Amplitude of a wave and general laws of physics

[swansont]

10 hours ago, Rob McEachern said:

Any experiment in which a multi-channel detector, responds to a mono-energy input (quanta) within each individual channel. That is what every Fourier transform's power spectrum describes. Thus, since wave-functions are mathematically described via Fourier transforms, it is true of every wave-function, whenever each "bin" in the transform receives only entities (quanta) with a single energy, E_photon ,which may differ from one channel to the next. That is the origin of the Born rule.

The Born rule applies to quantum states. 

A multi-channel detector that detects different energies is not an example of a probabilistic system. If one channel detects energies of 1.00 - 1.10 eV, and your photons have 1.05 eV, then they will all be detected in that one channel. Not the others.

10 hours ago, Rob McEachern said:

It happens with every experiment, that can be described mathematically via a Fourier transform's power spectrum (as are all wave-functions-squared), since it is a mathematical property of a Fourier transform, having nothing to do with physics. The only experimental condition that must be met, for the Born rule to be valid, is that each bin of the power-spectrum/histogram be obtained by a mono-chromatic integration of quanta. In other words E_photon must be the same for all the photons detected in a bin, but can differ from bin-to-bin. For example, a diffraction grating or prism, will deflect photons of different energies/frequencies at different angles. Consequently, detectors placed at different angles will each receive photons within a single, narrow band of energies, thus ensuring the validity of the Born rule.

No, that's not the Born rule. If your photon has a certain wavelength, the odds it will diffract at the angle predicted by formula is 100%. There's no wave function in this problem.

 

[Rob McEachern]

2 hours ago, swansont said:

A multi-channel detector that detects different energies is not an example of a probabilistic system

Thermodynamics. That is where Fourier analysis originated.

 

2 hours ago, swansont said:

There's no wave function in this problem

There is no wave function in any problem. The use of the term "wave" is a misnomer. The structure of Schrodinger's wave equation is unlike that of the classical wave equation. But it is identical to the classical heat equation; it would have been more appropriate to call it the Schrodinger heat equation, describing a process similar to heat-flow, rather than wave propagation.

[swansont]

1 hour ago, Rob McEachern said:

Thermodynamics. That is where Fourier analysis originated.

Albuquerque! See I can do it too. Snorkel!

(Last I checked, thermodynamics wasn't QM)

1 hour ago, Rob McEachern said:

There is no wave function in any problem. 

Please stop making stuff up.