Superconductivity at room temperature??

[johanfprins]

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

I do not dispute that this is the accepted dogma: In fact I have taught the same nonsense to my students for years. I am just pointing out that a wave packet can only form from a spectrum of waves, each of which must represent a single electron. How many electron-waves are according to you superposing in free space to form a wave-packet?

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.

I have also taught this gibberish to my students: If this is true Galileo's inertia is wrong, and so is then all physics since Galileo.

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 Physics.

Ptolemy's model of the universe also match experimental results and was believed to be the ultimate truth for 2000 years. But is it correct? And what do you mean with "non-relativistic" systems. There are no such systems. Relativity is always valid whether it is for fast relative speeds or for low relative speeds. Do you want to tell me that Schroedinger's equation is only valid when the electron is stationary? In fact, this is not far from the truth since it only moves when the boundary conditions force it to morph into a moving field. Furthermore, any theory that can be fudged by the mathematically invalid process called renormalisation can be patched up to fit data. It does not mean that the theory is correct.

Einstein's theory of gravity rests on the equivalence principle which is the weak (Galilean) equivalence principle and the Einstein equivalence principle.

All Einstein's arguments start with time-dilation and length-contraction, which according to him demands that space-time must be curved: I have recently met a physicist who used the Lorentz transformation to prove that length contraction does not happen at all, and that a moving clock does not actually slow down, it only seems from outside the inertial reference frame within which this clock is stationary AS IF the clock is slowing down. In fact his arguments are compelling that time is absolute, just like Newton has postulated.

He then took these arguments further and proved in terms of Einstein's rotating disc experiment, that a force caused by acceleration is not really equivalent to a gravity field. The reason why space and time is curved seems to require that all matter consist solely of waves. Just as Schroedinger had proposed when he postulated his approximate wave equation as if it is determined by a "particle-like" Hamilton-operator.

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.

The non-spreading wave packet is not a solution of the Schroedinger equation. I thought we are talking about the latter. Maybe YOU should visit that website and study it?

 

[Ziggurat]

I do not dispute that this is the accepted dogma: In fact I have taught the same nonsense to my students for years.

In other words, I was right: you object to standard quantum mechanics.

And yes, standard quantum mechanics requires that single-particle wave packet solutions be valid solutions.

I have recently met a physicist who used the Lorentz transformation to prove that length contraction does not happen at all

Oh goodie! Not only do you disagree with quantum mechanics, you disagree with relativity.

Stop bothering with this superconductivity stuff, you have more important matters to attend to. We eagerly await your overthrow of the entirety of 20th century physics.

 

[ben m]

I do not dispute that this is the accepted dogma: In fact I have taught the same nonsense to my students for years. I am just pointing out that a wave packet can only form from a spectrum of waves, each of which must represent a single electron. How many electron-waves are according to you superposing in free space to form a wave-packet?

How many energy eigenstates? Many. How many electrons? Just one, in a superposition.

This is not an abstract theory question, it's explicitly established by experiments. In the double-slit experiment, for example, Schrodinger's Equation suggests that the electron wavefunction is something like 1/sqrt(2) * (delta (x-x1) + delta(x-x2)). It's in a superposition of two position eigenstates. Is that, as you suggest, "really" two electrons---"every wave must represent a single electron"? Or is it, as the Born Interpretation says, just one electron, simultaneously in both states? Experiments say very clearly that it's just one electron.

The equivalent experiment for energy eigenstates is, say, Nature 459, 546-549 (28 May 2009) . The authors have a small resonant cavity which they can load with quantum superpositions of different amounts of energy. The energy eigenstates are states with energy 0, hw, 2hw, 3hw, etc.. The authors are able to put the cavity in arbitrary non-eigenstates which are superpositions of 0 and hw, 0 and 3hw, etc.. They demonstrate that the 1/(sqrt2)( |0> + |3> ) state is different than the 1/(sqrt2)( |0> + i*|3> ) state, but that both have a 50% chance of collapsing to the |0> state (and 50% to the |3> state) on measurement. Etc. They also demonstrate the |1> + delta*|3> + |6> three-way superposition.

Your "it's really two electrons" objection is nonsensical here; it's one cavity, a little stripline resonator on a chip. Preparing the |0>+|3> state does not mean you manufactured two cavities, one with the |0> and one with the |3>.

Again, Johan, your objection might have been worth looking twice at in 1926, when the Schrodinger Equation was a mysterious mathematical abstraction and the Born Rule didn't have all its ducks in a row. The subsequent 84 years of experiments have been unanimous in saying you're wrong.

 

[Reality Check]

I do not dispute that this is the accepted dogma: In fact I have taught the same nonsense to my students for years.

Wrong: it is a mathematic fact that you have being teaching.
The fact is that there is a wave packet solution of Schroedinger's equation for a single free particle.

I am just pointing out that a wave packet can only form from a spectrum of waves, each of which must represent a single electron.

Some basic quantum mechanics for you johanfprins:
For each particle in a system the wave function contains a separate set of position coordinates. If the wave function has a single set of position coordincates then it describes a single particle.
Now read and try to understand the quoted solution:

Originally Posted by sol invictus

, from which one obtains

See the single position coordinate (just x for 1D): CAn you count to 1, johanfprins ?

How many electron-waves are according to you superposing in free space to form a wave-packet?
Electron-waves do not exist so none.
The wave function quoted above is a supposition of an infinite number of plane waves each with a different k.

...
Relativity is always valid whether it is for fast relative speeds or for low relative speeds. Do you want to tell me that Schroedinger's equation is only valid when the electron is stationary?

Why are you mentioning relativity in the context of Schroedinger's equation? It makes you look so ignorant that you do not know that Schroedinger's equation is non-relativistic.

And it is obvious that Schroedinger's equation applies whatever the momentum of the particle. Look at the above solution. N.B. This is a solution to the time-dependent Schroedinger's equation (not that the wave function has time in it). You seem to be confusing this with the time independent or stationary equation.

...
All Einstein's arguments start with time-dilation and length-contraction, which according to him demands that space-time must be curved:
...snipped personal anecdote...

"All Einstein's arguments", i.e. the derivation of Einstein's theory of gravity , starts with the equivalence principle. It is this that which requires the introduction of curved space-time.
That space-time is then endowed with a pseudo-Riemannian metric in order to add Lorentz invariance, i.e. relativity.

...The reason why space and time is curved seems to require that all matter consist solely of waves.

No - all matter consists of waves even in non-relativistic theories (see Schroedinger's equation).

The non-spreading wave packet is not a solution of the Schroedinger equation. I thought we are talking about the latter. Maybe YOU should visit that website and study it?

The non-spreading wave packet is a solution of the Schroedinger equation.
There is no "latter" in your post.
What web site?

ETA: It just clicked: Maybe what you mean by "non-spreading" is nondispersive. The above solution is a dispersive solution. The width of the wave packet does spread with time.

 

[johanfprins]

Wrong: it is a mathematic fact that you have being teaching.
The fact is that there is a wave packet solution of Schroedinger's equation for a single free particle.

To derive a mathematical solution does not mean that it describes the physics you are trying to model.

Some basic quantum mechanics for you johanfprins:
For each particle in a system the wave function contains a separate set of position coordinates. If the wave function has a single set of position coordinates then it describes a single particle.

Correct! So why are you using a multitude of different waves, each having its own coordinate representation, and superpose them to form a wave packet which YOU then claim is a single electron wave?

Electron-waves do not exist so none.The wave function quoted above is a supposition of an infinite number of plane waves each with a different k.

You have just said that they do not exist? So you have a non-existent wave packet?

Why are you mentioning relativity in the context of Schroedinger's equation?

Your wave packet moves with a speed v<<c. So there must be an inertial refrence frame within which it is stationary.

It makes you look so ignorant that you do not know that Schroedinger's equation is non-relativistic.

A moving body with mass has nothing to do with relativity!? Really!

And it is obvious that Schroedinger's equation applies whatever the momentum of the particle. Look at the above solution. N.B. This is a solution to the time-dependent Schroedinger's equation (not that the wave function has time in it).

"Within" which stationary medium is this wave moving? It is not a light wave, and therefore it must have a stationary medium (a matter ether) within which it is moving for your statement to be correct.

You seem to be confusing this with the time independent or stationary equation.

I am not confusing anything.

"All Einstein's arguments", i.e. the derivation of Einstein's theory of gravity , starts with the equivalence principle. It is this that which requires the introduction of curved space-time.

Really!? I do not think so.

No - all matter consists of waves even in non-relativistic theories (see Schroedinger's equation).

I am glad you stated something I can agree with: So you also reject the existence of particles and Bohr's principle of complementarity?

The non-spreading wave packet is a solution of the Schroedinger equation.
There is no "latter" in your post.
What web site?

The WIKI one from which you quote for the wave-packet solution.

ETA: It just clicked: Maybe what you mean by "non-spreading" is nondispersive. The above solution is a dispersive solution. The width of the wave packet does spread with time.

Correct! And that is why a free electron cannot be a wave packet since it will spread so that the probability (a la Born) to find it will become zero. Experimentally we know that this is not so.

 

[sol invictus]

And that is why a free electron cannot be a wave packet since it will spread so that the probability (a la Born) to find it will become zero. Experimentally we know that this is not so.

No, experimentally we know it is so. The spreading of that wavepacket is a direct consequence of the uncertainty principle. The momentum isn't definite; therefore the uncertainty in position grows with time.

That effect is observed in thousands of experiments. A closely related phenomenon is the basis of scanning-tunneling microscopes, which are an off-the-shelf technology these days, and it's also a major issue for chip manufacturers (because electrons need to be very well localized to stay within a circuit element on a microchip).

This is all basic, basic QM - and QM is by far the most successful theory in the history of science. Its closest competitor is evolution, but QM is far better tested and perhaps even more broadly applicable.

 

[Ziggurat]

To derive a mathematical solution does not mean that it describes the physics you are trying to model.

Quite correct. If the theory is wrong, then the mathematical solution doesn't represent reality.

You think quantum mechanics is wrong. Good luck with that.

Correct! So why are you using a multitude of different waves, each having its own coordinate representation, and superpose them to form a wave packet which YOU then claim is a single electron wave?

One can "decompose" a wave packet by taking its Fourier transform, and one can reconstruct a wave packet by taking the inverse Fourier transform of the Fourier transform. But that's actually irrelevant. That's just a mathematical operation can be performed on any function, including single-particle bound state wave functions. The fact remains that the wave packet previously given solves the single-particle Schrodinger equation. A multiple-particle solution requires multiple coordinates, which the previous solution did not contain. So the previous solution obviously cannot represent multiple electrons.

 

[Reality Check]

To derive a mathematical solution does not mean that it describes the physics you are trying to model.

It turns out that this mathematical solution does decribe the physics of a free particle.

Correct! So why are you using a multitude of different waves, each having its own coordinate representation, and superpose them to form a wave packet which YOU then claim is a single electron wave?

Wrong! Every wave in the suposition has the same coordinate representation.

You have just said that they do not exist? So you have a non-existent wave packet?

I said the "electron-waves" do not exist. There are no waves of electrons in the solution. There is a single pa reticle.

Your wave packet moves with a speed v<<c.

The wave packet moves with any velocity that we like. It is a solution from a non-relativistic equation. The momentum and so speed is unlimited.

I am glad you stated something I can agree with: So you also reject the existence of particles and Bohr's principle of complementarity?

Wrong: All matter is particles and waves (wave/particle duality)

The WIKI one from which you quote for the wave-packet solution.
Correct! And that is why a free electron cannot be a wave packet since it will spread so that the probability (a la Born) to find it will become zero. Experimentally we know that this is not so.

Wrong!
The solution shows that measurements of momentum become more uncertain with position as expected from the uncertainty principle.

johanfprins: Can you confirm that you can understand that the solution is for a single particle?
As I pointed out this is obvious because the wave function has just 1 set of position coordinates (for that particle). A wave function for multiple particles has a set of position coordinates for each particle.

 

[johanfprins]

It turns out that this mathematical solution does decribe the physics of a free particle.

Show me an experiment on a single electron in free space that confirms it.

Wrong! Every wave in the suposition has the same coordinate representation.

The waves completely overlap but they must be different waves with different energies. This is what you have in an ideal metal: Every electron-wave has its own coordinate system but since they overlap one can solve for each of these waves as if each is determined by the SAME single particle wave. When there are N valence electron waves, only these N waves can superpose to form N wave-packets when you apply an electric-field so that a current can flow. If there were not N actual waves, you would not have had wave-packets forming within the metal: And yes, of course these wave packets can spread, since when switching off the electric field, they spread to again fill the whole metal as delocalised waves.

I said the "electron-waves" do not exist. There are no waves of electrons in the solution.

So what are you then superposing in free space? Nothing?

There is a single pa reticle.

What is a pa reticle? Even if you meant "particle": Please define what a "particle" is.

The wave packet moves with any velocity that we like. It is a solution from a non-relativistic equation. The momentum and so speed is unlimited.

In contrast to your wrong argument, an electron has mass and it only moves with a constant speed when it is observed from an inertial refrence frame which moves relative to the inertial refrence frame within which it is stationary. This defines all relativity! Thus your own words condemn your interpretation.

Wrong: All matter is particles and waves (wave/particle duality)

Nonsense! You can only state this if you can explain to me in which way a particle and a wave differs from each other. Please define what a "particle" is and why it is different from a localised stationary field that has a distribution in free space; and which can only be observed to move from another inertial refrence frame passing the stationary wave with a speed v.

Wrong!The solution shows that measurements of momentum become more uncertain with position as expected from the uncertainty principle.

The relationship (delta)k*(delta)x>g is valid for all harmonic waves and has nothing to do with such a wave's momentum.

johanfprins: Can you confirm that you can understand that the solution is for a single particle?

In free space it must be a non-spreading stationary wave within its own inertial refrence frame. If it is not the case, the solution violates Galileo's inertia and Newton's first law. In fact, it violates all physics.

As I pointed out this is obvious because the wave function has just 1 set of position coordinates (for that particle). A wave function for multiple particles has a set of position coordinates for each particle.

So has the solution for the N valence electron waves within an ideal metal. But there are still N waves involved.

I do not want to get into a discussion with Sol, Ben or Ziggurat, since they cannot follow any logic outside the dogma they want to believe. Nonetheless the following by Sol is hilarious: "That effect is observed in thousands of experiments. A closely related phenomenon is the basis of scanning-tunneling microscopes, which are an off-the-shelf technology these days, and it's also a major issue for chip manufacturers (because electrons need to be very well localized to stay within a circuit element on a microchip)."So a scanning-tunnelling microscope measures electrons in free space!! How deranged can you become. The scanning-tunnelling microscope measures wave packets on the surface and within material. In this case each charge-carrier is a wave packet that is formed by the superposition of N electron waves, and therefore they will obviously spread. But this does not happen for a single electron in free space.

 

[Reality Check]

Do you understand that a wave function has position coordinates for each particle

Show me an experiment on a single electron in free space that confirms it.

The double slit experiment when performed with single photons or electrons?
It certainly shows that the particle is a wave.

The waves completely overlap but they must be different waves with different energies.

That is almost right - the wave function is an integration over an infinite number of plane waves each with a different k.
The rest is just a rant about an ideal metal.

So what are you then superposing in free space?

Read the equation:

Originally Posted by sol invictus

, from which one obtains

What is integrates is an infinite number of plane waves each with a different k.

In contrast to your wrong argument, an electron has mass and it only moves with a constant speed
...snipped off topic relativity stuff...

That is nothing to do with what I said. The free particle can have any velocity. There is no restriction on momentum in the non-relativistic equation.
A specific particle (like an electron) in free space like an electron has mass and has a constant speed, i.e. a constant momentum. That constant momentum is the p₀ in:

​

Nonsense! You can only state this if you can explain to me in which way a particle and a wave differs from each other.

Wow - you are really displaying your ignorance

• particle
• wave
• wave/particle duality

...But there are still N waves involved.

Wrong: The wave function for a free particle has a infinite number of plane waves in it. The wave function for N free particles has an infinte number of plane waves in it.

You seem to be ignorant of what a wave packet is:

In physics, a wave packet (or wave train) is a short "burst" or "envelope" of wave action that travels as a unit. A wave packet can be analyzed into, or can be synthesized from, an infinite set of component sinusoidal waves of different wavenumbers, with phases and amplitudes such that they interfere constructively only over a small region of space, and destructively elsewhere.

And once more I will ask the question:
Do you understand the basic quantum mechanics that a wave function has a set of position coordinates for each particle?

The above wave function has 1 set of position coordinates. It describes 1 free particle.

 

[johanfprins]

The double slit experiment when performed with single photons or electrons?
It certainly shows that the particle is a wave.

Of course a free electron is a wave: It is never a particle and when it diffracts through a double slit it is also not a wave-packet. To obtain the clean diffraction patterns one obtains from electrons within an electron microscope, an electron wave must be coherent and this means that each electron-wave cannot be a wave-packet. Have you ever seen how a light diffraction pattern looks when you have incoherent waves: i.e. waves which are a superposition of different wavelengths?

That is almost right - the wave function is an integration over an infinite number of plane waves each with a different k.

Thus an electron moving with a speed v cannot have a single well-defined wavelength? Oh my poor de Broglie!

The rest is just a rant about an ideal metal.

It is not a rant since it is only under these conditions that electron-waves form wave-packets.

Read the equation:What is integrates is an infinite number of plane waves each with a different k.

Where do these plane waves come from if each of them is not an electron-wave in free space?

That is nothing to do with what I said. The free particle can have any velocity. There is no restriction on momentum in the non-relativistic equation.

There is no such thing as a "non-relativistic" description of a moving entity. If it moves, then what you observe must ALWAYS be determined by the Lorentz relativistic transformation equations. When the movement becomes slow these equations become the Galilean equations, but they are still valid. Thus any electron moving with any speed MUST have an inertial reference frame within which it is stationary: i.e. within which its momentum and position manifest simultaneouly. I ask you again to derive the solution from Schroedinger's equation for an electron that is stationary within its own inertial refrence frame. If you cannot then your physics is poppycock.

A specific particle (like an electron) in free space like an electron has mass and has a constant speed, i.e. a constant momentum.

Momentum is always a relativistic parameter. It depends on which inertial refrence frame you use to measure it in. You cannot ignore relativity. You might want to argue that Schroedinger's equation can only be valid for electrons moving with slow speeds: But this makes your statement that the Schroedinger equation is valid for any momentum invalid.

That constant momentum is the p₀ in:

[qimg]http://www.randi.org/latexrender/latex.php?$|\psi(x,t)|^2=\pi \frac{e^{-\frac{(x-2 p_0 t)^2}{2 \left(1+t^2\right)}} }{\sqrt{1+t^2}}$[/qimg]
​

Wow - you are really displaying your ignorance

• particle
• wave
• wave/particle duality

Please stop these pathetic references to dogma I have already learned more than 40 yeras ago. It is time that you try and argue logic by yourself. All you are doing is to quote cliches: This is not arguing physics!

Wrong: The wave function for a free particle has a infinite number of plane waves in it.

Coming from where?

The wave function for N free particles has an infinte number of plane waves in it.

Wrong! When you use periodic boundary conditions, which applies when you send a current through an ideal metal, each valence electron wave is a plane wave; and only these N plane waves superposse to form a wave packet.

And once more I will ask the question:
Do you understand the basic quantum mechanics that a wave function has a set of position coordinates for each particle?

Not true in all cases: For example the many-body electron wave within a metal, when no current is flowing, has no position coordinates for each particle. The two electrons forming a covalent bond has then same position coordinates. Similarly in the case of a REAL Bose-Einstein condensate: Here I am not talking about liquid helium which is not a real Bose-Einstein condensate; but about the real Bose-Einstein condensate that are formed a few millionths of a degree above absolute zero temperature.

The above wave function has 1 set of position coordinates. It describes 1 free particle.

No it does not: It describes a wave packet and a free electron can never be a wave packet.

 

[Ziggurat]

Of course a free electron is a wave: It is never a particle and when it diffracts through a double slit it is also not a wave-packet. To obtain the clean diffraction patterns one obtains from electrons within an electron microscope, an electron wave must be coherent and this means that each electron-wave cannot be a wave-packet. Have you ever seen how a light diffraction pattern looks when you have incoherent waves: i.e. waves which are a superposition of different wavelengths?

This is simply not true. A wave packet is at least slightly non-monochromatic, but the wavelength spread can be quite small. And for a small wavelength spread, the blurring of the diffraction pattern is correspondingly small. In fact, it can be arbitrarily small. So it's simply false to claim that you don't get a diffraction pattern from a wave packet. You do. You must, because we can obtain single-particle diffraction patterns, and (via both the Heisenberg and the energy-time uncertainty principles) it's physically impossible to create perfectly monochromatic single-particle beams.

I ask you again to derive the solution from Schroedinger's equation for an electron that is stationary within its own inertial refrence frame. If you cannot then your physics is poppycock.

This is standard textbook quantum mechanics (Griffiths, "Introduction to Quantum Mechanics, 2nd Ed", page 67):

psi(x,t) = (2a/pi)^1/4exp[-ax²/(1+2i*hbar*a*t)]/sqrt(1+2i*hbar*a*t)

This solves the Schrodinger equation. It has <p> = 0. And it's even a minimum uncertainty state at t=0.

You may object that this isn't stationary since it spreads out over time. This is a necessary consequence of the Heisenberg uncertainty principle, which requires that <p²> is NOT zero. And it's an experimental fact as well. You have mentioned that every electron is stationary in its own reference frame. This is true. So is the converse: every stationary electron is moving in some reference frame. So if you could make a stationary (meaning <p>=0) wave function which DOESN'T spread out over time, then you can also make a moving electron which doesn't spread out over time. Can we do that? No, we cannot. It's impossible to make a free electron beam with zero divergence. Which is exactly equivalent to saying that it's impossible for a stationary free electron to not spread out over time. The only way to prevent divergence is to put the electrons in a confining potential, at which point it's no longer a free electron.

So your arguments are a complete failure. Your claims match neither theory nor experiments.

Not true in all cases: For example the many-body electron wave within a metal, when no current is flowing, has no position coordinates for each particle. The two electrons forming a covalent bond has then same position coordinates.

That's simply not true. Wave functions may be symmetric (or antisymmetric) with respect to the various coordinates, but it quite definitely must HAVE multiple coordinates to describe multiple particles. If someone exploits symmetry to reduce the notation to one coordinate, that doesn't mean that the actual complete wave function doesn't have multiple coordinates. It does. That's standard quantum mechanics. You're free to disagree with standard quantum mechanics (but good luck), but you don't get to pretend that standard quantum mechanics says something it clearly doesn't say.

 

[ben m]

Have you ever seen how a light diffraction pattern looks when you have incoherent waves: i.e. waves which are a superposition of different wavelengths?

Yes I have. It is blurry depending on the amount of incoherence.

Indeed, have you ever seen a diffraction pattern that did not involve a light wavepacket? At the very least, you turned on your light source at time A, and you turned it off at time B. Sorry, Johan, under those circumstances it is impossible to describe the light pulse with a sine wave. Instead, it's a truncated sine wave, which---guess what?---is a wavepacket, localized in space (until dispersion takes over) and containing a range of frequencies.

Thus an electron moving with a speed v cannot have a single well-defined wavelength? Oh my poor de Broglie!

Where are you going to get "an electron moving at speed v"? From an infinitely-long apparatus? You've forgotten Heisenberg's Uncertainty Principle. dx dp = m dx dv > hbar/2. The only way to make an electron with no velocity uncertainty is to make it perfectly sinusoidal from one edge of the Universe to the other.

No, rather, any real electron has a range of values for v,p,lambda,E, and so on.

 

[sol invictus]

There is no such thing as a "non-relativistic" description of a moving entity. If it moves, then what you observe must ALWAYS be determined by the Lorentz relativistic transformation equations. When the movement becomes slow these equations become the Galilean equations, but they are still valid. Thus any electron moving with any speed MUST have an inertial reference frame within which it is stationary: i.e. within which its momentum and position manifest simultaneouly.

You've exhibited many times your total failure to comprehend the distinction between <p> and <p²>, or between <v> and <v²>, and here it is again.

The electron's rest frame is by definition the frame in which <v>=<p>=0, but in that frame neither <v²> nor <p²> can be zero.

I ask you again to derive the solution from Schroedinger's equation for an electron that is stationary within its own inertial refrence frame. If you cannot then your physics is poppycock.

Already done (for a free electron in 1D). It's the wavepacket solution I posted earlier, with p₀=0.

 

[ben m]

Just piling on, not that it will help:

http://www.nature.com/nphys/journal/v2/n5/full/nphys290.html

Electrons---yes, single electrons---can be prepared in wavepacket states.

a) The experiment guarantees that they're single electrons, because they're produced from a single argon atom by two sequential ionizing laser pulses.

b) The electrons are explicitly localized; they're emitted from a particular atom, in a narrow time window set by the laser pulse. These are not infinite plane waves.

c) The experiment explicitly detects the wide momentum range of the single electrons.

d) The experiment explicitly shows that the wide momentum range is present in each electron---as opposed to a classical statistical effect where the results are scattered due to an error---because each electron pair goes through the full all-momenta diffraction pattern. (An incoherent experiment, where each electron had one randomly-chosen momentum, would have no fringes visibility.)

There's no ambiguity here, Johan. There hasn't been ambiguity on these questions since the mid-1920s. Your statements to the contrary do not sound like "dissident physics tries a radical new approach", they sound like "you criticized my pet hypothesis, citing X, so I'm going to say X is wrong. What is this X? I don't care, the important thing is saying it's wrong."

 

[Reality Check]

Can I conclude that you have no idea what a wave function is

Thus an electron moving with a speed v cannot have a single well-defined wavelength?

So an electon moving with a speed v has have a single well-defined de Broglie wavelength.

Where do these plane waves come from if each of them is not an electron-wave in free space?

You ignorance continues: The plane waves come from each of the possible values of k in

This is a single particle (1 electron) wave function.

...
But this makes your statement that the Schroedinger equation is valid for any momentum invalid.

That is stupid. The Schroedinger equation is non-relativistic. Thus it is valid for any value of momentum.

You need to use quantum field theory to include Lorentz invariant (special relativity) in quantum theory.

Please stop these pathetic references to dogma I have already learned more than 40 yeras ago.

The links are to the scientific definition that you asked for.

Coming from where? Wrong! When you use periodic boundary conditions, ...

More idoicy anout ideal meatals.
We are talking about a free particle. There is no metal (ideal or otherwise). There are no periodic boundary conditions.

Not true in all cases: For example the many-body electron wave within a metal, when no current is flowing, has no position coordinates for each particle.

That is totally wrong: By definition the wave function describes all of the possible states of the system. That includes the positions, momentum and spins of each particle in the system.

Can I conclude from this that you have no idea what a wave function is?

No it does not: It describes a wave packet and a free electron can never be a wave packet.

Yes it does. It has 1 set of postion coordinates for the single free particle.
johanfprins, can you count to 1 ?

A free electron is never a wave packet. The expectation values for its state variables like position and momentum are wave packets. The width of the wave packet is a measure of the uncertainty in a measurement of the state variable.

 

[johanfprins]

So an electon moving with a speed v has have a single well-defined de Broglie wavelength.

What does "has had" mean? The fact remains that a photon has the same single frequency and value for k within all inertial reference frames: Thus it cannot be a wave packet. A free electron also has a single frequency and value for k within any inertial reference frame relative to which it is moving; but in this case it is not the same within all inertial refrence frames since it is directly proportional to the speed with which it is moving relative to the inertial refrence frame from which it is observed. Nonetheless, the wave vector k and frequency both again have definite values: Thus the free electron also, like the photon, cannot be a wave packet.

Your ignorance continues: The plane waves come from each of the possible values of k in
[qimg]http://www.randi.org/latexrender/latex.php?$\psi(x,t) = \int dk e^{i(kx-\omega t) -k^2 + i p_0 x}$[/qimg]
This is a single particle (1 electron) wave function.

It only exists when the boundary conditions are such that it can exist. You have concocted it without stating what the boundary conditions must be for it to exist. And then you assumed, without giving any logical physics-reason, that it represents an electron in free space. There is no experimental proof that this must be so. The only experimental proof we have of the wave-nature of a freely moving electron is through diffraction, and this gives a single value for k and the frequency when an electron moves with a speed v. This proves that a freely moving electron is coherent and therefore it cannot be a wave packet.

That is stupid. The Schroedinger equation is non-relativistic. Thus it is valid for any value of momentum.

It is clear that you do not have a clue what the theory of relativity entails.

You need to use quantum field theory to include Lorentz invariant (special relativity) in quantum theory.

Any theory which is based on a vacuum energy which is infinite and in which one has to subtract infinity from infinity to get a finite number must be pure poppycock.

The links are to the scientific definition that you asked for.

I did not ask for links since my background in physics is obviously far superior to yours. I am asking you to define what you are writing about without falling back on the mantra that what I am arguing must be wrong since it is not what the "accepted" interpretation is. The fact that an interpretation is "accepted" does not mean it is correct. The most basic principle on which physics must be based is that all "accepted" models must always be open to scrutiny, and that at any time an "acccepted" model can be proved wrong: If you do not follow this rule, then I hope you are not a physicist.

More idoicy anout ideal meatals.
We are talking about a free particle. There is no metal (ideal or otherwise). There are no periodic boundary conditions.

Yes we are talking about a free particle and you keep on claiming that a free electron is a wave packet which it cannot be since it has a unique de Broglie wavelength. And I keep pointing out to you that an electronic wave packet can only manifest within a conducting material as a pseudoparticle.

That is totally wrong: By definition the wave function describes all of the possible states of the system. That includes the positions, momentum and spins of each particle in the system.

That is the "accepted" dogma based on the assumption that there are particles. Electrons cannot be particles since particles can NEVER diffract. Furthermore, a free electron cannot be a wave packet, since a moving electron has a unique de Broglie wavelength: It does NOT consist of a superposition of wavelengths.

Can I conclude from this that you have no idea what a wave function is?

Can I conclude that you have no idea when a wave function models a physically real situation and when not?

Yes it does. It has 1 set of postion coordinates for the single free particle.
johanfprins, can you count to 1 ?

But it does not model a free solitary electron.

A free electron is never a wave packet. The expectation values for its state variables like position and momentum are wave packets. The width of the wave packet is a measure of the uncertainty in a measurement of the state variable.

This is the "accepted" dogma which must be wrong for many-many reasons which I discuss in detail in my book.

 

[sol invictus]

I'll tell you what, Johan - why don't you write down the solution to the Schrodinger equation that you think represents a single, stationary electron in empty space. I don't want words, I want an equation, like ψ(x,t)=...

Remember, |ψ(x,t)|² is the probability density for finding the particle at x, and therefore the integral of |ψ(x,t)|² over x had better equal 1.

 

[Reality Check]

What does "has had" mean? The fact remains that a photon has the same single frequency and value for k within all inertial reference frames

So an electon moving with a speed v has a single well-defined de Broglie wavelength.

No: a photon has the same measured single frequency and value for k.
The free-particle solution is integrated over all posisble k:

Thus it cannot be a wave packet.

Right: an electron is not a wave packet.
The expectation values for its state variables (measurable quantities) are wave packets. That is why the square of the magnitude of the wave function has been given:

This solution is one value of k (p₀)

You have concocted it without stating what the boundary conditions must be for it to exist.

How dumb: The boundary conditions for a free-particle is that there are no boundary conditions because there are no boundaries !

And then you assumed, without giving any logical physics-reason, that it represents an electron in free space.

The free-particle solution represents a free particle for the simple physical reasons that

1. There is no potential in the Hamiltonian (it is a Hamiltonian for a free particle).
2. The wave function has 1 set of position coordinates and so it is for a single particle.

As for experiment proof: Attosecond electron wave packet interferometry
Let us hope that you do not go the crank route and whine about the experiment not being on a single electron !

It is clear that you do not have a clue what the theory of relativity entails.

I know what the theory of relativity entails.

I can also read. I can read that the Schrödinger equation is a non-relativistic equation. That is something you seem unable to grasp.

Any theory which is based on a vacuum energy which is infinite and in which one has to subtract infinity from infinity to get a finite number must be pure poppycock.

Nothing to do with the theory of relativity. What you are refering to is quantum field theory and the need to renormalize.
You have it wrong:
Vacuum energy is not infinite , e.g. in QED it is 10¹⁰⁷ Joules per cubic centimeter.
There is no subtracting infinity from infinity. No one would be dumb enough to do that because infinity is not a number. The reduction of infinities in renormalization (regulization) is basically the repackaging of series of finite terms that each sum up to infinity into equivalent series of finite terms that do not sum up to infinity.

I did not ask for links since my background in physics is obviously far superior to yours.

Actually given your ignorance about what the wave function is and the rest of the misconceptions that you have, your knowledge of physics is certainly less than mine.
Our backgrounds in physics do not matter - if you are displaying misconceptions about such basic physics then you need to go back and relearn the physics.

I am asking you to define what you are writing.

I do not need to define such basic terms as wave or particle.

Yes we are talking about a free particle and you keep on claiming that a free electron is a wave packet which it cannot be since it has a unique de Broglie wavelength.

And there you go again.
I d not recall claiming that that a free particle is a wave packet.
A free particle is not a wave packet.
The expectation values for its state variables (measurable quantities) are wave packets. That is why the square of the magnitude of the wave function has been given.

And I keep pointing out to you that an electronic wave packet can only manifest within a conducting material as a pseudoparticle.

And I keep on pointing out that you are wrong.

That is the "accepted" dogma based on the assumption that there are particles. Electrons cannot be particles since particles can NEVER diffract.

That is ignorant. Electrons interact as waves and so ythey must be waves. Electrons interact as particles and so they must be particles. Eleactons have the proerties of both particle and waves.

Furthermore, a free electron cannot be a wave packet, since a moving electron has a unique de Broglie wavelength: It does NOT consist of a superposition of wavelengths.

Furthermore, a free electron is not a wave packet. A moving electron has a unique measured de Broglie wavelength.
The wave function of the free electron consists of a superposition of wavelengths. The expectation value of its momentum is a non-dispursive wave packet (in contrast to the expectation value of its position which is a dispursive wave packet).
A measurement of the momentum of a free electron collapses the wave function (in the Copenhagen interpretation) and you get a unique measured de Broglie wavelength.

Can I conclude that you have no idea when a wave function models a physically real situation and when not?

I have plenty of ideas when it does not and when it does.

Then you are wrong as has been show:

• There is a solution to the Schrödinger equation for a free particle.
• The result is supported by evidence: Attosecond electron wave packet interferometry

As a complete aside: It is the Dirac equation that more completely describes a free electron. But we are not interested in relativity or spin so the Schrödinger equation suffices. A pity because many years ago a third year pure math course at university was to write a paper on the development of a key concept in mathematics and I selected the Dirac equation. So I used to know a lot about it. Maybe in another thread.
Of course if we really wanted to discuss electrons in a metal as you seem to bring up all of the time than we should look at something I used in my thesis - the Wannier representation in WFT.

 

[Ziggurat]

The fact remains that a photon has the same single frequency and value for k within all inertial reference frames: Thus it cannot be a wave packet.

Well, that's rather obviously wrong. Haven't you ever heard of Doppler shift? The wavelength, frequency, and momentum of a photon ALL depend on reference frame. The velocity does not, but if you thought velocity and momentum were related for a photon, well, that would be monumentally stupid.

The fact that an interpretation is "accepted" does not mean it is correct.

True. But you have in the past denied that you disagree with basically all of 20th century physics. And it's quite clear that you do. The fact that you disagree doesn't prove you wrong (though plenty of provably false claims about experimentally verified facts, such as Doppler shift of light, do), but it does set a scale for what it is you would need to accomplish in order to prove that you're right. And honestly, you're just not up to the task. Not when you keep making mistake after mistake after mistake, a predictable litany of failures.

That is the "accepted" dogma based on the assumption that there are particles. Electrons cannot be particles since particles can NEVER diffract.

Oh goodie! A purely semantic objection!

If you define particles as things which never diffract, then sure, nothing is a particle. But there's no need to define particles that way. And modern physics doesn't. So your objection is based purely on your personal definition of the word, a definition that basically no one else in the field shares.