---
title: "Derivation of time independent Schrodinger equation"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:schrodinger-equation-derivation"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/schrodinger-equation-derivation
site: BCT Engineering
---

# Derivation of time independent Schrodinger equation

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2075 Asoj, 8 marks
- Video: [Time Independent Schrodinger Wave Equation Derivation AKTU BTech 1st Year Eng. Physics BAS101 BAS201](https://www.youtube.com/watch?v=YGzi6mN2ZBg) — EduGlue AKTU
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/schrodinger-equation-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

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.

## 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.

## Answer

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

#### Step 1: Classical wave
A wave travelling in the $+x$ direction is

$$
y = A\sin 2\pi\left(ft - \frac{x}{\lambda}\right)
$$

Differentiate with respect to $x$:

$$
\frac{\partial y}{\partial x} = -\frac{2\pi}{\lambda}A\cos2\pi\left(ft - \frac{x}{\lambda}\right)
$$

$$
\frac{\partial^2y}{\partial x^2} = -\frac{4\pi^2}{\lambda^2}A\sin2\pi\left(ft - \frac{x}{\lambda}\right) = -\frac{4\pi^2}{\lambda^2}\,y
$$

So

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

#### Step 2: Replace $y$ by the matter wave $\psi$
For a particle, the wave is described by the wave function $\psi$. Separating out the time part ($\psi(x,t) = \psi(x)\,e^{-j\omega t}$), the space part obeys the same equation:

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

#### Step 3: Use de Broglie's relation
$\lambda = \dfrac{h}{p}$, so $\dfrac{1}{\lambda^2} = \dfrac{p^2}{h^2}$.

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

$$
p^2 = 2m(E - V) \quad\Rightarrow\quad \frac{1}{\lambda^2} = \frac{2m(E - V)}{h^2}
$$

#### Step 4: Substitute in (2)

$$
\frac{d^2\psi}{dx^2} + \frac{4\pi^2\cdot2m(E - V)}{h^2}\psi = 0
$$

$$
\frac{d^2\psi}{dx^2} + \frac{8\pi^2m}{h^2}(E - V)\psi = 0
$$

Since $\hbar = h/2\pi$, $\dfrac{8\pi^2m}{h^2} = \dfrac{2m}{\hbar^2}$:

$$
\frac{d^2\psi}{dx^2} + \frac{2m}{\hbar^2}(E - V)\psi = 0
$$

or, in the usual form,

$$
-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi
$$

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

$$
\nabla^2\psi + \frac{2m}{\hbar^2}(E - V)\psi = 0
$$

Here $\psi$ is the wave function, $m$ the mass, $E$ the total energy, $V$ the potential energy and $\hbar = h/2\pi$. It is used for problems where $V$ does not depend on time, such as the potential well, the barrier (tunneling) and the hydrogen atom.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Differentiating y twice with respect to x gives d^2y/dx^2 equal to:
   A. (4 pi^2/lambda^2) y
   B. -(2 pi/lambda) y
   C. -(4 pi^2/lambda^2) y
   D. -(4 pi^2 f^2) y
   Answer: C. Each differentiation brings a factor of 2pi/lambda, and the sine returns with a minus sign. (Key point: Starting wave)

2. Which relation links the wavelength to the particle?
   A. lambda = h/p
   B. lambda = p/h
   C. lambda = hf
   D. lambda = E/p
   Answer: A. de Broglie: lambda = h/p. (Key point: de Broglie)

3. How is 1/lambda^2 expressed using the total energy?
   A. 2mE/h^2
   B. 2m(E + V)/h^2
   C. m(E - V)/h^2
   D. 2m(E - V)/h^2
   Answer: D. p^2 = 2m(E - V), so 1/lambda^2 = p^2/h^2 = 2m(E-V)/h^2. (Key point: Energy relation)

4. The time-independent Schrödinger equation in 1-D is:
   A. -hbar^2/2m d^2psi/dx^2 + V psi = E psi
   B. hbar^2/2m d^2psi/dx^2 + V psi = E psi
   C. -hbar^2/2m dpsi/dx + V psi = E psi
   D. d^2psi/dx^2 = 2m(E - V) psi
   Answer: A. d^2psi/dx^2 + (2m/hbar^2)(E-V)psi = 0 rearranges to this form. (Key point: Final form)

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