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

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Thayer Watkins
Silicon Valley
USA

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

 | 0 H^/h | | −H^/h 0 |

## 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

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.) .