Do higher spin particles obey Klein-Gordon or Dirac equations?

ndung200790

We know that 0-spin particles obey Klein-Gordon equation and 1/2spin particles obey Dirac equation.But I do not know whether higher integer spin particles obey Klein-Gordon equation or not.Similarly,do higher half integer spin particles obey Dirac equation?Because if we can not demontrate the higher spin particles obey Klein-Gordon(for Boson) or Dirac(for Fermion) equations,how can we say about the commutation relations of field operator and momentum operator of field for Bose fields and also how about the anticommutation relations for Fermi fields.Therefore,how can we make the quantization of fields, and how about the exclusion principle of Pauli....etc.
Thank you very much in advanced.

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kof9595995

That depends on what you really mean by "obey", solutions of Dirac eqn obey Klein-Gordon equation automatically, so is Maxwell's equation for photon. This is because Klein-Gordon really tells you nothing but that the theory is relativistic. In this sense all relativistic particles obey Klein-Gordon equation.
However Klein-Gordan does not describe electrons appropriately, although electrons "obey" Klein-Gordan.

dextercioby

Homework Helper
I think the analysis made by Gelfand and Yaglom in 1948 remains the best when it comes to differential equations for <objects> transforming under the Lorentz group. It is exposed in the book by Gelfand, Minlos and Shapiro, 2nd Chapter of Section II.

RedX

Spin 3/2 obeys the Rarita-Schwinger equation. The spin 3/2 field has a mixture of vector and spinor indices.

The spin 2 field should obey the Einstein equation.

ndung200790

So,the higher half integer spin may be considered as a mixture of vector indices and spinor indices.Therefore, all half integer spin particles obey Dirac equation?

RedX

So,the higher half integer spin may be considered as a mixture of vector indices and spinor indices.Therefore, all half integer spin particles obey Dirac equation?
I'm not that familiar with the spin 3/2 equation.

But does the electromagnetic potential obey the Klein-Gordan equation? I don't think it does unless you choose the Lorenz gauge or the Coloumb gauge.

it goes something like this:

$$\Box A^\mu-\partial^\mu \partial_\nu A^\nu=0$$

The Lorenz gauge obviously makes the 2nd term go away. The Coloumb gauge makes the 4-divergence produce the timed derivative of the scalar potential, which is zero (just set mu equal to zero in the equation to see that the scalar potential obeys Laplace's equation in free space).

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ndung200790

I think we need to set the gauge fix to eliminate redundant configurations.So photons still obey Klein-Gordon equation.

homology

I think the analysis made by Gelfand and Yaglom in 1948 remains the best when it comes to differential equations for <objects> transforming under the Lorentz group. It is exposed in the book by Gelfand, Minlos and Shapiro, 2nd Chapter of Section II.
Do you have the title of the text?

dextercioby

Homework Helper
Do you have the title of the text?
Sure, it's an old book from the 1960's (English translation): <Gelfand, Minlos, Shapiro - Representations of rotation and Lorentz groups>.

RedX

I think we need to set the gauge fix to eliminate redundant configurations.So photons still obey Klein-Gordon equation.
Doesn't it depend on which gauge you choose? What if I chose the axial gauge: A3=0 ?

Would:

$$\Box A^\mu-\partial^\mu \partial_\nu A^\nu=0$$

reduce to the Klein-Gordan equation?

kof9595995

Doesn't it depend on which gauge you choose? What if I chose the axial gauge: A3=0 ?

Would:

$$\Box A^\mu-\partial^\mu \partial_\nu A^\nu=0$$

reduce to the Klein-Gordan equation?
But the thing is, for massless Klein-Gordan eqn you'll also have the freedom of choosing a gauge, without changing any physical content. So I think in a loose sense it still obeys KG eqn.

RedX

But the thing is, for massless Klein-Gordan eqn you'll also have the freedom of choosing a gauge, without changing any physical content. So I think in a loose sense it still obeys KG eqn.
I was under the impression that discussion was over the free, non-charged KG equation.

The free, charged KG equation has a U(1) symmetry but not a gauge symmetry.

The solution to the EOMs of free KG fields are unique, so I don't believe there is any freedom to impose gauge conditions.

kof9595995

The solution to the EOMs of free KG fields are unique, so I don't believe there is any freedom to impose gauge conditions.
No, I meant we could change a gauge for KG equation, and the form of KG eqn will look different, however it's still KG eqn in a loose sense because nothing physical has changed, just like in electromagnetism, different gauge fixing gives different $$\Box A^\mu-\partial^\mu \partial_\nu A^\nu=0$$ but does not change the physics. (Actually I don't know formulate and prove " nothing physical has changed" in this quantum context, could somebody enlighten me?)

kof9595995

I was under the impression that discussion was over the free, non-charged KG equation.
Of course not necessarily non-charged, or else I wouldn't say Dirac field also obeys KG eqn. Anyway what I tried to say was simply asking what equation a field "obeys" is not enough, there is missing information in how the field transforms.
The free, charged KG equation has a U(1) symmetry but not a gauge symmetry.
I'm confused, why not U(1) a gauge symmetry group?

RedX

No, I meant we could change a gauge for KG equation, and the form of KG eqn will look different, however it's still KG eqn in a loose sense because nothing physical has changed, just like in electromagnetism, different gauge fixing gives different $$\Box A^\mu-\partial^\mu \partial_\nu A^\nu=0$$ but does not change the physics. (Actually I don't know formulate and prove " nothing physical has changed" in this quantum context, could somebody enlighten me?)
O I see what you're saying. So for the charged KG equation:

$$\Box \phi + m^2 \phi=0$$

and it's conjugate equation, you can add a term to get a new equation:

$$\Box \phi + m^2 \phi+\partial^\mu(\phi^*\partial_\mu\phi-\phi \partial_\mu \phi^*)=0$$

whose solution is the KG-field, since the term in parenthesis is zero since it is the conserved 4-current of the KG-equation without this extra term, so that a solution of the KG-field without this extra term is also a solution to the KG-equation with this extra term.

This only works for the charged KG-field though, or else that extra term is zero.