# Quantum Mechanics

## Schrodinger equation

The general case of the Schrodinger equation is written as

$\hat H \psi = E \psi$


There are various ways of expressing this equation and here are some common expressions:

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

$\hbar = \frac{h}{2 \pi}$
$h \,$ is Planck's constant
$m \,$ is the mass of the particle
$\psi\,$ is the (complex-valued) wavefunction that we want to find
$V\left(x\right)\,$ is a function describing the potential at each point x and
$E\,$ is the energy, a real number.

## The particle in a 1-dimensional box

• Particle in a box

For the 1-dimensional case in the x direction, the time-independent SchrÃ¶dinger equation can be written as:

$- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^2} \psi = E \psi$

The general solutions are:

$\psi = A e^{ikx} + B e ^{-ikx} \;\;\;\;\;\; E = \frac{k^2 \hbar^2}{2m}$

$\psi = C \sin kx + D \cos kx \;$ (exponential rewrite)

The presense of the walls of the box restricts the acceptable solutions to the wavefunction. At each wall :

$\psi = 0 \; \mathrm{at} \;\; x = 0,\; x = L$

Consider x = 0

• sin 0 = 0, cos 0 = 1. To satisfy $\psi = 0 \;$ D = 0 (cos term is removed)

Now Consider: $\psi = C \sin kx \;$

• at X=L, $\psi = C \sin kL \;$
• If C = 0 then $\psi =0 \;$ for all x and would conflict with Born interpretation
• therefore sin kL must be satisfied by
$kL = n \pi \;\;\;\; n = 1,2 \;$

By substituting k in the energy equation, we get:

$E_n = \frac {n^2 \hbar^2 \pi^2}{2mL} = \frac {n^2 h^2}{8mL^2}$