No, you're wrong. In quantum mechanics, kinetic energy is not "rate of change of time of the position of the center of mass". Kinetic energy is proportional to the 2nd derivative of the wavefunction, and the 2nd derivative of the wavefunction is not zero.
I know that this is the Copenhagen interpretation of the Schroedinger's wave function: But it is of course nonsense. If an electron has kinetic energy it must move relative to an inertial reference frame, and if it has a wave "associated" with it by "wave-particle duality" then even this wave must also move with the electron-particle. A standing Schroedinger wave does NOTHING of the sort; and therefore it cannot have kinetic energy, and neither can its "dual" electron.
Let me give you an example where the physics is as obvious as possible. Take one laser beam and shoot a pulse to the left. Take another laser and shoot a pulse to the right. Let the pulses fly through space, at some point, cross and overlap. The region of space where both, counterpropagating pulses are present has---uh-oh---a photon wavefunction with exactly the same properties as an ensemble of photons in a cavity.
Again more of your BS. A laser beam does NOT consist of separately identifiable photon-waves. The two laser beams are EACH a single coherent macro-wave which is modelled directly by Maxwell's equations. Furthermore, when the beams superpose the resultant wave does not remain stationary like a standing wave within a cavity does.
In other words, it looks in every detail like a standing wave; it has zero NET momentum;
Only momentarily while the two,laser pules move through each other.
the center-of-momentum isn't moving anywhere; etc.
the two waves superpose and stay two separate waves each having its own "centre of momentum". Within a cavity the wave is not a superposition of two ACTUAL coherent waves running in opposite directions. The standing wave is a single non-superposed entity.
(In fact, if you were standing in this region you would be unable to tell whether the light you're seeing is passing-in-the-night or trapped-in-a-cavity.
Nonsense!!! In the cavity the intensity of the standing wave becomes ZERO at the walls of the cavity. In your example of two laser pulses in space this is NOT the case. The end points of the two laser pulses are not stationary relative to you.
Go ahead, do the math.) But you must agree that this region of space contains photons with ordinary kinetic energy---50% of it, on average, heading left and 50% heading right.
There are no photons just two coherent laser waves. But suppose that the two laser pulses are two photons You then have two photons moving through one another without any boundaries to pin them; and you want to argue that this is the same as a photon standing wave within a cavity with boundaries. But within a cavity you can have a SINGLE PHOTON wave that is stationary: YOU DO NOT NEED TWO PHOTON WAVES MOVING ALONG BOTH DIRECTIONS. IN FACT IT IS PHYSICALLY IMPOSSIBLE FOR ANY RUNNING WAVES TO EXIST WITHIN A CAVITY.
The laws of quantum mechanics---i.e., KE = (p-hat)^2/2m---will look at this wavefunction and get that fact right. (It'll say p = {left or right}, <p> = 0, <p^2> != 0, KE != 0, etc.) Your heuristic----"this is the wavefunction of a standing wave, so let's ignore Schrodinger and think about relativity"---will not.
Thus acording to you a standing light wave cannot have less energy tha 2*h*(nu). You have really lost it have you not?
Johan has been denying that massive particles, in bound states, have kinetic energy. I diverted him onto photons only briefly.
There are no "massive particles in bound states" since the bound states are waves
not particles.
The rest of your post is the usual BS you expound: So I will ignore it and try to redirect the discussion to superconduction at room temperature.
