The Variational Principle

Also known as the Ritz theorem.

From \(H|n\rangle = E_n|n\rangle\). We have \(|\psi\rangle = \sum_n c_n|n\rangle\). Thus, \(\langle \psi|H|\psi\rangle = \sum_n |c_n|^2E_n\) and \(\langle \psi|\psi\rangle = \sum_n |c_n|^2\). The weighted average is then, \(\langle H\rangle = \frac{\langle \psi|H|\psi\rangle}{\langle\psi|\psi\rangle} = \frac{\sum_n|c_n|^2E_n}{\sum_n|c_n|^2}\). Note that, \(\langle \psi|H|\psi\rangle\geq E_0\sum_n|c_n|^2\) since that would assume all energies are the ground state at best and there exist larger energies at worst. Thus, \(\langle H\rangle \geq E_0\).

  1. Choose a trial ket \(|\psi(\alpha)\rangle\), \(\alpha\) is called a Ritz parameter
  2. Compute \(\langle H\rangle(\alpha)\)
  3. Minimize with respect to \(\alpha\), \(\left.\frac{\partial \langle H\rangle}{\partial \alpha}\right|_{\alpha=\alpha_0}=0\).
  4. Thus, \(\langle H\rangle(\alpha_0)\) is the lowest upper bound for the energy of the ground state for that trial function.

Example: Simple Harmonic Oscillator

\(H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2x^2\)

  1. Let \(\psi_\alpha(x) = \exp(-\alpha x^2)\).
  2. Then, \(\langle \psi_\alpha| H|\psi_\alpha\rangle = \int_\mathbb{R}\exp(-\alpha x^2)\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{m\omega^2 x^2}{2}\right)\exp(-\alpha x^2)dx = \left(\frac{\hbar^2}{2m}\alpha + \frac{m\omega^2}{8\alpha}\right)\sqrt{\frac{\pi}{2\alpha}}\). \(\langle \psi_\alpha|\psi_\alpha\rangle = \sqrt{\frac{\pi}{2\alpha}}\).

Thus, \(\langle H\rangle(\alpha) = \frac{\hbar^2}{2m}\alpha + \frac{m\omega^2}{8\alpha}\).

  1. \(\frac{\partial \langle H\rangle}{\partial \alpha} = \frac{\hbar^2}{2m} - \frac{m\omega^2}{8\alpha_0^2} = 0\) implies \(\alpha_0^2 = \frac{m^2\omega^2}{4\hbar^2} = \left(\frac{m\omega}{2\hbar}\right)^2\).
  2. The lowest upper bound is then, \(\langle H\rangle = \frac{\hbar\omega}{2}\).

Example: SHO with a different Trial Function

  1. \(\psi_\alpha = \frac{1}{x^2+\alpha}\) with \(\alpha>0\).
  2. Then, \(\langle \psi_\alpha|H|\psi_\alpha\rangle = \frac{\hbar^2}{8m}\frac{\pi}{\alpha^{5/2}} + \frac{m\omega^2\pi}{4\sqrt{\alpha}}\). \(\langle \psi_\alpha|\psi_\alpha\rangle = \frac{\pi}{2\alpha\sqrt{\alpha}}\). \(\langle H\rangle(\alpha) = \frac{\hbar^2}{4m\alpha} + \frac{m\omega^2\alpha}{2}\).
  3. \(\partial_\alpha \langle H\rangle = -\frac{\hbar^2}{4m\alpha_0^2} + \frac{m\omega^2}{2} = 0\Rightarrow\alpha_0^2 = \frac{\hbar^2}{2m\omega}\).
  4. \(\frac{\hbar\omega}{\sqrt{2}}\).

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:17