Superconductivity at room temperature??

[Ziggurat]

1. Do you agree that the charge-carriers within a superconductor must all have the same energy at any instant of time whether a current is flowing or not?

Within a single superconductor, we can assume that for DC situations. That need not be the case even for DC case if we're examining two different superconductors. Of course, the two sides of the Josephson junction are not the same superconductor, because (surprise) they're separated by something which does not superconduct.

2. Do you agree that without a voltage over the Josephson junction the charge-carriers on both sides of the junction must have the same energy?

3. Do you agree that when applying a voltage V to the Josephson junction the charge carriers on one side is forced to be at an energy qV lower than the charge carriers on the other side whether a current is flowing or not?

4. Do you agree that a charge-carrier on the high energy side has an energy qV higher than a charge-carrier on the other side, even when it does not move through the junction?

5. Do you agree that a charge-carrier moving from the high energy side to the low energy side will be accelerated to gain energy qV.

Good so far. But you're completely wrong on this next bit:

6. Do you agree that a charge carrier starting off with an energy qV higher than it must have on the other side, and then being accelerated to gain energy qV when it moves to the other side must get rid of its initial higher energy qV PLUS the energy it gained while moving to the other side: ALSO qV: i.e. THAT IS 2qV.

Nope, it's not 2qV. It's just qV. The charge carrier has already lost the potential energy difference by the time it emerges from the Josephson junction. That's where the kinetic energy came from. There is no second qV.

Can you not count at all. I already knew when I was two years old that 1 plus 1 is two!

And one plus zero is still one. The problem is not that we can't add, the problem is that you pulled that second one out of nowhere.

You violated conservation of energy. Conservation of energy must apply regardless of the details of the system. An analysis which treats the total energies directly shows a violation of energy conservation. You have yet to show that this analysis is wrong. You have, instead, simply claimed that one cannot perform such a generalized analysis. But we can. In fact, we MUST be able to, or energy conservation is meaningless.

 

[ben m]

you have not! You have said no matter what is in the box the charge carier will emerge with an energy qV higher. So stop lying please.

Nope. "the voltage is V higher on the left" means exactly, precisely, identically the same thing as "a charge carrier on the left starts with qV potential energy".

So why did you not accept that this increase happens in addition to the higher energy that the charge carrier has before moving through the junction?

Because it doesn't. The meaning of "the charge carrier has potential energy" is that "we've kept track of how its energy will change in a vector force field". The +qV energy gain in electric fields in the junction is the conversion of the potential energy ("the potential is V higher on the left") into kinetic energy.

Johan, do you agree with the statement U = -integral(F.dx), potential energy change is the integral of the force over a path?

how many times must I tell you that we are NOT talking about a resistor within which the charge-carriers are accelerated and scattered all the time! Please, stop displaying your ignorance.

I'm sorry, Johan---you seem to be refusing to do any energy calculations at all. I am trying to suggest simpler and simpler circuits until we find something that you're willing to do a calculation with. Resistors are easy and (like it or not) they embody the physics error that I think you're falling into. (If I'm wrong about your error, maybe they don't embody your error. If so, I'd like to find out. How can one tell if you won't calculate anything?)

Listen---is there anything whatsoever for which you're willing to complete Zig's energy-budget calculation? You pick.

You get to pick the circuit element. SIS junction, resistor, vacuum tube, Penning trap, LED, whatever you want. Whatever's handy for you.

You get to draw the circuit diagram. Put the element itself in a black box. Put all of the work-performing engines (batteries, current supplies, Wimshurst generators---your choice) are outside of the box. You can run wires into and out of the box (as many as you like) as long as you're prepared to keep track of the current in each.

Then please actually do the energy budget calculation. How much work is done by the machines outside the box? How much work is done on the device inside the box?

 

[sol invictus]

So you deny that you have maintained that even when there is "anything" in your "box" an injected charge-carrier must emerge with an energy qV higher than it entered with? This is impossible if the wires going into the box are connected to a superconductor!

My previous posts are all there in black and white, for anyone to read.

I will try and be patient: It is exactly at this point that you demonstrate that you do not know what superconduction is all about. So let me try and force you to answer some questions which you refused to respond to above:

1. Do you agree that the charge-carriers within a superconductor must all have the same energy at any instant of time whether a current is flowing or not?

2. Do you agree that without a voltage over the Josephson junction the charge-carriers on both sides of the junction must have the same energy?

3. Do you agree that when applying a voltage V to the Josephson junction the charge carriers on one side is forced to be at an energy qV lower than the charge carriers on the other side whether a current is flowing or not?

4. Do you agree that a charge-carrier on the high energy side has an energy qV higher than a charge-carrier on the other side, even when it does not move through the junction?

5. Do you agree that a charge-carrier moving from the high energy side to the low energy side will be accelerated to gain energy qV.

For the sake of argument, I agree to all the above.

6. Do you agree that a charge carrier starting off with an energy qV higher than it must have on the other side, and then being accelerated to gain energy qV when it moves to the other side must get rid of its initial higher energy qV PLUS the energy it gained while moving to the other side: ALSO qV: i.e. THAT IS 2qV. Can you not count at all. I already knew when I was two years old that 1 plus 1 is two!

Of course I don't agree.

Look - let's model the charges as little balls with definite position and definite momentum and charge q. OK? Then the initial state is a little ball with charge q on the "high" side of the junction. If it's at rest, it has energy qV (zero kinetic energy, qV of potential energy). So E_initial=qV, just as you said in item 4 above.

Next, it wanders (very slowly) a little too close to the junction, gets sucked in by the electric field, and flies out the other side.

What's its energy now? Assuming it hasn't radiated anything yet, by conservation of energy E_final=E_initial=qV. Since its potential energy is now 0 (it's on the low side of the junction), its kinetic energy is qV.

What next? Well, it can radiate a photon - with energy qV - and return to a state with zero kinetic energy. If it radiated a photon with energy 2qV, it would either have negative kinetic energy (-qV), which is impossible, negative potential energy (-qV), which is impossible since we know this piece of superconductor is at V=0, or violate conservation of energy.

Yes I would very much like to see how you prove that 1 plus 1 is equal to 1. Fascinating new algebra!

Very well. Work is force integrated over distance. Begin with a charge q on one side of the junction. We want to compute

[latex]$W=\int_{x_i}^{x_f} \vec F \cdot d \vec l$[/latex].

where the limits on the integral are the initial and final points of the path along which we move the charge. Next, use the Lorentz force law

[latex]$\vec F=q \vec E = -q \vec \nabla V$[/latex].

Substituting this the integral and using the fundamental theorem of calculus gives

[latex]$W=-q [V(x_f)-V(x_i)]=\pm qV$[/latex],

where the sign depends whether we move across the junction with or against the electric field (and whether we compute the work done on the charge or by the charge). Notice that the result does not depend on the path, and it is not equal to 2qV.

 

[sol invictus]

As I have said time and again, a superconductor is not a resistor within which the charge carriers are moving with an average drift speed because they are accelerated and scattered all then time. Because in a resistor the charge-carriers are accelerated and scattered all the time. Jeez, do you not even understand high school physics? Let me rather sign off before I lose my temper.

You're the one struggling with high school physics, Johan. Why do you think you haven't been able to convince any other physicists that you're right on this, after years and years of trying? Do you really believe that everyone else is so much stupider than you? Isn't it more probable that you, a single individual, might have made a mistake, than that all the other physicists in the world have been wrong for the last 50 years?

The only significant difference between the situation we are considering - where the charge carrier passes through a Josephson junction with voltage drop qV and radiates a single photon - and the situation in a normal resistor with voltage drop qV, is that in a normal resistor the charge bleeds away energy qV in lots of little steps as it bounces around inside the resistor and warms it up, rather than all in one single photon as in the superconductor. But the total energy that charge radiates is exactly qV in both cases, because anything else violates conservation of energy.

 

[davefoc]

Exactly what is going on in the barrier?

Does an electron enter the barrier with a net zero energy, gain qV energy as it passes through the barrier, lose that energy just as it passes out of the barrier through an electromagnetic emission and enter the other side of the junction with exactly zero energy?

Does the electron have a non-zero energy when it enters the barrier, gain qV energy in the barrier, lose that energy just as it passes out of the barrier through an electromagnetic emission and exit the barrier on average with the same non-zero energy that it entered the barrier with?

Or is my understanding here so screwed up that my question makes no sense?

 

[Skwinty]

Isn't it important to first decide on some basic parameters.

1. What type of material is being used as a superconductor.
eg, metal, ceramic or organic. The mechanism of each is surely different.

I understand that there are different classses of superconductors.
Conventional, unconventional and half conventional.

2. Which mechanism is being discussed.
eg electron-phonon, soliton, bisoliton or exciton.

BCS theory (electron-phonon) cooper pairs applies to low temperature superconductors but not room or high temperature superconductors.
BCS theory also does not predict superconductivity in different materials and alternate theories have been proposed.

I have also read that
"Zabusky and Kruskal found that the solitary waves are not changed in collisions, like rigid bodies, and on passing through each other, two solitary waves accelerate."

Could this explain the additional energy that Johan is talking about?

ETA:

Mathematically, there is a difference between

​

​

​

​

“solitons” and “solitary waves.”

Solitons are localized solutions of integrable equations, while solitary waves are localized solutions of non-integrable equations. Another characteristic feature of solitons is that they are solitary waves that are not deformed after collision with other solitons. Thus the variety of solitary waves is much wider than the variety of the “true” solitons. In fact, real systems do not carry exact soliton solutions in the strict mathematical sense (which implies an infinite life-time and an infinity of conservation laws) but quasi-solitons which have most of the features of true solitons. In particular, although they do not have an infinite life-time, quasi-solitons are generally so long-lived that their effect on the properties of the system is almost the same as that of true solitons. This is why physicists often use the word “soliton” in a relaxed way which does not agree with mathematical rigor.​

 

[marplots]

I think it must be very difficult psychologically to know you are right and find that when you present your ideas to what should be peers, your ideas are universally rejected. I am glad that I am able to reject my own ideas as they occur.

 

[johanfprins]

I think it must be very difficult psychologically to know you are right and find that when you present your ideas to what should be peers, your ideas are universally rejected. I am glad that I am able to reject my own ideas as they occur.

Not just psychologically but physically (meaning physics-logic) also. It is difficult to convince people that the earth is not stationary after they have observed for thousands of years that it is everything else that is moving, not the Earth.

You probably all agree that it will be an impossible task to answer all the comments that have been posted since I last posted yesterday. So I read them and tried to understand why I cannot explain that the Earth is not stationary. I came to the conclusion that it is since you, just like every other scientist to date, do not understand what a superconductor really is. So I am going to start afresh, and try and explain superconduction in terms of an analogy that we can imagine and see.

Consider a Chinese Checkers board: I assume that we all know what this is. And assume that all the dents of equal depth are filled with exactly the same marbles. Now this array of marbles is analogous to a superconducting phase. To simplify the discussion even more, let us assume a "useless" Chinese Checkers board consisting of a single row of identical marbles within identical dents all spaced at the same distance from each other. Now an external source injects an extra marble at one end. It has not got its own dent, but is cheeky enough to move to the first marble within a dent and threaten this marble to move on.

The latter marble prays and says "Oh God if you do not get me out of this pit I am a goner!" And God responds and says there is a Cinderella solution which has been discovered by Werner Heisenberg that allows Me to give to you an energy (delta)E to move out of the dent of energy depth (delta)E_S and also have kinetic energy 1/2m v^2 so that you can move to the next marble and replace this marble within a time interval (delta)t. Of course, after this time interval the magic will be gone and you will not have this energy anymore.

So the next marble prays and says: "But Oh God but now I am a goner". But God responds: "No I can now lend you the same amount of energy so that you can move on and so forth until the last marble along the row can leave the row. Thus the injected marble causes a relay race until the last marble is ejected: Thus a marble appears on the other side without any of you becoming a goner: No net energy change is required. And since you are special marbles each with a charge q, this means that an injected charge q appears as an ejected charge q on the other side without any of you having to be accelerated by an electric force from your dent to the next dent. You move without having to resist the movement in any way: Is it not SUPER?"

While preparing another linear Chinese Checkers board, the guy who drilled the dents was called away for tea when he was half way through, and when he returned, he drilled the remaining dents deeper so that now a marble must borrow Cinderella energy (delta)E_SS + the kinetic energy to reach the other side. Thus if the distances between these deeper dents are the same as between the initial shallower dents, they will not be able to borrow enough Cinderella-energy to move on. The super relay race cannot occur any more unless the deeper marbles can be accelerated from their dents. This will be hurtful to these marbles.

Fortunately, however, the dent-driller had a bit of whiskey in his tea, and he did not just drill the dents deeper but also nearer to each other, so that the relay race was still possible; the marbles could still move in this SUPER manner from one end to the other. However, the marble jumping from the last shallow dent into the adjacent deeper dent, does now not end up within the bottom of the deeper dent. After its Cinderella-jump it must have the same energy than with what it started off, so that it has to lose extra energy (delta)E_DD=(delta)E_SS-(delta)E_Sto reach the bottom of the deeper dent. This energy is then radiated away. Thus the super-movement is not so purely super than it had been before the driller made the mistake.

The driller after discovering his mistake, thought now this is interesting: "What would happen when I put a higher barrier with energy (delta)E_B between two adjacent dents of equal depth?" Of course he found that the larger he made (delta)E_B he could still have a Cinderella jump across this barrier when this barrier is thin enough. Thus the super-motion of the marbles is not affected in any way as long as the barrier is thin enough. But if the barrier is too thick this super motion cannot occur anymore. He thought: "I will call the barrier which is thin enough a Josephson-junction"

So he argued: "OK let me see what happens when I now drill the holes deeper on the other side of the Josephson barrier." And of course he still observed that the difference in the depths of the dents (delta)E_DD =(delta)E_SS-(delta)E_S must now be radiated away. So he decided for his next experiment he is not going to use a flat-topped Josephson barrier, but give it a slope so that the barrier height is still (delta)E_B on the side of the shallower dent, but at an energy (delta)E_DD lower on the other side. So while the marble is making the Cinderella jump to end up at an energy that is the same as the energy it has started off with, it also rolls down the slope on top of the barrier to gain in addition an energy (delta)E_DD. Since with the flat top barrier it has to radiate away energy (delta)E_DD, the sloped top requires that it now radiates away twice this energy.

 

[johanfprins]

Isn't it important to first decide on some basic parameters.

1. What type of material is being used as a superconductor.
eg, metal, ceramic or organic. The mechanism of each is surely different.

Not so! The mechanism is exactly the same for each material. You need an array of stationary orbitals (marbles). Therefore in the metals you must go through a metal-insulator transition to obtain these "marbles" or so-called "localized electron-states". In the ceramics these states form between the crystallographic layers; and you will find that all superconductors EVER discovered (except the phase that I extracted from a diamond) have such arrays of localized electron states which need not be doubly-charged; and are not doubly charged for most, if not all, the superconductors discovered to date.

1. I understand that there are different classses of superconductors.
Conventional, unconventional and half conventional.

A misconception that was planted in ignorance by Brian Pippard at a superconduction conference.

2. Which mechanism is being discussed. eg electron-phonon, soliton, bisoliton or exciton.

Not one of these since not one of them can explain superconduction.

BCS theory (electron-phonon) cooper pairs applies to low temperature superconductors

It does not apply to any superconductor since Cooper Pairs is a figment of Cooper's imagination.

"Zabusky and Kruskal found that the solitary waves are not changed in collisions, like rigid bodies, and on passing through each other, two solitary waves accelerate."

There is a lot of confusion when talking about "solitons" which are non-harmonic waves and a harmonic solitary wave, like an electron.

Could this explain the additional energy that Johan is talking about?

No the loaned energy is available to any harmonic solitary matter-wave for a limited time interval (delta)t

 

[sol invictus]

Exactly what is going on in the barrier?

Does an electron enter the barrier with a net zero energy, gain qV energy as it passes through the barrier, lose that energy just as it passes out of the barrier through an electromagnetic emission and enter the other side of the junction with exactly zero energy?

Does the electron have a non-zero energy when it enters the barrier, gain qV energy in the barrier, lose that energy just as it passes out of the barrier through an electromagnetic emission and exit the barrier on average with the same non-zero energy that it entered the barrier with?

Or is my understanding here so screwed up that my question makes no sense?

The process by which the electrons pass through the barrier is quantum - they tunnel through it. As a result you can't really ask about their energy "while" they are inside it. Naively their kinetic energy would be negative, since total energy is conserved and the barrier is high. And indeed, there is an approximation (the WKB approximation) in which one uses that to compute the rate at which the tunneling occurs.

But more accurately there is no sense in which the kinetic energy is ever negative. No measurement will ever reveal it to be so, and in particular, no photon emitted by the electron will have measured energy greater than qV.

 

[sol invictus]

You probably all agree that it will be an impossible task to answer all the comments that have been posted since I last posted yesterday.

I doubt anyone agrees with you on that either, Johan - unless "impossible" means "impossible without admitting I am wrong".

The driller after discovering his mistake, thought now this is interesting: "What would happen when I put a higher barrier with energy (delta)E_B between two adjacent dents of equal depth?" Of course he found that the larger he made (delta)E_B he could still have a Cinderella jump across this barrier when this barrier is thin enough. Thus the super-motion of the marbles is not affected in any way as long as the barrier is thin enough. But if the barrier is too thick this super motion cannot occur anymore. He thought: "I will call the barrier which is thin enough a Josephson-junction"

So he argued: "OK let me see what happens when I now drill the holes deeper on the other side of the Josephson barrier." And of course he still observed that the difference in the depths of the dents (delta)E_DD =(delta)E_SS-(delta)E_S must now be radiated away. So he decided for his next experiment he is not going to use a flat-topped Josephson barrier, but give it a slope so that the barrier height is still (delta)E_B on the side of the shallower dent, but at an energy (delta)E_DD lower on the other side. So while the marble is making the Cinderella jump to end up at an energy that is the same as the energy it has started off with, it also rolls down the slope on top of the barrier to gain in addition an energy (delta)E_DD. Since with the flat top barrier it has to radiate away energy (delta)E_DD, the sloped top requires that it now radiates away twice this energy.

So the latest claim is that an electron tunneling through a barrier like this:

[FONT="Courier New"]
    |-----|   
    |     |
----|     |
          |
          |------
[/FONT]

will radiate a photon with a larger energy than an electron tunneling through a barrier like this?

[FONT="Courier New"]
      /\   
     /  \
----/    \
          \
           \------
[/FONT]

Is that what you are claiming now?

 

[johanfprins]

Exactly what is going on in the barrier?

The postulate that an electron can move "through" an energy barrier, and thus "tunnel" through the barrier is false. It also violates Pauli's exclusion principle since the barrier also consists of matter waves. What really happens during the process which is called by the misnomer of tunneling, is that the electron-wave takes the Cinderella option, it borrows enough energy to jump over the energy barrier and reach the other side with the same energy that it had before it jumped the barrier.

Consider a "quantum" tennis ball approaching a wall. According to the accepted crackpot interpretation, the tennis ball moves through the wall "having negative energy" and proceeds. According to my realistic interpretation, the ball reaches the wall, take the Cinderella option, jumps over the wall and then moves on as if the wall were not even there.

Does an electron enter the barrier with a net zero energy, gain qV energy as it passes through the barrier, lose that energy just as it passes out of the barrier through an electromagnetic emission and enter the other side of the junction with exactly zero energy?

No, the electron takes the Cinderella option, borrows the energy to jump across the barrier to reach the other side at exactly the same energy that it has had before jumping. If the orbital on the other side is at a lower energy, then the electron has to radiate away this difference in energy before it can replace the "waiting orbital". If during the Cinderella jump, it encounters an electric field that accelerates it, it will in addition also have to radiate away this energy.

Putting an electric field across a Josephson junction, it lowers the orbitals on the opposite side relative to the orbitals jumping the barrier and this difference in energy qV must then be radiated away. But it also causes the jumping orbital to be accelerated and gain additional energy during the Cinderella jump, and this also has to be radiated away. Thus the energy radiated must be 2qV; and the energy measured is 2eV, so that the jumping orbitals each has a single electron charge.

 

[johanfprins]

The latest claim is that an electron tunneling through a barrier like this:

[FONT="Courier New"]
    |-----|   
    |     |
----|     |
          |
          |------
[/FONT]

will radiate a photon with a larger energy than an electron tunneling through a barrier like this?

[FONT="Courier New"]
      /\   
     /  \
----/    \
          \
           \------
[/FONT]

Is that what you are claiming now?

Your second drawing is not correct since the barrier does not have a peak in voltage at its center when you place a voltage across it. Furthermore I have not made "a latest claim", I have just tried to explain in a Kindergarten manner what I have said all along with the hope that you will now understand it.

Oh and by the way, if you can arrange your first diagram to be valid without an electric field over the junction, it will radiate away less energy than in the case where there is also an electric-field.

 

[johanfprins]

The process by which the electrons pass through the barrier is quantum - they tunnel through it. As a result you can't really ask about their energy "while" they are inside it.

I see you still believe in fairy tales.

Naively their kinetic energy would be negative, since total energy is conserved and the barrier is high. And indeed, there is an approximation (the WKB approximation) in which one uses that to compute the rate at which the tunneling occurs.

"Naively" is the understatement of the century. Let us call it what it really is: It is pure unadulterated BS. You cannot have an electron moving "through an energy barrier". Only if you are totally cuckoo will you believe this Copenhagen nonsense.

 

[sol invictus]

No, the electron takes the Cinderella option, borrows the energy to jump across the barrier to reach the other side at exactly the same energy that it has had before jumping. If the orbital on the other side is at a lower energy, then the electron has to radiate away this difference in energy before it can replace the "waiting orbital". If during the Cinderella jump, it encounters an electric field that accelerates it, it will in addition also have to radiate away this energy.

This, of course, is exactly the same basic mistake you've been making all along. Showing you that was the point of all the posts you've just ignored. The electric field is the reason there is a difference in energy on the two sides of the barrier in the first place - that there is a potential energy difference is simply another way of saying, in a slightly different language, that there is a field.

Your second drawing is not correct since the barrier does not have a peak in voltage at its center when you place a voltage across it.

Then draw your own, please. Good luck.

 

[sol invictus]

I see you still believe in fairy tales.
"Naively" is the understatement of the century. Let us call it what it really is: It is pure unadulterated BS. You cannot have an electron moving "through an energy barrier". Only if you are totally cuckoo will you believe this Copenhagen nonsense.

Yes, quantum mechanics - the most successful theory in the history of science, the theory that explains everything from metals to Bose-Einstein condensates to emission line spectra to the sun to particle physics to the origin of galaxies with unprecedented precision - is obviously "pure unadulterated BS".

I'm sorry Johan, but the (totally unsupported) assertion of someone that repeatedly makes a basic, simple physics mistake - a mistake that any competent high school physics student would recognize - doesn't carry much weight.

 

[johanfprins]

I doubt anyone agrees with you on that either, Johan - unless "impossible" means "impossible without admitting I am wrong".

From my past experience with the main stream crackpots over the past 10 years, and the way in which you twist my words to implicate that I am a charlatan, I expect that you will never admit that you are wrong even after it struck you that you must be wrong. It is uncanny how people still do not see mountains on the moon if they want to believe that there cannot be mountains on the moon.

 

[johanfprins]

Yes, quantum mechanics - the most successful theory in the history of science, the theory that explains everything from metals to Bose-Einstein condensates to emission line spectra to the sun to particle physics to the origin of galaxies with unprecedented precision - is obviously "pure unadulterated BS".

You cannot even see how silly you are. I did not say quantum mechanics is BS; but it suits your murky agenda to throw around such false accusations. What I said is that the Copenhagen interpretation that "tunneling" is a probability process is unadulterated BS. And I can assure you that you do not even know what a "Bose-Einstein" condensate is. You probably naively think that such a condensate is built up of "particles"; even though you are not even able to define what a "particle" is. I really whish you would grow up. As a matter of interest, how old are you?

I'm sorry Johan, but the (totally unsupported) assertion of someone that repeatedly makes a basic, simple physics mistake -

I have not made any mistake in my reasoning. If you cannot understand that my model is fully supported by quantum mechanics sans the crackpot probability interpretation, then the fault lies with you!

 

[sol invictus]

I did not say quantum mechanics is BS

Yes, you did.

but it suits your murky agenda to throw around such false accusations.

"Murky agenda", Goebbels - you're paranoid, Johan.

I have not made any mistake in my reasoning.

You have made an extremely simple, extremely basic mistake. It's become quite clear to me that no matter what we say we will not be able convince you of that. So, I'm afraid all the comments on the first page of this thread were spot-on.

How's that diagram of the barrier coming, Johan?

 

[johanfprins]

This, of course, is exactly the same basic mistake you've been making all along. Showing you that was the point of all the posts you've just ignored. The electric field is the reason there is a difference in energy on the two sides of the barrier in the first place -

I have never disputed that and have pointed out over and over and over again that this shift will require the radiation qV when an orbital jumps over the barrier by borrowing and returning energy; which is allowed by quantum mechanics. This will happen even when the electron is not accelerated during the jump. It does not require acceleration since the energy to scale the barrier is available until it has to be returned a time (delta)t later.

I have then logically pointed out that if during the jump which is allowed by the borrowing and returning of energy, the electron is in addition accelerated, as it must be owing to the presence of the voltage V, it cannot also give this energy back and thus has to also radiate this energy; hence 2qV. You consistently claimed that this voltage cannot accelerate the jumping electron to have energy above the energy with which it started off with. My high school text books taught me in no uncertain terms that when you start off with an energy E(0) and accelerate through a potential energy V(p) you energy at the end will be E(0) +V(p). You maintain that it must still be E(0). I think you are in need to reread your elementary physics books!

Then draw your own, please. Good luck.

....................
|\
| \
| \ qV
| \
|....... \..........
| |
| |
| |
____|.........|..........
|
| qV
|
|......................




I tried but could not get it to reproduce well. If you can make out what I tried to draw you will see that there are two energy regions qV: One owing to the shift of the level on the other side to be lower by energy qV, and the same energy difference qv which accelerate the jumping electron.