Superconductivity at room temperature??

[sol invictus]

Basically a wave-packet can only be a solution of a many-body wave equation: i.e. the superposition of many waves each of which is in essence an electron-wave and thus has a rest-mass. When you solve the single particle wave equation, you do get a spectrum of possible solutions: But only one of them can be a solution for a single particle wave equation. It makes no sense to assume that the other possible solutions can be superposed to obtain a wave-packet since they do not really exist unless they are each representing an additional electron.

Just FYI RC, this is total nonsense. All solutions to the Schrodinger equation of the type we've been discussing represent a single electron. The only condition they need to satisfy is that they be normalizable (that the integral over space of the square of the absolute value of the wavefunction be equal to 1). Any such wavefunction represents a state with total charge e, never 2e or 3e.

A two-electron wavefunction looks completely different - it is a function of two sets of spatial variables (x1,y1,z1) and (x2, y2, z2) (one for each electron). It satisfies a different equation - the SE but with derivatives taken with respect to both sets of variables, not just one. It satisfies a different normalizability condition. And it must have certain properties that don't even make sense for a single particle wavefunction - by the spin-statistics/Pauli exclusion principle for example the two-electron WF must be odd under interchange of the two sets of coordinates.

As for the energy: the SE is the non-relativistic limit of a relativistic theory in which electrons and positrons can be created and destroyed. In the relativistic theory therefore, the number of particles is something that can vary depending on the state. But in the NR limit, the particles have a kinetic energy which is much less than their rest mass energy mc^2. Therefore there is not enough energy to create an electron-positron pair and change the particle number, and so NR wavefunctions represent a strictly fixed number of particles (which is one in the case of the WFs discussed in this thread). As such, the total rest mass energy is an absolutely fixed constant. One can include it in the energy, but it has no effect at all (constants added to the energy do not affect non-gravitational physics).

Bear in mind that all of this is a basic part of the best-tested theory in the history of humanity. What Johan is saying is absolutely ruled out by a century of the most precise and careful experiments ever performed, done in an incredible variety of circumstances.

 

[johanfprins]

Rest mass energy = m(0)*c^2

Kinetic energy is T=m*c^2-m(0)*c^2

Kinetic energy is ONLY possible when m*c^2 is larger that m(0)*c^2: Only a certifiable idiot will argue otherwise. When an electron is bound, one has that m*c^2 is less than
m(0)*c^2 and this means that it cannot have kinetic energy and it cannot have momentum. QED.

 

[Ziggurat]

Rest mass energy = m(0)*c^2

Kinetic energy is T=m*c^2-m(0)*c^2

Kinetic energy is ONLY possible when m*c^2 is larger that m(0)*c^2: Only a certifiable idiot will argue otherwise. When an electron is bound, one has that m*c^2 is less than
m(0)*c^2 and this means that it cannot have kinetic energy and it cannot have momentum. QED.

Not so. That merely means that the potential energy is negative (and hence reduces the atom's mass). It's a well-known result from BOTH classical mechanics and quantum mechanics that for a 1/r potential <V> = -2<T>. <T> is positive, but <V> is negative and larger, leading to a negative total binding energy. It's a result of the virial theorem.

Furthermore, you seem to be confusing <p> = 0 with <p²> = 0. The former does not imply the latter. That's not even quantum mechanics, that's basic statistics. But the Heisenberg uncertainty principle further requires that <p²> > <p>². Kinetic energy depends on <p²>, not <p>². You will find that bound state solutions have nonzero <p²>, and hence nonzero kinetic energy.

This is basic quantum mechanics. Did you forget this stuff, or did you never learn it?

 

[Cuddles]

Rest mass energy = m(0)*c^2

Kinetic energy is T=m*c^2-m(0)*c^2

Kinetic energy is ONLY possible when m*c^2 is larger that m(0)*c^2: Only a certifiable idiot will argue otherwise. When an electron is bound, one has that m*c^2 is less than
m(0)*c^2 and this means that it cannot have kinetic energy and it cannot have momentum. QED.

No, that would mean it has negative energy. This just keeps getting better and better. What are we up to now? All quantum physics is wrong, all relativity is wrong, all classical waves are wrong, all physicists are wrong, all textbooks are wrong, all maths is irrelevant. Have I missed anything?

Seriously Johan, Ziggurat is right. There's absolutely no point in you making claims about or trying to discuss superconductivity, since you disagree with absolutely everything that would be required to discuss it. You'd be much better off if you instead spend some time trying to come up with a coherent explanation of what your view of physics actually is, and then discussing that. As it is, it's simply impossible to have a coherent discussion with you because you don't appear to accept any part of physics from the last 200 years or so.

 

[Reality Check]

Where is the mistake in the derivation of the wave packet solution, johanfprins

Basically a wave-packet can only be a solution of a many-body wave equation:
...snipped...

You are wrong johanfprins: The wave packet solution to the Schrodinger equations comes from a single-body wave equation. There is only one set of variables pertaining to that single particle (as sol invictus points out).

See the wave packet solution of Schrodingers equation for a single free particle.
Where is the mistake in the derivation of the wave packet solution, johanfprins?

 

[sol invictus]

Rest mass energy = m(0)*c^2

Kinetic energy is T=m*c^2-m(0)*c^2

Kinetic energy is ONLY possible when m*c^2 is larger that m(0)*c^2: Only a certifiable idiot will argue otherwise. When an electron is bound, one has that m*c^2 is less than
m(0)*c^2 and this means that it cannot have kinetic energy and it cannot have momentum. QED.

So according to Prinsphysics the earth has zero kinetic energy and zero momentum, despite the fact that it hurtles around the sun at 100,000 km/hour.

Brilliant, Johan, brilliant.

 

[johanfprins]

You are wrong johanfprins: The wave packet solution to the Schrodinger equations comes from a single-body wave equation. There is only one set of variables pertaining to that single particle (as sol invictus points out).

Do you agree that a wave packet is a superposition of waves with different energies? Do you agree that the single-electron Schroedinger equation is an eigenvalue equation which gives you a set of possible solutions for a single electron. The electron can thus only be one of these solutions; not all of them at the same time.

To have waves with different energies superpose, each wave must be an electron. This means that you must have more than one electron to form a wave packet. It is simple physics. Have you ever seen a circular, coherent wave on the surface of water that suddenly changes into a wave-packet. You must remember that we are dealing with physics here not just mathematical hop-skotch.

See the wave packet solution of Schrodingers equation for a single free particle.
Where is the mistake in the derivation of the wave packet solution, johanfprins?

Unfortunately I have to meet a deadline. I will look at this derivation as soon as I have time and I promise that I will then get back to you as soon as possible.

 

[sol invictus]

Do you agree that a wave packet is a superposition of waves with different energies?

Yes.

Do you agree that the single-electron Schroedinger equation is an eigenvalue equation which gives you a set of possible solutions for a single electron.

No, not really. The SE is not an eigenvalue equation at all. It can be written as one if you first assume the solution separates - but that's an assumption, and it's not valid in general.

The electron can thus only be one of these solutions; not all of them at the same time.

To have waves with different energies superpose, each wave must be an electron.

And this pretty much puts the cap on it. The single most essential feature of QM - the feature that makes it quantum, that distinguishes it from classical physics - is that a single particle can be in a state that is a superposition of states with definite values for X, where X can be any observable (for example spin, or energy, or position). Quantum states form a linear vector space (called a Hilbert space). Requiring that all particles must be in energy eigenstates (and not superpositions of them) makes as much sense as requiring that all vectors must point due north, due east, or straight up, but never in any direction in between.

The last century of physics experiments - most notably Stern-Gerlach, double-slit type experiments, Mach-Zender interferometers, EPR type experiments, but actually including every experiment in atomic physics, quantum optics, particle physics - all of them demonstrate conclusively that single particles can be in superpositions of energy eigenstates.

That's the whole point of QM. Why are you trying to revolutionize superconductivity, Johan? First, you need to re-write every single QM text, and convince every single physicist and physics student that the entire basis of their field (one that is the best-tested and most successful theory in the history of science) is totally wrong, and all those classic experiments were flukes.

 

[sol invictus]

One other comment on Johan's latest assertion - for a free particle, a state of definite energy is also a state of definite momentum (or to be precise, of definite momentum squared). But as every school kid knows, the Heisenberg uncertainty principle tells us that states of definite momentum are not allowed - if the uncertainty in momentum is zero, the uncertainty in position is infinite, which means the particle can be found anywhere with equal probability and the state cannot be normalizable.

So not only are single particle states that are superpositions of energy eigenstates (such as wave packets) allowed in QM, they are essential for the simplest possible case of the free particle in infinite volume - all physical states are superpositions. Energy eigenstates for the free particle are unphysical and impossible, they essentially are the only states that the particle cannot be in.

 

[Ziggurat]

No, not really. The SE is not an eigenvalue equation at all. It can be written as one if you first assume the solution separates - but that's an assumption, and it's not valid in general.

Furthermore, the fact that single-particle wave functions can be in superposition states is explicitly verified by experiments such as Stern-Gerlach.

 

[ben m]

Do you agree that a wave packet is a superposition of waves with different energies?

Yes.

Do you agree that the single-electron Schroedinger equation is an eigenvalue equation which gives you a set of possible solutions for a single electron. The electron can thus only be one of these solutions; not all of them at the same time.

No. No no no no. A single electron can be in a superposition of energy eigenstates, under exactly the same rules as it can be in a superposition of position eigenstates (like the double-slit experiment) or angular momentum (like Stern-Gerlach) or whatever. Indeed, except in the limit of infinitely-long-lived states, every electron is in a superposition of energy eigenstates.

To have waves with different energies superpose, each wave must be an electron.

Are you stating this under the impression that this is standard quantum mechanics? Or are you stating this as one of the tenets of your quantum-mechanics-textbooks-are-all-wrong argument?

This means that you must have more than one electron to form a wave packet. It is simple physics. Have you ever seen a circular, coherent wave on the surface of water that suddenly changes into a wave-packet.

Non sequitur. First: I've never seen a classical wave collapse onto a position eigenstate, either, but this has no bearing whatsoever on the truth of quantum mechanics. Second: what? Whose pro-quantum argument assumed or implied that coherent waves spontaneously "suddenly change" into wavepackets?

 

[Reality Check]

Do you agree that a wave packet is a superposition of waves with different energies? Do you agree that the single-electron Schroedinger equation is an eigenvalue equation which gives you a set of possible solutions for a single electron. The electron can thus only be one of these solutions; not all of them at the same time.

As sol invictus and ben m have stated, the answers are:

• Yes.
• Not quite right - the SE is not an eigenvalue equation.
• Wrong: A fundemental part of quantum mechanics is the superposition of states. The electron will be in all of the possible states.

 

[johanfprins]

See the wave packet solution of Schrodingers equation for a single free particle.
Where is the mistake in the derivation of the wave packet solution, johanfprins?

I have had time to look at this webpage today: Obviously I did not find "a mistake" in the mathematics at all. What I find unacceptable is that a wave-packet is derived by using an arbitrary superposition of trigonometric functions, and it is then claimed, without any physical reason, that such a wave-packet represents a free electron. I can create millions of superpositions by summing or integrating over Fourier-space, but it does not mean that each of these solutions is physically possible. A solution is only possible when one can give a physics-reason (find the actual boundary conditions) why it is possible.

The other aspect is that in the case of the Schroedinger equation, this wave-packet solution, if it is physically possible (which I doubt), can only be possible within a single, unique inertial reference frame. Classical waves, except for light waves, are only possible within a single unique inertial refrence frame defined by the medium within which the waves are moving. Light waves are their own medium: i.e. one can choose ANY inertial reference frame and obtain the same wave.

For electron waves this is not the situation: Thus the wave-packet derived in your example for the Schroedinger equation can only exist within a single, unique inertial refrence frame; and if you cannot give a rule to decide which one of the infinite number of inertial reference frames it is, the solution is meaningless. After all the de Broglie wavelength is a relativistic parameter and therefore the solution for an electron wave must be different when the same electron is viewed from different inertial reference frames. The wave-function must transform according to the Galilean transformation at low speeds and the Lorentz transformation at high speeds.

Then comes the other problem: Within all inertial reference frames, a moving electron diffracts and interfere as if each electron is a coherent wave which cannot consist of a superposition of waves with different values for the k-vector: i.e. the wave can thus not be a spreading wave-packet when it impinges into a diffraction grating; AND the wavelength with which it diffracts is solely determined by the relative speed with which it moves relative to the diffraction grating: In other words, the applicable value of k is different for different difrraction gratings which are stationary within different inertial refrence frames.

Thus to grab an arbitrary superposition of states out of the air to create a wave-packet within a single unique inertial reference frame, cannot be the correct description for the same electron which will be observed within different inertial reference frames to move with different speeds.

Furthermore, it is assumed that the potential energy of an electron in "free space" is constant (chosen as zero). This is probably wrong if Einstein's theory of gravity is correct. An electron has mass and mass curves space around it: It is thus unlikely that one can describe curved space in terms of a zero potential energy.

One expects that, in order to satisfy Galileo's inertia and Einstein's theory of gravity, an electron wave must be a stationary, non-spreading, wave within the unique reference frame within which the electron is stationary. That there must be such an inertial reference frame is mandated by Galileo's inertia. Thus to claim that the centre-of-mass of an electron has "uncertainties" in position and momentum violates Galileo's inertia and all physics ever postulated: Including Schroedinger's equation. It is also the reason why it will never be possible to unite gravity with quantum mechanics. The latter will only become possible when the concepts of "uncertainty" and and "probability waves" are dropped. Goodbye Hiiggs!

 

[sol invictus]

I have had time to look at this webpage today: Obviously I did not find "a mistake" in the mathematics at all.

That is indeed obvious.

What I find unacceptable is that a wave-packet is derived by using an arbitrary superposition of trigonometric functions, and it is then claimed, without any physical reason, that such a wave-packet represents a free electron. I can create millions of superpositions by summing or integrating over Fourier-space, but it does not mean that each of these solutions is physically possible. A solution is only possible when one can give a physics-reason (find the actual boundary conditions) why it is possible.

Except that the description of particles in terms of wave packets is a foundational element of the best-tested theory in the history of the human race. But as we already know, experiments mean nothing to you.

The other aspect is that in the case of the Schroedinger equation, this wave-packet solution, if it is physically possible (which I doubt), can only be possible within a single, unique inertial reference frame. Classical waves, except for light waves, are only possible within a single unique inertial refrence frame defined by the medium within which the waves are moving. Light waves are their own medium: i.e. one can choose ANY inertial reference frame and obtain the same wave.

For electron waves this is not the situation: Thus the wave-packet derived in your example for the Schroedinger equation can only exist within a single, unique inertial refrence frame; and if you cannot give a rule to decide which one of the infinite number of inertial reference frames it is, the solution is meaningless. After all the de Broglie wavelength is a relativistic parameter and therefore the solution for an electron wave must be different when the same electron is viewed from different inertial reference frames. The wave-function must transform according to the Galilean transformation at low speeds and the Lorentz transformation at high speeds.

All of those questions are answered in any graduate level QM course. The Schrodinger equation is the non-relativistic limit of a relativistic equation. Its solutions do of course transform as expected under Galilean transforms, there is nothing even slightly mysterious about it. To see Galileo turn into Lorentz, you need a relativistic form, which of course is readily available.

Then comes the other problem: Within all inertial reference frames, a moving electron diffracts and interfere as if each electron is a coherent wave which cannot consist of a superposition of waves with different values for the k-vector: i.e. the wave can thus not be a spreading wave-packet when it impinges into a diffraction grating; AND the wavelength with which it diffracts is solely determined by the relative speed with which it moves relative to the diffraction grating: In other words, the applicable value of k is different for different difrraction gratings which are stationary within different inertial refrence frames.

Wrong.

Furthermore, it is assumed that the potential energy of an electron in "free space" is constant (chosen as zero). This is probably wrong if Einstein's theory of gravity is correct. An electron has mass and mass curves space around it: It is thus unlikely that one can describe curved space in terms of a zero potential energy.

The curvature of space due to a single electron is miniscule. The effects you refer to are far smaller than (special) relativistic corrections.

One expects that, in order to satisfy Galileo's inertia and Einstein's theory of gravity, an electron wave must be a stationary, non-spreading, wave within the unique reference frame within which the electron is stationary. That there must be such an inertial reference frame is mandated by Galileo's inertia. Thus to claim that the centre-of-mass of an electron has "uncertainties" in position and momentum violates Galileo's inertia and all physics ever postulated: Including Schroedinger's equation. It is also the reason why it will never be possible to unite gravity with quantum mechanics. The latter will only become possible when the concepts of "uncertainty" and and "probability waves" are dropped. Goodbye Hiiggs!

Yes, Johan, all of physics is wrong. Only you know the truth, and the computer you're using works by coincidence.

 

[ben m]

What I find unacceptable is that a wave-packet is derived by using an arbitrary superposition of trigonometric functions, and it is then claimed, without any physical reason, that such a wave-packet represents a free electron. ... A solution is only possible when one can give a physics-reason (find the actual boundary conditions) why it is possible.

So think of an experiment. Use that to set some boundary conditions. There, now you've got a non-arbitrary superposition of electron energy eigenstates.

(Your statement is sort of like, "I find the Pythagorean theorem unacceptable. It describes completely arbitrary right triangles. A right triangle is only possible when you actually build one and measure two of its sides.")

The other aspect is that in the case of the Schroedinger equation, this wave-packet solution, if it is physically possible (which I doubt), can only be possible within a single, unique inertial reference frame. Classical waves, except for light waves, are only possible within a single unique inertial refrence frame defined by the medium within which the waves are moving.

Nonsense. Take any wavefunction; transform all of its coordinates via the Lorentz transform. Now you've got the same wavefunction as seen from a different inertial frame. It's still a valid wavefunction, and it does exactly what you'd expect such a transformed wavefunction to do.

Light waves are their own medium: i.e. one can choose ANY inertial reference frame and obtain the same wave.

Not true. A light wave's energy, direction, etc., depends on the observer. Just like an electron's does.

For electron waves this is not the situation: Thus the wave-packet derived in your example for the Schroedinger equation can only exist within a single, unique inertial refrence frame; and if you cannot give a rule to decide which one of the infinite number of inertial reference frames it is, the solution is meaningless. After all the de Broglie wavelength is a relativistic parameter and therefore the solution for an electron wave must be different when the same electron is viewed from different inertial reference frames. The wave-function must transform according to the Galilean transformation at low speeds and the Lorentz transformation at high speeds.

Gibberish.

 

[Reality Check]

I have had time to look at this webpage today: Obviously I did not find "a mistake" in the mathematics at all. What I find unacceptable is that a wave-packet is derived by using an arbitrary superposition of trigonometric functions, and it is then claimed, without any physical reason, that such a wave-packet represents a free electron.
...the rest of your post is addressed by sol invictus and ben m...

If there is no mistake in the mathematics then the mathematics is correct and there is a wave packet solution of Schroedinger's equation for a single free particle.

The choice is not arbitrary. The choice of the wave function is a wave function that is a solution to the Schroedinger's equation for a single free particle. The expectation value for that wave function is a wave packet.

The physical reason why we expect something like a wave packet is simple: The particle has to exist somewhere !
The Heisenberg uncertainty principle tells us that the particle cannot exist at an exact measurable position because then it will always have no measurable momentum. The particle's position will be spread out in some way. Thus a wave packet is an appropriate function to describe a free particle.

 

[johanfprins]

If there is no mistake in the mathematics then the mathematics is correct and there is a wave packet solution of Schroedinger's equation for a single free particle.

I have not disputed this. The fact is that any differential wave equation can give one solutions which cannot manifest. Spreading wave packets do seem to exist within a conductor when one applies an electric-field to generate an electric current. They form since the application of the electric-field changes the boundary conditions. All the otherwise delocalised valence-electron waves then superpose to form these wave packets. It is to my mind highly unlikely that one will have a wave-packet in free space consisting of the superposition of possible solutions for a single free electron. To superpose you need a multitude of electrons and the required boundary conditions for this to happen.

The choice is not arbitrary. The choice of the wave function is a wave function that is a solution to the Schroedinger's equation for a single free particle. The expectation value for that wave function is a wave packet.

As I have pointed out above, I am able to mathematically superpose Fourier components to my heart's delight and generate a multitude of solutions which cannot be generated in reality. The solutions which are possible are determined by the boundary conditions. For the wave packet in free space it is assumed that there are no boundary conditions, and that one can superpose possible solutions to generate a solution which is not determined by boundary conditions.This is not physics!

The physical reason why we expect something like a wave packet is simple: The particle has to exist somewhere !

I am glad you realise this important fact. But is a spreading wave packet the wave describing a stationary electron within its inertial reference frame? I doubt it very much.

The Heisenberg uncertainty principle tells us that the particle cannot exist at an exact measurable position because then it will always have no measurable momentum. The particle's position will be spread out in some way. Thus a wave packet is an appropriate function to describe a free particle.

I know this is what the Copenhagen interpretation claims. Heisenberg in a desperate atempt to waylay Schroedinger created this red herring. All harmonic waves exist within both position-space and reciprocal k-space. This has NOTHING to do with uncertainty in the position and momentum of an electron. According to Galileo's concept of inertia, as quantified by Newton, a body with rest mass must be stationary within an inertial reference frame when there are no forces acting on it. An electron has a rest mass. This means that position and momentum manifests with 100% accuracy at the same time. If this is not so, then all physics, including Heisenberg's matrix mechanics and Schroedinger's wave equation must be wrong.

Also note that Einstein's theory of gravity rests on Galileo's inertia. And then the "experts" are amazed that they cannot unify Einstein's theory of gravity with quantum mechanics! As long as Heisenberg's relationship for position and momentum is interpreted as "uncertainties" in the position and momentum of an electron, for so long will these two fields not unify. On the other hand, when returning to Schroedinger's first intent that an electron is a wave and not a particle, Einstein's gravity and Schroedinger's wave mechanics do become unifyable.

This implies that the wave representing an electron within its own inertial refrence frame must be a non-spreading localised wave with its intensity proportional to its mass and curvature around mass. Such a wave has a centre-of-mass and this centre of mass is stationary within its own inertial reference frame for as long as the electron does not interact with other matter.

 

[Reality Check]

I have not disputed this.
...snipped conductor rant...

The fact is this is the wave packet solution of Schroedinger's equation for a single free particle. There is no conductor.

...This means that position and momentum manifests with 100% accuracy at the same time. If this is not so, then all physics, including Heisenberg's matrix mechanics and Schroedinger's wave equation must be wrong.

This is blatent ignorance: Heisenberg's matrix mechanics and Schroedinger's wave equation means that position and momentum never manifests with 100% accuracy at the same time.
The match with experimental results says that Schroedinger's wave equation is right (for non-relativistic systems).
The match with experimental results says that quantum field theory is right, e.g. QED is one of the most precisely tested theory in [hysics.

Also note that Einstein's theory of gravity rests on Galileo's inertia.
...snipped gibberish...

No.
Einstein's theory of gravity rests on the equivalence principle which is the weak (Galilean) equivalence principle and the Einstein equivalence principle.

ETA

But is a spreading wave packet the wave describing a stationary electron within its inertial reference frame? I doubt it very much.

Then you are showing your ignorance yet again. The wave packet above that sol invictus cites is not spreading. It is moving. It does not describe a stationary particle.

 

[dafydd]

A wave with a specific frequency must always be an energy eigen state. A coherent running wave must thus be an eigen state. A coherent running wave around a ring, if it can form which it cannot, must thus be an eigen state and not a wave packet.

Until you confirm that the solution of Schroedinger's time-dependent wave equation must be time dependent, and that such a solution, to be coherent, must have a single, non-zero, frequency, I will also not respond to your postings anymore. I hope there are people on this forum who are able to understand physics and who will thus post relevant discussion material. I will gladly respond to them: Not anymore to flat-earthers.

You have no understanding of physics. Your misplaced hubris is very entertaining.

 

[dafydd]

I have not disputed this. The fact is that any differential wave equation can give one solutions which cannot manifest. Spreading wave packets do seem to exist within a conductor when one applies an electric-field to generate an electric current. They form since the application of the electric-field changes the boundary conditions. All the otherwise delocalised valence-electron waves then superpose to form these wave packets. It is to my mind highly unlikely that one will have a wave-packet in free space consisting of the superposition of possible solutions for a single free electron. To superpose you need a multitude of electrons and the required boundary conditions for this to happen.
As I have pointed out above, I am able to mathematically superpose Fourier components to my heart's delight and generate a multitude of solutions which cannot be generated in reality. The solutions which are possible are determined by the boundary conditions. For the wave packet in free space it is assumed that there are no boundary conditions, and that one can superpose possible solutions to generate a solution which is not determined by boundary conditions.This is not physics!
I am glad you realise this important fact. But is a spreading wave packet the wave describing a stationary electron within its inertial reference frame? I doubt it very much.
I know this is what the Copenhagen interpretation claims. Heisenberg in a desperate atempt to waylay Schroedinger created this red herring. All harmonic waves exist within both position-space and reciprocal k-space. This has NOTHING to do with uncertainty in the position and momentum of an electron. According to Galileo's concept of inertia, as quantified by Newton, a body with rest mass must be stationary within an inertial reference frame when there are no forces acting on it. An electron has a rest mass. This means that position and momentum manifests with 100% accuracy at the same time. If this is not so, then all physics, including Heisenberg's matrix mechanics and Schroedinger's wave equation must be wrong.

Also note that Einstein's theory of gravity rests on Galileo's inertia. And then the "experts" are amazed that they cannot unify Einstein's theory of gravity with quantum mechanics! As long as Heisenberg's relationship for position and momentum is interpreted as "uncertainties" in the position and momentum of an electron, for so long will these two fields not unify. On the other hand, when returning to Schroedinger's first intent that an electron is a wave and not a particle, Einstein's gravity and Schroedinger's wave mechanics do become unifyable.

This implies that the wave representing an electron within its own inertial refrence frame must be a non-spreading localised wave with its intensity proportional to its mass and curvature around mass. Such a wave has a centre-of-mass and this centre of mass is stationary within its own inertial reference frame for as long as the electron does not interact with other matter.

Way to go for the Nobel prize! When can we expect to see you onstage in the full soup and fish in Oslo?