Question

The Klein-Gordon Equation is given as follows, using natural units (ℏ→1,c→1

):−∂2Ψ∂t2=−∇2Ψ+m2Ψ

The way I tried to derive this is as follows, starting from the time-dependent Schrodinger equation: i∂Ψ∂t=H^Ψ
where H^=p^22m and p^=i∇
Now, we are going to make the following substitution:

H^=p^22m→H^2=p^2+m2
Squaring the time-dependent Schrodinger equation gives:

−∂2Ψ∂t2=H^2Ψ=(p^2+m2)Ψ=(−∇2+m2)Ψ=−∇2Ψ+m2Ψ

as stated in the beginning.
However, is my derivation valid? For example, I squared the time-dependent Schrodinger equation, but this feels off to me, because I am not squaring the wave function but only the partial derivative and the H^
operator. Are there any flaws in this?

Answer 1

The attempted derivation of the Klein-Gordon Equation by squaring the Schrödinger equation is flawed. To correctly derive the Klein-Gordon Equation, one should use the relativistic energy-momentum relation with appropriate quantum mechanical substitutions.

Step-by-step explanation:

Your derivation of the Klein-Gordon Equation using the time-dependent Schrödinger equation involves a clever substitution, but there is indeed a flaw in your reasoning when you square the equation. The Schrödinger equation which you started from is a first-order differential equation in time.

However, the Klein-Gordon Equation is a second-order relativistic wave equation that derives from the relativistic energy-momentum relation E2 = p2c2 + m2c4. It incorporates both the time and space derivatives to second order, whereas in non-relativistic quantum mechanics, the Schrödinger equation involves only the first-order time derivative.

To correctly derive the Klein-Gordon equation in natural units, we should begin from the relativistic energy-momentum relation E2 = p2 + m2 (after setting ħ = 1 and c = 1), and use the quantum mechanical substitutions E → i ∂/∂t and π → -i
abla. This leads directly to the Klein-Gordon equation without the need to square the Schrödinger equation.