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Derivation of the Schrödinger Equation 📂Quantum Mechanics

Derivation of the Schrödinger Equation

Overview

  • Time-independent Schrodinger equation

$$ H\psi=\left(-\frac{\hbar^{2}}{2m}\frac{ d ^{2} }{ d x^{2} }+V\right)\psi=E\psi \\ H\psi=\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+V\right)\psi=E\psi $$

  • Time-dependent Schrodinger equation

$$ i\hbar\frac{ \partial \psi}{ \partial t}=\left(-\frac{\hbar^{2}}{2m}\frac{ \partial ^{2} }{\partial x^{2} }+V\right)\psi \\ i\hbar\frac{ \partial \psi}{ \partial t}=\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+V\right)\psi $$

The Schrodinger equation is a partial differential equation related to the energy, position, and time of a complex wave function. In simpler terms, it’s like the following in classical mechanics:

$$ F=ma $$

Using this, we can calculate the wave function and energy of the wave function in various potential situations. First, the one-dimensional wave function concerning time and position, where the wave number is $k$ and the angular frequency is $\omega$, is as follows:

$$ \psi (x,t)=e^{i(kx-\omega t)} \tag{1} $$

To simplify the equation, we omit the preceding constants. The De Broglie relation is as follows:

$$ \lambda=\frac{h}{p} $$

$$ k=\frac{p}{\hbar} \tag{2} $$

From Planck’s black-body radiation and Einstein’s photoelectric effect, the following relationship is obtained:

$$ E=h\nu=\hbar \omega \tag{3} $$

$\nu=\frac{\omega}{2\pi}$ is the frequency of the particle. Quantum mechanics is described through wave functions, operators, and eigenvalue equations, so we will derive the Schrodinger equation using these.

Time-independent Schrodinger Equation

The goal is to obtain the energy operator $E_{op}$, which has the wave function $\psi$ as its eigenfunction and the energy $E$ of $\psi$ as its eigenvalue. Since the energy of the particle is kinetic energy + potential energy,

$$ E=\frac{p^{2}}{2m}+V $$

Due to the De Broglie relation $(2)$, it follows that $p=k\hbar$

$$ E=\frac{\hbar^{2}k^{2}}{2m}+V $$

Multiplying both sides by the wave function $\psi$,

$$ \frac{\hbar^{2}k^{2}}{2m}\psi+V\psi=E\psi \tag{4} $$

Since the wave function is $(1)$,

$$ \frac{d^{2}\psi }{dx^{2} }=-k^{2}\psi\quad \implies\quad -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi }{dx^{2} }=\frac{\hbar^{2}k^{2}}{2m}\psi $$

Thus, $(4)$ is

$$ \begin{align*} &&-\frac{\hbar^{2}}{2m}\frac{ d ^{2}\psi}{ dx^{2} }+V\psi=E\psi \\ \implies &&\left(-\frac{\hbar^{2}}{2m}\frac{ d ^{2}}{ dx^{2} }+V\right)\psi=E\psi \end{align*} $$

This equation is called the time-independent Schrodinger equation. Additionally, the energy operator that obtains energy is simply denoted as $H$ and is called the Hamiltonian. In three dimensions, the Hamiltonian and Schrodinger equation are as follows:

$$ H=-\frac{\hbar^{2}}{2m}\nabla^{2}+V $$

$$ \left(-\frac{\hbar^{2}}{2m}\nabla^{2}+V\right)\psi=E\psi \tag{5} $$

Using $H$, the time-independent Schrodinger equation can be simplified as follows:

$$ H\psi=E\psi $$

Time-dependent Schrodinger Equation

According to $(3)$, the energy of the particle is expressed by the angular frequency $\omega$ and Planck constant $\hbar$. The angular frequency can be obtained when the wave function $(1)$ is differentiated with respect to time. $$ \frac{ \partial \psi}{ \partial t }=-i\omega\psi $$ Therefore, $$ E\psi=\hbar \omega \psi=i\hbar\frac{ \partial \psi}{ \partial t } $$ By substituting this into $(5)$, the time-dependent Schrodinger equation is obtained. $$ i\hbar\frac{ \partial \psi}{ \partial t}=\left(-\frac{\hbar^{2}}{2m}\frac{ \partial ^{2} }{ \partial x^{2} }+V\right)\psi \\ i\hbar\frac{ \partial \psi}{ \partial t}=\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+V\right)\psi $$