insists there is a similar relationship between energy and time; the shorter amount of time the greater the random variation from a zero value there is.

In quantum mechanics energy and the time per unit change of a variable are conjugate variables. So they satisfy an Heisenberg uncertainty relation, often written $\Delta E \Delta t \geq \hbar$ . This is sloppy though and not quite right. What is right is given any operator $A$ and the Hamiltonian $H$ defining the time evolution of $A$, then $\Delta A \Delta H \geq \frac{1}{2} \hbar [d<A>/dt]$ . In this case I don't see what is the time per unit change in the expected value of the energy density between the plates?  The plates are assumed stationary.

Brent

In the time-energy form of the HUP, what is the role of time as an operator? What does time per unit change of a variable mean? Which variable is referenced? About virtual particles; aren't they elements of a perturbation expansion and thus not to be considered real since those terms violate conservation of energy? TY, AG

That's why I include the equations (although I see they didn't get converted to display).  It can be any variable whose change is encoded by the Hamiltonian, A and H respectively in the equation.  It doesn't have anything to do with how you might solve the equations; which is where perturbation expansions and virtual particles enter.

Brent

Can you give some examples of what A could be, and mustn't A be an operator, not a variable, that commutes with H? If your claim is correct, ISTM that Clark cannot apply the Time-Enegy form of the HUP to make his claim about the Casmir Effect. Do you agree? TY, AG 

[Brent Meeker]

Yes, A is an operator, but it doesn't commute with H.  That would imply the variable measured by A is constant in time.  The time per unit change in the expected value of the variable is the inverse of [d<A>/dt].

If your claim is correct, ISTM that Clark cannot apply the Time-Enegy form of the HUP to make his claim about the Casmir Effect. Do you agree?

I don't know.  I don't understand the proposed role of time.  It doesn't seem to have anything to do with conducting plates.  Perhaps he means the period of EM fields filling the gap between the plates.  Those of long period being excluded from between the plates would thereby remove their repulsive pressure and leave an unbalanced compressive pressure.

Brent

TY, AG 

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[Alan Grayson]

On Thursday, May 1, 2025 at 8:42:45 PM UTC-6 Brent Meeker wrote:

On 5/1/2025 7:25 AM, Alan Grayson wrote:

      On Wednesday, April 30, 2025 at 11:06:07 PM UTC-6 Brent Meeker wrote:

             On 4/30/2025 5:54 PM, Alan Grayson wrote:

                    On Wednesday, April 30, 2025 at 2:41:43 PM UTC-6 Brent Meeker wrote:

                            On 4/30/2025 4:29 AM, John Clark wrote:

                                  On Tue, Apr 29, 2025 at 8:09 PM Brent Meeker <meeke...@gmail.com> wrote:

                                       >> If you place two macroscopic conductive plates close to each other the Casimir Effect will                                               cause the two plates to attract each other; this occurs regardless of if you make any                                                           measurements or not. It happens because there are fewer virtual particles between the two                                             plates than there are outside the plates. And virtual particles exist because it's impossible                                               for the energy in the electromagnetic field to be exactly zero for any arbitrary length of                                                     time;  and the shorter the time the greater the deviation from zero it's likely to be.  JC

>That's why the qualification about measure like interactions.  The two conductive plates exclude longer wavelengths. 

Yes.

 > I don't recall that the effect depended on duration. 

Heisenberg's uncertainty principle is not just about the relationship between momentum and position, it also insists there is a similar relationship between energy and time; the shorter amount of time the greater the random variation from a zero value there is.

In quantum mechanics energy and the time per unit change of a variable are conjugate variables. So they satisfy an Heisenberg uncertainty relation, often written $\Delta E \Delta t \geq \hbar$ . This is sloppy though and not quite right. What is right is given any operator $A$ and the Hamiltonian $H$ defining the time evolution of $A$, then $\Delta A \Delta H \geq \frac{1}{2} \hbar [d<A>/dt]$ . In this case I don't see what is the time per unit change in the expected value of the energy density between the plates?  The plates are assumed stationary.

Brent

In the time-energy form of the HUP, what is the role of time as an operator? What does time per unit change of a variable mean? Which variable is referenced? About virtual particles; aren't they elements of a perturbation expansion and thus not to be considered real since those terms violate conservation of energy? TY, AG

That's why I include the equations (although I see they didn't get converted to display).  It can be any variable whose change is encoded by the Hamiltonian, A and H respectively in the equation.  It doesn't have anything to do with how you might solve the equations; which is where perturbation expansions and virtual particles enter.

Brent

Can you give some examples of what A could be, and mustn't A be an operator, not a variable, that commutes with H?

Yes, A is an operator, but it doesn't commute with H.  That would imply the variable measured by A is constant in time.  The time per unit change in the expected value of the variable is the inverse of [d<A>/dt].

What is d? Can you give one or two specific examples of A? I thought the HUP is applicable only for non-computing operators. Am I mistaken? AG

If your claim is correct, ISTM that Clark cannot apply the Time-Enegy form of the HUP to make his claim about the Casmir Effect. Do you agree? AG

I don't know.  I don't understand the proposed role of time. 

Same here, and more generally, so I find applying the time-energy form of the HUP dubious at best, but this is how the Casmir Effect is presumably established in quantum EM theory.  If you recall, Bruce Kellet, an excellent physicist IMO, claimed the Casmir Effect can fully accounted for classically. He also vehemently denied that virtual particles are real due to energy considerations, given that virtual particles violate energy conservation. AG

It doesn't seem to have anything to do with conducting plates.  Perhaps he means the period of EM fields filling the gap between the plates.  Those of long period being excluded from between the plates would thereby remove their repulsive pressure and leave an unbalanced compressive pressure.

Hopefully, Clark can explain what he meant. AG

Brent

[Alan Grayson]

On Thursday, May 1, 2025 at 10:17:19 PM UTC-6 Alan Grayson wrote:

On Thursday, May 1, 2025 at 8:42:45 PM UTC-6 Brent Meeker wrote:

On 5/1/2025 7:25 AM, Alan Grayson wrote:

      On Wednesday, April 30, 2025 at 11:06:07 PM UTC-6 Brent Meeker wrote:

             On 4/30/2025 5:54 PM, Alan Grayson wrote:

                    On Wednesday, April 30, 2025 at 2:41:43 PM UTC-6 Brent Meeker wrote:

                            On 4/30/2025 4:29 AM, John Clark wrote:

                                  On Tue, Apr 29, 2025 at 8:09 PM Brent Meeker <meeke...@gmail.com> wrote:

                                       >> If you place two macroscopic conductive plates close to each other the Casimir Effect will                                               cause the two plates to attract each other; this occurs regardless of if you make any                                                           measurements or not. It happens because there are fewer virtual particles between the two                                             plates than there are outside the plates. And virtual particles exist because it's impossible                                               for the energy in the electromagnetic field to be exactly zero for any arbitrary length of                                                     time;  and the shorter the time the greater the deviation from zero it's likely to be.  JC

>That's why the qualification about measure like interactions.  The two conductive plates exclude longer wavelengths. 

Yes.

 > I don't recall that the effect depended on duration. 

Heisenberg's uncertainty principle is not just about the relationship between momentum and position, it also insists there is a similar relationship between energy and time; the shorter amount of time the greater the random variation from a zero value there is.

In quantum mechanics energy and the time per unit change of a variable are conjugate variables. So they satisfy an Heisenberg uncertainty relation, often written $\Delta E \Delta t \geq \hbar$ . This is sloppy though and not quite right. What is right is given any operator $A$ and the Hamiltonian $H$ defining the time evolution of $A$, then $\Delta A \Delta H \geq \frac{1}{2} \hbar [d<A>/dt]$ . In this case I don't see what is the time per unit change in the expected value of the energy density between the plates?  The plates are assumed stationary.

Brent

In the time-energy form of the HUP, what is the role of time as an operator? What does time per unit change of a variable mean? Which variable is referenced? About virtual particles; aren't they elements of a perturbation expansion and thus not to be considered real since those terms violate conservation of energy? TY, AG

That's why I include the equations (although I see they didn't get converted to display).  It can be any variable whose change is encoded by the Hamiltonian, A and H respectively in the equation.  It doesn't have anything to do with how you might solve the equations; which is where perturbation expansions and virtual particles enter.

Brent

Can you give some examples of what A could be, and mustn't A be an operator, not a variable, that commutes with H?

Sorry, I meant that the operator A does NOT commute with H. AG 

[Cosmin Visan]

Respect my religious belief in operators!

[Alan Grayson]

On Friday, May 2, 2025 at 12:34:27 AM UTC-6 Cosmin Visan wrote:

Respect my religious belief in operators!

Asshole. You're definitely conscious, but not sufficiently so to have the basic self-realization that your repetitive one-note comments prove beyond doubt to other conscious entities here, that you're an immature dumb schmuck who craves attention, and has nothing of substance to offer. IOW, STFU. We're NOT interested in your stupid rantings! AG 

[Cosmin Visan]

How about the epicycles ?!!!

[Alan Grayson]

On Friday, May 2, 2025 at 6:50:45 AM UTC-6 Cosmin Visan wrote:

How about the epicycles ?!!!

They were more accurate in their predictions than your incessant mantra, "consciousness".  The time for you to cease being an asshole is long past. AG

[Cosmin Visan]

The theory of consciousness predicts that you have eternal life. What more do you want ?

[Brent Meeker]

On 5/1/2025 9:17 PM, Alan Grayson wrote:
Yes, A is an operator, but it doesn't commute with H.  That would imply the variable measured by A is constant in time.  The time per unit change in the expected value of the variable is the inverse of [d<A>/dt].

What is d?

d/dt is the derivative operator.

Can you give one or two specific examples of A? I thought the HUP is applicable only for non-computing operators. Am I mistaken? AG

"Non-communting"  That's right.  But any time varying variable fails to commute with the Hamiltonian.

Brent

[Cosmin Visan]

@Brent, you're just regurgitating your Sunday school memes. At any point an experiment can turn up in which your cherished memes dont apply.