The time-dependent Schroedinger equation is

ih(∂ψ/∂t) = H^ψ

where the wave function ψ is complex.

The Cartesian Representation
of the Wave Function

Let ψ(z, t)=φ(z, t)+iγ(z, t), where z denote the location vector and t is time, Then

ih(∂ψ/∂t) = ih(∂φ/∂t) − h(∂γ/∂t) = H^φ + iH^γ

The separation of the real and imaginary terms gives

h(∂φ/∂t) = H^γ
h(γ/∂t) = − H^φ

Let Ψ denote the column vector with components φ and γ. The above system can be symbolically represented in matrix form as

(∂Ψ/∂t) = MΨ

where M is the matrix of operators

Polar Representation of
the Wave Function ψ

Let ψ=r·exp(iθ). Then

(∂ψ/∂t) = (∂r/∂t)exp(iθ) + r·exp(iθ)(i∂θ/∂t))
= exp(iθ)[(∂r/∂t) + ir(∂θ/∂t)]

The RHS of Schroedinger's equation, H^(r·exp(iθ), cannot be separated at the general level because H^ usually involves the Laplacian operator which for a product of functions is complicated; i.e.,

∇²(f·g) = (∇²f)g + 2(∇f·∇g) +f(∇²g)

Consider the case of a particle of mass m in a potential field V(z). Its Hamiltonian operator is

H^ = (h²/(2m))∇² + V

For one spatial dimension x

∇² = (∂²/∂x²)

Then

∇²(r·exp(iθ)) = (∂(∂(r·exp(iθ)/∂x)/∂x)
= ((∂((∂r/∂x)exp(iθ) + r·exp(iθ)i(∂θ/∂x))/∂x
= (∂²r/∂x²)exp(iθ) + 2i (∂r/∂x)(∂θ/∂x)exp(iθ) − (∂²θ/∂x²)exp(iθ)

Note that each term has a factor of exp(iθ). When the substitutions into the Schroedinger equation are made and the common factor of exp(iθ) cancelled the result is

ih[(∂r/∂t) + ir(∂θ/∂t)] = (h²/(2m))[(∂²r/∂x²)+ 2i (∂r/∂x)(∂θ/∂x) − (∂²θ/∂x²)] + Vr

A separation of real and imaginary terms gives

−hr(∂θ/∂t) = (h²/(2m))[(∂²r/∂x²) − (∂²θ/∂x²)] + Vr
h[(∂r/∂t) = 2(h²/(2m))(∂r/∂x)(∂θ/∂x)
which reduce to
(∂θ/∂t) = (h/(2rm))[(∂²r/∂x²) − (∂²θ/∂x²)] + V/h
(∂r/∂t) = (h/m)(∂r/∂x)(∂θ/∂x)

Now consider θ=Ω/h. This replacement results in the previous two equations becoming

(∂Ω/∂t) = (h²/(2rm))[(∂²r/∂x²) − (∂²Ω/∂x²)/h²] + V
(∂r/∂t) = (1/m)(∂r/∂x)(∂Ω/∂x)

The first of these reduces to

(∂Ω/∂t) = (h²/(2rm))(∂²r/∂x²) − (1/(2rm))(∂²Ω/∂x²) + V

A nice touch is to let r=√ρ. Then ψψ*=ρ and thus ρ is the probability density function for the physical system. With this substitution

the equation
(∂r/∂t) = (1/m)(∂r/∂x)(∂Ω/∂x)
becomes
(1/2√ρ)(∂rρ∂t) = (1/(2m√ρ))(∂ρ/∂x)(∂Ω/∂x)
which reduces to
(∂ρ/∂t) = (1/m)(∂ρ/∂x)(∂Ω/∂x)

(To be continued.) .