That would be a violation of conservation of energy. The only way the electron can radiate anything at all is if it is accelerated during the jump. If both sides of the barrier are at the same height, the electron will not and cannot radiate (as in the DC Josephson effect).
I have just typed this answer and then lost it when posting it
. So here we go again: Let us first see if we can agree on quantum mechanics. Do you agree that a localized orbital can through a quantum fluctuation attain an increase in energy which allows it to jump a barrier and arrive at the other side with the same energy. If not, and you think that an electron can only tunnel through a barrier, then we are wasting each other's time since then, according to my opinion you are living in Alice's Wonderland. I will now go ahead by assuming that you agree that such a scaling or "barrier-jumping" as I propose here is what is actually happening during a "tunneling event".
We seem to agree that when a SC charge-carrier moves from one side to the other under the conditions where the initial energy of the charge-carrier is the same as the energy of the charge carrier it has to replace on the other side, there will be no EM-radiation at all. Now consider an interface between two superconductors where the binding energy of the localized charge carriers are lower in one than in the other. A charge-carrier jumping from the the one SC into the other, by borrowing and returning energy, will then end up at a higher energy than the energy of the charge-carrier it has to replace. In order to replace this charge carrier it will have to radiate away this energy even though it experienced
no electric-field while it jumped.
It is only accelerated by the potential energy difference between its starting energy and the final energy it has to be at, after it has reached the other side.
Now in the case of a Josephson junction with a voltage V across it, the latter energy difference is still responsible for radiating the offset qV between the two energy levels after the electron has jumped across;
even if the electron, during its jump, ignored the electric field within the junction. But it cannot ignore this field, so it is in addition accelerated. Thus instead of just ending up at a potential energy qV above its final energy which must be radiated away by acceleration, it also has kinetic energy qV which must also be emitted as EM radiation.
our high school texts did not teach you that - you made that mistake all by yourself. When you start with energy E(0) and move through a region with a potential energy gradient, you end with exactly the same energy you started with. That's called conservation of energy. The only effect of the acceleration is that potential energy is exchanged for an exactly equal amount of kinetic energy.
I agree, in the heat of the moment I wrote nonsense by not distinguishing for the starting energy between energy that can be dissipated and energy that must first be converted into kinetic energy before it can be dissipated. Sorry about that.
But it does not change my logic on this thread by a single iota! The fact is that since the charge carrier borrows the energy to reach the other side it will end up at energy qV above the energy level it has to end up at, after returning the borrowed energy. If however, during this jump it is in addition accelerated during the jump it will still have the potential energy qV it must radiate away by being accelerated PLUS in addition the kinetic energy that it gained during the jump owing to the presence of the electric field within the barrier.
I'll try to reproduce your diagram here:
Code:
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| \
| \ qV
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|....... \..........
| |
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____|.........|..........
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| qV
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Thanks you did a sterling job
Your diagram (like mine) shows that an electron at rest on the right side of the barrier has an energy that's less than an electron at rest on the left side by qV.
correct I never had any argument with this.
Therefore, by conservation of energy, if the initial and final states are an electron at rest on the left and right sides of the barrier, exactly qV of energy must have been radiated away.
Only if the movement of the electron across the barrier does not require an extra increase in energy.
The shape of the barrier is totally irrelevant, as I proved in one of the posts you ignored (after you requested the proof).
As far as the borrowed energy is concerned the shape of the barrier is irrelevant since it allows to electron to move across the barrier and ends up at an energy qV above the energy it must end up at after the borrowed energy is returned. From where it is accelerated to release energy qV: However the shape is important as to whether the jumping electron is accelerated or not during the jump. If the junction could be flat topped, only the energy qV would need to be dissipated; but in this case it is not flat topped, so that during the jump the electron gains kinetic energy qV in addition to the energy qV it has to dissipate by further acceleration
after it has reached the other side.
