Ciro Santilli $$Sponsor Ciro$$ 中国独裁统治 China Dictatorship 新疆改造中心、六四事件、法轮功、郝海东、709大抓捕、2015巴拿马文件 邓家贵、低端人口、西藏骚乱

# Derivation of the Klein-Gordon equation

The Klein-Gordon equation directly uses a more naive relativistic energy guess of squared.
But since this is quantum mechanics, we feel like making 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: $$H=∇2+m2 (3)$$
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: $$Hψ=i∂t∂ψ​ (4)$$ taking the Hamiltonian twice leads to: $$H2ψ=−∂2t∂2ψ​ (5)$$
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.