# Klein-Gordon equation for electro-magnetic field?

## Main Question or Discussion Point

Can we imagine electro-magnetic field in vakuum as a massless particle that respects
Klein-Gordon equation (instead of Electromagnetic wave equation)?
It seems to me that both equations are the same, except that the electro-magnetic field can have 2 possible polarizations (then we count them as 2 different particles?).

## Answers and Replies

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pam
Yes. The EM wave equation for A^\mu is the KG equation of a massless particle.
Only having two polarizations is due to the masslessness.

However, note that because we do not observe longitudinal modes (which are allowed by the KG eqns) you have to include a constraint into the solution.

but the Klein-Gordon equation is a scalar equation. Are the two modes not coupled? They propagate independently each satisfying the (massless) K-G equation?

As a comparison, there is the Dirac equation. The 4 components each individually satisfy the K-G equation as well (for free particles, that is) but they are coupled by the Dirac equation and not independent.

but the Klein-Gordon equation is a scalar equation. Are the two modes not coupled? They propagate independently each satisfying the (massless) K-G equation?

As a comparison, there is the Dirac equation. The 4 components each individually satisfy the K-G equation as well (for free particles, that is) but they are coupled by the Dirac equation and not independent.
All good and true... but you can decouple the Dirac equation by finding it's eigenspinors. This is what you do, e.g., when you perform a mode expansion of the field.

Every free relativistic field will satisfy the KG equation, as it merely expresses $$p^2 = m^2$$. Note that $$\mathbf{E}$$ and $$\mathbf{B}$$ also obey the massless Klein Gordon equation. I think to get the KG equation for $$A$$ you need to be in the Lorenz gauge.

Ok. I get it. The components of A are in general coupled by the Maxwell equations. It is just by selecting a special gauge they are de-coupled.

Can we imagine electro-magnetic field in vakuum as a massless particle that respects
Klein-Gordon equation (instead of Electromagnetic wave equation)?
It seems to me that both equations are the same, except that the electro-magnetic field can have 2 possible polarizations (then we count them as 2 different particles?).
Out from the dunce corner with my weekly dumb question. Looked up Klein-Gordon equations in wikipedia--nothing. Looked in my McGraw-Hill encyclopedia of Physics--nothing. Kindly generalize in words, no acronyms please, what those(or that) equation(s) is about.

klien gordon is kind of a relativistic schodinger equation. it uses energy squared is equal to the (momentum times the speed of light in vaccuum) squared plus the mass squared times the speed of light in vacuum to the fourth power, i think. and turns everything into waves and operators.

the electro magnetic equations can be found by comparing maxwells equations to eachother. something like the laplacian of E equals the E double dot, where E is the electric field.

or something

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Out from the dunce corner with my weekly dumb question. Looked up Klein-Gordon equations in wikipedia--nothing. Looked in my McGraw-Hill encyclopedia of Physics--nothing. Kindly generalize in words, no acronyms please, what those(or that) equation(s) is about.
$$\partial^\mu \partial_\mu \psi + m \psi^2 == 0$$

Mentz114
Gold Member
Genneth made a small typo - it should read

$$\partial^\mu \partial_\mu \psi + m^2 \psi == 0$$

or

$$(\square^2 + \mu^2)\psi = 0$$

[nrqed- you're too quick - I'm still editing - d*mn latex. It depends on your definition of $$\square$$]

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nrqed
Homework Helper
Gold Member
Genneth made a small typo - it should read

$$\partial^\mu \partial_\mu \psi + m^2 \psi == 0$$

or

$$(\Box^2 + \mu^2)\psi = 0$$
Mentz made a very minor typo: there is no exponent of two on $$\Box$$

Mentz114
Gold Member
nrqed - for me

$$\square^2 \equiv -\frac{d^2}{dt^2} + \nabla^2$$

nrqed
Homework Helper
Gold Member
nrqed - for me

$$\square^2 \equiv -\frac{d^2}{dt^2} + \nabla^2$$
Ok. It's just that I have never seen this definition used in any book so it's not the standard definition, as far as I know. But I may be wrond and if you know of any book using that notation I would be interested to know.

Regards

Mentz114
Gold Member
nrqed,

it's true it is used differently in various texts. Itzykson&Zuber use your definition, but the Wiki page uses mine. I guess I&Z are more authoritative. Best to define just what one means by it, I guess.

M

pam
I think Gordan spelled his name that way, although Google votes 100-1 the other way.

pam
nrqed - for me

$$\square^2 \equiv -\frac{d^2}{dt^2} + \nabla^2$$
Both signs are used in books.
The opposite sign
$$\square^2 \equiv +\frac{d^2}{dt^2} - \nabla^2$$
is winning out recently, because it corresponds to the metric (1,-1,-1,-1).

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I think Gordan spelled his name that way, although Google votes 100-1 the other way.
The Gordon in Klein-Gordon and the Gordan in Clebsch-Gordan are two different people.

Is relativistic quantum mechanic also supposed to describe the dynamics of photons? Or are the similarities between equations concidental?

Both signs are used in books.
The opposite sign
$$\square^2 \equiv +\frac{d^2}{dt^2} - \nabla^2$$
is winning out recently, because it corresponds to the metric (1,-1,-1,-1).
Thanks for sending the details of that operator but I've been seeing such equations for years. Are there any physical/intuitive ideas associated with this one?

Mentz114
Gold Member
Normouse,
the square is called the D'Alembertian operator. More info here -

en.wikipedia.org/wiki/D'Alembertian

It should strictly be written like this ( note the factor c^2)

$$\square^2 \equiv +\frac{1}{c^2}\frac{d^2}{dt^2} - \nabla^2$$

in which case, (as has been stated above)

$$(\frac{1}{c^2}\frac{d^2}{dt^2} - \nabla^2)\psi = 0$$ means that $$\psi$$ is a field whose quanta travel at c.

Hans de Vries
Gold Member
Thanks for sending the details of that operator but I've been seeing such equations for years. Are there any physical/intuitive ideas associated with this one?
You might say so

The whole first part of my book is devoted to this equation....
for instance chapter 1:

http://physics-quest.org/Book_Chapter_EM_basic.pdf

Regards, Hans

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pam
The Gordon in Klein-Gordon and the Gordan in Clebsch-Gordan are two different people.
Thank you. I now recall that I knew that once.

pam