Wave functions and Schrödinger equation

Exploring Quantum Physics – Week 1 Lecture 1

Early Experiments

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.

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


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s