Derive Schrödinger's time-independent wave equation for matter waves. (asked in MU June 2025 KT)

Answer: \(\nabla^2\psi+\dfrac{8\pi^2m}{h^2}(E-V)\psi=0\) (in one dimension \(\dfrac{d^2\psi}{dx^2}+\dfrac{8\pi^2m}{h^2}(E-V)\psi=0\)).

Step 1 – Classical wave equation. Let \(\psi(x,y,z,t)\) be the wave function of the matter wave travelling with velocity \(v\). It obeys the general wave equation

\[\nabla^2\psi=\frac1{v^2}\frac{\partial^2\psi}{\partial t^2}\qquad\ldots(1)\]

Step 2 – Separate the time part. For a wave of fixed frequency \(\nu\), write \(\psi=\psi_0(x,y,z)\,e^{-i\omega t}\) with \(\omega=2\pi\nu\). Then

\[\frac{\partial^2\psi}{\partial t^2}=-\omega^2\psi\qquad\ldots(2)\]

Putting (2) in (1): \(\nabla^2\psi=-\dfrac{\omega^2}{v^2}\psi\). Since \(\dfrac{\omega}{v}=\dfrac{2\pi\nu}{\nu\lambda}=\dfrac{2\pi}{\lambda}\),

\[\nabla^2\psi+\frac{4\pi^2}{\lambda^2}\psi=0\qquad\ldots(3)\]

Step 3 – Bring in de Broglie's relation. \(\lambda=\dfrac{h}{mv}\Rightarrow\dfrac1{\lambda^2}=\dfrac{m^2v^2}{h^2}\).

Total energy \(E=\) kinetic \(+\) potential \(=\frac12mv^2+V\), so \(m^2v^2=2m(E-V)\) and

\[\frac1{\lambda^2}=\frac{2m(E-V)}{h^2}\qquad\ldots(4)\]

Step 4 – Substitute (4) in (3):

\[\nabla^2\psi+\frac{4\pi^2\cdot2m(E-V)}{h^2}\psi=0\ \Rightarrow\ \boxed{\nabla^2\psi+\frac{8\pi^2m}{h^2}(E-V)\psi=0}\]

With \(\hbar=h/2\pi\) this is \(\nabla^2\psi+\dfrac{2m}{\hbar^2}(E-V)\psi=0\). It does not contain time, so it is the time-independent (steady-state) Schrödinger equation, used for problems such as a particle in a box where \(E\) and \(V\) do not change with time.