Can photons or other particles form a wave-function again after an "observation" or "measurement"?

[Mordred]

Well one problem I see is your comparing pentacenes with photons. There is significant differences between the two. Those differences goes beyond probability functions.

A single photon for example has transverse and longitudinal wavefunctions determined by the EM field which is defined through the Maxwell equations. Those wavefunctions are not the probability functions. You still have those wavefunctions after measurement.

I'm sure you can recognize significant differences between quantum elementary particles to a multiparticle molecule such as pentacene 

Edited December 20, 2019 by Mordred

[Enthalpy]

To my understanding, the EM field is a macroscopic illusion, like the magnetisation vector is one for the orientation of the magnetic moment of the nuclei, which is not a vector but a binary value in any direction.

In NMR, one can compute a magnetic  "vector" as is done macroscopically, then deduce for one hydrogen nucleus the probability for each state. And even: most people do it like this, because it's concrete, while the precession matrices are not.

But for one photon, writing an EM field and deducing a wavefunction of a probability isn't always possible.

Take a hydrogenoid 3s->2p transition: it can emit a photon in any direction with equal probability and with any polarisation. This can not be written as an EM field, as I already pointed out.
https://en.wikipedia.org/wiki/Hairy_ball_theorem
But as a scalar wavefunction it's easy.

That is, computing first an EM field, then deduce a photon wavefunction, fails sometimes. And the EM field is not the photon's wavefunction.

Edited December 20, 2019 by Enthalpy

[Mordred]

The EM field is a macroscopic illusion ??? better clarity on that bit please

 

 

Edited December 21, 2019 by Mordred

[swansont]

12 hours ago, Enthalpy said:

 

But for one photon, writing an EM field and deducing a wavefunction of a probability isn't always possible.

Take a hydrogenoid 3s->2p transition: it can emit a photon in any direction with equal probability and with any polarisation. This can not be written as an EM field, as I already pointed out.

An atomic transition is described by the wave function of the atom, not the wave function of a photon, and isn't presented as an EM field.

 

On 12/19/2019 at 5:20 PM, Enthalpy said:

True, that was already answered. What the pentacene pictures bring new is that the wavefunction can be observed and then released intact - in some cases.

Not sure how you conclude that. You don't know what the wave function was before the measurement, and I don't know what "released intact" means.

It's dubious to say an AFM image is the measurement of the wave function. It's a measurement of electron position.

[Mordred]

Ok lets get some better details here is the QM methodology on photon transverse and longitudinal wavefunctions. The paper also discusses the electrons.

https://www.nist.gov/system/files/documents/pml/div684/fcdc/photon-wave.pdf

You will see how the Maxwell equations are involved including the Schrodinger equations, though as the photon is relativistic the Klien Gordon equation is a better option however the Dirac equations are Lorentz invariant. Which the paper checks for Lorentz invariance. I chose this method via the Scrodinger as it will become important in regards to decoherence that was mentioned earlier this thread.

Decoherence also involves the Schrodinger equation.

 

Edited December 21, 2019 by Mordred

[Edgard Neuman]

On 12/19/2019 at 11:20 PM, Enthalpy said:

I haven't grasped from you text where past/present/future makes a difference.

 

Picture the family tree of a specie (that's an analogy)... at any point in time you have several versions of the prototypical individual (each version of the DNA of the specie) living concurrently.. but as time pass, every branches eventually die off. Until you get to a point where only one ancestor is still represented by its descendant.. (the last common ancestor). At each point in time there are several individuals simultaneously, but far enough in the past, only one of them is needed to explain the present.

Now replace "individuals", by "alternate universes" (concurrent versions of the Feynman diagram).
I love this analogy, because it also understand how branches that died off are still necessary to explain what happened in the past, just as you need all the concurrent Feynman scenari to compute the probability of the final state..

Edited January 20, 2020 by Edgard Neuman