# 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. $$\frac{\hbar\omega}{\sqrt{2}}$$.

Created: 2024-05-30 Thu 21:17

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