I've read that every relativistic scalar field (and in some sense, any field) satisfies the Klein-Gordon equation. Is the reasoning for this just based on the quantum mechanical substitution of $E\to i\,\partial_t\,,p\,\to -i\nabla$ and special relativity's $E^2=p^2+m^2\,?$ This would seem to essentially make it a postulate, is this true?

Though I am a bit confused because special relativity's equations are originally for particles, not fields. I understand that the Klein-Gordon operator is the simplest linear differential operator acting on scalar fields, but it seems like one can easily define a Lorentz invariant scalar field on spacetime that doesn't satisfy the Klein-Gordon equation, in the same way that one can easily define a scalar field on $\mathbb{R}^3$ that doesn't satisfy Laplace's equation.


1 Answer 1


Not all fields obey the Klein-Gordon equation - all free fields do, since the Klein-Gordon equation is the equation of motion for the Lagrangian density

$$ \mathscr{L}[\phi,\partial_\mu \phi] = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2 \phi^2$$

containing no interaction terms.

  • $\begingroup$ What's the definition of a free field? Is it defined to be a field with that Lagrangian density? This is how I've seen it defined. Which would make it a tautology... $\endgroup$
    – JLA
    Commented Feb 1, 2015 at 21:57
  • $\begingroup$ @JLA: Jep, a free field is by definition a field that satisfies Klein-Gordon/has this Lagrangian density. It's called "free" because it is precisely the Lagrangian of a particle without forces/of a field without (self-)interactions. $\endgroup$
    – ACuriousMind
    Commented Feb 1, 2015 at 22:02
  • $\begingroup$ Without forces? The analogous classical mechanics Lagrangian is $\frac{\dot{x}^2}{2}-kx^2\,,$ equivalently $F=kx\,,$ so doesn't the term involving $m$ result in a force of some kind? Is the definition of noninteracting "linear"? $\endgroup$
    – JLA
    Commented Feb 1, 2015 at 22:09
  • $\begingroup$ @JLA: Ooops, sorry, forget the bit about that particle. The definition of "non-interacting" is that you do not get interaction vertices for the Feynman diagrams. If you view it classically as a theory on a lattice, it means that you have a simple lattice of harmonic oscillators damped by the mass $m$. $\endgroup$
    – ACuriousMind
    Commented Feb 1, 2015 at 22:29
  • $\begingroup$ Ok thanks. One more question: Is there a sense in which quantum fields satisfy the Klein-Gordon equation? I mean since they are operators. I've read that particles satisfy the Klein-Gordon equation, something about their mass being on-shell, but it's not really clear to me how to go from a field to a particle, some confusion resulting by the presence of real particles and virtual particles. Though I know that light (or electromagnetic waves) satisfies the Klein-Gordon equation with $m=0\,.$ $\endgroup$
    – JLA
    Commented Feb 2, 2015 at 0:41

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