Incredibly huge inconsistency of modern physics

[Reality Check]

I always thought that the classical and QM notion of wave couldn't be mixed;.

They are the same thing and so can be "mixed".
A wave in classical physics is the same as a wave in QM. For example the Schrödinger solution for a hydrogen atom uses the mathematics for waves worked out decades before QM existed.
But I suspect you are persisting with ignorance about wave equations. Maybe

The wave function is not a wave and has no wavelength !

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).

needs to be added to: 20 November 2015: Some science you should learn, TheAdversary?

 

[Perpetual Student]

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?

The existence of electromagnetic waves (and the notion of wavelength) is a concept that precedes QM, and was predicted by Maxwell and later demonstrated experimentally by Hertz. When the concept of light Quanta (photons) was postulated by Einstein and demonstrated, the notion of the wavelength of a photon was an obvious consequence. This is how QM began. The energy and momentum associated with the photon's wavelength was also deduced by Einstein.
This happened many years before de Broglie, who first postulated the concept of wavelength for massive particles.
Today, the theoretical bases for all the characteristics of the photon are derived through quantum field theory, or more specifically, quantum electrodynamics.

 

[TheAdversary]

Ok then tell me this. Since all information about the system needs to be contained by the wave-function, how is the information for the wavelength stored therein.
And how can I retrieve it.
And stop your smug educational attitude. I might be ignorant in some details, but you already ran into TWO reductio ad absurdums. Let's be fair here!
A Quantum Mechanics expert that fails to realize the importance of conservation of probability, or conservation laws in general, and even calls them fantasies!
And I'm quite sure you can't answer this question and you're going to find another way to beat around the bush and making it look like I'm stupid.
How is the information for the wavelength of a photon stored in the quantum model and how can I retrieve it? Surely I don't need the Maxwell Equations for that?
All the information about the system must be contained within the wave-function, right? Including the wavelength for a photon ...

 

[TheAdversary]

I know how the answer is going to look : You're so stupid. You need to learn Science. Here let me add this to the list.
And you fail to answer the question I'm actually asking. Yup, great educational skills there ... you're obfuscating, not explaining, mainly.

 

[TheAdversary]

This is the problem with modern physicists : You can't reason with them, because you're ignorant, and then they present to you a theory that violates basic
thermodynamics, after layers and layers of obfuscation.
QM starts to look like the Vietnam of physics; You don't know; You were not there.
But I'm quite sure you can't create energy from nothing, which is what would happen if conservation of probability was a 'fantasy'.

 

[TheAdversary]

I'll repeat this from Wikipedia :

https://en.wikipedia.org/wiki/Free_particle

Schrodinger equation. Solution for a free particle is a complex plane wave having an amplitude and a frequency (simplified).
Probability density function is absolute square of wave-function (amplitude), momentum can be calculated from the frequency of the wave-function using de Broglie.
After all, same wave function, as I already thought.

 

[TheAdversary]

So, according to the information on Wikipedia, this statement was already entirely correct :

'I'm talking about a single photon. Energy and momentum are related to frequency/wavelength via the de Broglie hypothesis. The absolute square of the
amplitude forms a probability distribution function which when integrated over an interval gives you the probability of measuring the photon in that interval.
So if the probability is higher, you'll detect more photons, all of which have an energy and momentum determined by the frequency/wavelenght.'

Informal? Maybe. Non-science? I don't think so.

This is the reason I kept bugging you about the wavelength.

 

[TheAdversary]

You might say, but it's non-relativistic and you came from a relativistic perspective, but that would just be pedantry.

 

[TheAdversary]

'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.'

This turned out to be false. And you intermingle with QM interpretation which is implicitly suspicious.

'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.

Wave packet is a more complicated construct, which is why I asked for a single photon. That's a complex plane wave, with amplitude and frequency,
allowing you to calculate to probability and momentum. I'm not interested in 'interpretations' of QM. It's PDFs and measurements to me,
not 'particles', whatever that means.

'This is the de Broglie equation and has nothing to do with probability.'

Except for the fact that it is central to the theory which relies on probability. But you can isolate it like that and be pedantic about it if you like ...

'Where "waves" appear in QM is the wave function which is not the de Broglie equation.'

I meant what is mentioned on this page : https://en.wikipedia.org/wiki/Free_particle
You can see that they are part of the same theory. Pedantry.

'There are many interpretations of QM.'

And all of them are nonsense because it is trivial to proof that you'll only end up trying to define pure chance, which is insane.

'Some of these interpretations are deterministic, i.e. no probability distribution functions.'

They will very soon all be proven false, if I'm not mistaken.

'One interpretation of QM is the Copenhagen interpretation where the wave function.'

Yep, that's the one that uses PDFs and measurements. Not 'many worlds' or similar nonsense.

'Maxwell's equations describe electromagnetic waves with wave lengths. A photon is the quantum of electromagnetic waves. Thus photons have wavelengths.'

A photon doesn't exist in the Maxwell model. Which is why the PDFs where necessary in the first place, because our treatment of 'particles'
depends on it. Sure, you can plug in the quantities, and a wave is a wave, but the model is very different. So your conclusion doesn't follow.

'The de Broglie hypothesis is that matter also has wave-like properties, e.g. you can assign a wavelength to massive particles.'

Hey, I agree with this!

 

[fagin]

Who are you talking to?

 

[TheAdversary]

To those who have eyes to see and ears to hear

 

[fagin]

Carry on then.

 

[schrodingasdawg]

You should second-quantize the electromagnetic field instead of talking about the wave function of a single photon, really.

So, in classical electrodynamics, you can decompose the wave solutions for the EM field into normal modes: have an integral over all wave numbers, and a sum over polarizations, of something like a(k,s)*e(k,s)*exp(-ik.x) + complex conjugate. A number a(k,s) for the amplitude of the mode, a polarization vector e(k,s), and a periodic factor. When you pass from classical to quantum electrodynamics, you replace the mode amplitude a(k,s) by an annihilation operator, and its complex conjugate by a creation operator. You say that [a(k,s), a^t(p,z)] = delta(s,z) * delta(k-p) (Kronecker and Dirac delta, respectively), times some other factors depending on how you define the momentum integration measure.

In classical electrodynamics, the amplitudes for the normal modes can take on any values: there are no photons. You get photons in quantum electrodynamics because the number operator N(k,s) = a^t(k,s) a(k,s) has a discrete spectrum. However, in both the classical and quantum theories, each mode has a wavelength 2*pi/k.

 

[Dancing David]

 

[Perpetual Student]

This is the problem with modern physicists : You can't reason with them, because you're ignorant, and then they present to you a theory that violates basic
thermodynamics, after layers and layers of obfuscation.
QM starts to look like the Vietnam of physics; You don't know; You were not there.
But I'm quite sure you can't create energy from nothing, which is what would happen if conservation of probability was a 'fantasy'.

Your ignorance of (or failure to understand) a theory does not make it wrong.

 

[Darwin123]

Ok then tell me this. Since all information about the system needs to be contained by the wave-function, how is the information for the wavelength stored therein.
And how can I retrieve it.
....

Experimenters can answer these questions with a small level of detail as follows.

The wavelength is 'stored' in a wave function by the spatial variation of the field strength at any one moment of time as defined in the reference frame of the observer.

The observer can retrieve (i.e., measure) the wave function several ways. Classically, the wave length can be measured by a dispersive device such as a diffraction grating or prism. A detector counts the number of particles associated with each angle of diffraction or refraction.

Therefore, the observer retrieves the wavelength by overlapping the wave with a material that has 'ruled' divisions of length. The interaction between the wave and the shaped material determines the wavelength.

Quantum mechanics provides other means to 'measure' the wavelength. The de Broglie relations show that the momentum of the particle is stored in the wavelength of the wave function. So one can retrieve the wavelength of wave by measuring the momentum of the single particle.

The momentum is measured by looking at the motion of materials that are overlapped with the wave function. The associated particle will scatter particles in the material, transferring momentum.

Again, the interaction of the quantum particles with the particles in the material is what allows one to 'measure' the wavelength.

Basically, what allows one to measure energy or momentum is an interaction between the wave being examined and a material.

Your literal questions were easy to answer. I suspect that you are asking another question in an indirect manner. I suggest that you ask a more detailed question as your literal question was easy to answer.

No doubt you will point out how stupid my answers are

 

[Perpetual Student]

...

A Quantum Mechanics expert that fails to realize the importance of conservation of probability, or conservation laws in general, and even calls them fantasies!

...

I seriously doubt that there is anyone here who fails to realize the importance of conservation laws.
Many of us also know that the wave function represents the probability amplitude for a given observable.
However, I would like to understand why you believe probability conservation is specifically neglected by commentators here.
Simply stated, probability conservation is consistent with any symmetry of the system. Consequently, symmetry transformations are implemented by operators which are known to conserve scalar products, hence probabilities. Here is a straightforward demonstration:
LINK

 

[TheAdversary]

+schrodingasdawg Your explanation is beyond my comprehension at this time, though I have some knowledge of basis functions. I've used spherical harmonics before
for image compression. But the answer is what I expected : It requires Quantum Electro Dynamics.
You see, he was making a logic error that I explain here :

'Maxwell's equations describe electromagnetic waves with wave lengths. A photon is the quantum of electromagnetic waves. Thus photons have wavelengths.'

A photon doesn't exist in the Maxwell model. Which is why the PDFs where necessary in the first place, because our treatment of 'particles'
depends on it. Sure, you can plug in the quantities, and a wave is a wave, but the model is very different. So your conclusion doesn't follow.

This is why we needed Quantum Electro Dynamics in the first place! That is, his expression contains a very small and subtle a priory assumption ...

 

[TheAdversary]

+Darwin123 Yes, I was asking another question in disguise Your answer is also what I expected, not stupid

 

[TheAdversary]

+Perpetual Student At one point in the discussion, the conservation of probability principle was called into question. I just took off on a tangent a little
because I was trying to trick him into a reductio ad absurdum. Thanks for the link; It looks beautiful.