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 Schrodinger 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:

$H=β_{2}+m_{2}$

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 Schrodinger equation which looks like:
taking the Hamiltonian twice leads to:

$HΟ=iβtβΟβ$

$H_{2}Ο=ββ_{2}tβ_{2}Οβ$

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.

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