Derivation of the Klein-Gordon equation

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 = \nabla^{2} + m^{2}

(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:

H ψ = i \frac{\partial ψ}{\partial t}

(4)

taking the Hamiltonian twice leads to:

H^{2} ψ = - \frac{\partial ^{2} ψ}{\partial ^{2} t}

(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.

Derivation of the Klein-Gordon equation

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