Superconductivity at room temperature??

[johanfprins]

I find it rather telling that the one thing you are NOT willing to do here is actually calculate the answer based on the wave function Sol gave you. All the necessary information is right there, but you can't do it, can you? You've lost the ability to do simple calculus.

Have YOU ever understood calculus and the conservation of energy? Obviously not!

I can obviously calculate the mathematical answer but this answer is given relative to a zero-energy baseline which does not make any physics-sense: In text books it is assumed that the mathematical answer is the full answer and that therefore the energy is "kinetic energy" measured from a undefined baseline at which the kinetic energy is supposedly zero. In other words it is stupidly assumed that you can add any potential energy factor without changing the physics.

Although the latter is mostly possible in classical mechanics, this is NOT the case in quantum mechanics; since the energy of a stationary electron-wave MUST be its mass energy, and this can only be correctly interpreted physically when the base-line energy from which it is measured is the rest mass energy of the electron.

When you insert this baseline into the Schroedinger equation, you find that the energy of the electron given by Sol's wave, is less than the rest mass energy of an electron: THUS THE ENERGY OF THIS WAVE CANNOT BE KINETIC ENERGY EVER!!!!! It can only have a kinetic-energy component if the energy of an electron wave is larger than m*c^2. Have YOU ever heard of Einstein and his Special Theory of Relativity?

 

[ben m]

Johan, let me simplify it for you.

Look at an electromagnetic wave in a high-Q resonant cavity. It's a standing wave, with boundary conditions, with <v>=0. It has all of the properties you attribute to an electron wave in a metal.

If I slice one end off the cavity---changing the boundary conditions---does the wave stay in the cavity, unable to cruise out "because <v>=0"? Obviously not.

 

[ben m]

Although the latter is mostly possible in classical mechanics, this is NOT the case in quantum mechanics; since the energy of a stationary electron-wave MUST be its mass energy, and this can only be correctly interpreted physically when the base-line energy from which it is measured is the rest mass energy of the electron.

What's the energy of an electron bound to a stationary atom? It's not 510.9989 keV, it's less. In a ground-state hydrogen atom, for example, the electron's total energy is 510.9852 keV --- it's 13.7 eV below the rest mass. Its kinetic energy has a nonzero, positive expectation value. Its potential energy has a nonzero, negative expectation value.

 

[johanfprins]

Johan, let me simplify it for you.

Look at an electromagnetic wave in a high-Q resonant cavity. It's a standing wave, with boundary conditions, with <v>=0. It has all of the properties you attribute to an electron wave in a metal.

Thanks, you are at last starting to talk physics. You are correct, the electromagnetic wave in this case has NO KINETIC ENERGY since it is a standing wave.

If I slice one end off the cavity---changing the boundary conditions---does the wave stay in the cavity, unable to cruise out "because <v>=0"? Obviously not.

Exactly!!! Now it can cruise out, since it is not a standing wave anymore.

In the metal it is similar: You have standing waves when not making contacts to the material and applying a voltage. As soon as you do this, you have wave-packets which can move through the contacts: And only now you have kinetic energy at the Fermi-level energy position.

 

[Ziggurat]

Have YOU ever understood calculus and the conservation of energy? Obviously not!

And yet, you were the one who couldn't get conservation of energy right back when we were talking about Josephson junctions and you claimed that each electron would give off 2eV of energy when crossing, something which would violate conservation of energy really obviously.

I can obviously calculate the mathematical answer

I don't believe you can. You've never demonstrated any ability to use math beyond arithmetic here (and even that is quite spotty, per above).

but this answer is given relative to a zero-energy baseline which does not make any physics-sense: In text books it is assumed that the mathematical answer is the full answer and that therefore the energy is "kinetic energy" measured from a undefined baseline at which the kinetic energy is supposedly zero. In other words it is stupidly assumed that you can add any potential energy factor without changing the physics.

Like I said before, you fundamentally disagree with standard quantum mechanics.

Although the latter is mostly possible in classical mechanics, this is NOT the case in quantum mechanics; since the energy of a stationary electron-wave MUST be its mass energy

Nope. Not in non-relativistic quantum mechanics. And remember, YOU asked us to use non-relativistic quantum mechanics. Oh, wait, I'm sorry, you can't remember things, so you probably already forgot that.

When you insert this baseline into the Schroedinger equation, you find that the energy of the electron given by Sol's wave, is less than the rest mass energy of an electron: THUS THE ENERGY OF THIS WAVE CANNOT BE KINETIC ENERGY EVER!!!!!

Really? That would be quite remarkable, if you could actually do this. Please, demonstrate.

But you won't. Because you can't.

It can only have a kinetic-energy component if the energy of an electron wave is larger than m*c^2. Have YOU ever heard of Einstein and his Special Theory of Relativity?

Have YOU ever heard that the Schrodinger equation is explicitly non-relativistic? And did you hear that YOU are the one who asked us to use it?

 

[Ziggurat]

What's the energy of an electron bound to a stationary atom? It's not 510.9989 keV, it's less. In a ground-state hydrogen atom, for example, the electron's total energy is 510.9852 keV --- it's 13.7 eV below the rest mass. Its kinetic energy has a nonzero, positive expectation value. Its potential energy has a nonzero, negative expectation value.

We can go a step further, in fact, and actually quantify those contributions. Its kinetic energy is +13.7 eV and its potential energy is -27.4 eV.

 

[johanfprins]

What's the energy of an electron bound to a stationary atom? It's not 510.9989 keV, it's less. In a ground-state hydrogen atom, for example, the electron's total energy is 510.9852 keV --- it's 13.7 eV below the rest mass.

EXACTLY!! ITS ENERGY IS LESS THAN THE ENERGY IT MUST HAVE TO HAVE KINETIC ENERGY!!

Its kinetic energy has a nonzero, positive expectation value.

How is this possible if its TOTAL energy is less than its rest mass energy???

Its potential energy has a nonzero, negative expectation value.

NO!! Its lowest potential energy is m*c^2 MINUS V. where V in this case goes to MINUS infinity at r=0. Its lowest allowed energy is m*c^2 MINUS 13.6 eV, which CANNOT be kinetic energy SINCE it is lower than m*c^2.

 

[ben m]

Since there is a voltage difference, there MUST be an electric-field between the points A and B, and this electric field MUST attract such a pseudo-electron from one position to the other position. If there is no scattering, such a pseudo-electron MUST be accelerated from one position to the other.

Still no (emphatically). You are still, wrongly, pretending that the Fermi energy is exactly the same thing as electric potential energy. This is still wrong. The Fermi energy is the sum of (a) an energy attributable to electric potential (which you'll find electric fields to explain) and (b) an energy attributable to Pauli-exclusion-of-lower-bound-states in the material.

In case it helps, which I doubt, let me challenge you with another thought experiment. Consider a gold block and a sodium block; neutral and NOT in contact. It is clear that their Fermi levels are different---if I want to take an electron out of the gold into the vacuum, I have to expend 5.1 eV; if I lower it this electron into the sodium, I only gain 2.3 eV. According to your mental picture, this difference must represent an electric field. Where is the field? In the vacuum between the blocks? No it isn't, the field here is zero. (It's a vacuum, I can use Gauss's Law.) At the surface of the metal? No it isn't; if I drop a proton or a positron onto either block, it doesn't feel the Fermi level difference at all.

 

[Ziggurat]

Thanks, you are at last starting to talk physics. You are correct, the electromagnetic wave in this case has NO KINETIC ENERGY since it is a standing wave.

Exactly!!! Now it can cruise out, since it is not a standing wave anymore.

And... you walked right into ben's trap. If it's got kinetic energy when inside the cavity, and you just remove a wall (an action which DOES NOT provide the photon with more energy), then why does it suddenly now have kinetic energy? You have violated conservation of energy, johan. This was the obvious setup of ben's question, and you missed it, because you simply have no clue.

 

[johanfprins]

We can go a step further, in fact, and actually quantify those contributions. Its kinetic energy is +13.7 eV and its potential energy is -27.4 eV.

LOL

 

[johanfprins]

And... you walked right into ben's trap. If it's got kinetic energy when inside the cavity, and you just remove a wall (an action which DOES NOT provide the photon with more energy), then why does it suddenly now have kinetic energy? You have violated conservation of energy, johan. This was the obvious setup of ben's question, and you missed it, because you simply have no clue.

You are really slow: When you remove one wall, you do not change the energy of the wave but you now allow this energy to propagate as kinetic energy. This also happens when an atomic electron "orbital" releases light energy: Before releasing this energy it is purely STATIONARY mass energy. When released this SAME energy is purely kinetic light energy. The energy is TRANSFORMED, and energy conservation is not violated; as you have done all along in your arguments on this thread.

 

[Ziggurat]

LOL

Go on, try it. Take the ground-state wave function of the hydrogen atom, and actually calculate the kinetic energy and the potential energy. Tell us what you get. This is standard introductory textbook quantum mechanics, it's not really hard.

But you won't do that, perhaps because you can't, or perhaps because you don't actually want to know the answer. I'm still leaning towards the former, but feel free to prove me wrong about both.

Alternatively, since I'm quite confident that you can't do the calculation, you could admit that and ask one of us to do the calculation for you, and show our work. I can do it, and I'm sure Sol and ben can do it too. And I'm quite confident that we will all agree on what the answer is, too.

 

[johanfprins]

In case it helps, which I doubt, let me challenge you with another thought experiment. Consider a gold block and a sodium block; neutral and NOT in contact. It is clear that their Fermi levels are different---if I want to take an electron out of the gold into the vacuum, I have to expend 5.1 eV; if I lower it this electron into the sodium, I only gain 2.3 eV. According to your mental picture, this difference must represent an electric field. Where is the field? In the vacuum between the blocks? No it isn't, the field here is zero. (It's a vacuum, I can use Gauss's Law.) At the surface of the metal? No it isn't; if I drop a proton or a positron onto either block, it doesn't feel the Fermi level difference at all.

This such a silly argument: But it requires a long anaswer. It has been said that a fool can ask more questions within 10 minutes than a wise man can answer in a lifetime! It is late in South Africa: So I will return to this aspect tomorrow if I have the time.

 

[Ziggurat]

You are really slow: When you remove one wall, you do not change the energy of the wave but you now allow this energy to propagate as kinetic energy.

What energy? There's no potential within the cavity. If your energy isn't potential energy, then it's kinetic energy. It sure as hell isn't rest mass energy, because photons have no rest mass.

 

[ben m]

You are really slow: When you remove one wall, you do not change the energy of the wave but you now allow this energy to propagate as kinetic energy. This also happens when an atomic electron "orbital" releases light energy: Before releasing this energy it is purely STATIONARY mass energy. When released this SAME energy is purely kinetic light energy. The energy is TRANSFORMED, and energy conservation is not violated; as you have done all along in your arguments on this thread.

Even if you believe this, you ought to see that it conflicts with your statements about contact potentials. Your false understanding of standing waves disagrees with your false understanding of Fermi levels.

How is an electron-in-sodium different than a photon-in-a-cavity? You say they're both standing waves. But if I chop one of the boundary conditions off of the cavity, you recognize that the photon gets out. If I chop one of the boundary conditions off the sodium, by bringing in the gold, you somehow think that the electron can't get out without a "force" to accelerate it out. Why the difference, Johan?

 

[ben m]

EXACTLY!! ITS ENERGY IS LESS THAN THE ENERGY IT MUST HAVE TO HAVE KINETIC ENERGY!! How is this possible if its TOTAL energy is less than its rest mass energy??? NO!! Its lowest potential energy is m*c^2 MINUS V. where V in this case goes to MINUS infinity at r=0. Its lowest allowed energy is m*c^2 MINUS 13.6 eV, which CANNOT be kinetic energy SINCE it is lower than m*c^2.

Johan, let me ask you for the 12th time: do you know an actual, in-person scientist of any sort? Do you have a niece or nephew with a bachelor's in chemistry or physics? A neighbor who teaches at a community college? Please ask someone to walk you through the physics of hydrogen and check your arguments. Do this in person, on pencil and paper with a textbook in front of you.

You have nothing to lose. If you're right, and if Zig and I are morons, surely your random acquaintance will back you up---maybe they'll suggest a devastating proof that will put us in our place.

If Zig and I are right, you'll learn something and stop being humiliated.

 

[johanfprins]

I find it rather telling that the one thing you are NOT willing to do here is actually calculate the answer based on the wave function Sol gave you. All the necessary information is right there, but you can't do it, can you? You've lost the ability to do simple calculus.

OK let us derive the answer correctly.

Let us derive Sol's standing wave using the rest mass energy m(e)*c^2 as the baseline-energy at which we know that there is no kinetic energy. Thus, for the energy E that must be inserted into the Schroedinger equation, we must use E=m*c^2 where m is different from m(e). Since we are using m(e)*c^2 as the baseline, we must also measure the potential energy V of the electron from this baseline: i.e. choose V=0 on this base line.

When the electron has V=0 and E=m(e)*c^2, then it must be at the vacuum level and can thus not be a bound electron. Sol's wave function is that of a bound electron in a "box": Thus it must find itself at a potential energy that is less than m(e)*c^2; AND its E must also be less than the energy it has at the vacuum level outside the box: What you thus have to insert into the Schroedinger equation for E and V, are:

E=m(e)*c^2-(modulus)((delta)E)

where (delta)E is the energy of the electron as measured from the baseline m(e)*c^2

AND
V=m(e)*c^2-(modulus)((delta)V)

Where (delta)V is the potential energy of the electron as measured from the baseline m(e)*c^2.

When inserting these expression into the Schroedinger equation one finds that one can only obtain Sol's solution when (modulus)((delta)V))>(modulus)((delta)E)). In other words if you set (delta)V equal to zero; as is done in text books, you derive a mathematical solution which cannot physically manifest. Not that it seems to matter, since this has been par for the course for nearly 100 years when modern day theoretical physicists are allowed to model physics: They prefer to produce, what I call Voodoo, paranormal metaphysical models.

When you solve the Schroedinger equation with the correct expressions using the baseline m(e)*c^2, one obtains the following for the energy-offset:

(mod)((delta)E)=(mod)((delta)V)-[n*(hbar)^2*(pi)^2]/[2*m(e)*L^2] where L is the length of the standing wave.

Thus, we have that

E=m*c^2=m(e)*c^2-(mod)((delta)E) which is less than m(e)*c^2:

AND THEREFORE THE ENERGY OF SOL'S STANDING WAVE CANNOT BE KINETIC ENERGY AT ALL. If it is kinetic energy, it violates Einstein's Special Theory of Relativity which mandates that KE=m*c^2-m(e)*c^2 where m must be larger than m(e).

 

[Ziggurat]

OK let us derive the answer correctly.

Let us derive Sol's standing wave using the rest mass energy m(e)*c^2 as the baseline-energy at which we know that there is no kinetic energy.

1) You're assuming something which the Heisenberg uncertainty principle says cannot be true, namely that the ground state has no kinetic energy.

2) This is explicitly non-relativistic quantum mechanics, so rest mass energy is irrelevant, just like it is irrelevant in Newtonian mechanics. I don't care if that's wrong, that's still the theory, and YOU told us to use non-relativistic mechanics.

3) We don't have to derive the solution, we can simply look it up in any undergrad quantum mechanics textbook. Hell, Sol even did that part for us. Our task is merely to calculate the kinetic energy of that solution.

Thus, for the energy E that must be inserted into the Schroedinger equation, we must use E=m*c^2 where m is different from m(e). Since we are using m(e)*c^2 as the baseline, we must also measure the potential energy V of the electron from this baseline: i.e. choose V=0 on this base line.

I don't care what you want to insert into the Schrodinger equation (though I note that you haven't actually done what you say needs doing), we already HAVE the solution. We don't need to insert the total energy in anywhere in order to calculate the kinetic energy.

Sol's wave function is that of a bound electron in a "box": Thus it must find itself at a potential energy that is less than m(e)*c^2; AND its E must also be less than the energy it has at the vacuum level outside the box: What you thus have to insert into the Schroedinger equation for E and V, are:

E=m(e)*c^2-(modulus)((delta)E)

I don't care what the potential is. If I have the wave function (and Sol already gave it), I don't need to know anything about the potential in order to calculate the kinetic energy. Again, this is basic quantum mechanics, and you can't do it.

When inserting these expression into the Schroedinger equation one finds that one can only obtain Sol's solution when (modulus)((delta)V))>(modulus)((delta)E)). In other words if you set (delta)V equal to zero; as is done in text books, you derive a mathematical solution which cannot physically manifest.

In other words, you object to standard quantum mechanics. You think the theory is wrong from top to bottom.

When you solve the Schroedinger equation with the correct expressions using the baseline m(e)*c^2, one obtains the following for the energy-offset:

(mod)((delta)E)=(mod)((delta)V)-[n*(hbar)^2*(pi)^2]/[2*m(e)*L^2] where L is the length of the standing wave.

I don't care what you think the total energy is, that's not what I want to find. I ONLY want to find the kinetic energy. And for that, I can ignore the potential as long as you give me the actual wave function. All I need to do to calculate the kinetic energy is to calculate
integral[psi^*(p²/2m)psi dx]
At no point do I need to insert ANYTHING about the potential or the total energy, all I need is the wave function itself

Furthermore, you object to Sol's wave function, but I notice that you can't actually provide a wave function of your own. Hell, you can't even tell me what your deltaV's are quantitatively.

AND THEREFORE THE ENERGY OF SOL'S STANDING WAVE CANNOT BE KINETIC ENERGY AT ALL. If it is kinetic energy, it violates Einstein's Special Theory of Relativity which mandates that KE=m*c^2-m(e)*c^2 where m must be larger than m(e).

But we already know that the Schrodinger equation violates special relativity. It is explicitly non-relativistic. So is the equation KE = (1/2)mv², yet that still works pretty damned well, doesn't it? So too with the Schrodinger equation, WITHOUT resorting to rest mass. Your simultaneous appeals to the Schrodinger equation AND special relativity reveal that you are deeply ignorant of at least one (and probably both) of those theories.

And look, we don't even have to do a particle in a box. We can do the hydrogen atom instead. That's a very physical system, if you think the particle in a box system is too artificial.

 

[johanfprins]

I decided to jump ahead to this post of Ziggurat since it demonstrates the superstition and Voodoo that Heisenberg deliberately introduced in order to keep the concept of a "particle" alive; after Schroedinger's equation rendered incontrovertible proof that an electron is always a wave-entity.

1) You're assuming something which the Heisenberg uncertainty principle says cannot be true, namely that the ground state has no kinetic energy.

I know that Heisenberg's Voodo-interpretation of this relationship can be used to argue such nonsense, and has been used far too long to argue such nonsense.

All this relationship for "momentum" and position is saying is that a harmonic wave "lives" in both position-space and reciprocal space: Just as each (and every harmonic wave)has always done. This is shown by the fact that Planck's constant cancels out to give you the correct relationship between position space and reciprocal space (delta)x*(delta)k>1/2. A stationary harmonic wave HAS NO MOMENTUM AT ALL and therefore the De Broglie relationship CANNOT define a wavelengh for it. The De Broglie relationship is only valid when the momentum IS NOT ZERO; and it has only been experimentally measured WHEN THE MOMENTUM IS NOT ZERO.

To state that there is an uncertainty in position and momentum of the centre-of-mass of an electron-entity, violates Galileo's concept of inertia. Furthermore,if at absolute zero the position and momentum do not manifest with 100% accuracy, the electron will have kinetic energy at absolute zero: This violates the concept of absolute zero temperature.

2) This is explicitly non-relativistic quantum mechanics, so rest mass energy is irrelevant, just like it is irrelevant in Newtonian mechanics.

Here I again disagree. Even with Newtonian mechanics you have a rest mass. This fact smoothly dovetails with Einstein's Speacial Theory of Relativity. Since Schroedinger uses rest mass in his equation, this equation should also smoothly dovetail with Special Relativity.

In fact it does when you accept the obvious, and that is that the wave-energy of a solitary electron within its own inertial reference frame within which it is stationary (i.e. (delta)p=(delta)x=0), must be its rest mass energy. This means that the correct wave equation should give a solution for a solitary electron that is a localised stationary wave within its own inertial reference frame.

Both Schroedinger and Dirac's equtions do not give such a solution for a stationary electron. So they must both be inadequate for energies at, and above the rest mass energy.

s not treq I don't care if that's wrong, that's still the theory,

And the standard interpretation of the theory MUST BE HOLY DOGMA to you Sol and Ben. I feel sorry for you guys.

and YOU told us to use non-relativistic mechanics.

BTW: There is no such thing as "non-relativistic" mechanics. Newton's mechanics is fully commensurtae with Einstein's Special Theory of Relativity at low speeds.

3) We don't have to derive the solution, we can simply look it up in any undergrad quantum mechanics textbook. Hell, Sol even did that part for us. Our task is merely to calculate the kinetic energy of that solution.

I know that this is what you are doing since none of you can do the mathematics. The fact is still that if you want to claim kinetic energy for an electron, you must define it relative to the energy at which it does not have kinetic energy and above which it starts to have kinetic energy. This energy level is the rest mass energy of the electron, which is also valid for Newton' mechanics.

I don't care what you want to insert into the Schrodinger equation (though I note that you haven't actually done what you say needs doing), we already HAVE the solution. We don't need to insert the total energy in anywhere in order to calculate the kinetic energy.

Oh yes you do or else you do not have an energy reference above which kinetic energy is possible. You think that you can take any energy as a reference energy. There is only ONE reference energy for an electron and it is its rest mass energy. Only above this energy can an electron have kinetic energy. Below this energy the electron acts as a glue so that its mass energy is lower than it rest mass energy: And therefore it CANNOT have kinetic energy. Or do you want to argue that quntum mechanical bonding does not involve a change in mass energy? Have you ever heard of Hiroshima and Nagasaki?

I don't care what the potential is. If I have the wave function (and Sol already gave it), I don't need to know anything about the potential in order to calculate the kinetic energy. Again, this is basic quantum mechanics, and you can't do it.

No it is not: You cannot calculate a wave-function without specifying its potential energy. And you cannot calculate a kinetic energy if the highest energy of the electrons within a metal (or a box) is at an energy W (the workfunction) below the electron's rest mass energy; unless you believe in Voodoo, paranormal metaphysics. Which seems to be your, Ben's and Sol's specialty.

In other words, you object to standard quantum mechanics. You think the theory is wrong from top to bottom.

Nope, I have NEVER claimed this. But I do claim that the Copenhagen interpretation of quantum mechanics is wrong. Many other sdientists have been and still are claiming that the copenhagen interpretation must be wrong. They are not being accused of "objecting to standard quantum mechanics".

All I am pointing out is that if you reject the notion that the intensity of an electron wave is a probability ditribution, but rather assume that it is the electron's mass and gravitational energy, everything falls into place; since the theory then becomes totally causal; and starts to dovetail with Einstein's model for gravity. But of course our modern day theoretical physicists do not want physics to be causal: It must be Voodoo: They rather advocate ideas that: 1. Our universe must split up into zillions of universes every instant in time; 2. What we do now must affect the past; 3. Diffraction only occurs when neither you or the electron knows through which slit it has moved. etc. etc. etc. For God's sake man can you not see that these conclusions are nothing else paranormal metaphysics?

I don't care what you think the total energy is, that's not what I want to find. I ONLY want to find the kinetic energy.

Keep on searching since it is NOT there where you are looking for it, and cannot be there for as long as the electron has less energy than its rest mass energy.

And for that, I can ignore the potential as long as you give me the actual wave function.

So you are throwing energy conservation out of the window; and then telling me that I am the one who is doing this?

All I need to do to calculate the kinetic energy is to calculate
integral[psi^*(p²/2m)psi dx]
At no point do I need to insert ANYTHING about the potential or the total energy, all I need is the wave function itself

A stationary electron wave has NO MOMENTUM: So you are calculating an energy which cannot exist except in Alice's wonderland.

Furthermore, you object to Sol's wave function, but I notice that you can't actually provide a wave function of your own. Hell, you can't even tell me what your deltaV's are quantitatively.

I have not "objected to the wavefunction", but to the interpretation that its energy is kinetic energy. If you could have handled calculus, which you obviously cannot, you could have derived for yourself that I get exactly the same wave-function but that the energy is different since in my case the relationship with the k-number is:

k^2=[(2*m(e))/(hbar)²*[(mod)(delta)V-(mod)(delta)E]

And this solution is only valid when (mod)(delta)V>(mod)(delta)E. Different metals have different (delta)V's and the same wave (same qunatum number)within two different metals do not necessarily have the same energy E=m(e)*c^2-(mod)(delta)E

already know that the Schrodinger equation violates special relativity.

Like Newton's laws "violate" Special Relativity? They do not violate anything at low speeds. neither shoudl Sxchroedinger's equation if it is relevant; which it is.

So is the equation KE = (1/2)mv², yet that still works pretty damned well, doesn't it?

At low speeds it does, but it does not violate the Special theory of Relativity. It still means that when an electron has an energy in excess of its rest mass energy, which is small since the speed is low, then you still have that KE=m*c²-m(e)*c²= (1/2)*m(e)*v²; provided that m>m(e): If however m<m(e) as in the case of all bound electrons, these electrons CANNOT have ANY kinetic energy AT ALL!!!

So too with the Schrodinger equation, WITHOUT resorting to rest mass.

You cannot ignore rest mass at all in the Schroedinger equation if you want to determine whether the energy of a wave is kinetic energy or not. If you do this you are doing Voodoo, paranormal metaphysics.

Your simultaneous appeals to the Schrodinger equation AND special relativity reveal that you are deeply ignorant of at least one (and probably both) of those theories.

Just the opposite: Your appeal that an electron with an energy less than its rest mass energy can have kinetic energy as long as it is "non-relativistic" still violates Eisntein's Special Theory of Relativity NO MATTER AT WHAT SPEED THE ELECTRON IS MOVING.

And look, we don't even have to do a particle in a box. We can do the hydrogen atom instead. That's a very physical system, if you think the particle in a box system is too artificial.

Exactly: And also in this case the bound electron energies are all below the rest mass energy and can therefore not have a kinetic-energy component. Only in the case where an electron has energy that is larger than its rest mass energy, can it have momentum and kinetic energy and can it therefore move inn the space around the nucleus. Such states are known as Rydberg states: They continuously decay owing to the acceleration that the electron experiences: Such a state thus has to be "pumped" with energy to keep it going. Orbitals below the rest mass energy do NOT have momentum, and therefore also not kinetic energy: There is thus no acceleration involved: For this reason they can remain stable for long times.

 

[sol invictus]

What's <p²> in the wavefunction I gave, Johan? And I don't want an answer like "It MUST be zero because [fill in nonsense here]", I want a calculation.

Good luck, you'll need it.