San José State University

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The Transformation of the Time-Dependent
Schroedinger Equation into Two Linked Equations
Involving Only Real Dependent Variables

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^ = (h ²/(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)] = (h ²/(2m))[(∂²r/∂x²)+ 2i (∂r/∂x)(∂θ/∂x) − (∂²θ/∂x²)]
+ Vr
A separation of real and imaginary terms gives

−h r(∂θ/∂t) = (h ²/(2m))[(∂²r/∂x²) − (∂²θ/∂x²)] + Vr
h [(∂r/∂t) = 2(h ²/(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) = (h ²/(2rm))[(∂²r/∂x²) − (∂²Ω/∂x²)/h ²] + V
(∂r/∂t) = (1/m)(∂r/∂x)(∂Ω/∂x)
The first of these reduces to

(∂Ω/∂t) = (h ²/(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.)
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