Johan, you have corrected your diagram (in the erratum) to show a
potential energy drop of 2qV.
Code:
ground --- power supply ----------- junction ---------- ground
___
| |
| |
| |
__________________| | . . . . . . . V
/ |
________________/ | .............. 0
|
| __________ -V
You show the charge dropping from the "V" level to the "-V" level, a change of 2qV of potential energy. But this charge has to get pushed back up to the V level ... which won't happen with the one-V battery of the diagram above. The battery needs to do 2qV of work to get move this charge from the -V level back to the +V level ... which means that it's a battery whose voltage is 2V to begin with.
In other words, the above voltage scenario---a battery of voltage V pushing electrons through a circuit where they can drop by 2V---again does not conserve energy. Let me fix it for you:
Code:
ground --- power supply ----------- junction ---------- ground
___
| |
battery | |
has voltage 2V | |
__________________| | . . . . . . . V
/ |
/ | .............. 0
/ |
_____________/ |_______________ -V
That's the voltage landscape that corresponds to your
corrected fig. 28-6b. This voltage picture does conserve energy, finally. But look, it's gone back to describing the
mainstream Josephson picture:
The energy of the photon emitted is equal to, not double, the net battery voltage times the charge of the charge-carrier.
The V is getting confusing now, let's add a clearer variable name. The above circuit has a battery of voltage B. When a charge q moves through the junction, it emits a photon of energy E = qB. Just like Josephson said. The fact that we observe photons at E = 2eB tells us that q = 2e and therefore that conduction comes from electron pairs.
Your quantity V = B/2 is not the battery voltage, nor the photon energy, nor does it appear to be observable in any way. You seem to be saying "there's V worth of polarization potential plus V worth of Fermi energy", but one way or another this energy needs to be supplied by the battery.