Variational method (quantum mechanics)

Suppose we are given a Hilbert space and a Hermitian operator over it called the Hamiltonian H. Ignoring complications about continuous spectra, we look at the discrete spectrum of H and the corresponding eigenspaces of each eigenvalue λ (see spectral theorem for Hermitian operators for the mathematical background):

${\displaystyle \langle \psi _{\lambda _{1}}\mid \psi _{\lambda _{2}}\rangle =\delta _{\lambda _{1}\lambda _{2}}}$

where ${\displaystyle \delta _{i,j}}$ is the Kronecker delta

${\displaystyle \delta _{ij}={\begin{cases}0&{\text{if }}i\neq j,\\1&{\text{if }}i=j.\end{cases}}}$

and the Hamiltonian is related to λ through the typical eigenvalue relation

${\displaystyle {\hat {H}}\left|\psi _{\lambda }\right\rangle =\lambda \left|\psi _{\lambda }\right\rangle .}$

Physical states are normalized, meaning that their norm is equal to 1. Once again ignoring complications involved with a continuous spectrum of H, suppose it is bounded from below and that its greatest lower bound is E₀. Suppose also that we know the corresponding state |ψ⟩. The expectation value of H is then

${\displaystyle {\begin{aligned}\left\langle \psi \mid H\mid \psi \right\rangle &=\sum _{\lambda _{1},\lambda _{2}\in \mathrm {Spec} (H)}\left\langle \psi |\psi _{\lambda _{1}}\right\rangle \left\langle \psi _{\lambda _{1}}|H|\psi _{\lambda _{2}}\right\rangle \left\langle \psi _{\lambda _{2}}|\psi \right\rangle \\&=\sum _{\lambda \in \mathrm {Spec} (H)}\lambda \left|\left\langle \psi _{\lambda }\mid \psi \right\rangle \right|^{2}\geq \sum _{\lambda \in \mathrm {Spec} (H)}E_{0}\left|\left\langle \psi _{\lambda }\mid \psi \right\rangle \right|^{2}=E_{0}\end{aligned}}}$

Obviously, if we were to vary over all possible states with norm 1 trying to minimize the expectation value of H, the lowest value would be E₀ and the corresponding state would be an eigenstate of E₀. Varying over the entire Hilbert space is usually too complicated for physical calculations, and a subspace of the entire Hilbert space is chosen, parametrized by some (real) differentiable parameters α_i (i = 1, 2, ..., N). The choice of the subspace is called the ansatz. Some choices of ansatzes lead to better approximations than others, therefore the choice of ansatz is important.

Let's assume there is some overlap between the ansatz and the ground state (otherwise, it's a bad ansatz). We still wish to normalize the ansatz, so we have the constraints

${\displaystyle \left\langle \psi (\mathbf {\alpha } )\mid \psi (\mathbf {\alpha } )\right\rangle =1}$

and we wish to minimize

${\displaystyle \varepsilon (\mathbf {\alpha } )=\left\langle \psi (\mathbf {\alpha } )|H|\psi (\mathbf {\alpha } )\right\rangle .}$

This, in general, is not an easy task, since we are looking for a global minimum and finding the zeroes of the partial derivatives of ε over all α_i is not sufficient. If ψ (α) is expressed as a linear combination of other functions (α_i being the coefficients), as in the Ritz method, there is only one minimum and the problem is straightforward. There are other, non-linear methods, however, such as the Hartree–Fock method, that are also not characterized by a multitude of minima and are therefore comfortable in calculations.

There is an additional complication in the calculations described. As ε tends toward E₀ in minimization calculations, there is no guarantee that the corresponding trial wavefunctions will tend to the actual wavefunction. This has been demonstrated by calculations using a modified harmonic oscillator as a model system, in which an exactly solvable system is approached using the variational method. A wavefunction different from the exact one is obtained by use of the method described above.

Although usually limited to calculations of the ground state energy, this method can be applied in certain cases to calculations of excited states as well. If the ground state wavefunction is known, either by the method of variation or by direct calculation, a subset of the Hilbert space can be chosen which is orthogonal to the ground state wavefunction.

${\displaystyle \left|\psi \right\rangle =\left|\psi _{\text{test}}\right\rangle -\left\langle \psi _{\mathrm {gr} }\mid \psi _{\text{test}}\right\rangle \left|\psi _{\text{gr}}\right\rangle }$

The resulting minimum is usually not as accurate as for the ground state, as any difference between the true ground state and ${\displaystyle \psi _{\text{gr}}}$ results in a lower excited energy. This defect is worsened with each higher excited state.

In another formulation:

${\displaystyle E_{\text{ground}}\leq \left\langle \phi |H|\phi \right\rangle .}$

This holds for any trial φ since, by definition, the ground state wavefunction has the lowest energy, and any trial wavefunction will have energy greater than or equal to it.

Proof: φ can be expanded as a linear combination of the actual eigenfunctions of the Hamiltonian (which we assume to be normalized and orthogonal):

${\displaystyle \phi =\sum _{n}c_{n}\psi _{n}.\,}$

Then, to find the expectation value of the Hamiltonian:

${\displaystyle {\begin{aligned}&\left\langle \phi |H|\phi \right\rangle \\={}&\left\langle \sum _{n}c_{n}\psi _{n}|H|\sum _{m}c_{m}\psi _{m}\right\rangle \\={}&\sum _{n}\sum _{m}\left\langle c_{n}^{*}\psi _{n}|E_{m}|c_{m}\psi _{m}\right\rangle \\={}&\sum _{n}\sum _{m}c_{n}^{*}c_{m}E_{m}\left\langle \psi _{n}\mid \psi _{m}\right\rangle \\={}&\sum _{n}|c_{n}|^{2}E_{n}.\end{aligned}}}$

Now, the ground state energy is the lowest energy possible, i.e. ${\displaystyle E_{n}\geq E_{\text{ground}}}$. Therefore, if the guessed wave function φ is normalized:

${\displaystyle \left\langle \phi |H|\phi \right\rangle \geq E_{\text{ground}}\sum _{n}|c_{n}|^{2}=E_{\text{ground}}.\,}$

The helium atom consists of two electrons with mass m and electric charge −e, around an essentially fixed nucleus of mass M ≫ m and charge +2e. The Hamiltonian for it, neglecting the fine structure, is:

${\displaystyle H=-{\frac {\hbar ^{2}}{2m}}(\nabla _{1}^{2}+\nabla _{2}^{2})-{\frac {e^{2}}{4\pi \varepsilon _{0}}}\left({\frac {2}{r_{1}}}+{\frac {2}{r_{2}}}-{\frac {1}{|\mathbf {r} _{1}-\mathbf {r} _{2}|}}\right)}$

where ħ is the reduced Planck constant, ε₀ is the vacuum permittivity, r_i (for i = 1, 2) is the distance of the i-th electron from the nucleus, and |r₁ − r₂| is the distance between the two electrons.

If the term V_ee = e²/(4πε₀|r₁ − r₂|), representing the repulsion between the two electrons, were excluded, the Hamiltonian would become the sum of two hydrogen-like atom Hamiltonians with nuclear charge +2e. The ground state energy would then be 8E₁ = −109 eV, where E₁ is the Rydberg constant, and its ground state wavefunction would be the product of two wavefunctions for the ground state of hydrogen-like atoms:[5]

${\displaystyle \psi (\mathbf {r} _{1},\mathbf {r} _{2})={\frac {Z^{3}}{\pi a_{0}^{3}}}e^{-Z(r_{1}+r_{2})/a_{0}}.}$

where a₀ is the Bohr radius and Z = 2, helium's nuclear charge. The expectation value of the total Hamiltonian H (including the term V_ee) in the state described by ψ₀ will be an upper bound for its ground state energy. <V_ee> is −5E₁/2 = 34 eV, so <H> is 8E₁ − 5E₁/2 = −75 eV.

A tighter upper bound can be found by using a better trial wavefunction with 'tunable' parameters. Each electron can be thought to see the nuclear charge partially "shielded" by the other electron, so we can use a trial wavefunction equal with an "effective" nuclear charge Z < 2: The expectation value of H in this state is:

${\displaystyle \langle H\rangle =\left[-2Z^{2}+{\frac {27}{4}}Z\right]E_{1}}$

This is minimal for Z = 27/16 implying shielding reduces the effective charge to ~1.69. Substituting this value of Z into the expression for H yields 729E₁/128 = −77.5 eV, within 2% of the experimental value, −78.975 eV.[6]

Even closer estimations of this energy have been found using more complicated trial wave functions with more parameters. This is done in physical chemistry via variational Monte Carlo.