Incredibly huge inconsistency of modern physics

The absolute square of the amplitude of the wave-function, which has an intrinsic wavelength, is a PDF, right? So in the end I get one scalar quantity : the probability to
measure something there. In this case, I detect a photon. How is the wavelength of this photon, from which I can obtain the momentum, connected to the wavelength of the
wave-function, from which I obtain the probability? Or is there no connection? Is that number then randomly assigned?

Virtual particles are calculation mechanisms that in the end do not cause a net gain or loss of anything that you can eventually actually detect as a conserved quantity.
And creation/destruction of particles is something else than actual removal or addition of a conserved quantity. No physics theory ever says anything about that.
You assume there is stuff, and stuff never gets lost, though it might change form. And in the end it just ends up being photons.

Conservation of probability is a natural consequence of keeping conserved quantities, well, conserved; Even if the details are nasty, as many papers will show you.

I'm quite sure that experimentally you cannot change the actual conserved quantities like energy. You'd be creating energy from nothing, then. That's perpetual motion.
I'm quite sure therefore, even if I have to repeat myself, that the laws of QM do not allow creation of energy from nothing.
It's not an argument from ignorance to be suspicious of someone who claims that he can create energy from nothing.
 
TheAdversary said:
The absolute square of the amplitude of the wave-function, which has an intrinsic wavelength, is a PDF, right?
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No. 20 November 2015: Some science you should learn, TheAdversary.

The wave function is not a wave and has no wavelength :eye-poppi!
A wave function in quantum mechanics describes the quantum state of an isolated system of one or more particles. There is one wave function containing all the information about the entire system, not a separate wave function for each particle in the system. Its interpretation is that of a probability amplitude. Quantities associated with measurements, such as the average momentum of a particle, can be derived from the wave function. It is a central entity in quantum mechanics and is important in all modern theories, like quantum field theory incorporating quantum mechanics, while its interpretation may differ. The most common symbols for a wave function are the Greek letters ψ or Ψ (lower-case and capital psi).
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When you plug a wave function into a wave equation such as the Schrödinger equation, you get wave solutions, e.g. the solution for a free particle is a wave packet. The Schrödinger solution for a hydrogen atom contains spherical harmonics (basically 3D wave equations).

Some incoherent stuff but you get virtual particles right
In physics, a virtual particle is an explanatory conceptual entity that is found in mathematical calculations about quantum field theory. It refers to mathematical terms that have some appearance of representing particles inside a subatomic process such as a collision. Virtual particles, however, do not appear directly amongst the observable and detectable input and output quantities of those calculations, which refer only to actual, as distinct from virtual, particles.
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Virtual particles exist because we can see their effects on measurements but they are virtual so we will never detect them directly.

That some things (charge, energy, etc.) are conserved does not mean that a phrase like "conservation of probability " makes sense without a coherent explanation.
 
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You see, the problem I'm having is this : You say Maxwell's equations give the photon its wavelength. But the photon doesn't even exist in the Maxwell equations.
Do I have to temporarily introduce it somewhere then? And how does that mix?
In layman's terms; Just conceptual.
 
TheAdversary said:
You see, the problem I'm having is this : You say Maxwell's equations give the photon its wavelength.
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I said that Maxwell's equations give an electromagnetic wave that has a wave length. I assumed (wrongly it seems) that you knew that photons are electromagnetic waves as well as particles.
 
You say the solution for a free particle is a wave packet. But a wave packet is just the de Broglie hypothesis that every particle can be seen as a wave packet
where the momentum is given by Planck's constant divided by the wavelength ... I assume the wavelength that's associated with the concept of wave packet.
Where could it otherwise come from? Correct me if I'm wrong, but I'm not interested in the time evolution or anything complicated. Just a single wave packet.
How do you calculate the momentum of a wave packet?
 
TheAdversary said:
Just no is not an argument. Some logic you should learn.
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Just not learning about the science that you are talking about and have been supplied will not educate you. Some logic you should learn, TheAdversary.
There is no such thing as a de Broglie wave or a de Broglie hypothesis wave packet:
  1. There is no such thing as a " de Broglie wave". There is no actual wave just the assignment of a wavelength to the momentum of particles. Particles get "wave-like" properties. Particles do not become waves.
  2. There is no such thing as a " de Broglie wave packet". There is no actual wave packet just the assignment of a wavelength to the momentum of particles. Particles get "wave-like" properties (not "wave packet-like" properties). Particles do not become wave packets.
  3. This is the de Broglie equation and has nothing to do with probability.
  4. Where "waves" appear in QM is the wave function which is not the de Broglie equation.
  5. There are many interpretations of QM.
    Some of these interpretations are deterministic, i.e. no probability distribution functions.
  6. One interpretation of QM is the Copenhagen interpretation where the wave function
  7. Maxwell's equations describe electromagnetic waves with wave lengths. A photon is the quantum of electromagnetic waves. Thus photons have wavelengths.
    The de Broglie hypothesis is that matter also has wave-like properties, e.g. you can assign a wavelength to massive particles.
 
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Look, Mr. Great Educator, I'm just asking a simple question; Where does the wavelength of the photon come from, at its core, without over-complicating the matter.
I want to calculate the momentum of a photon. This requires the wavelength. Where the F does that come from? It can't come from the Maxwell equations.
 
Is it safe to assume, according to you, that the concept of the wavelength of a photon ultimately derives from the de Broglie hypothesis and not the Maxwell equations?
 
'Particles do not become waves.'
PDF's become measurements. The notion of particle already implies a philosophical commitment. This is why your story is so confusing.
 
Oh, and local hidden variable theories are nonsense, and non-local hidden variable theories always end up with a preferred frame of reference. I know modern day
physicists don't care about this but I think a preferred frame of reference in a physics theory is absurd. Like creating energy from nothing.
 
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I asked where the wavelength comes from! I know how to calculate the momentum from the wavelenght.
WAVELENGTH!!!
 
I always thought that the classical and QM notion of wave couldn't be mixed; That they were fundamentally different things.
 
TheAdversary: Please show that non-local hidden variable theories always up with a...

TheAdversary said:
Oh, ...
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Local hidden variable theories are scientific theories. They are very probably wrong since the Bell inequality experiments are getting closer and closer to "loophole-less" results.
20 November 2015 TheAdversary: Please show that non-local hidden variable theories always up with a preferred frame of reference

FYI: Classical QM (Schrödinger equation) is the quantum extension of classical Newtonian mechanics. Quantum field theory is the quantum extension of special relativity.
 
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