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Every Solution of the Dirac Equation is also a solution of the Klein-Gordon equation.

So the K-G equation does not necessarily represent particles with non-zero spin.

Would it be incorrect to conclude that KG equation relates solely to zero-spin particles?

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KG equation written in the momentum space is just a mass-shell condition, so it holds for any particles that are on-shell (e.g. gauge-fixed Maxwell equations in the vacuum are $\square A_{\mu}=0$, which is just KG equation for the massless field).

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  • $\begingroup$ If you put m=0 in the KG equation you simply get the standard wave equation. Incidentally photons have spin=1 . You are trying to extend my arguments with further examples I believe $\endgroup$
    – Anamitra Palit
    Commented Aug 9, 2012 at 7:38
  • $\begingroup$ The argument is just about the mass shell condition being independent on spin -- that's all, example is just for illustration. $\endgroup$
    – M.Dedushenko
    Commented Aug 9, 2012 at 8:06
  • $\begingroup$ Also should add that although it's indeed just a terminological question, people usually say "it satisfies KG equation" when discussing fields of arbitrary spin (see S.Weinberg's book on QFT, section 5.1 for example) $\endgroup$
    – M.Dedushenko
    Commented Aug 9, 2012 at 8:24
  • $\begingroup$ The mass shell equation is simply the energy equation :$E^2=p^2c^2+m_{0}^2c^4$Putting in the appropriate operators we get the KG equation. For going back to the momentum space we are doing the reverse action.We get back the energy equation which is valid real particles.But if you insert the operators for E and p is the same equation supposed to be valid for all real particles? $\endgroup$
    – Anamitra Palit
    Commented Aug 9, 2012 at 9:11
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There is a simple terminological confusion here. What you refer to the "Klein-Gordon equation" is known as the "wave equation" or "d'Alembert equation" with a mass term. Fields of any spin may be the variables in such an equation. Of course it's useful to realize the similarities between the equations governing the scalar field and equations governing other fields. The essence is always the same.

However, the Klein-Gordon equation is, by definition, the wave equation for a spinless, scalar field. In particular, the Klein-Gordon field has to be a scalar field, by definition. So the answer to your question is No. If the equation contains a field with a non-zero spin, then it is not the Klein-Gordon equation. This claim is no different from the claim that the formula $s=vt$ for the distance as a product of the velocity and the time isn't called the Ohm's law. Ohm's law has the same mathematical form – something is the product of other two things, $U=RI$ – but it actually requires the objects that enter the product to have some particular properties. That's why $s=vt$ isn't Ohm's law.

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  • $\begingroup$ $K-G{\;}Eqn \times I_{4\times 4}=0$ can take you out of the scalar field $\endgroup$
    – Anamitra Palit
    Commented Aug 9, 2012 at 8:20
  • $\begingroup$ $(i\gamma^\mu \partial_\nu-m)\psi=0$ Or,$(-i\gamma^\mu \partial_\nu-m)(i\gamma^\mu \partial_\nu-m)\psi=0$ Or,$(1/2{{}\gamma^\mu \gamma^\nu{}}\partial_\mu \partial{\nu}+m^2)\psi=0$Or,$I_{4 \times 4}(\partial^2+m^2)\psi=0$ Or,$(\partial^2+m^2)\psi=0$ Here $\psi$ itself is a column vector. $\endgroup$
    – Anamitra Palit
    Commented Aug 9, 2012 at 8:31
  • $\begingroup$ $\gamma^\mu \gamma^\nu $ should be enclosed by curly brackets in the last comment $\endgroup$
    – Anamitra Palit
    Commented Aug 9, 2012 at 8:38
  • $\begingroup$ You may consider the equation: $(\partial^2+m_{0}^2)\psi=0$. If $\psi$ is taken to be a $4\times 1$ column vector, each of the four component equations should obey the scalar KG equation. These solutions are neither zero spin nor half spin particles.(The coefficients in the Dirac equation contain quantities like energy and momentum components)What is the possibility of their existence and detection in nature? $\endgroup$
    – Anamitra Palit
    Commented Aug 9, 2012 at 9:32
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    $\begingroup$ Dear @Anamitra, a spinor is something else than a collection of four scalars. So the equation with $\psi$ being a 4-component spinor may look similar, and indeed mathematically equivalent, to an equation in which $\psi$ is a column of 4 scalars, these two equations must be considered physically different equations because the objects in them have different transformation rules under rotations. In physics, we only call it "Klein-Gordon equation" if the fields in the equations transform as scalars i.e. don't transform at all. $\endgroup$
    – Luboš Motl
    Commented Aug 10, 2012 at 7:34
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In addition to the answers above, I would like to add what goes wrong if you try to use Klein Gordon equation for higher spin. In this case, you always get ghost particles, you get particles whose contribution to the probability is negative.

This issue is best explained by trying this trick with a vector. Consider a four-component vector, with each component separately obeying the KG equation. In this case, you get

$$ S = \int \partial_\mu A_\nu \partial_\mu A_\nu - {m^2\over 2} A_\mu A\mu d^4x$$

and the kinetic term on the time component of A is opposite sign to all the space components. This means you have a timelike ghost field, or three spacelike ghost fields, and the equation is not describing a reasonable free field theory.

Nevertheless, this thing (in the massless case) is what you get in QED when you impose Feynman gauge (or in t'Hooft gauge for the nonabelian theory). The ghost time component doesn't contribute to physical processes by gauge invariance, and this means that the negative propagator states are not propagating degrees of freedom.

This disease affects the KG equation with any spin other than 0, so the other wave equations are the proper ghost-free trucations of the Klein Gordon equation, which is best viewed as only part of the equation of motion--- the mass shell projection.

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Each free particle or field (each component of corresponding field or wave-function) must satisfy the Klein-Gordon equation, because it corresponds to relativistic energy-momentum relation (or, more formally, refers to the Casimir operator $p^{\mu}p_{\mu}$ of the Poincare group).

But each free particle or field must satisfy relation $W^{\mu}W_{\mu} \Psi = -ms(s + 1)\Psi$, where $W_{\mu}W^{\mu}$ is the second Casimir operator (refers to the spin degrees of freedom). There is $2s + 1$ spin degrees of freedom (energy is equal for corresponding states). So you need to add some relations to your function (it may refer to the spinor $\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{m}}$ or, in particular cases, to 4-tensor $A_{\mu_{1}...\mu_{n}}$), which left only $2s + 1$ independent components (or only one for massless particles). For the free particles almost all other charges lose significance, because they are important only in an interaction cases. So we usually say that free particle has mass $m$ and spin (helicity) $s$.

For example, let's have spin-s massive particle. This case corresponds to s-rank 4-tensor $A_{\mu_{1}...\mu_{s}}$ with Klein-Gordon equation for each component $$ \tag 1 (\partial^{2} + m^{2})A_{\mu_{1}...\mu_{s}} = 0 $$ and conditions $$ \tag 2 \partial_{\mu_{k}}A^{\mu_{1}...\mu_{k}...\mu_{s}} = 0,\quad k = 1, ..., s, $$ $$ \tag 3 A^{\mu_{1}...\mu_{k}...\mu_{l}...\mu_{s}} = A^{\mu_{1}...\mu_{l}...\mu_{k}...\mu_{s}}, \quad k = 1, ..., s, $$ $$ \tag 4 A_{\mu_{k}}^{\quad \mu_{1}...\mu_{k}...\mu_{s}} = 0, \quad k = 1, ..., s. $$

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