The Klein-Gordon equation

$$\left(\frac{\partial ^2}{\partial t^2} - |\nabla|^2 + m^2\right)\phi = 0\tag{1}$$

should describe non interacting particles without spin. So what particles in the standard model are described by it? And what about famous non-elementary particles? For example: is the Higgs boson described by (1)? It is indeed a spinless, chargeless particle; but I am not sure.

Also we know that the time component of the conserved four current:

$$J_\mu=\phi^*\partial _\mu \phi -\phi \partial_\mu \phi^* \tag{2}$$

is not positive definite, so we cannot interpret it as a probability current as we normally do in QM, this is not a problem for the K-G dynamics since we perform second quantisation, right? We don't need to interpret $J_0$ as a probability.

  • $\begingroup$ Be careful. You could add interactions. Also, should the Schrödinger equation only describe spinless particles? $\endgroup$
    – my2cts
    Feb 26 at 21:24

2 Answers 2


A real (hermitean) Klein-Gordon field describes chargeless spin $0$ bosons. Examples: $\pi^0$, Higgs.

A complex (non-hermitean) Klein-Gordon field describes charged spin $0$ bosons, where the "charge" is not necessarily of electromagnetic nature. Examples: $\pi^\pm$, $K^\pm$ (both with electromagnetic charge) or $K^0, \bar{K}^0$ (electromagnetically neutral with strangeness "charge").

In quantum field theory, $J^\mu =i[\phi^\ast \partial^\mu \phi- (\partial^\mu \phi^\ast) \phi]$ represents the 4-current-density operator with the associated charge operator given by $Q=\int \! d^3x \, J^0(x)$.


The Klein-Gordon equation is the field version of the special relativistic energy momentum relation. When one abandons the restriction that it should only be used for scalar fields it becomes clear that it applies to all free matter, just like the special relativistic energy momentum relation. For example solutions of the free Dirac equation solve the Klein-Gordon equation. The Maxwell equations can be written as a wave equation, that is, a massless Klein-Gordon equation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.