Is psi-function really a cloud of probability?

[altsci2]

No, it is not! Actually it is something to do with the scalar wave equation. Let us assume that a solution of the scalar wave equation ( Ω ) is a steady state with amplitude ψ and the oscillating factor exp(-tω), Ω=ψ(r)exp(-itω(r)). Usually the frequency ω assumed to be a constant. Now, suppose it depends of coordinates: ω(r). The kinetic energy, velocity, and De-Broglie frequency (f=cmv/h) of the electron that is orbiting the atom will depend of radius. We can express De-Broglie's frequency of the electron through its kinetic energy. Now we can request that the local frequency of the Ω be equal to De-Broglie's frequency of the electron at the same location: ω(r)=f. Substituting Ω to the wave equation we will find the equation for ψ which will be exactly the Schrodinger Equation.
That means that ψ-function of Quantum Theory is actually the amplitude of a standing wave of some scalar wave equation!

 

[bhobba]

That means that ψ-function of Quantum Theory is actually the amplitude of a standing wave of some scalar wave equation!

Of course it is - and as you note it's called Schroedinger's Equation - although standing waves is not what I would necessarily call the solutions.

And if you want to understand its true basis see Chapter 3 - Ballentine - QM - A Modern Development:
https://www.amazon.com/dp/9810241054/?tag=pfamazon01-20

Believe it or not its a requirement of symmetry - that the QM probabilities are frame independent.

Thanks
Bill

 

[Nugatory]

That means that ψ-function of Quantum Theory is actually the amplitude of a superposition of standing waves of [STRIKE]some[/STRIKE] a particular scalar wave equation!

The equation in question being Schrodinger's equation.

However, it is correct to interpret $\psi^*\psi$ as a probability density, which I think is what you mean by "cloud of probability".

 

[altsci2]

Clarification: The wave equation is not a Schrodinger Equation. We can derive the Schrodinger Equation from the wave equation by requiring that the solution of the wave equation has a form of standing wave with amplitude ψ (independent of time) and phase with time dependence. The Schrodinger Equation for ψ follows from the Wave Equation for Ω . Now the Schrodinger Equation does not need to be postulated -- it can be derived

 

[bhobba]

Now the Schrodinger Equation does not need to be postulated -- it can be derived

Without commenting of what you said before (to be blunt, as far as I can see, its basically nonsense - you will need to provide a lot more detail), its derived from the Principle Of Relativity in the reference I gave before - its not postulated.

It has been known since Wigner its true basis is symmetry:
https://noppa.aalto.fi/noppa/kurssi/rak-54.1300/materiaali/Rak-54_1300_hyva_lukea_-_symmetry_in_physics_--_wigner_e._1.pdf

Thanks
Bill

 

[ShayanJ]

The requirement that the time evolution of the state be unitary, gives a much better derivation of Schrodinger's equation. At least I like it more than other derivations, including the one suggested here. Other derivations are just too far from Quantum Mechanics!
Of course by other derivations, I mean the ones using things from outside QM. Of course if QM could be derived from CM, then CM was right in the microscopic world and we wouldn't need QM in the first place!
So to me, it seems nonsense to struggle with classical equations and concepts to get Schrodinger's equation!
Schrodinger equation can't be derived without extra assumptions on old things, and of course without a fundamental change in our viewpoint!

 

[bhobba]

The requirement that the time evolution of the state be unitary, gives a much better derivation of Schrodinger's equation.

That is required by Wigners Theorem.

It is a major ingredient, but something else like the frame independence of probabilities is also required.

Insofar as I can make sense out of what the OP wrote its intermixing all sorts of concepts such as de Broglies which have their own issues eg if you go to a frame where the particle is at rest what is its phase velocity.

Thanks
Bill

 

[ShayanJ]

That is required by Wigners Theorem.

It is a major ingredient, but something else like the frame independence of probabilities is also required.

Insofar as I can make sense out of what the OP wrote its intermixing all sorts of concepts such as de Broglies which have their own issues eg if you go to a frame where the particle is at rest what is its phase velocity.

Thanks
Bill

In page 12 of this, you can find a derivation assuming only unitary evolution!

My problem with OP's argument is that, like many others, it seems to be insisting on the coherence between quantum mechanics and classical mechanics.

 

[bhobba]

In page 12 of this, you can find a derivation assuming only unitary evolution!

See equation 2.34 on the form of the Hamilitonian. That is an assumption.

Ballentine makes no such assumption and derives that form from frame invariance.

My problem with OP's argument is that, like many others, it seems to be insisting on the coherence between quantum mechanics and classical mechanics.

My issue is it's very vague what he is doing, and what he does spell out is full of holes and semi explicit assumptions.

Exactly how you can derive it from Classical analogies has been done to death eg:
http://arxiv.org/pdf/1204.0653.pdf

Thanks
Bill

 

[mal4mac]

No, it is not!

Yes, it is!

According to an article published in Nature:

http://www.scientificamerican.com/article/quantum-theorys-wavefunction/

 

[bhobba]

Yes, it is!

That's the PBR result, and Scientific American got it wrong - that's not what it shows at all:
https://www.physicsforums.com/showthread.php?t=767669

Thanks
Bill

 

[ShayanJ]

See equation 2.34 on the form of the Hamilitonian. That is an assumption.

Yeah, that's an assumption but there are 2 points:

1- That doesn't seem to be related to frame invariance of probabilities.
2- That's not used in deriving the equation $i \hbar \frac{\partial}{\partial t}|\Psi,t\rangle=H(t)|\Psi,t\rangle$ and because different systems can have different Hamiltonians than 2.34, we should accept that when we talk about Schrodinger's equation, we're not talking about the form of the Hamiltonian and so 2.34 isn't used in the proof.

 

[bhobba]

That doesn't seem to be related to frame invariance of probabilities.

The interesting thing is it is. The proof's not easy, but the detail can be found in Chapter 3 of Ballentine. Its one of the most beautiful bits of math I have seen - but does require attention - I spent a whole day going through it step by step.

That's not used in deriving the equation $i \hbar \frac{\partial}{\partial t}|\Psi,t\rangle=H(t)|\Psi,t\rangle$ and because different systems can have different Hamiltonians than 2.34, we should accept that when we talk about Schrodinger's equation, we're not talking about the form of the Hamiltonian and so 2.34 isn't used in the proof.

Its perfectly fine. Its just that form of the Hamiltonian can be reduced to a simpler more general axiom - the POR.

Pedantically its important because if you start with the Principle Of Least Action (PLA) in classical mechanics you can show similar things via the POR - the detail can be found in Landau - Mechanics.

This means the real basis for dynamics is the symmetries of the POR - you get classical or quantum depending on what you apply it to.

Thanks
Bill

 

[altsci2]

Somebody said:

"(to be blunt, as far as I can see, its basically nonsense - you will need to provide a lot more detail),

I answered to that yesterday by referring to my article (attached). But the post was lost, so I repeat it here

 

[Nugatory]

Somebody said:

"(to be blunt, as far as I can see, its basically nonsense - you will need to provide a lot more detail),

I answered to that yesterday by referring to my article (attached). But the post was lost, so I repeat it here

This would be a good time for you to take another look at the physics forum rules about personal theories and posting non-peer- reviewed papers.

 

[berkeman]

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