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

#### referframe

Gold Member
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?

Last edited:
Related Quantum Physics News on Phys.org

#### dextercioby

Homework Helper
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?

#### bhobba

Mentor
Last edited by a moderator:

#### referframe

Gold Member
Did you read a book about the scalar diffraction theory in optics?
Not yet. Can you recommend a good one?

#### olgerm

Gold Member
Why $\Psi$ becomes a vector if m=0?

#### Hans de Vries

Gold Member
.
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.

#### olgerm

Gold Member
What is the correct value for m? Every source that I found said that restmass of photon is 0.

#### bhobba

Mentor
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 cant think of others though - but they may exist - I certainly am no expert on the standard model):

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:

#### olgerm

Gold Member
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:

#### olgerm

Gold Member
$\vec{\Psi}=\vec{E}+i*\vec{B}$ 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?

#### vanhees71

Gold Member
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.

#### olgerm

Gold Member
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 cant 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: