Part VI, Chapter 3 | Page 1 of 3

The Variational Principle

Upper bounds on the ground state energy through optimized trial wave functions

The variational method is a powerful approximation technique that works even when perturbation theory fails. It requires no small parameter and provides a rigorous upper bound on the true ground state energy. The key insight is that any trial wave function, no matter how crude, gives an energy that is guaranteed to be at least as high as the true ground state energy.

The Variational Theorem

For any normalized trial state $|\psi\rangle$, the expectation value of the Hamiltonian provides an upper bound on the ground state energy:

$$\boxed{\frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle} \geq E_0}$$

where $E_0$ is the true ground state energy. Equality holds if and only if $|\psi\rangle$ is the exact ground state. This deceptively simple result is one of the most useful tools in quantum mechanics.

Proof from Completeness

The proof is elegant and instructive. Expand the trial state in the exact energy eigenstates $|n\rangle$:

$$|\psi\rangle = \sum_n c_n |n\rangle, \quad \hat{H}|n\rangle = E_n|n\rangle, \quad E_0 \leq E_1 \leq E_2 \leq \cdots$$

Compute the expectation value:

$$\langle\psi|\hat{H}|\psi\rangle = \sum_n |c_n|^2 E_n$$

Since every $E_n \geq E_0$, we can bound each term:

$$\langle\psi|\hat{H}|\psi\rangle = \sum_n |c_n|^2 E_n \geq E_0 \sum_n |c_n|^2 = E_0$$

The last step uses normalization: $\sum_n |c_n|^2 = \langle\psi|\psi\rangle = 1$. The inequality is strict unless $c_n = 0$ for all $n \neq 0$, meaning $|\psi\rangle$ is exactly the ground state.

The Variational Method in Practice

The practical recipe is:

Step 1: Choose a trial wave function with adjustable parameters

$$|\psi(\alpha_1, \alpha_2, \ldots, \alpha_k)\rangle$$

Step 2: Compute the energy functional

$$E(\alpha_1, \ldots, \alpha_k) = \frac{\langle\psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}$$

Step 3: Minimize with respect to all parameters

$$\frac{\partial E}{\partial \alpha_i} = 0 \quad \text{for } i = 1, 2, \ldots, k$$

Step 4: The minimum gives the best variational estimate

$$E_{\min} = E(\alpha_1^*, \ldots, \alpha_k^*) \geq E_0$$

The quality of the estimate depends entirely on how well the functional form of the trial wave function can approximate the true ground state.

Example: Hydrogen Ground State with Gaussian Trial

To illustrate the method, consider the hydrogen atom with a Gaussian trial function instead of the exact exponential:

$$\psi_{\text{trial}}(r) = A e^{-\alpha r^2}$$

Normalization: $A = (2\alpha/\pi)^{3/4}$. Now compute the kinetic and potential energy expectation values.

Kinetic energy:

$$\langle T\rangle = -\frac{\hbar^2}{2m}\langle\psi|\nabla^2|\psi\rangle = \frac{3\hbar^2\alpha}{2m}$$

Potential energy:

$$\langle V\rangle = -\frac{e^2}{4\pi\epsilon_0}\left\langle\frac{1}{r}\right\rangle = -\frac{e^2}{4\pi\epsilon_0}\sqrt{\frac{2\alpha}{\pi}} \cdot 2 = -\frac{e^2}{4\pi\epsilon_0}\frac{2}{\sqrt{\pi}}\sqrt{2\alpha}$$

Total energy functional:

$$E(\alpha) = \frac{3\hbar^2\alpha}{2m} - \frac{e^2}{4\pi\epsilon_0}\sqrt{\frac{8\alpha}{\pi}}$$

Minimization:

$$\frac{dE}{d\alpha} = 0 \quad \Rightarrow \quad \alpha^* = \frac{8me^4}{9\pi^2\hbar^4(4\pi\epsilon_0)^2}$$

Result:

$$E_{\min} = -\frac{4}{3\pi}\frac{me^4}{2\hbar^2(4\pi\epsilon_0)^2} = -\frac{8}{3\pi} \times 13.6\text{ eV} \approx -11.5\text{ eV}$$

The exact answer is $-13.6$ eV. Our Gaussian trial gives $-11.5$ eV, about 85% of the correct value. The variational bound is respected ($-11.5 > -13.6$), and the error comes from the Gaussian's inability to reproduce the cusp at $r=0$ and the exponential tail of the true wave function.

Example: Harmonic Oscillator with Gaussian

For contrast, try a Gaussian for the harmonic oscillator:

$$\psi(x) = Ae^{-\alpha x^2}$$

The energy functional evaluates to:

$$E(\alpha) = \frac{\hbar^2\alpha}{2m} + \frac{m\omega^2}{8\alpha}$$

Minimizing:

$$\frac{dE}{d\alpha} = \frac{\hbar^2}{2m} - \frac{m\omega^2}{8\alpha^2} = 0 \quad \Rightarrow \quad \alpha^* = \frac{m\omega}{2\hbar}$$

$$E_{\min} = \frac{\hbar\omega}{2}$$

Exact result! This is not a coincidence -- the true ground state of the harmonic oscillator is a Gaussian. When the trial function family includes the exact solution, the variational method finds it.

Choosing Good Trial Functions

A good trial function should:

Respect the symmetries of the problem (parity, angular momentum, etc.)
Satisfy the boundary conditions (vanish where required)
Have correct asymptotic behavior (decay at infinity for bound states)
Be flexible enough to approximate the true wave function well
Be simple enough that the integrals can be evaluated analytically or efficiently

Common trial function families:

Slater-type orbitals: $r^n e^{-\zeta r}$ (good for atoms, correct cusp)
Gaussian-type orbitals: $r^n e^{-\alpha r^2}$ (easier integrals, widely used in quantum chemistry)
Jastrow factors: $e^{-u(r_{12})}$ (explicit electron-electron correlation)
Polynomial times exponential: $(1 + \beta r + \gamma r^2)e^{-\alpha r}$ (systematic improvement)

Looking Ahead

On the next page, we tackle the helium atom -- the simplest two-electron system and one of the most important applications of the variational method. We will see how a single variational parameter (effective nuclear charge) captures the essential physics of electron-electron repulsion.

← Degenerate PT Helium Ground State →

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