{c}

The Klein-Gordon equation directly uses a more naive <relativistic energy> guess of $p^2 + m^2$ squared.

But since this is <quantum mechanics>, we feel like making $p$ into the "<momentum operator>", just like in the <Schrödinger equation>.

But we don't really know how to apply the momentum operator twice, because it is a <gradient>, so the first application goes from a scalar field to the vector field, and the second one...

So we just cheat and try to use the <laplace operator> instead because there's some squares on it:
$$
H = \laplacian{} + m^2
$$

But then, we have to avoid taking the square root to reach a first derivative in time, because we don't know how to take the square root of that operator expression.

So the Klein-Gordon equation just takes the approach of using this squared Hamiltonian instead.

Since it is a Hamiltonian, and comparing it to the <Schrödinger equation> which looks like:
$$
H \psi = i \pdv{\psi}{t}
$$
taking the Hamiltonian twice leads to:
$$
H^2 \psi = - \pdv{^2 \psi}{^2 t}
$$

We can contrast this with the <Dirac equation>, which instead attempts to explicitly construct an operator which squared coincides with the relativistic formula: <derivation of the Dirac equation>.


Derivation of the Klein-Gordon equation