Superconductivity at room temperature??

[sol invictus]

And for the free particle, here's another animation

http://demonstrations.wolfram.com/TimeEvolutionOfAQuantumFreeParticleIn2D/

 

[johanfprins]

I guess we need to back up even more, then.

To patronize you even more:

(-3) Do you agree that the QM wavefunction ψ is a complex function of the real variables x,y,z,t, where (x,y,z) are Cartesian coordinates on space, and t is time - so ψ=ψ(x,y,z,t)?

Yes.

(-2) Do you agree that the QM wavefunction ψ has a real part and an imaginary part, ψ=Re(ψ)+iIm(ψ)?

Yes.

(-1) Do you agree that given (-2) and (-1), Re(ψ) and Im(ψ) are themselves functions of x,y,z, and t?

Yes.

(0) Do you agree that Re(ψ)=x and Im(ψ)=y solves the Schrodinger equation for a free particle?

No! Only when x=y=0: The latter is not a solution at all. Furthermore, Schroedinger's equation cannot be used to solve for a "free particle" by merely setting the potential energy equal to zero. A "free particle's" energy is its rest mass within its inertial reference frame. Schroedinger's equation uses this mass as input and it can therefore not give it as the answer as it must. Thus, it cannot be solved to obtain the energy of a "free particle" at all until it is reformulated not to have the "particle's" rest mass as an input.

 

[johanfprins]

What happened in your teacup? Did the water sit still and flat?

(Remember, you are moving the cup, not spinning it around its axis)

No the tea did move but it did not move as spokes around the centre-point of the cup. It is obviously another of your nonsensical arguments. We all know that you will obtain ripples at the interface when a liquid moves along a surface, since there is a viscous interaction. This has NOTHING to do with the argument of whether one can have a coherent wave of wave fronts moving around a circle. So stop being childish and wasting everbody's time on this forum!

 

[sol invictus]

T
No! Only when x=y=0: The latter is not a solution at all.

The Schrodinger equation is

[latex]$-\nabla^2 \psi = i \hbar \partial_t \psi$[/latex], where I've set some constants to 1 for simplicity and [latex]$\nabla^2 = \partial_x^2 + \partial_y^2 + \partial_z^2$[/latex]

Since [latex]$\nabla^2 x = \nabla^2 y = 0, \nabla^2 \psi = \nabla^2 (x+iy)=\partial_t(x+iy) = 0$[/latex] and the free Schrodinger equation is satisfied for all x and y and t.

You are wrong yet again, ψ=x+iy does in fact solve the free particle Schrodinger equation.

Would you like another chance at answering questions 1-4 now?

 

[W.D.Clinger]

No I do not. There is no gradient of S in the Argand plane: There is an Argand plane at each point within the intensity of a wave, and within each separate Argand plane, S is a constant, and a constant does not have a gradient. You can only have a gradient of S in real space...

Egad.

johanfprins has just proved that every complex-valued function is constant.

That's going to revolutionize mathematics as well as physics.

Really, and when you do this you see crests and troughs along the circumference (like wheel spokes turning around the centre of the cup) or do you see crest and troughs along the radial direction (as rings around the centre-point of your cup)? If you are able to do the first and create spokes, there awaits a lucrative career for you as an illusionist.

Recipe for omelets, serves four:

1. Break a dozen eggs into a large mixing bowl.
2. Break the yolks with a fork.
3. Stir rapidly until the yolks blend with the whites, taking care not to slosh eggs over the side of the bowl.
4. ...

When johanfprins sloshes eggs over the side of the bowl, he sees the same degree of mess outside every point on the rim of the bowl.

 

[johanfprins]

Any solution with <p>=/=0

can you tell me what this expression means? Is <p> zero or is it not zero?

Of course in that case <p> is a function of time, since there is a potential. But it clearly shows peaks and troughs moving, unless you choose special initial conditions.

<p> is only a function of time when the potential energy is a function of time. And even in this case you do not have crests and troughs driving the wave along a direction that is perpendicular to parallel wave fronts moving freely along in space. But let us not discuss this now since it will only serve to obscure the original topic (this is probably your intent).

Lets us first stick to the case where the potential energy is not changing with time as we have talked about right from the start.

 

[ben m]

No the tea did move but it did not move as spokes around the centre-point of the cup.

Was there a high point (a crest, you might call it) that moved around and around? Was the high point directly across the cup from a low point (or trough?) Did the trough also move?

So there was a "spoke". Just one spoke, not many like a wagon wheel, but that's the sort of limitation you encounter when you've abandoned the fancy-schmancy idea that the wave equation describes waves.

 

[johanfprins]

The Schrodinger equation is

[latex]$-\nabla^2 \psi = i \hbar \partial_t \psi$[/latex], where I've set some constants to 1 for simplicity and [latex]$\nabla^2 = \partial_x^2 + \partial_y^2 + \partial_z^2$[/latex]

Since [latex]$\nabla^2 x = \nabla^2 y = 0, \nabla^2 \psi = \nabla^2 (x+iy)=\partial_t(x+iy) = 0$[/latex] and the free Schrodinger equation is satisfied for all x and y and t.

You are wrong yet again, ψ=x+iy does in fact solve the free particle Schrodinger equation.

Would you like another chance at answering questions 1-4 now?

Now I am convinced that you have a screw loose: You differentiate with time while ignoring the fact that a harmonic wave has a frequency (omega). Thus the wave which you have to substitute when you use the time derivative is PSI=(x+iy)*exp(-i(omega)). When you now take the partial time derivative you obtain -i(omega)*PSI, and this leads you to the time independent Schroedinger equation which is:

(del)^2*(x+iy) + k^2*(x+iy)=0.

Now do your partial space differentiation and the first term becomes zero, so that the second term gives x+iy=0. Thus, the only solution possible is when x=y=0: As I have already posted. Are you well? Or are you writng from the LOOney Bin: Another one that flew over the Cuckoo's nest!

 

[johanfprins]

Was there a high point (a crest, you might call it) that moved around and around? Was the high point directly across the cup from a low point (or trough?) Did the trough also move?

I did not see this at all. Maybe I am not doing it right. But in either case it is irrelevant to our discussion.

So there was a "spoke". Just one spoke, not many like a wagon wheel, but that's the sort of limitation you encounter when you've abandoned the fancy-schmancy idea that the wave equation describes waves.

I have not seen a spoke going round and round: An even when this is possible the motion is not a coherent, harmonic wave at all. Or do you argue that a superconducting wave does NOT move coherently?

 

[Ziggurat]

I suppose you are talking about Rydberg atoms. They are not solutions of Schroedinger's equation since Schroedinger's equation cannot model running waves.

No, I'm not talking about Rydberg atoms. I'm specifically talking about the explicit solutions to the Schrodinger equation, solutions which do exactly what you claim cannot be done. Crack open just about any intro quantum mechanics textbook, look up the solutions to the hydrogen atom, find the non-zero m solutions, and you'll find "running wave" solutions. Check them if you don't believe me: they solve the Schrodinger equation.

You either disagree with standard quantum mechanics, or you don't know what standard quantum mechanics IS.

There is no standard solution of the Schroedinger equation that gives you a running wave circling the nuicleus.

Yes there is. The n=2, l=1, m=1 solution is a wave circling the nucleus.

Are you now talking of stationary Schroedinger waves or Bohr's obsolete and physically impossible model?

You can call it stationary if you like, but it's got angular momentum around the nucleus. It's circling the nucleus.

There is not much use to argue with a person who believes that Schroedinger's equation can give a solution that is a coherent wave moving along a ring; or moving along any direction for that matter.

Like I said: you disagree with standard quantum mechanics.

These waves are not coherent since the ring now acts as a waveguide.

That doesn't make the waves incoherent. What on earth made you think it did?

Schroedinger's equation cannot be solved to obtain any coherently moving wave fronts in ANY direction: Not even a linear direction.

How can you be so ignorant of standard undergraduate quantum mechanics?

 

[sol invictus]

can you tell me what this expression means? Is <p> zero or is it not zero?

<p> not equal to zero.

<p> is only a function of time when the potential energy is a function of time.

Wrong, as always. <p> is conserved if the potential is independent of space, not time. If the potential is constant in space but depends on time, <p> is constant but the energy is not.

I notice you ignored (as usual) the examples I posted which falsify your previous (absurd) claim (that the Schrodinger equation does not have solutions with moving waves).

 

[sol invictus]

Now I am convinced that you have a screw loose: You differentiate with time while ignoring the fact that a harmonic wave has a frequency (omega). Thus the wave which you have to substitute when you use the time derivative is PSI=(x+iy)*exp(-i(omega)).

Johan, I asked you if the wavefunction PSI=(x+iy) satisfies the free Schrdinger equation. You said no, so I proved you wrong.

Your response: "Well sure, that wavefunction satisfies the free Schrodinger equation, but here's another one of my invention that doesn't."

Congratulations, you are wrong on absolutely every point. Go play with your grandkids, Johan.

 

[Ziggurat]

Now do your partial space differentiation and the first term becomes zero, so that the second term gives x+iy=0. Thus, the only solution possible is when x=y=0: As I have already posted. Are you well? Or are you writng from the LOOney Bin: Another one that flew over the Cuckoo's nest!

Not so. The second term is k²(x+iy)=0. Which simply means k = 0. That also means that omega = 0, and this solution has no time dependence. So Sol is right, and you are wrong.

 

[johanfprins]

No, I'm not talking about Rydberg atoms. I'm specifically talking about the explicit solutions to the Schrodinger equation, solutions which do exactly what you claim cannot be done. Crack open just about any intro quantum mechanics textbook, look up the solutions to the hydrogen atom, find the non-zero m solutions, and you'll find "running wave" solutions. Check them if you don't believe me: they solve the Schrodinger equation.

OK if you want to interpret it in that manner: But you are wrong. If it is a running wave, then its intensity will change with time. Differentiate the intensity of any Schroedinger wave around a nucleus an you will find it does NOT change at all. The intensity is time-independent for any value of the quantum numbers. A travelling matter wave must have a change in positioon of its centre-of-mass. You do not have this for the stationary waves around the nucleus.

You either disagree with standard quantum mechanics, or you don't know what standard quantum mechanics IS.

I agree with Schrodinger's equation as being a good approximation of a more general wave equation which do not require the rest mass of the electron as an input. I do not agree that any of the standing waves around a nucleus has momentum since for each such a wave one has that <p>=0. If you want to interpret this as "not agreeing with standard qunatum mechanics" then so be it. Yes I do not agree with the Copenhagen interpretation if you want to call this "standard" qunatum mechanics. But so does Bohm, de Broglie, Einstein etc. I have asked you define what you mean by "standard quantum mechanics" but you did not respond. This proves that you do not know what you yourself mean by "standard" quantum mechanics. In either case there is not a single theory or model in physics that should be acalled "standard": Physics does not generate dogmas but theories and models that are always subject to revision. Clearly you do not know very much about the rules of physics.

Yes there is. The n=2, l=1, m=1 solution is a wave circling the nucleus.

No it is not since even in this case you have <p>=0. So how can a wave circle the nucleus when its momentum is zero?

You can call it stationary if you like, but it's got angular momentum around the nucleus. It's circling the nucleus.

Prove this by experiment please.

Like I said: you disagree with standard quantum mechanics.

Like I said your statement is empty since you cannot even define what you mean by "standard" quantum mechanics.

That doesn't make the waves incoherent. What on earth made you think it did?

A coherent wave has a single frequency and its moving wave fronts are parrallel. This is NOT possible around a circle. Or define to me what you understand as a coherent wave.

How can you be so ignorant of standard undergraduate quantum mechanics?

You are the one who is ignorant since you want to argue about issues which you are too stupid to define.

 

[johanfprins]

<p> not equal to zero.

Wrong, as always. <p> is conserved if the potential is independent of space, not time. If the potential is constant in space but depends on time, <p> is constant but the energy is not.

For once you are making a good point; but you are still missing the effect of relativity. Please solve the Schroedinger equation to get a wave-intensity that is moving past you. If you do not have a wave-intensity moving past you you do not have momentum. The Schroedinger equation does not give such a solution; even when you solve for the so-called free electron by putting V=0: In this case the intensity is the same everywhere and it fills infinite space, so the wave cannot move. In fact, the Schroedinger equation only give solutions for electron waves which are each stationary within its own inertial refrence frame, and within this reference frame one MUST have that <p> is always zero.

The only possibility to see <p> as non-zero is to view the stationary wave from another inertial reference frame moving relative to the reference frame in which the electron wave is stationary: In other words you must use the Lorentz transformation to a see such a wave moving with a momentum <p> that is not zero. And it is only under these conditions that the de Broglie wavelength come into play: Not when the wave is stationary. Obviously a stationary electron-wave cannot have a de Broglie wavelength, since such a wavelength will have to be inifinite.

I notice you ignored (as usual) the examples I posted which falsify your previous (absurd) claim (that the Schrodinger equation does not have solutions with moving waves).

Which examples? I have not seen any such example.

 

[johanfprins]

Johan, I asked you if the wavefunction PSI=(x+iy) satisfies the free Schrdinger equation. You said no, so I proved you wrong.

You did not since a wave-function MUST always have a frequency when it is a solution of the time-dependent Schroedinger equation. You stupidly used the time-dependent Schroedinger equation and ignored the time dependence that every Schroedinger wave MUST have. Are you really so stupid or are you trying to play games?

Your response: "Well sure, that wavefunction satisfies the free Schrodinger equation, but here's another one of my invention that doesn't."

That was not my response: You claimed a time-independent wave function which I inserted into the time-independent Schroedinger equation and proved that it cannot be a wave function at all.

Congratulations, you are wrong on absolutely every point. Go play with your grandkids, Johan.

I do play with my grandchildren. May I advise that you enrol at a school for the mentally retarded? You are in dire need of getting help.

 

[johanfprins]

Not so. The second term is k²(x+iy)=0. Which simply means k = 0. That also means that omega = 0, and this solution has no time dependence. So Sol is right, and you are wrong.

It is clear that you claim that you know "standard" quantum mechanics while not even knowing that k cannot be zero for any wave ever. If it is zero for a free wave, the energy of the wave is zero. Great solution for a free electron: It has no energy!! Ha! Ha! How stupid can a person really be? And you call yourself a physicist?

 

[sol invictus]

For once you are making a good point; but you are still missing the effect of relativity. Please solve the Schroedinger equation to get a wave-intensity that is moving past you.

I linked to animations of solutions of the SE that show that clearly. You ignored them. Generally, nearly any solution that is the sum of at least two energy eigenstates has that property.

In fact, the Schroedinger equation only give solutions for electron waves which are each stationary within its own inertial refrence frame, and within this reference frame one MUST have that <p> is always zero.

Complete and utter nonsense. See either of the animations, open any QM text, or simply superpose some free particle stationary solutions. Here's an explicit analytic example in 1D:

[latex]$\psi(x,t) = \int dk e^{i(kx-\omega t) -k^2 + i p_0 x}$[/latex], from which one obtains

[latex]$|\psi(x,t)|^2=\pi \frac{e^{-\frac{(x-2 p_0 t)^2}{2 \left(1+t^2\right)}} }{\sqrt{1+t^2}}$[/latex].

That is a moving wave packet with momentum p₀, and it solves the free-particle Schrodinger equation.

 

[sol invictus]

You did not since a wave-function MUST always have a frequency when it is a solution of the time-dependent Schroedinger equation. You stupidly used the time-dependent Schroedinger equation and ignored the time dependence that every Schroedinger wave MUST have. Are you really so stupid or are you trying to play games?

No Johan, I am not stupid. The wavefunction I posted does not have a time-dependence, and yet as I proved mathematically, it is nevertheless a solution to the time-dependent free-particle SE. The reason for that is obvious to everyone else reading this thread. I picked that solution because it is the simplest counterexample to one of your myriad wrong assertions.

As always, you cannot understand simple facts.

 

[Ziggurat]

OK if you want to interpret it in that manner: But you are wrong. If it is a running wave, then its intensity will change with time.

Your objection has no basis. The average momentum of a stationary state is indeed zero, but the angular momentum is not. And angular momentum is rather more relevant here. Your objection is equivalent to stating that a rotating object is motionless since its net linear momentum is zero. But of course, it's not motionless.

Differentiate the intensity of any Schroedinger wave around a nucleus an you will find it does NOT change at all.

Except that's not even true. The linear combination Psi₁₀₀+Psi₂₁₁ (where the indices are nlm) is a solution to the Schrodinger equation with an intensity which DOES vary with azimuthal angle. In fact, it does more than that: the net LINEAR momentum of this state is also nonzero, and the center of mass moves in a circle about the origin. So even if we restrict ourselves to YOUR definition of motion (which contradicts standard quantum mechanics), we still find solutions which do exactly what you say cannot be done.

I agree with Schrodinger's equation as being a good approximation of a more general wave equation which do not require the rest mass of the electron as an input. I do not agree that any of the standing waves around a nucleus has momentum since for each such a wave one has that <p>=0.

Net linear momentum isn't relevant. Angular momentum is. And they very much have nonzero angular momentum. Plus, as above, there's no reason we need to restrict ourselves to stationary state solutions.

I have asked you define what you mean by "standard quantum mechanics" but you did not respond.

What you find in basically any modern introductory quantum mechanics textbook. For example, Griffiths' "Introduction to Quantum Mechanics".

This proves that you do not know what you yourself mean by "standard" quantum mechanics.

No it doesn't. I simply don't recall you ever asking this before. Quite possibly I simply missed it, but there's no proof of anything here. Given your own repeated failures to justify such elementary aspects of your own theory as energy conservation, you're hardly in a position to criticize me about having overlooked one rather unimportant question whose answer should have been obvious from the start.

No it is not since even in this case you have <p>=0. So how can a wave circle the nucleus when its momentum is zero?

How can a classical object spin when p = 0? Quite easily: L (angular momentum) does NOT equal zero.

How can a wave function with nonzero L NOT circle the nucleus?

A coherent wave has a single frequency and its moving wave fronts are parrallel. This is NOT possible around a circle.

Sure it is. Like I said, the standard classical vibrational modes for a solid ring include exactly those sorts of waves.

It is clear that you claim that you know "standard" quantum mechanics while not even knowing that k cannot be zero for any wave ever.

I don't care if you refuse to call it a wave, it's still a solution to the Schrodinger equation. What do you think k is, anyways?

If it is zero for a free wave, the energy of the wave is zero.

That is correct. So what?

Great solution for a free electron: It has no energy!! Ha! Ha! How stupid can a person really be? And you call yourself a physicist?

I never called myself a physicist.

And Sol never claimed that electrons existed in this particular state. He claimed that it's a mathematical solution to the Schrodinger equation, which has mathematical properties which you claimed were not mathematically possible. On that point he is right, and you are wrong.

In terms of physical manifestation, there's a rather more obvious problem here than zero energy, a problem which should have jumped out at you the moment the solution was written down. The fact that it wasn't obvious is rather telling about your own lack of experience actually doing quantum mechanics.