Quantum Mechanics

From Chemistry

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 Schrödinger 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}