What is the PDE for photons proposed by Bialynicki-Birula?

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SUMMARY

The discussion centers on the absence of a Schrödinger equation for photons, which are massless particles. Participants clarify that the wave equation for the four-potential Aμ serves as the equivalent Schrödinger equation for photons and is derived from the zero mass limit of the Klein-Gordon equation. The Schrödinger equation is deemed non-relativistic and unsuitable for photons, while a relativistic Schrödinger equation can be formulated but faces limitations regarding locality and causality. Bialynicki-Birula's work is referenced as an attempt to establish a partial differential equation (PDE) for photons.

PREREQUISITES
  • Understanding of the Schrödinger equation and its limitations for massless particles
  • Familiarity with the Klein-Gordon equation and its implications in quantum mechanics
  • Knowledge of relativistic quantum theory and its challenges
  • Basic grasp of electromagnetic theory, particularly Maxwell's equations
NEXT STEPS
  • Study the wave equation for the four-potential Aμ in quantum electrodynamics
  • Explore the implications of the Klein-Gordon equation for massless particles
  • Investigate Bialynicki-Birula's publications on the PDE for photons
  • Learn about the limitations of relativistic quantum mechanics regarding locality and causality
USEFUL FOR

This discussion is beneficial for theoretical physicists, quantum mechanics researchers, and students interested in the behavior of massless particles and the foundations of quantum field theory.

birulami
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Neither in Sam Treiman's http://books.google.de/books?id=e7fmufgvE-kC" I was able to find the Schrödinger equation for a photon, i.e. a particle without rest mass. The Schrödinger equation straight from Treiman's book (typos are mine, if any)

-\frac{\hbar^2}{2m}\Delta\Psi + V\Psi = i\hbar\frac{\partial\Psi}{\partial t}​

with potential V and the Laplace-Operator \Delta applied for all coordinates except t, does contain the (rest)mass m in the denominator, so I guess this won't work for the photon.

How then does the equation look like to cover massless particles? Or does it not apply?

Thanks,
Harald.
 
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You are not going to find the Shrodinger equation for photon anywhere in literature. Check my responses #95 and #96 in the thread What really is a photon?
 
The wave equation for A^\mu is the equivalent SE for a photon, with A^\mu being considered the wave function of the photon. It is the zero mass limit of the Klein Gordon equation. This interpretation is in the literature.
 
birulami said:
The Schrödinger equation [...] does contain the (rest)mass m in the denominator, so I guess this won't work for the photon.

Note that the photon is highly relativistic, and the Schrödinger equation is non-relativistic. The SE contains the classical definition of kinetic energy, translated into operator form:

K = \frac{1}{2}mv^2 = \frac{p^2}{2m} \rightarrow \frac {1}{2m} \left( -i \hbar \frac{\partial}{\partial x} \right)^2 = - \frac{\hbar^2}{2m} \frac {\partial^2}{\partial x^2}

(in one dimension)
 
jostpuur said:
You are not going to find the Shrodinger equation for photon anywhere in literature.

Autsch. I wondered already why in one part of a book I find Maxwell's equations and a description of the photon, and in other parts of a books I find the Schrödinger equation. But a connection between the two is suspiciously absent.

Too bad.
Harald.
 
birulami said:
Autsch. I wondered already why in one part of a book I find Maxwell's equations and a description of the photon, and in other parts of a books I find the Schrödinger equation. But a connection between the two is suspiciously absent.

Come on now. We can certainly write down a Schrödinger equation for a system composed of matter and photons (in Columb gauge to be specific).


\frac{i\partial}{\partial t}|\Psi> = H |\Psi>

H = H_{\rm matter} + H_{\rm free photon} + H_{\rm interaction}

H_{\rm free photon} = \sum_{k,j}a^{\dagger}_{k,j}a_{k,j} c \hbar k

where c is the speed of light and k is the wave-vector and j is one of two polarizations. And \Psi is some god-awful wavefunction.

(sorry for not putting TeX wrappers around the TeX... my TeX comes out completely different from what I wrote when I do that. Anyone else have this problem?)
 
pam said:
The wave equation for A^\mu is the equivalent SE for a photon, with A^\mu being considered the wave function of the photon. It is the zero mass limit of the Klein Gordon equation. This interpretation is in the literature.

Klein-Gordon equation is not a consistent generalization of the non-relativistic Shrodinger equation. The second order (in respect to time) PDE does not become first order PDE in the low energy limit. Classical fields do not become quantum mechanical wave functions either, since they usually have different dimensions.

With the photon these problems are not so obvious because there is no non-relativistic limit anyway, but the fundamental problem with relativistic quantum theory is still there.

We can write down a relativistic SE easily as

<br /> i\partial_t \Psi = \sqrt{-\nabla^2 + m^2}\Psi<br />

and it is easy to substitute m=0 also and get

<br /> i\partial_t \Psi = \sqrt{-\nabla^2}\Psi<br />

but the mainstream story goes on so, that having found the relativistic SE, we next notice that it cannot be used for some locality and causality related reasons.

olgranpappy said:
Come on now. We can certainly write down a Schrödinger equation for a system composed of matter and photons (in Columb gauge to be specific).


\frac{i\partial}{\partial t}|\Psi> = H |\Psi>

H = H_{\rm matter} + H_{\rm free photon} + H_{\rm interaction}

H_{\rm free photon} = \sum_{k,j}a^{\dagger}_{k,j}a_{k,j} c \hbar k

where c is the speed of light and k is the wave-vector and j is one of two polarizations. And \Psi is some god-awful wavefunction.

(sorry for not putting TeX wrappers around the TeX... my TeX comes out completely different from what I wrote when I do that. Anyone else have this problem?)

Working in momentum space is closer to hiding the problem than solving it. You wouldn't bother transforming this into the position representation?
 
olgranpappy said:
(sorry for not putting TeX wrappers around the TeX... my TeX comes out completely different from what I wrote when I do that. Anyone else have this problem?)

Not now, but I remember having some problem like this once. I think I just edited it until it started working. Like through deleting and rewriting.

hmhmh... or did this happen before I had understood that one has to reload the page once after editing the equations in order to see the correctly? I'm not sure...
 
jostpuur said:
Working in momentum space is closer to hiding the problem than solving it. You wouldn't bother transforming this into the position representation?

Nah, I'm happy with that Hamiltonian the way it is.

I'm interested in practical things such as calculating scattering amplitudes for x-ray experiments, so I do just fine in momentum space, thank you very much.
 
  • #10
I saw an interesting attempt to find PDE for photons made by Polish physicist Bialynicki-Birula:

http://www.cft.edu.pl/~birula/publ/APPPwf.pdf
http://www.cft.edu.pl/~birula/publ/reconstr.tex
http://www.cft.edu.pl/~birula/publ/CQO7.pdf
 
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