Lesson 26: Schrodinger’s Equation « No Free Lunch

Posted by: musubk | July 3, 2009

Lesson 26: Schrodinger’s Equation

Schrödinger’s equation describes the quantum state (or wavefunction) of a system as a function of time. To derive Schrödinger’s equation we start with the assumption that the total energy of a particle is its kinetic energy plus its potential energy:

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

We specify the wavefunction as a complex plane wave (in one dimension):

$\Psi = Ae^{-i\omega(t-\frac{x}{v})}$

Where $\omega = 2 \pi \nu$ and $v = \lambda \nu$ so that:

$\Psi = Ae^{-2 \pi i (\nu t - \frac{x}{\lambda})}$

We also assume:

$E = 2 \pi \hbar \nu$

$\lambda = \frac{2 \pi \hbar}{p}$

So that we get:

$\Psi = Ae^{-\frac{i(Et - px)}{\hbar}}$

We also take the partial in time and the second partial in space:

$\frac{\partial \Psi}{\partial t} = \frac{-iE}{\hbar}\Psi$

$\frac{\partial^2 \Psi}{\partial x^2} = -\frac{p^2}{\hbar^2}\Psi$

Therefore:

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

$E \Psi = -\frac{\hbar}{i} \frac{\partial \Psi}{\partial t}$

$p^2 \Psi = -\hbar^2 \frac{\partial^2 \Psi}{\partial x^2}$

Bringing all this together:

$i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} (\frac{\partial^2 \Psi}{\partial x^2}) + V \Psi$

In three dimensions we simply have three partials in space, which can be written as:

$i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m} \nabla^2 \Psi + V \Psi$

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