# Wave functions and Schrödinger equation

Exploring Quantum Physics – Week 1 Lecture 1

## Early Experiments

• Photo electric effect: beam of light hitting charged plates with a potential across them creates current.
• Expected: amount of current is independent of frequency of the light, dependent on the intensity.
• Actual: current depends on frequency of light (no current below frequency threshold), independent of intensity.
• Einstein proposed photons (nobel prize)
• Electron diffraction: (Davisson-Germer experiment)
• Crystalline structure acts as a diffraction grating, electron beam demonstrates wave interference like light.
• Experiment supports DeBroglie hypothesis:
$\lambda = h / p$, $E = \frac{hc}{\lambda} = \hbar \omega$
• For a derivation of the DeBroglie wave length, see
http://www.chip-architect.com/physics/deBroglie.pdf

## Wave Equation

Wave equation:
$\displaystyle \begin{array}{ll} u(x, t) &= A_0 \sin(kx - \omega t + \phi_0) \\ &= A_0 \Im e^{i(kx - \omega t + \phi_0)} \end{array}$
Where k is the wave number

$\displaystyle \lambda = \frac{2 \pi}{k} \qquad \omega = c k$
Where c is the wave velocity

We can represent particles as a wave (but not the other way around) via the Fourier transform:

$\displaystyle f(x) = \int dk A_k e^{ikx}$
Where $A_k$ is the “fourier harmonics” (amplitudes)

Represent a particle as a Gaussian wave with a sharp peak:

Particle as a Gaussian wave.

## The Schrödinger wave equation

$\displaystyle \Psi(x,t) = C e^{i(kx - \omega t)}$
Where $\Psi$ is the wave function, and $C$ is the constant of proportionality, ignore for now.

Since $p = \hbar k$ and $E = \hbar \omega$, can write as:
$\displaystyle \Psi(x,t) = C e^{\frac{i}{\hbar}(px - Et)}$

Taking the derivative:
$\displaystyle \frac{\partial}{\partial t} \Psi = \frac{-i}{\hbar} E \Psi \qquad \frac{\partial}{\partial x} \Psi = \frac{-i}{\hbar} p \Psi$
(or for a momentum vector: $\nabla \Psi = \frac{i}{\hbar} \vec{p} \Psi$)

For a free particle (not in a potential field), $E = \mathrm{k.e.} = \frac{mv^2}{2} = \frac{p^2}{2m}$, so:
$\displaystyle E \Psi = i \hbar \frac{\partial}{\partial t} \Psi = \frac{\vec{p}^2}{2m} \Psi = \frac{(-i \hbar \nabla)^2}{2m} \Psi = \frac{\hbar^2 \nabla^2}{2m} \Psi$

Giving the free schrödinger equation:

$\displaystyle \left[ i \hbar \frac{\partial}{\partial t} + \frac{\hbar^2 \nabla^2}{2m} \right] \Psi(\vec{r}, t) = 0$

$\displaystyle \hat{H} = \mathrm{Hamiltonian} = \mathrm{K.E.} + \mathrm{Potential} = \frac{\vec{p}^2}{2m} + V(\vec{r})$

Using the Hamiltonian for energy gives the fundamental equation of quantum physics:

$\displaystyle i \hbar \frac{\partial \Psi}{\partial t} = \hat{H} \Psi$

## Particle delocalization

Take a Gaussian wave-packet $\Psi(x, 0) = A e^{-x^2 / 2d^2}$
Solve schrödinger with this gives
$\displaystyle \left| \Psi(x,t) \right|^2 \propto \exp \left[ -\frac{x^2}{d^2 (1+t^2/\tau^2)} \right]$
where $\tau = md^2 / \hbar =$ delocalization time

$\displaystyle \Psi(x, 0) = Ae^{-\frac{x^2}{2d^2}} = \int \frac{dp}{h} \phi_p e^{\frac{i}{\hbar} px}$

Recognize the Gaussian integral: $\int_{-\infty}^\infty e^{-\alpha x^2 + \beta x} dx = \sqrt{\frac{\pi}{\alpha}} e^{\frac{\beta^2}{4\alpha}}$

Gives wave-function in momentum space:
$\phi_p = \int dx \Psi(x, 0)e^{-\frac{i}{\hbar}px} \propto e^{\frac{p^2d^2}{2\hbar^2}}$

Notice the uncertainty $\delta x \sim d$, but $\delta p \sim \hbar / d$. This is a manifestation of the Heisenberg uncertainty principle:
$\displaystyle \delta x \cdot \delta p \geq \hbar$