schrodinger-equation-solutions

CASE 1: Infinite square well (Particle in a box)

Choosing a 1D “box” as the potential where

$$V(r)=\{\begin{array}{ll}0& when0\le x\le a\\ \infty & otherwise\end{array}$$

Outside the box, $\psi (x,t)=0$

We solve the differential equation with $V(x)=0$ between $x=0$ and $x=a$ for the eigenfunctions and eigenvalues of $\mathbb{H}$

Time independent Schrodinger equation for infinite square well =

$$-\frac{{\hslash }^2}{2m}\frac{{\partial }^2\psi (x)}{\partial x^2}=E\psi (x)$$

Solutions for the infinite square well

Discrete Energy values (eigenvalues of $\mathbb{H}$) =

$$E_n=\frac{n^2{\psi }^2{\hslash }^2}{2m{a}^2}$$

Allowed states (eigenfunctions of $\mathbb{H}$) =

$${\varphi }_n(x)=\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi }{a}x\right)$$

Stationary states = The solutions to the time independent Schrodinger equation can be used to write the time dependent stationary wave function =

$$\psi (x,t)=\sum _{n=1}^{\infty }{c}_n\sqrt{\frac{2}{a}}\sin \left(\frac{n\pi }{a}x\right)e^{-i(n^2{\pi }^2{\hslash }^2/2m{a}^2)t}$$

where

$$c_n=\sqrt{\frac{2}{a}}{\int }_0^a\sin \left(\frac{n\pi }{a}x\right)\psi (x,0)dx$$

method of solving = Differential equation

$$\frac{{\partial }^2\psi }{\partial x^2}=-k^2\psi$$

where $\varphi (n)=A\sin (kx)B\cos (kx)$

general solution =

$$\varphi (x)=A\sin \left(\frac{n\pi }{a}x\right)$$

stationary states =

$$\psi (x,t)=\sum _{n=1}^{\infty }{c}_m{\varphi }_m(x)e^{-iEt/\hslash }$$

where

$$c_m=\sum _n{c}_n⟨m|n⟩=⟨m|\psi (0)⟩$$

CASE 2: Harmonic oscillator

The harmonic oscillator is represented by the force equation $F=-kx$ which gives the potential energy $V(x)=-k{x}^2/2$

Many potentials reduce to this form for small values of x, so many phenomena can be understood using the harmonic potential

For quantum systems, we want to solve the Schrodinger equation using the potential

$$V(x)=\frac{1}{2}m{\omega }^2{X}^2,\omega =\sqrt{\frac{k}{m}}$$

Time independent Schrodinger equation for harmonic oscillator =

$$-\frac{{\hslash }^2}{2m}\frac{{\partial }^2\psi (x)}{\partial x^2}+\frac{1}{2}m{\omega }^2{X}^2\psi (x)=E\psi (x)$$

This can be solved using the same method as for CASE 1 Infinite square well (Particle in a box) but can be solved quicker using harmonic-oscillator-ladder-operators

Solutions to the harmonic oscillator

The equation has an infinite number of discrete solutions

ground state wave function =

$${\psi }_0(x)={\frac{m\omega }{\pi \hslash }}^{1/4}{e}^{-(m\omega /2\hslash )x^2}$$

ground state energy =

$$E_0=\frac{1}{2}\hslash \omega$$

High energy solutions =

$${\psi }_n(x)=A_n(a_{+}{)}^n{\psi }_0(x)$$

where $A_n$ is the normalization constant

$$E_n=\left(n+\frac{1}{2}\right)\hslash \omega$$