Klein-Gordan's equation: free particle

The idea here is to take into account the relativistic effect on a fast moving free particle, in the Schrodinger's equation. Klein-Gordan's equation is just the time-dependent relativistic Schrodinger's equation.

The classical expression of the total energy E of a free particle of mass at rest m and momentum p is:

E = p²/2m

The relativistic expression of the total energy E of a free particle of mass at rest m_o and momentum p is:

E² = p²c²+ m_o²c⁴

The quantum expression of the momentum is:

P = - i ħ ∇

The time-dependent non-relativistic Schrodinger's equation for this particle is:

(p²/2m) ψ = E ψ = iħ∂ψ/∂t

We replace the momentum p by its quantum mechanical operator P = - i ħ ∇, and E by its relativistic expression : [p²c²+ m_o²c⁴]^1/2, and get:

iħ∂ψ/∂t = [(- i ħ ∇)²c²+ m_o²c⁴]^1/2 ψ

To get rid of the square root operation, we apply the operator iħ∂ψ/∂t two times and get:

[iħ∂ /∂t]² ψ = [(- i ħ ∇)²c²+ m_o²c⁴] ψ

That is:
- ħ²∂²ψ/∂t²ψ = - ħ² ∇²c²ψ + m_o²c⁴ψ
or
∂²ψ/∂t² - ∇²c²ψ + (m_o²c⁴/ħ²)ψ = 0

(1/c²)∂² /∂t² ψ - ∇²ψ + (m_o²c²/ħ²)ψ = 0

The operator ☐ = (1/c²)∂²/∂t²- ∇²
is called d'Alembertian operator.

The above expression can be written as:

☐ψ + (m_o²c²/ħ²)ψ = 0

Klein-Gordan equation is:

(1/c²)∂²/∂t² ψ - ∇²ψ + (m_o²c²/ħ²)ψ = 0

or

☐ψ + (m_oc/ħ)²ψ = 0

with

☐ = (1/c²)∂²/∂t²- ∇²