We can derive the Klein-Gordon equation using the relativistic energy-momentum relation, and here, I have no problem, it is an easy thing to do. However, I found by research that it applies only to spinless, spin 0, particles, but why is that. How can I deduce this just by looking at the equation.

Basically my question generalizes to: "how can I deduce, which kind of particles a particular wave equation describes, by looking at the equation?", since I would also like to know why I can say that the Dirac equation describes massive spin-1/2 particles and that Einstein's field equations describes a massless spin-2 field.


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  • $\begingroup$ Note that the Klein-Gordon equation also applies to each component of the Dirac Equation, which describes spin-1/2 particles. $\endgroup$ – Physics_Plasma Aug 14 '16 at 16:53
  • $\begingroup$ Please do not vandalize your own posts. $\endgroup$ – HDE 226868 Jan 15 '17 at 19:53

My total guess, which I hope other answers can correct is that, if you remember back to the Schroedinger Equation, it was based on the classical, non-relativistic version of energy. It does not incorporate spin and is not Lorentz invariant.

The Klein Gordon equation does take into account SR, by using E$^2$ = $p^2c^2 + m^2c^2$, but it can be interpreted as a scalar field, producing spin 0 particles, because it transforms as a scalar field.

How can I deduce, which kind of particles a particular wave equation describes, by looking at the equation?

The glib answer is: don't just look, follow through on finding a solution(s) and when the second quantization is performed, you will know if you have a vector field, which produces particles with spin.

  • $\begingroup$ Oh ok so I would need to look at how the equation transforms and at the nature of one or more solutions to deduce that, right? $\endgroup$ – abcabc123 Aug 14 '16 at 16:23
  • $\begingroup$ Can I suggest you wait for other answers to confirm my answer, ( or laugh at it, as I am a newbie to QFT)? But even without transformation, working through a solution is 99 percent of figuring out what it represents, as D.E.s are easy to set up, compared to solving them. The Dirac Equation is a classic in this regard, not making sense even to Dirac, until the "right" interpretation was found. $\endgroup$ – user108787 Aug 14 '16 at 16:30

It's not true that only spin-$0$ species can satisfy something that looks like the Klein-Gordon equation. For example, from the Dirac equation we can prove that a Dirac spinor $\psi$ satisfies $\square \psi=\pm M^2\psi$ (the sign depends on your relativity conventions), just like a spin-0 Klein-Gordon field.) Indeed, the Dirac equation was first dreamed up to "square root" the Klein-Gordon equation by obtaining a more stringent result which, Dirac hoped, would lack solutions of the Klein-Gordon equation that were physically troublesome. (A simple analogy is "square-rooting" $\partial_x^2 u = a^2 u$ to $\partial_x u = a u$, a strictly stronger equation.) It was only when Dirac experimented with the details that he realised that, among other things, $\psi$ must have at least 4 components for his new equation to work in 4D space. It turns out this predicts the electron's spin; a $2S+1$-fold spin degeneracy is doubled to include antimatter, so we solve $4S+2=4$ to give $S=1/2$.

More generally, given a differential equation in physics one can deduce from it a higher-order differential equation that has all the same solutions, but in general a number of others too. All Dirac solutions are also Klein-Gordon solutions, but not vice versa. Physically, we're interested in the most demanding constraint we can place, which will have as low a degree as we can muster. What's especially natural about these is that they are the equations of motion of the Lagrangian we use for the whole theory. For example, the Dirac equation is an equation of motion of the Dirac Lagrangian. The Klein-Gordon "squared" variant, however, is just a consequence rather than an equation of motion of the same Lagrangian.

So the real question is not which spins allow a field to satisfy an equation, but which are consistent with a Lagrangian from which it emerges as the strictest applicable constraint. Physical Lagrangians typically incorporate multiple types of field (some necessitated by difficulties with others), whether or not we include interaction terms for them. For example, the theory of electromagnetism has both the spin-$\tfrac{1}{2}$ electron field $\psi$ and the spin-$1$ photon $A^\mu$ (the latter can be motivated by a desire to generalise a global invariance of spinless fields to a local invariance). Bosons' spins are easily inspected; they are the number of Greek indices on the field. (For example, $g_{\mu\nu}$ is the spin-$2$ graviton.) Fermions can usually be inferred as above to be spin-$\tfrac{1}{2}$. Hypothetically, spin-$\tfrac{3}{2}$ species will obey a cousin of the Dirac equation such as this one, though there are some theoretical headaches such as faster-than-light propagation. Luckily, the only known spin-$\tfrac{3}{2}$ particles are non-elementary.

  • $\begingroup$ -1: It's not as simple as the number of Greek indices being the spin (see the example from Green, Schwarz and Witten's Superstring Theory involving a Fock vacuum and a theory defined on a Khaler manifold). $\endgroup$ – JamalS Jan 15 '17 at 21:16
  • $\begingroup$ @JamalS I'm having trouble finding in said textbook an example of a spin-$S$ boson on which the number of spacetime indices differs from $S$. Could you please narrow it down somewhat? The only discussions I can find of spin in the context of Fock vacuums & Kähler manifolds is of the $SO\left( 2N\right),\,SU\left( N\right)$ representations. $\endgroup$ – J.G. Jan 16 '17 at 6:35
  • $\begingroup$ One example of a field with more indices than spin states is a higher gauge 2-form $A_{\mu\nu}$ in four dimensions: Its free higher Yang-Mills theory is equivalent to the theory of a massless scalar field, since the dual potential to a 2-form in four dimensions is a scalar. $\endgroup$ – ACuriousMind Jan 19 '17 at 18:47

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