Superconductivity at room temperature??

[johanfprins]

Slippery little customer is the electron. Heck, even a surface wave can be difficult to grasp. I rather like the third picture down on the wiki wind wave article because it gets across the way a wave is an "extended entity".

It is late in South Africa. I will comment tomorrow on this and the rest of your very interesting and relevant comments. As I have indicated, this will require an objective engagement between open minds. I suspect that this is possible between us, but I also suspect that this thread is not the best venue for it. But for now: See you later

 

[Farsight]

As a starting point, I must first explain my experiment and why I claim that it is superconduction at room temperature. Although this experiment caused me to doubt the Copenhagen interpretation, and is esaier to explain when accepting upfront that this interpretation is wrong, I believe that I can also prove my conclusion by staying within the paradigm of accepted Solid State Physics and Solid Sate Electronics: The latter fields have been the most spectacularly succesful ones after 1927. I am thus restating my first two statements. If we cannot find any fault in them in terms of Solid State Physics and Electronics, I will move on to the third statement: If again no fault, the fourth statement etc. I hope you find this procedure acceptable.

Sure, but you don't need to prove it to me, just tell me.

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.

No problem. I know a bit about semiconductors.

Statement 2: If the density of donors within the n-type semiconductor is sufficiently high, an equilibrium dipole will manifest when the Fermi-level at the metal-dipole interface has the same energy as the Fermi-level at the semiconductor-dipole interface.

Sounds good. Carry on.

 

[ben m]

I have NEVER stated that <p>=0 implies that <p²> is zero. PLEASE STOP LYING SO BLATANTLY!!

Although it is correct that it does not have any momentum, it DOES NOT have <p>=0 but only that <k>=0. And as any first year studemt should know k has NOTHING to do with the momentum p when you have a stationary wave. PLEASE WAKE UP out of your stupor!

Are we still in "all quantum textbooks are wrong" land? If so, then stop saying "as any first year student should know".

Because, according to standard physics, you're still wrong. It is true that the bound states we're talking about have <p>=0. It's also true that p and k are related (p = k/hbar). It's also true that the p and k operators can be applied---equally easily and equally sensibly---both to momentum eigenstates like a plane wave, and also to bound states which are not eigenfunctions of momentum. If you really think that this is standard physics, please cite a textbook---any textbook---that uses it. I happen to have Eisberg & Resnick, 2nd ed., in front of me and page 148-149 contradict you explicitly.

ETA:

Example 5-10. Use the "particle in a box" wave function treated in example 5-9 to evaluate the expectation values of x, p, x^2, and p^2 ... thus we obtain <p> = 0 ... physically, the expectation value of the momentum of the particle is zero because, if the particle is confined to the region between -a/2 and a/2, and moving with total energy E, it must be bouncing back and forth between the ends of the region ... that is, the magnitude of the momentum must be such that p^2/2m = E, but since it is equally probably that the sign of the momentum will be either positive or negative, measurements of this quantity will average out to zero.

 

[W.D.Clinger]

I'm quoting the whole thing so there shouldn't be any question about the context of the part I've highlighted:

Good, so at least you understand that <p> and <p²> are different quantities, and that <p>=0 does not imply <p²>=0.

I have NEVER stated that <p>=0 implies that <p²> is zero. PLEASE STOP LYING SO BLATANTLY!!

The wavefunction I posted has <p>=0. It does not have <p²>=0.

Although it is correct that it does not have any momentum, it DOES NOT have <p>=0 but only that <k>=0. And as any first year student should know k has NOTHING to do with the momentum p when you have a stationary wave. PLEASE WAKE UP from your stupor!

Dr Prins is saying <p>=0 does not hold, which implies <p> is nonzero, which implies p cannot be identically zero, which directly contradicts Dr Prins's claim that "it does not have any momentum".

That's just basic calculus, logic, and the definition of the expected value <p>. If p were identically zero, then the expected/expectation value <p> would also be zero.

That Dr Prins would make such an obvious mistake while telling others to "PLEASE WAKE UP" from their stupor is...typical.

 

[Farsight]

I don't know what you're referring to. The question in my most recent post wasn't about waves and doesn't require relativity (you can take p<<mc), although the answers don't change if you include relativistic corrections.

Johan was talking about a wave in a cavity shorter than its wavelength, and you started talking about a massive particle in a cavity. You're missing the point that the wave in the cavity is simulaneously moving this way → and this way →. It doesn't matter at which time you measure the momentum, it's always +1-1=0.

As for the wavefunction I wrote earlier, it describes a massive particle such as an electron confined to a one-dimensional box.

I know. That's why I said you were talking about a wave in a box that's moving back and forth inside another box. You need to take it down a level and try to think of the electron as a photon in a box. Do read the van der Mark ’t Hooft article. It is good.

Again, relativity is not relevant.

It is relevant, because we were talking about momentum and mass. Trap a massless photon in a box and its momentum now appears as inertia. Inertia is just the flip side of momentum when you say it's you moving instead of the photon. When you push the box you have to change the photon momentum from +1 and -1 to +1.1 and -1. It's all rather simpler than you might think. Sometimes this kind of thing seems to be too simple.

To describe a box full of photons, you cannot use a single-particle wavefunction like that (although again, the answers to the questions I asked wouldn't change qualitatively in that situation)

OK. But the trick of this is to think about what might be happening in pair production, and make the leap to the electron wavefunction being a standing-wave version of the 511keV photon wavefunction.

Physicsworld: The secret lives of photons revealed

 

[sol invictus]

And as any first year student should know k has NOTHING to do with the momentum p when you have a stationary wave.

How much do you want to bet on that, Johan? Would you like me to ask some students?

Johan was talking about a wave in a cavity shorter than its wavelength, and you started talking about a massive particle in a cavity. You're missing the point that the wave in the cavity is simulaneously moving this way → and this way →. It doesn't matter at which time you measure the momentum, it's always +1-1=0.

I know. That's why I said you were talking about a wave in a box that's moving back and forth inside another box. You need to take it down a level and try to think of the electron as a photon in a box. Do read the van der Mark ’t Hooft article. It is good.

It is relevant, because we were talking about momentum and mass. Trap a massless photon in a box and its momentum now appears as inertia. Inertia is just the flip side of momentum when you say it's you moving instead of the photon. When you push the box you have to change the photon momentum from +1 and -1 to +1.1 and -1. It's all rather simpler than you might think. Sometimes this kind of thing seems to be too simple.

OK. But the trick of this is to think about what might be happening in pair production, and make the leap to the electron wavefunction being a standing-wave version of the 511keV photon wavefunction.

[qimg]http://images.iop.org/objects/phw/news/thumb/15/6/4/double1.jpg[/qimg]
Physicsworld: The secret lives of photons revealed

I'm sorry Farsight, but I have no idea what you're talking about.

 

[johanfprins]

Slippery little customer is the electron. Heck, even a surface wave can be difficult to grasp. I rather like the third picture down on the wiki wind wave article because it gets across the way a wave is an "extended entity".

It is really wonderful to find another person who is not scared to think outside that accepted mainstream paradigm.

Physics is all about creating paradigm shifts, and how will one discover a paradigm shift by not thinking, as John Bell would have said, "the unspeakable". As you can see on this thread, the defenders of the mainstream paradigm are acting like a man who searches for his lost wallet under a street lamp, even though he lost it somewhere in the dark: Why? Since there is more light under the street lamp.

What is extremely worrying is that the mentality displayed by people like Sol, Ben, Ziggurat etc. on this thread is obstructing new discoveries in physics. AND this is happening at a time when any honest physicist MUST agree that there is something gravely wrong with physics. Even the revered physicists of the 20thn century has consistently admitted that they do not understand quantum mechanics: For example Bohr: “If a person thinks that he/she understands quantum mechanics than he/she does not understand it”; Feynman: “Nobody understands quantum mechanics” and “Yes! Physics has given up. We do not known how to predict what would happen in a given circumstance; and we believe now that it is impossible.”

Only Einstein stuck to his guns by proclaiming that the purpose of physics is to find out “how God did it”. If you accept that the latter is impossible, then you should stop doing physics; but admit that we can only understand the universe through religious belief. “Albert, stop telling God what he can or cannot do." Copenhagen thus steered physics to become a religious dogma; and in all cases that this has happened, any thinking outside this dogma is soon ridiculed. Inquisitors, like Sol, Ben, Ziggurat, etc. soon consider it their duty to preserve the “holy dogma” at all costs.

I will, however, not be intimidated to not think outside this “holy dogma”; no matter how insulting the inquisitors become. If I do not search for my lost wallet in the dark, I will never find it. I believe, like Einstein, that it is our purpose as physicists to try and figure out “how God did it”. After all, if you do not pursue this route, you have to believe in miracles and in a bizarre nature. I detest it when I read in scientific journals that “quantum mechanics is bizarre”. I believe that when physical behaviour seems “bizarre” it proves that we DO NOT understand the physics involved; and that the correct physics must then lie outside the mainstream dogma in the “dark”. To wander there is of course dangerous; but not to do so, means your life as a physicist is totally futile.

By wandering outside the cosy light of mainstream dogma I have found that the “bizarre” behaviour in quantum mechanics falls away when one makes a few simple postulates most of which (except the concept of entanglement) all follow logically from classical mechanics. Whether these postulates are totally correct must eventually be determined by experiment: And if they are proved to be wrong, it does not make me a “crackpot”. A summary of the postulates that in my opinion should be further investigated is the following:

1.The intensity of all harmonic waves is proportional to the energy of the wave.
2.A harmonic wave can exist as a superposition of other harmonic waves, where the other harmonic waves keep their separate identities.
3. A harmonic wave can exist as an entanglement of other harmonic waves, where the other harmonic waves have totally lost their separate identities.
4. An entangled harmonic wave is a holistic entity which is in immediate contact with itself within the whole region of space that it occupies.
5. A photon is the lowest energy entangled light wave that can exist and an electron is the lowest energy entangled matter wave that can exist.
6. Entangled waves morph in shape and size when the boundary conditions change: Morphing also includes forming separate fractions in space: Where these fractions still stay in instantaneous contact with one another.: For example, when an s-wave morphs into a p-wave, the latter wave now consists of two separate regions. For d-waves, the fractions are even more than for p-waves.
7. When a photon is absorbed by an electron wave, they entangle to form an electron-wave with a higher mass energy, which then has to morph in shape and size to form a higher energy electron-wave (this is a quantum jump).
8. When a photon is emitted by an electron wave it disentangles from the existing electron wave and speeds away at speed c. This forces the electron wave to morph into a lower energy electron wave (the latter is also a quantum jump).

At this point I first want to stop temporarily before coming to the rest of your interesting post. But I have to conclude by responding to your metaphor above of wind waves, since it is appropriate when you generate a current through a metal by applying an electric field. Before applying the electric field the electrons are standing waves having no momentum and thus no kinetic energy: Ignoring temperature fluctuations, there are NO “particles” that move with momentum p.

When applying an electric field, the boundary conditions are changed, and a situation arrives where the applied electric field acts like a wind on the surface of water. The standing waves, just before applying the electric field, now morph into running waves which superpose to form wave-packets. These wave packets, which are not there when there is no electric field, act as the charge-carriers. The charge-carriers are pseudo-electrons as anybody who ever did Solid State Physics should know. There are no “free electrons” within a material since free-electrons must have energies that are larger than their respective rest mass energies: Such electrons CANNOT bond atoms together.

The rest of your comments on this post will require quite a bit of discussion. Maybe it should be postponed for now so that I rather go ahead with my statements about my experiment?

 

[johanfprins]

Sure, but you don't need to prove it to me, just tell me.

No problem. I know a bit about semiconductors.

Sounds good. Carry on.

Statement 3: When reverse-biasing the Schottky barrier, the Fermi-level on the metal side is pulled down in energy relative to the Fermi-level on the n-type semiconductor side: If the donor density is high enough, the dipole width will increase to also eliminate this offset in Fermi-levels. No current can thus flow through the dipole since there is no net electric-field within the dipole. By further increasing the reverse bias, a situation wll be reached at which the dipole cannot adjust anymore, and a breakdown current will flow through the dipole.

 

[johanfprins]

Statement 4

I decided to alo post statement 4 since it flows logically from astatement 3:

Statement 4: The voltage at which reverse breakdown occurs is determined by the density of the donors so that there are different types of breakdown: i.e.

(i) Low donor density: Dielectric breakdown: AKA avalanche breakdown.

(ii) Intermediate donor density: Zener breakdown where at the breakdown voltage electrons start to tunnel through the depletion layer.

(iii) Very high donor density: Immediate breakdown; since even though the dipole adapts to align the Fermi-levels, the dipole layer is so thin that charge-carriers immediately start to tunnel through the dipole-layer. It is said that an "ohmic-contact" has formed.

 

[Small Bones]

Statement 3: When reverse-biasing the Schottky barrier, the Fermi-level on the metal side is pulled down in energy relative to the Fermi-level on the n-type semiconductor side: If the donor density is high enough, the dipole width will increase to also eliminate this offset in Fermi-levels. No current can thus flow through the dipole since there is no net electric-field within the dipole. By further increasing the reverse bias, a situation wll be reached at which the dipole cannot adjust anymore, and a breakdown current will flow through the dipole.

Johan,

Before you throw insults and tell me to sue my nursery school teacher could you say whether your above statement is based on existing text-book physics or whether you are disregarding some elements of currently accepted physics? The above statement doesn't agreement with text-book descriptions of a Schottky barrier that I have seen. You say that no current can flow because there is no net electric field when reverse biasing the junction. A text book interpretation would say that, as the reverse bias increases, the E-field across the depletion region increases and the depletion region increases in width. At some point the field is large enough to cause breakdown. You say the dipole increases in width so I am assuming you are using dipole where a text book would use 'depletion region'.
A key point I have a problem with in your description is that you say "No current can thus flow through the dipole since there is no net electric-field within the dipole" but a reverse leakage current always flows in a reverse biased diode and this is usually explained by the E-field caused by the reverse bias. Essentially, the text book explanation is that the depletion region that exists with zero bias will increase in width with increasing reverse bias. Are there parts of the text book explanation that you think are incorrect?

 

[johanfprins]

Johan,

Before you throw insults and tell me to sue my nursery school teacher could you say whether your above statement is based on existing text-book physics or whether you are disregarding some elements of currently accepted physics? The above statement doesn't agreement with text-book descriptions of a Schottky barrier that I have seen. You say that no current can flow because there is no net electric field when reverse biasing the junction.

An excellent question: This I the quality of questions we need on this thread. Well done.

A text book interpretation would say that, as the reverse bias increases, the E-field across the depletion region increases and the depletion region increases in width. At some point the field is large enough to cause breakdown.

The latter is correct, but what the text book left out is to take the increase in polarisation field caused by the increase in both the width of the depletion region and the increase in the number of negative charges at the surface of the metal into account.

When the density of donors is high enough in the depletion region, this polarisation field opposes the applied electric field, and will cancel the applied electric field within the dipole for as long as it can keep the Fermi-levels aligned on both sides of the dipole. Only once the dipole cannot adjust to keep the Fermi-levels at the same energy anymore, can the applied electric field increase within the dipole and thus the depletion layer. Breakdown then occurs thereafter.

I should have explained better why the breakdown voltage is determined by the density of the donors. Thus the text book explanation is not wrong, but only sloppy, since for completeness they should have explained that the voltage only increases within the depletion layer after the layer cannot increase by enough to keep the Fermi-levels on both sides of the dipole at the same energy level. According to the book only avalanche breakdown occurs. This is not so since Zener breakdown occurs while the applied electric field is still cancelled within the dipole.

You say the dipole increases in width so I am assuming you are using dipole where a text book would use 'depletion region'.

NO: Dipole means the depletion layer AND the negative charges that accumulate at and near the surface of the metal. The depletion layer is only one half of the dipole. Even though the applied electric field is also cancelled within this region.

A key point I have a problem with in your description is that you say "No current can thus flow through the dipole since there is no net electric-field within the dipole" but a reverse leakage current always flows in a reverse biased diode and this is usually explained by the E-field caused by the reverse bias.

A leakage current is different from a full current. The charge-carriers forming the leakage current are usually those charge-carriers that enter the dipole layer with enough ballistic energy to move right through it.

In this respect you have raised another important point which should be included when doing a thorough analysis of the Schottky diode. Even though there is no net electric field within the dipole, a charge entering the dipole disturbs the equilibrium charge distribution that cancels the applied electric field. A force then acts on the charge-carrier to pull it back. But as already explained above, if the charge-carrier enters the dipole with a high ballistic speed it can reach the other side before it is turned around; and this causes the leakage current.

Essentially, the text book explanation is that the depletion region that exists with zero bias will increase in width with increasing reverse bias.

This is exactly what I am also saying.

Are there parts of the text book explanation that you think are incorrect?

No, not incorrect, but incomplete since it does not explain how the density of the donors can affect the reverse breakdown. I should have added it already in statement 3; and have added it now in statement4.

But in a text book these aspects are not really required and thus glossed over. You must remember that it is also assumed in text books that there are zero electric field within the metal and the semiconductor even when you bias the device. This is obviously not correct, but in normal devices these fields are so small, that they can be neglected. As you will see, these fields cannot be neglected when working with a superconducting element.

Thank you again for raising these issues. It feels good to be actually talking physics!

 

[ben m]

No current can thus flow through the dipole since there is no net electric-field within the dipole.

Since your whole superconductivity-theory, Johan, has to do with whether or not there is an electric field in a certain place, it behooves you to get this right.

Statement 3 is wrong and contains a basic misunderstanding of physics.

"equal Fermi levels" does NOT mean no electric field, Johan. The reason the Fermi level is flat is that the electrostatic potential (due to the field) has cancelled out the difference in binding energies (the electron chemical potential) between the two materials. Remember? In reverse bias, the Fermi levels are not even equal and there's an extra-large electric field.

A nice illustration of this is shown http://en.wikipedia.org/wiki/File:Pn-junction-equilibrium-graphs.png in the context of a p-n junction; again, a Schottky diode is qualitatively exactly the same. Another diagram showing the reverse-bias case explicitly is http://hyperphysics.phy-astr.gsu.edu/hbase/solids/pnjun2.html

Given that you've spent the past eight years boasting of your great theoretical certainty where there are and are not electric fields, and that you're so certain of yourself that you can use the result to (a) claim HTS, (b) overturn quantum mechanics, and (c) insult everyone who disagrees ... well, here's your problem. "no change in Fermi level" means "no change in electrochemical potential", which does NOT mean "no electric field".

 

[johanfprins]

here's your problem. "no change in Fermi level" means "no change in electrochemical potential", which does NOT mean "no electric field".

Show me any material with two contacts across which you applied a voltage V (and thus an electric field between the contacts), which has that the Fermi-levels at the contacts are at the same energy. Stop posting BS.

Maybe you have never noticed or learned it (have you ever done a course in elementary physics?): But a voltage is also a potential, and in electrochemical studies voltages are applied to change electrochemical potentials.

If there is an electric field between two contacts in a material, the Fermi-levels at the two contacts MUST be at different energies: When there is no electric field, the Fermi-levels MUST be at the same energy within the materials under discussion here.

 

[ben m]

Show me any material with two contacts across which you applied a voltage V (and thus an electric field between the contacts), which has that the Fermi-levels at the contacts are at the same energy.

Sure. Any two materials with different work functions can have this property. The "Volta potential" http://en.wikipedia.org/wiki/Volta_potential is the experimental consequence and has been known for at least a century.

(ETA: And you have to be careful about "apply a voltage". If you apply a voltage by injecting electrons from a power supply, the power supply wire is made of some metal with its own work function, and it's easy to get the a loop of contact potentials to cancel out---obviously they can't drive a loop current, for example---but an electrometer outside the wires can easily see these differences. I have been avoiding the word "voltage" in this discussion in favor of the unambiguous words "Fermi level" and "electrostatic potential")

If there is an electric field between two contacts in a material, the Fermi-levels at the two contacts MUST be at different energies: When there is no electric field, the Fermi-levels MUST be at the same energy within the materials under discussion here.

Nope. You're ignoring the things that influence electrons other than electric potential, like the Pauli exclusion principle.

 

[johanfprins]

Sure. Any two materials with different work functions can have this property. The "Volta potential" http://en.wikipedia.org/wiki/Volta_potential is the experimental consequence and has been known for at least a century.

Are you really so stupid? There is NOTHING in this reference that does not agree with what I have been stating. Similarly in your references above. Instead of quoting wikipedia, why do you not try and think for yourself? Ah, I should not have asked the impossible from you.

(ETA: And you have to be careful about "apply a voltage". If you apply a voltage by injecting electrons from a power supply, the power supply wire is made of some metal with its own work function, and it's easy to get the a loop of contact potentials to cancel out---obviously they can't drive a loop current, for example---but an electrometer outside the wires can easily see these differences. I have been avoiding the word "voltage" in this discussion in favor of the unambiguous words "Fermi level" and "electrostatic potential")

Again trying to be a smart ass. Please stop wasting our time on this thread.

Nope. You're ignoring the things that influence electrons other than electric potential, like the Pauli exclusion principle.

How does the Pauli exclusion principle cause an electric field to be zero between two contacts at which the Fermi-level has the same energy? Or does it cause in some miraculous manner that there is an electric field between the two contacts when the Fermi-level has the same energy at the two contacts?

Please get a first year student in Electrical Engineering to instruct you. Not that it will help at all. You are clueless!

 

[Farsight]

Physics is all about creating paradigm shifts, and how will one discover a paradigm shift by not thinking, as John Bell would have said, "the unspeakable"...

I agree with this. There's a kind of schism between learning and thinking. People who've been taught a lot and cherish their expertise seem to have a real problem with thinking for themselves, and resent others who do. There's some powerful psychology at work there, with people thumping their good book even in the face of patent scientific evidence. But let's not dwell on it.

Only Einstein stuck to his guns...

Agreed. I'm something of an Einstein "fan" I suppose, even though I know he wasn't perfect. That's another one for another day.

I will, however, not be intimidated to not think outside this “holy dogma”; no matter how insulting the inquisitors become...

I wish there were more like you.

By wandering outside the cosy light of mainstream dogma I have found that the “bizarre” behaviour in quantum mechanics falls away...

Me too. Only I didn't wander outside, I'm a creature of the night. I'm not a professional physicist, I've got a Computer Science degree, that's why I know a bit about semiconductors and can get under the maths. I got there my analysing the terms in common expressions, things like E, m, c and so on. I was motivated by the demolition of the Manchester University Maths tower and our two teenage children giving up all their science subjects.

1.The intensity of all harmonic waves is proportional to the energy of the wave.
2.A harmonic wave can exist as a superposition of other harmonic waves, where the other harmonic waves keep their separate identities.
3. A harmonic wave can exist as an entanglement of other harmonic waves, where the other harmonic waves have totally lost their separate identities.
4. An entangled harmonic wave is a holistic entity which is in immediate contact with itself within the whole region of space that it occupies.
5. A photon is the lowest energy entangled light wave that can exist and an electron is the lowest energy entangled matter wave that can exist.
6. Entangled waves morph in shape and size when the boundary conditions change: Morphing also includes forming separate fractions in space: Where these fractions still stay in instantaneous contact with one another.: For example, when an s-wave morphs into a p-wave, the latter wave now consists of two separate regions. For d-waves, the fractions are even more than for p-waves.
7. When a photon is absorbed by an electron wave, they entangle to form an electron-wave with a higher mass energy, which then has to morph in shape and size to form a higher energy electron-wave (this is a quantum jump).
8. When a photon is emitted by an electron wave it disentangles from the existing electron wave and speeds away at speed c. This forces the electron wave to morph into a lower energy electron wave (the latter is also a quantum jump).

That gels well with me. There's something you might add to that - the electron wave is a particular photon-wave configuration. And I've just realised there was a typo in something I said earlier: pair production is where we convert a photon into two quasi-spherical-harmonic λ=4π/nc^1½ standing-wave configurations. I missed out the 1 from the 1½. Try it on your calculator. Here's another one: 3π/c^½.

...Before applying the electric field the electrons are standing waves having no momentum and thus no kinetic energy: Ignoring temperature fluctuations, there are NO “particles” that move with momentum p.

Agreed. There's an angular momentum in there, but they ain't going anywhere.

When applying an electric field, the boundary conditions are changed, and a situation arrives where the applied electric field acts like a wind on the surface of water. The standing waves, just before applying the electric field, now morph into running waves which superpose to form wave-packets.

I got you. The standing waves are undulating.

These wave packets, which are not there when there is no electric field, act as the charge-carriers. The charge-carriers are pseudo-electrons as anybody who ever did Solid State Physics should know. There are no “free electrons” within a material since free-electrons must have energies that are larger than their respective rest mass energies: Such electrons CANNOT bond atoms together.

I didn't know that. Interesting. But that's what's so good about this kind of conversation. I learn things.

The rest of your comments on this post will require quite a bit of discussion. Maybe it should be postponed for now so that I rather go ahead with my statements about my experiment?

Yes. Don't feel the need to reply to this post.

 

[Ziggurat]

How does the Pauli exclusion principle cause an electric field to be zero between two contacts at which the Fermi-level has the same energy?

It doesn't. You have completely misunderstood ben. Fermi exclusion helps create Fermi energy level differences even in the absence of an electric field. An electric field is necessary to equalize Fermi energies for metals with different work functions.

Which goes back to your complete ignorance of basic quantum mechanics. You have claimed that stationary states can't have any kinetic energy. Of course, this contradicts standard quantum mechanics, for reasons you can't understand because you don't understand quantum mechanics, but no matter. Perhaps the stumbling block is purely semantic. So let me try a slightly different question. Going back to Sol's wave function for a particle in a box, we have

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

These states have an energy of
[latex]E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}[/latex]

The energy is different for different n's. And if we stick multiple particles in our box, then Pauli exclusion forces some electrons into the higher energy states. Yet the potential energy of all these states is exactly the same. Standard quantum mechanics tells us that the energy difference is kinetic energy. You insist that it's not. Fine, we'll play your game: the energy different is "johan energy". The important point here is that it's NOT a potential energy difference.

The conclusion that follows should be obvious, but I know I need to spell it out for you, and you still won't accept it because it's inconvenient to your delusions: the Fermi energy is NOT simply the potential energy. Differences in Fermi energy are a combination of differences in voltage and differences in something else (which standard quantum mechanics calls "kinetic energy" but we'll call "johan energy"). If those other things are different (and that's easy enough to determine by examining the work function), then equalizing the Fermi energy requires an electric field.

Every objection you have provided contradicts standard quantum mechanics. You may protest all you want to that standard quantum mechanics is wrong, but there's two problems with that argument. First and most glaring is that YOU told us to use it. And second, your objections, if correct, would break quantum mechanics so fundamentally that the theory should be completely useless. But right or wrong, the one thing we know about quantum mechanics is that it's not useless.

 

[Farsight]

Statement 3: When reverse-biasing the Schottky barrier, the Fermi-level on the metal side is pulled down in energy relative to the Fermi-level on the n-type semiconductor side: If the donor density is high enough, the dipole width will increase to also eliminate this offset in Fermi-levels. No current can thus flow through the dipole since there is no net electric-field within the dipole. By further increasing the reverse bias, a situation wll be reached at which the dipole cannot adjust anymore, and a breakdown current will flow through the dipole.

Sounds good. Anybody reading can check up on this on say wiki. I see ben has raised a point of objection, I'll look more closely at that. Maybe it will be related to no electric field outside a solenoid in the Aharonov-Bohm effect.

Statement 4: The voltage at which reverse breakdown occurs is determined by the density of the donors so that there are different types of breakdown: i.e.

(i) Low donor density: Dielectric breakdown: AKA avalanche breakdown.

(ii) Intermediate donor density: Zener breakdown where at the breakdown voltage electrons start to tunnel through the depletion layer.

(iii) Very high donor density: Immediate breakdown; since even though the dipole adapts to align the Fermi-levels, the dipole layer is so thin that charge-carriers immediately start to tunnel through the dipole-layer. It is said that an "ohmic-contact" has formed.

It's all reasonable semiconductor stuff. I looked a few things up on wiki and can't see anything wrong with it.

 

[ben m]

These states have an energy of
[latex]E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}[/latex]

The energy is different for different n's. And if we stick multiple particles in our box, then Pauli exclusion forces some electrons into the higher energy states. Yet the potential energy of all these states is exactly the same. Standard quantum mechanics tells us that the energy difference is kinetic energy. You insist that it's not. Fine, we'll play your game: the energy different is "johan energy". The important point here is that it's NOT a potential energy difference.

Yep. This is why I asked Johan to consider a gold block and a sodium block not in contact. Suppose they're both neutral; you have to conclude that there is no electric field between them.

Johan's "logic"---that Fermi level differences are always due to electric fields---tells us that these disconnected metals must have the same Fermi level. But this is incorrect; the photoelectric effect tells us that it requires an energy input to move an electron from the gold to the sodium, and that you can gain energy by moving an electron from the sodium to the gold. That energy-difference shows that the Fermi levels are not equal in this case, and the difference is purely due to quantum mechanics (i.e. binding energy differences) and not due to electrostatics.

 

[CriticalThanking]

Farsight, I have no issue with discussions of philosophy of science and attitudes toward "dogma". But Sol, Zig, ben m, et al are supplying experimental and mathematical evidence to support what the textbook is saying. That is not dogma - that is evidence. A valid response must provide the same level of evidence.

Every objection you have provided contradicts standard quantum mechanics. You may protest all you want to that standard quantum mechanics is wrong, but there's two problems with that argument. First and most glaring is that YOU told us to use it. And second, your objections, if correct, would break quantum mechanics so fundamentally that the theory should be completely useless. But right or wrong, the one thing we know about quantum mechanics is that it's not useless.

I think Zig's comment says it quite well. Why can QM make testable predictions that pass experimentally if they are wrong?