When can the Klein-Gordon Equation be used for a photon?

In summary: The Riemann-Silberstein vector ##\vec{\Psi}## is not a wavefunction in the traditional sense. It is a mathematical construct used to represent the massless spin-1 representation of the Poincare group. It does not have the same properties as a wavefunction and should not be interpreted in the same way.In addition, the scenario you described with photons decaying on a surface is not physically realistic. Photons do not decay in the same way that massive particles do. They can only be absorbed or emitted by other particles. And even then, the concept of position for a photon is not well-defined due to its wave-particle duality. It is better to describe a photon in terms of its momentum
  • #1
LarryS
Gold Member
345
33
Consider the double-slit experiment done with photons from a laser. If one was interested only in computing position (vertical) probability amplitudes and did not care about spin/helicity, could the Klein-Gordon Equation (with mass set to zero) be used?

Thanks in advance.
 
Last edited:
Physics news on Phys.org
  • #2
The Klein-Gordon equation for a (wave)function reads: ##(\square +m^2) \phi (x) =0 ## and applies for a massive photon, too (massive electromagnetic field). Drop the mass term and you've got the D'Alembert equation (the e-m potential satisfies this equation in the covariant Lorenz gauge).

Did you read a book about the scalar diffraction theory in optics?
 
  • #3
Last edited by a moderator:
  • #4
dextercioby said:
Did you read a book about the scalar diffraction theory in optics?

Not yet. Can you recommend a good one?
 
  • #5
Why ##\Psi## becomes a vector if m=0?
 
  • #6
.
olgerm said:
Why ##\Psi## becomes a vector if m=0?

The equation for photons with mass is really the Proca equation

##\partial_\mu(\partial^\mu B^\nu - \partial^\nu B^\mu)+\left(\frac{mc}{\hbar}\right)^2 A^\nu=0##

which is equivalent to a Klein Gordon equation for each of the four individual components of ##A^\nu##

##\left[\partial_\mu \partial^\mu+ \left(\frac{mc}{\hbar}\right)^2\right]A^\nu=0~~~~## with (in the massive case) ##~~\partial_\mu A^\mu=0##

Using the single component mass-less Klein Gordon equations is just a simplification which is mostly valid in some cases but not in others.
 
  • Like
Likes bhobba and Heikki Tuuri
  • #7
What is the correct value for m? Every source that I found said that restmass of photon is 0.
 
  • #8
olgerm said:
What is the correct value for m? Every source that I found said that restmass of photon is 0.

The Proca equation is the general equation of spin one particles. If the mass is zero you get photons and perhaps gluons - but the general equation can have mass, and indeed in the standard model some spin one particles do (ie the W and Z boson have rest mass. I can't think of others though - but they may exist - I certainly am no expert on the standard model):
https://en.wikipedia.org/wiki/Proca_action
However, the actual equations for the W and Z in the Standard Model are much more complicated because they start as mass-less fields and, if massive, acquire mass via the Higgs mechanism, and in addition, they have interactions with other fields, which the Proca equation does not incorporate.

Interestingly it is thought the gluons of the strong force have no rest mass but I do not think it has been directly measured. Puzzle for somebody more knowledgeable than me (as I said I am no expert on the Standard Model) what then makes up the mass of nucleons? Is it some relativistic effect?

Thanks
Bill
 
Last edited:
  • #9
How does it explain why ##\Psi## becomes a vector if m=0. And components of new ##\vec{\Psi}## are not copmonents of ##\vec{A}## (electromagnetic 4-potential).
 
Last edited:
  • #10
##\vec{\Psi}=\vec{E}+i*\vec{B}## is very different from wavefunction, because ##\Psi## may have many particles position as argument, but E and B are functions of position. What if positions of 2 photons are entangeled?
 
  • #11
There's no wave function for a photon to begin with (strictly speaking there's no wave function for any relativistic particle either though the non-relativistic approximation is valid for a broad range of systems, including lighter atoms, molecules, condensed matter).

A photon is described by a massless quantum field of spin one. There are different ways to represent the massless spin-1 representation of the proper orthochronous Poincare group. The standard one is using massless vector-fields ##\hat{A}^{\mu}##, which are necessarily gauge fields, because otherwise you'd have continuous spin-like intrinsic quantum numbers, which are not observed when dealing with photons. Thus the photons have only the two intrinsic spin-like degrees of freedom, called polarization states with the observable associate to them called helicity, which is the projection of the total angular momentum (and only total angular moementum is well defined not a split into spin and orbital angular momentum) to the direction of its momentum and can take the values ##\pm 1##.

Another representation of this irrep. of the Poincare group is based on the Riemann-Silberstein vector ##\hat{\vec{E}}+\mathrm{i} \hat{\vec{B}}##. I'm not sure, whether this possibility has ever been worked out at the same level as standard QED.
 
  • #12
How can ##\vec{\Psi}=\vec{E}+i*\vec{B}## describe a situation, where parameters of photons are dependnt of each other. For example if electron and positron decay on surface x=0 at time t=0:
probability that ##p_{x\ photon1}>0## is 0.5 and probability that ##p_{x\ photon2}>0## ,but
probability that ##p_{x\ photon1}>0## and ##p_{x\ photon2}>0## is not 0.25 ,but 0.

with wavefunction ##\Psi(x_{photon\ 1},y_{photon\ 1},z_{photon\ 1},x_{photon\ 2},y_{photon\ 2},z_{photon\ 2})## it can be expressed as
##\int_0^\infty(dx_1*\int_{-\infty}^\infty(dx_2*\int_{-\infty}^\infty(dx_3*\int_{-\infty}^\infty(dx_4*\int_{-\infty}^\infty(dx_5*\int_{0}^\infty(dx_6*|\Psi(x_1,x_2,x_3,x_4,x_5,x_6)|^2)))))=0.5##

##\int_{-\infty}^\infty(dx_1*\int_{-\infty}^\infty(dx_2*\int_{-\infty}^\infty(dx_3*\int_{-\infty}^\infty(dx_4*\int_{0}^\infty(dx_5*\int_{-\infty}^\infty(dx_6*|\Psi(x_1,x_2,x_3,x_4,x_5,x_6)|^2)))))=0.5##

##\int_0^\infty(dx_1*\int_{-\infty}^\infty(dx_2*\int_{-\infty}^\infty(dx_3*\int_{-\infty}^\infty(dx_4*\int_{0}^\infty(dx_5*\int_{-\infty}^\infty(dx_6*|\Psi(x_1,x_2,x_3,x_4,x_5,x_6)|^2)))))=0\neq 0.25##

, but ##\vec{\Psi}(t,x,y,z)=\vec{E}(t,x,y,z)+i*\vec{B}(t,x,y,z)## only function of space and time and can't describe such situation.

Given situation may be impossible because uncercanty principle, but question is still up: How to describe parameters of photon that are releated with each other that way.
 
Last edited:

1. Can the Klein-Gordon Equation be used for a photon in all situations?

No, the Klein-Gordon Equation can only be used for photons in certain situations. This equation is typically applied to massive particles, such as electrons, and cannot accurately describe the behavior of massless particles like photons. However, in certain scenarios where the photon behaves like a particle with mass (such as in a medium with high refractive index), the Klein-Gordon Equation can be used.

2. What is the difference between the Klein-Gordon Equation and Maxwell's Equations for describing photons?

The Klein-Gordon Equation and Maxwell's Equations are two different mathematical equations used to describe the behavior of photons. The Klein-Gordon Equation is based on the principles of quantum mechanics and describes the behavior of particles with mass, while Maxwell's Equations are based on classical electromagnetism and describe the behavior of electromagnetic waves, including photons. Maxwell's Equations are more accurate for describing the behavior of photons in most situations.

3. When can the Klein-Gordon Equation be used instead of Maxwell's Equations for photons?

As mentioned before, the Klein-Gordon Equation can be used for photons in certain situations where they behave like particles with mass. This includes scenarios such as in a medium with high refractive index or in interactions with other particles. However, in most cases, Maxwell's Equations are more accurate for describing the behavior of photons.

4. Can the Klein-Gordon Equation account for the wave-particle duality of photons?

Yes, the Klein-Gordon Equation can account for the wave-particle duality of photons. This equation is derived from the principles of quantum mechanics, which describe particles as having both wave-like and particle-like properties. The Klein-Gordon Equation can be used to describe the wave-like behavior of photons in certain situations.

5. Is the Klein-Gordon Equation limited to only photons?

No, the Klein-Gordon Equation is not limited to only photons. While it is most commonly used to describe the behavior of photons, this equation can also be applied to other particles with mass, such as electrons. In fact, the Klein-Gordon Equation was originally developed to describe the behavior of spin-0 particles, which include both photons and other particles.

Similar threads

Replies
1
Views
777
Replies
9
Views
761
  • Quantum Physics
Replies
2
Views
2K
  • Quantum Physics
Replies
2
Views
1K
Replies
41
Views
4K
  • Quantum Physics
Replies
2
Views
1K
  • Quantum Physics
Replies
3
Views
2K
Replies
113
Views
10K
  • Quantum Physics
2
Replies
40
Views
3K
  • Quantum Physics
Replies
21
Views
6K
Back
Top