Superconductivity at room temperature??

[Reality Check]

Plasmas only form under very special conditions which are jointly determined by both the pressure (vacuum quality) and the applied electric field. When switching off the electric field the plasma collapses. In my case my superconducting phase does not, and even when you then also collapse the vacuum, the phase still does not collapse as it MUST if it were a plasma.

That sounds strange - how did you meaure your "superconducting" phase without an electric field?

My experiment is the first that supplies the best impeccable evidence EVER, that there cannot be an electric field present (within the phase that I have discovered) while a current is transferred through this phase.

Good - can you cite your measurements that the electric field is zero?
Perhaps a link to your paper and the page(s) where you measured this?

He assumed ...

You remain mistaken about what he stated. He claimed that the measurements that you took can not demonstrate superconduction:
Superunlikely a letter? in New Sceintist by Archie Cambell

If the gap were made of copper the resistance would be about 0.04 micro-ohms, so the only conclusion that can be drawn is that the resistivity of the electron gas is no greater than around 10^11 times that of copper. This cannot be regarded as evidence of superconductivity.

Your measuments give an upper limit to the resistivity of the electron gas. That upper limit is around 10^11 times that of copper. This is not superconductivity.

Maybe they have tried but were just to incompetent?
They do not have the equipment to do so and have other programs and their future to worry about than to spend thousands of dollars replicating my results.

Guess what, johanfprins - you are not the only competent scientist in the world !
The rest is a a bit of paranoia ("their future to worry about") together with a real problem for your results. If there is no equipment anywhere in the world that can reproduce your experiment then your experiment will never be reproduced !

From what little I know about experimental solid state physics, your experiment looks easy to replicate.

The point remains that fact that your experiment has been ignored (except by Prof Doyle) for 8 years now. This means that your reported results have failed to convince any other scientist who has read your paper to make the effort to replicate it.

 

[Ziggurat]

Many times, but on this thread you, Sol, Ziggurat, Reality Check and others set the trend and I like to respond in kind.

Did I set the trend? No, I don't think I did. Let's go back to the early history of this thread. Do you recall that early history? This is one of our first interactions, where you responded to my post by saying,

A typical crackpot remark.

Post #123 of this thread. I hadn't insulted you, and you insulted me. You set the trend, not me. If you don't like the tone of our exchanges, you have nobody to blame but yourself.

 

[ben m]

Apropos of Zig's trip down memory lane, I wonder how the following early post is compatible with Johan's claim to have "many, excellent" physicists supporting him anonymously. (ArXiV endorsements are entirely confidential, BTW.)

Not true: You need to get somebody who has already published on ArXive to "recommend" (I do not know whether this is the correct word) that it can be published. I could not find such a person and then wrote to ArXive telling them that I am isolated. They then allowed me to publish these two manuscripts subject to certain restrictions.

Did I set the trend? No, I don't think I did. Let's go back to the early history of this thread. Do you recall that early history? This is one of our first interactions, where you responded to my post by saying,

Remember that the thread began, not with Johan himself posting, but with !Kaggen's stumbling across some random website (cathodixx) claiming to have discovered superconductivity. I did not hesitate to call that claim, and the many related claims in various papers on that website, crackpottery. So did several other people. (Ziggurat, you did not.)

When Dr. Prins actually showed up, I was much more polite. There's a big difference between talking to a person and talking about his website. But it didn't help! He doesn't seem to draw a distinction between statements like "you are a crackpot" (a rude insult), "this is a crackpot paper" (a harsh judgement), "You have misplaced a factor of 2" (a normal judgement), and "by X, do you mean Y or Z? Because Z is wrong" (a normal question).

 

[johanfprins]

How do I know what is or is not required? You're making a statement that you seem to intend to use eventually to "prove" your idiosyncratic theory of superconductivity.

Not idiosyncratic since my logic is based on elementary physics which you can find in any text book on Solid State Electronics.

I have no way of knowing what parts of this statement are important. How do I know what error, ambiguity, or mis-definition you might eventually rely on in your "proof" that BCS theory is wrong? Since I don't know how much detail you're going to squeeze out of this stupid "statement", the only thing I can do with it is ask for it to be actually correct.

It has become a simple statement about foming a Schottky barrier with an n-type semiconductor. Nothing more nothing less. If you want to argue that I have made an "error, ambiguity, or mis-definition" then do so whaen I post such a statement. In statement 1, as it stands now, there are no "error, ambiguity or mis-definition" in the descriptioon of the formation of a Schottky barrier.

I have a better idea, though. Get to the point.

I have tried that many times on this thread but then you Sol and Ziggurat duck and dived, by playing semantics and going off on tangents outside the topic that is being discussed.

The only option left to me is to go statement by statement through my logic. Even when you then duck and dive and deliberately try and confuse the issue, I can then reformulate the statement until you are satisfied before proceeding to the next statement. Until I get an agreement that I can proceed to the second statement, I will just keep on posting the first statement, and will not respond to issues outside the topic as to whom insulted whom the most on this thread: Neither will I waste time anymore to respond to people like Reality Check, who thinks that he did physics by programming Wannier functions and whose only recognition in physics is a citation by his supervisor. Here is statement 1 again:

Statement 1: When creating an interface between an n-type semiconductor and a metal, where the latter has a Fermi level at a lower energy than the n-type semiconductor, and when neither of the two materials are charged up before generating the interface, then, after creating the interface, a dipole layer forms across the interface, since negative charge-carriers move across the interface from the n-type semiconductor into the metal, thus leaving a positively charged depletion-layer behind (in contact with the interface), and forming a negatively charged layer (in contact with the interface) by accumulation within the metal. Yes or No?

 

[Reality Check]

Neither will I waste time anymore to respond to people like Reality Check, who thinks that he did physics by programming Wannier functions and whose only recognition in physics is a citation by his supervisor.

How idiotic: I "did physics" by learning physics, remembering physics and refreshing my memory when needed.
I did not "program Wannier functions". You would know how ridiculous that statement was if you knew any solid state physics. My thesis was expressed in terms of the Wannier representation (look up the one-band Wolff model).
An example of the representation : Wannier Representation for Tight-Binding Hamiltonian of a Periodic Chain with N Sites


You may have "did physics" but your statements in this thread show that you have forgotten some of it, e.g.

• that the Dirac Equation is correct, relativistic description of the electron.
• the law of conservation of energy (an electron that gains an energy of qV can only emit photons up to a total energy of qV, not 2qV as you claimed before).
• The Schroedinger equation for a single particle is .... an equation for a single particle!

P.S. I have also forgotton some of the physics I have learned. But I am aware of this, know that I will make mistakes and are able to take corrections from posters in the forum.
It is a pity that you are going to ignore me since I have an answer to your statement 1.

 

[ben m]

It has become a simple statement about foming a Schottky barrier with an n-type semiconductor. Nothing more nothing less.

For crying out loud. Why didn't you say so to begin with? I thought you were trying to say something general about interfaces. Why not make Statement 1 "a Schottky barrier, which forms when X, consists of Y ... "

(Anyway, your statement is, as far as I can tell, an accurate description of a Schottky barrier. Finally.)

 

[Reality Check]

Missed this:

Many people erroneously believe that the Meissner effect provides this evidence. THIS IS NOT SO: Let me quote the following recent remark from a person who is revered as an expert on superconduction:

""Further notes on the centenary of superconductivity Physics World

When Grant suggests that the Meissner effect is the ideal test for superconductivity, he continues a misunderstanding that has led to some very poor papers on new superconductors. There are two reasons why it is a very bad test. First, the best superconductors, such as niobium-titanium-tin, do not show a significant Meissner effect, because “flux pinning” prevents the field from being expelled. Second, even if an apparent diamagnetic signal is seen, it is very easily confused with a reduction in paramagnetism.

Archie Campbell, University of Cambridge"

So when Archie Cambell makes a comment about some new superconductors, he is right.
But when this person who is "revered as an expert on superconduction" states
Superunlikely a letter? in New Sceintist by Archie Cambell

The example of room-temperature superconductivity that you report is unconvincing (12 April, p 17). It is clear from the original paper by Johan Prins that it would be quite impossible to detect superconductivity with his measurements.

he is magically wrong just because you say so !

 

[johanfprins]

Anyway, your statement is, as far as I can tell, an accurate description of a Schottky barrier.

Thanks!

Statement 2: Just after the initial contact between a metal with a large work-function and an n-type semiconductor in order to generate a Schottky barrier, the Fermi-level within the metal is at a lower energy than the Fermi-level within the n-type semiconductor. Therefore there is an electric-field across the interface which accelerates negative charge-carriers (pseudo-electrons) out of the semiconductor into the metal where these entities lose energy and accumulate at the interface within the metal since they are attrracted back to the interface by the positively-charged depletion layer they are leaving behind within the semiconductor. Yes or no?

 

[ben m]

Thanks!

Statement 2: Just after the initial contact between a metal with a large work-function and an n-type semiconductor in order to generate a Schottky barrier, the Fermi-level within the metal is at a lower energy than the Fermi-level within the n-type semiconductor. Therefore there is an electric-field across the interface which accelerates negative charge-carriers (pseudo-electrons) out of the semiconductor into the metal

NO. Emphatically no. The Fermi level is the sum of (A) an electron chemical potential---a potential created by Pauli exclusion alone, and relevant to electrons alone---and (B) an actual electrostatic potential, relevant to any charged particle. In this case, the pre-contact Fermi levels are different purely due to chemical potential.

The "force" that rearranges electrons, in the picosecond-after-contact, about-to-forming-the-barrier moment you describe, is diffusion, not electrostatic acceleration. If you shot a muon (or a pion or a positron) through the system at this moment, it would NOT feel an electric field, because there isn't one there.

ETA: Perhaps this is a terminology subtlety that most people don't have to deal with. I'm one of the ones who does, because I do a lot of high-precision vacuum electronics. If you want, you can take the gradient of the voltage (i.e. of the Fermi level) and call it a pseudo-field; that's enough like an electric field to make simple energy-conservation and direction-of-charge-flow arguments. But it's not an electric field, it doesn't exert Newton-like forces, it doesn't obey Gauss's or Coulomb's law, and you basically shouldn't do it.

 

[johanfprins]

NO. Emphatically no. The Fermi level is the sum of (A) an electron chemical potential---a potential created by Pauli exclusion alone, and relevant to electrons alone---and (B) an actual electrostatic potential, relevant to any charged particle. In this case, the pre-contact Fermi levels are different purely due to chemical potential.

So you are arguing that when there is a difference in the energy of the Fermi-level at two different points this does not mean that there is an electric-field between these two points? Why does the Fermi-energy then slope when there is an applied electric field present in a material, if it is not a fact that that there is an actual net electric field between two points at which the Fermi-energy are not the same?

The "force" that rearranges electrons, in the picosecond-after-contact, about-to-forming-the-barrier moment you describe, is diffusion, not electrostatic acceleration.

Diffusion? Diffusion is slip-stick motion, and the "particles" that "diffuse" still move from a higher electric-energy state to a lower electric-energy position. You must also remember that the initial electric field is over a much much smaller distance than the diffusion length of an electron within the n-type semiconductor. Will an electron at the interface then "diffuse" across the interface instead of being accelerated acros the interface? But no matter how it moves from the n-type semiconductor into the metal you are, as usual, missing the point.

You are arguing that when a charged-entity is at higher electric energy-potential at a position A, than its electric potential-energy will be at a position B, there is no electric-field between these two positions: You should sue your primary school teacher who introduced you to physics. There MUST be an electric-force in such a case atracting the conduction-charges from A to B. The fact remains that such a charge-carrier MUST BE attracted by an electric field from position A to position B NO MATTER HOW IT MOVES FROM A TO B. The latter is irrelevant in this argument. An energy offset of the Fermi-level between two separated positions mandates the presence of an electric field between the two positions.

Note added: I do not want to go off on a tangent, but it is quite hilarious that you can now post that: "call it a pseudo-field; that's enough like an electric field to make simple energy-conservation and direction-of-charge-flow arguments. But it's not an electric field, it doesn't exert Newton-like forces, it doesn't obey Gauss's or Coulomb's law, and you basically shouldn't do it." In our prvious arguments you were adamant that an applied electric field within a material DOES obey Gauss's and Coulomb's laws. Now it does not suit you to argue this anymore? And now you want to call the applied electric field that slopes the Fermi-level a "pseudo-field". Sorry my friend: Whatever you want to call this field, it is still the field that accelerates charge-carriers within a material. I do, agree that Gauss's law does not apply to this field, since I have consistently argued above that it does not. I am glad you seem to have come to your senses.

 

[ben m]

So you are arguing that when there is a difference in the energy of the Fermi-level at two different points this does not mean that there is an electric-field between these two points? Why does the Fermi-energy then slope when there is an applied electric field present in a material, if it is not a fact that that there is an actual net electric field between two points at which the Fermi-energy are not the same?

Read carefully. The Fermi energy is the sum of the electrostatic potential energy (traditionally "U") and a chemical potential (electron kinetic energy, sustained due to the fact that low-kinetic-energy states are filled already). If there's a slope in the electrostatic potential energy within a single material, that makes a slope in the Fermi energy. If there's a gradient in the allowed-kinetic-energies, between two neutral materials at the same electrostatic potential, that ALSO makes a slope in the Fermi energy.

Your statement-2 "Schottky diode, just assembled" is an example of the latter. Your new example (an applied electric field) is an example of the latter. Get it?

Diffusion? Diffusion is slip-stick motion, and the "particles" that "diffuse" still move from a higher electric-energy state to a lower electric-energy position.

Diffusion is (I think) the standard term, even if it doesn't have exactly the properties you think of for (e.g. gaseous) diffusion. Perhaps "effusion" is a better thing to visualize.

A neutral, high-Fermi-level material (call it gold) has high momentum electrons within it. A neutral, low-Fermi-level material (call it sodium) has only low-momentum electrons in it. If you put them in contact, there is a net flow from the high to the low---a high-energy electron from the gold, which happens to reach the gold/sodium interface, will simply keep going at a constant speed, right past the interface. (Thereafter, it may scatter, and now that it's in sodium there are plenty of lower-momentum states to end up in after each scatter.) A low-energy electron from the sodium, hitting the same barrier, must get reflected back into the sodium because its momentum state is Pauli-excluded from the gold.

(For an analogy with the same energetics, minus the quantum mechanics: Consider a box of high-temperature gas, with a small hole poke in the wall. Gas molecules will, on average, leave the box. Does that mean there's a force that accelerates them through the hole? No---the atom isn't "accelerating out of the box", it's just "leaving when it gets a chance.")

You are arguing that when a charged-entity is at higher electric energy-potential at a position A, than its electric potential-energy will be at a position B, there is no electric-field between these two positions:

Electric potential energy is not the only form of potential energy. In the case you chose to present, the total energy of the electrons has both an electrostatic component---which is the same at A and B---and a chemical component (which is actually a form of kinetic energy). A valence electron at A has a different chemical/kinetic energy than a valence electron at B, so, yes, it can move from A to B without experiencing an electric field.

The electric field obeys Gauss's Law, and will only appear when there is a charge excess somewhere. Such a charge excess will be present after some electrons have been displaced by the effusion.

ETA: I want to emphasize, Johan, that it is appalling that you haven't at least heard of what I'm talking about. Even if you disagree with it, even if you've set out to prove it wrong, everything I'm saying ought to be familiar. Rather than "go back to school you idiot", a competent dissident-physicist might say, "Aha, I see you're following the Landau & Lifschitz view, which is inappropriate because X". When Einstein wrote a paper that discredited aether-based theories of light, he didn't do so having never heard of the aether, and thinking aether theorists were confused morons..

 

[ben m]

In our prvious arguments you were adamant that an applied electric field within a material DOES obey Gauss's and Coulomb's laws. Now it does not suit you to argue this anymore?

It does. The Fermi level does not obey Gauss's Law. The electric field does obey Gauss's Law. The Fermi level includes both the (Gauss's-Law-obeying) electrostatic potential and the (material-dependent) electron chemical potential.

In the old Josephson junction discussion, we were in a case where the material was the same on both sides of the junction, so the Fermi-level-change was indeed due to electric fields.

In the new about-to-become-a-Schottky junction discussion, we are in a case where the material is different on both sides, and the Fermi level changes for a reason other than the electric field.

(Note that: after a picosecond or so, when the Schottky junction has its depletion layer set up, the Fermi level will be constant and there will be an electric field. How is this possible in your world, Johan, if Fermi level is the same thing as electric potential? In my world, it's possible because the electric field's electrostatic potential, and the material-dependent electron chemical potential, will be equal and opposite and will cancel out.)

On the plus side, I'm glad to see that you remember discussing Josephson junctions. Perhaps you remember having some distinct difficulties with them. It's worth noting, in case it comes up, that a Cooper pair is a boson, not a fermion, and needs to be treated as such. Pauli exclusion, and the band structure imposed by it, does not apply to them.

 

[Ziggurat]

An energy offset of the Fermi-level between two separated positions mandates the presence of an electric field between the two positions.

No it doesn't. You see, Fermi levels are the energy levels for electrons in specific quantum states. They do NOT represent the energy of an arbitrary electron which is merely located in some particular region. And in quantum mechanics, the potential energy is only one part of the total energy. Electrons have kinetic energy as well based upon their quantum state, and two bound electrons in different locations can have very different kinetic energies (and thus different total energies) even if there is no electric field (and thus no voltage difference) between the two states. Since different materials will have different quantum states with different kinetic energies, we will get different Fermi levels for different materials even with absolutely no electric field at all.

Your track record of consistent failure is impressive.

 

[ben m]

No it doesn't. You see, Fermi levels are the energy levels for electrons in specific quantum states. They do NOT represent the energy of an arbitrary electron which is merely located in some particular region.

That's a good way of putting it.

The Fermi level doesn't tell you an electron's energy. It tells you the energy of electrons which happen to be at the top of the Fermi sea. It's possible to have electrons way above that, it's possible to have holes way below that.

By the way, remember that electron that I said would "fly freely" from gold into sodium, without changing energy? Swap the Au and Na---I had them backwards. Electrons will effuse from Na to Au. Darn you, Ben Franklin!

So the effusion from Na to Au will take with electrons in Na at Na's Fermi energy, let them move into Au without changing energy; they then find themselves with lots of unoccupied states between them and Au's low Fermi energy. (How do I know they don't change energy? Because there is no electric field to make them do so, because Gauss's Law is true.)

 

[mike3]

Guess what, johanfprins - you are not the only competent scientist in the world !
The rest is a a bit of paranoia ("their future to worry about") together with a real problem for your results. If there is no equipment anywhere in the world that can reproduce your experiment then your experiment will never be reproduced !

However, if it is ignored, as you mention below, then that is irrelevant: even if the equipment exists, someone has to first choose to replicate it.

From what little I know about experimental solid state physics, your experiment looks easy to replicate.

The point remains that fact that your experiment has been ignored (except by Prof Doyle) for 8 years now. This means that your reported results have failed to convince any other scientist who has read your paper to make the effort to replicate it.

 

[johanfprins]

Read carefully. The Fermi energy is the sum of the electrostatic potential energy (traditionally "U") and a chemical potential (electron kinetic energy, sustained due to the fact that low-kinetic-energy states are filled already).

There are NO LOW KINETIC ENERGY STATES, since within a material, when not applying an electric-field, the electron-energy states are all standing waves. To convince yourself, solve the Schroedinger equation when there is no electric-field being applied. The boundary conditions, defined by the surfaces of the material, then mandate STANDING ELECTRON WAVES. A standing wave has no momentum, and NEVER WILL HAVE momentum, and THEREFORE also not kinetic energy.

For this reason, the highest energy of a standing electron wave (ignoring temperature fluctuations) which has no kinetic energy defines the Fermi-level when there is no electric field being applied. Thus, to talk about a kinetic energy at the Fermi-level without an electric field within the material, is obviously nonsensical paranormal metaphysics. It is again BCS without the C

If there is an applied electric field, there is a slope in the Fermi energy.

Correct, and this slope is directly proportional to the applied electric field; and it is this applied electric field that changes the boundary conditions and is, in this way, responsible for the generation of wave packets that can be accelerated AND the actual acceleration of these wave packets: The latter entities were not there before the electric field has been applied: Only now you can talk about kinetic energy of the charge-carriers since the charge carriers are generated by the applied electric field and accelerated by the applied electric field. The Fermi-level now defines the zero kinetic energy state of the wave-packets (pseudo-electrons): The latter fact can be confirmed by measuring the Hall effect.

If there's a gradient in the allowed-kinetic-energies, between two neutral materials at the same electrostatic potential, that ALSO makes a slope in the Fermi energy.

The gradient is in the potential energy which is defined by the sloping Fermi-level.

Your statement-2 "Schottky diode, just assembled" is an example of the latter. Your new example (an applied electric field) is an example of the latter. Get it?

No, I am not so stupid to believe in paranormal physics. The incontrovertible experimental fact is that when there is a difference in the Fermi-levels at a surface, there is a potential-energy difference e*V across the interface where V is a voltage. This can be easily verified experimentally by using a voltage probe. THIS incontrovertible experimental fact demands that, when the distance between the Fermi-levels across the surface is d, there MUST be an electric field V/d that will accelerate charge-carriers through the surface: And since d is approximately zero, the impulse causing this acceleration is enormous.

I am thus not going to answer all the nonsense on this post by yourself and other similar metaphysical posts on this same issue that were posted yesterday, in a follow-up to your post. I realise that I have overestimated your abilities to connect the points between one of my statements and the next statement. I will thus advance more slowly from one statement to the next by now posting an alternative statement 2; that will be an intermediary statement.

New Satement 2: When the Fermi-level within a piece of matter, which can consist of different materials that are intimately connected to one another across interfaces, has different energy values E(A) at position A, and E(B) at position B, there will be a difference in energy for a negatively charged pseudo-electron (at the Fermi-level) at these two positions given by e*V=E*(A)-E(B) where -e is the charge of the pseudo-electron and V is a voltage difference. Since there is a voltage difference, there MUST be an electric-field between the points A and B, and this electric field MUST attract such a pseudo-electron from one position to the other position. If there is no scattering, such a pseudo-electron MUST be accelerated from one position to the other. Thus when the distance between A and B is less than the average scattering length of the pseudo-electrons, such a pseudo electron MUST be accelerated from one position to the other so that it gains kinetic energy equal to e*V. Yes or no?

 

[sol invictus]

There are NO LOW KINETIC ENERGY STATES, since within a material, when not applying an electric-field, the electron-energy states are all standing waves. To convince yourself, solve the Schroedinger equation when there is no electric-field being applied. The boundary conditions, defined by the surfaces of the material, then mandate STANDING ELECTRON WAVES. A standing wave has no momentum, and NEVER WILL HAVE momentum, and THEREFORE also not kinetic energy.

One electron in a well of width L. Solving the time-independent Schrodinger equation says

[latex]$\psi(x) =\sqrt{2 \over L} \sin \left( {n \pi x \over L} \right),~n \in {\cal Z}$[/latex]

What's the energy of that state, Johan? How much of that is kinetic?

 

[johanfprins]

One electron in a well of width L. Solving the time-independent Schrodinger equation says

[latex]$\psi(x) =\sqrt{2 \over L} \sin \left( {n \pi x \over L} \right),~n \in {\cal Z}$[/latex]

What's the energy of that state, Johan? How much of that is kinetic?

Assume its energy is E. Since this is a "single electron" standing wave within a well, its energy E is less than that of a "free electron" outside the well. The latter's minimum energy is m*c^2; where m is the rest mass of the free electron wave. As determined by Einstein's Special Theory of Relativity such an entity has NO kinetic energy.

Since the wave solution you have written down must have an energy E that is less than m*c^2, it cannot have kinetic energy at all, unless negative kinetic energy exists: i.e. unless an imaginary momentum can exist within our three dimensional space. The latter does not exist within our three dimensional space at all.

The only way your wave can have kinetic energy is for it to absorb light-energy so that its energy increases from E to Ep where Ep>m*c^2. This is what happens when a photo-electron wave is generated by the absorption of (entanglement with) a photon-wave.

No "single electron" wave within a material can have kinetic energy, since each one of these waves has an energy E that is lower than m*c^2. Single-electron waves can thus not act as charge-carriers within a conductor. The "free-electron" model assuming a "single electron gas" with kinetic energy, is thus an oxymoron concept.

 

[Ziggurat]

There are NO LOW KINETIC ENERGY STATES, since within a material, when not applying an electric-field, the electron-energy states are all standing waves. To convince yourself, solve the Schroedinger equation when there is no electric-field being applied. The boundary conditions, defined by the surfaces of the material, then mandate STANDING ELECTRON WAVES. A standing wave has no momentum, and NEVER WILL HAVE momentum, and THEREFORE also not kinetic energy.

Why are you asking us to solve the Schrodinger equation when you apparently don't accept quantum mechanics? Really, this is incredibly basic stuff here. <p>=0 doesn't mean <p²> = 0, and kinetic energy comes from <p²>, not <p>². If you don't accept this, then there's really no justification for using the Schrodinger equation to begin with. And if you do, then that means you're deeply ignorant of probably the simplest example of a solution to the Schrodinger equation, the very first one most students ever solve, which Sol just gave you above.

Thus, to talk about a kinetic energy at the Fermi-level without an electric field within the material, is obviously nonsensical paranormal metaphysics.

No, johan. It's bog-standard introductory quantum mechanics. Which you claim to accept but in fact never do. No bound state can ever have zero kinetic energy. That would violate the Heisenberg uncertainty principle.

No, I am not so stupid to believe in paranormal physics.

Or standard quantum mechanics, apparently. Your disagreement with the physics community at large is so basic that it's strange you're stuck on your whole superconductivity trip. You don't believe in the very foundations of quantum mechanics. Why aren't you spending your time challenging that?

 

[Ziggurat]

Since the wave solution you have written down must have an energy E that is less than m*c^2, it cannot have kinetic energy at all, unless negative kinetic energy exists: i.e. unless an imaginary momentum can exist within our three dimensional space. The latter does not exist within our three dimensional space at all.

I find it rather telling that the one thing you are NOT willing to do here is actually calculate the answer based on the wave function Sol gave you. All the necessary information is right there, but you can't do it, can you? You've lost the ability to do simple calculus.