Skip to main content

Derivation of time independent Schrodinger equation

Short answer

Begin with y = A sin 2π(ft - x/λ), derive d²y/dx² + (4π²/λ²)y = 0. Replace y with matter wave ψ as spatial part, use de Broglie λ = h/p and total energy E = p²/(2m) + V, to obtain the time-independent equation -ħ²/(2m) d²ψ/dx² + V ψ = E ψ.

Derive the time independent Schrodinger's equation, starting with classical wave equation, y = A sin 2π(ft − x/λ), where notations have their usual meanings.

The time-independent Schrödinger equation is obtained by combining the classical wave equation with de Broglie's relation λ=h/p\lambda = h/p. It describes the spatial part ψ(x)\psi(x) of a particle's wave function in a potential V(x)V(x).

Step 1: Classical wave

A wave travelling in the +x+x direction is

y=Asin⁡2π(ft−xλ)y = A\sin 2\pi\left(ft - \frac{x}{\lambda}\right)

Differentiate with respect to xx:

∂y∂x=−2πλAcos⁡2π(ft−xλ)\frac{\partial y}{\partial x} = -\frac{2\pi}{\lambda}A\cos2\pi\left(ft - \frac{x}{\lambda}\right) ∂2y∂x2=−4π2λ2Asin⁡2π(ft−xλ)=−4π2λ2 y\frac{\partial^2y}{\partial x^2} = -\frac{4\pi^2}{\lambda^2}A\sin2\pi\left(ft - \frac{x}{\lambda}\right) = -\frac{4\pi^2}{\lambda^2}\,y

So

∂2y∂x2+4π2λ2y=0(1)\frac{\partial^2y}{\partial x^2} + \frac{4\pi^2}{\lambda^2}y = 0 \qquad (1)

Step 2: Replace yy by the matter wave ψ\psi

For a particle, the wave is described by the wave function ψ\psi. Separating out the time part (ψ(x,t)=ψ(x) e−jωt\psi(x,t) = \psi(x)\,e^{-j\omega t}), the space part obeys the same equation:

d2ψdx2+4π2λ2ψ=0(2)\frac{d^2\psi}{dx^2} + \frac{4\pi^2}{\lambda^2}\psi = 0 \qquad (2)

Step 3: Use de Broglie's relation

λ=hp\lambda = \dfrac{h}{p}, so 1λ2=p2h2\dfrac{1}{\lambda^2} = \dfrac{p^2}{h^2}.

The total energy is E=p22m+VE = \dfrac{p^2}{2m} + V, so

p2=2m(E−V)⇒1λ2=2m(E−V)h2p^2 = 2m(E - V) \quad\Rightarrow\quad \frac{1}{\lambda^2} = \frac{2m(E - V)}{h^2}

Step 4: Substitute in (2)

d2ψdx2+4π2⋅2m(E−V)h2ψ=0\frac{d^2\psi}{dx^2} + \frac{4\pi^2\cdot2m(E - V)}{h^2}\psi = 0 d2ψdx2+8π2mh2(E−V)ψ=0\frac{d^2\psi}{dx^2} + \frac{8\pi^2m}{h^2}(E - V)\psi = 0

Since ℏ=h/2π\hbar = h/2\pi, 8π2mh2=2mℏ2\dfrac{8\pi^2m}{h^2} = \dfrac{2m}{\hbar^2}:

d2ψdx2+2mℏ2(E−V)ψ=0\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E - V)\psi = 0

or, in the usual form,

−ℏ22md2ψdx2+V(x)ψ=Eψ-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi

This is the time-independent Schrödinger equation in one dimension. In three dimensions d2/dx2d^2/dx^2 is replaced by ∇2\nabla^2:

∇2ψ+2mℏ2(E−V)ψ=0\nabla^2\psi + \frac{2m}{\hbar^2}(E - V)\psi = 0

Here ψ\psi is the wave function, mm the mass, EE the total energy, VV the potential energy and ℏ=h/2π\hbar = h/2\pi. It is used for problems where VV does not depend on time, such as the potential well, the barrier (tunneling) and the hydrogen atom.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Electrical Engineering Material chapter 1 questions →