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The wave associated with a particle moving along x can be written as

\[ \psi(x,t)=A\,e^{i(kx-\omega t)}=\psi_0(x)\,e^{-i\omega t} \]

where \(\psi_0(x)=Ae^{ikx}\).

Differentiating twice with respect to x:

\[ \frac{\partial^2\psi}{\partial x^2}=-k^2\psi=-\frac{4\pi^2}{\lambda^2}\psi \quad (k=2\pi/\lambda) \]

so \(\dfrac{1}{\lambda^2}=-\dfrac{1}{4\pi^2\psi}\dfrac{\partial^2\psi}{\partial x^2}\) ... (1)

de Broglie: \(\lambda=\dfrac{h}{mv}\), so \(\dfrac{1}{\lambda^2}=\dfrac{m^2v^2}{h^2}\) ... (2)

Total energy \(E=\tfrac12mv^2+V\), so \(m^2v^2=2m(E-V)\) ... (3)

From (1), (2) and (3):

\[ -\frac{1}{4\pi^2\psi}\frac{\partial^2\psi}{\partial x^2}=\frac{2m(E-V)}{h^2} \]

\[ \boxed{\frac{\partial^2\psi}{\partial x^2}+\frac{8\pi^2m}{h^2}(E-V)\psi=0} \quad\text{or}\quad \frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V)\psi=0 \]

This is the one-dimensional time-independent Schrödinger equation. In three dimensions: \(\nabla^2\psi+\dfrac{2m}{\hbar^2}(E-V)\psi=0\). For a free particle (V = 0): \(\nabla^2\psi+\dfrac{2mE}{\hbar^2}\psi=0\).

It applies to stationary states, where the potential and energy do not depend on time. ψ must be single-valued, finite and continuous, and \(|\psi|^2\) gives the probability density.

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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