Variational Estimate for the Helium Atom

The Helium Atom is the classic first serious test of the variational method. The exact two-electron problem is not separable because of electron-electron repulsion, but a simple screened-charge trial wavefunction already gives a surprisingly good ground-state energy. Here we derive the effective-charge estimate for a point nucleus of infinite mass in nonrelativistic Coulomb theory. The atomic discussion adds the spectrum, singlet–triplet structure and precision corrections.

Required background. The Variational Principle gives the bound; Trial Wavefunctions and Tensor Products explain the ansatz.

This page uses atomic units:

$$\hbar=m_e=e=4\pi\epsilon_0=1.$$

Energies are in Hartree. For helium, $Z=2$ and

$$H = - \frac12\nabla_1^2 - \frac12\nabla_2^2 - \frac{Z}{r_1} - \frac{Z}{r_2} + \frac{1}{r_{12}},$$

where

$$r_{12}=|\mathbf r_1-\mathbf r_2|.$$

Independent-Electron Trial State

Use a product of two normalized hydrogenic $1s$ orbitals with an adjustable effective charge $\zeta$:

$$\phi_\zeta(r) = \left(\frac{\zeta^3}{\pi}\right)^{1/2} e^{-\zeta r}.$$

The spatial trial state is

$$\Psi_\zeta(\mathbf r_1,\mathbf r_2) = \phi_\zeta(r_1)\phi_\zeta(r_2).$$

For the physical helium ground state, this spatial state is paired with an antisymmetric spin singlet so that the total two-electron state is antisymmetric, as required by the Pauli exclusion principle. The energy expectation below concerns only the spatial Hamiltonian.

Effective Nuclear Charge

The parameter $\zeta$ is interpreted as an effective nuclear charge. If electron-electron repulsion were ignored, each electron would see the full charge $Z=2$. Repulsion screens the nucleus, so one expects

$$1<\zeta<2.$$

For one electron in $\phi_\zeta$,

$$\left\langle - \frac12\nabla^2 \right\rangle = \frac{\zeta^2}{2}, \qquad \left\langle - \frac{Z}{r} \right\rangle = -Z\zeta.$$

For two independent electrons, the one-body contribution is

$$E_{\mathrm{one-body}}(\zeta) = \zeta^2-2Z\zeta.$$

For helium, this is

$$E_{\mathrm{one-body}}(\zeta) = \zeta^2-4\zeta.$$

Electron-Electron Repulsion

The repulsion expectation in this trial state is the standard integral

$$\left\langle \frac{1}{r_{12}} \right\rangle = \int d^3r_1d^3r_2\, \frac{ |\phi_\zeta(r_1)|^2 |\phi_\zeta(r_2)|^2 }{ |\mathbf r_1-\mathbf r_2| }.$$

A useful reduction makes the coefficient transparent. The angular average of $1/|\mathbf r_1-\mathbf r_2|$ is $1/\max(r_1,r_2)$. With radial probability $p_\zeta(r)=4\zeta^3r^2e^{-2\zeta r}$, symmetry between the two radii gives

$$\left\langle\frac1{r_{12}}\right\rangle =2\int_0^\infty\frac{p_\zeta(r)}r\,dr \int_0^r p_\zeta(s)\,ds.$$

The inner integral is $1-e^{-2\zeta r}(1+2\zeta r+2\zeta^2r^2)$. Integrating the remaining exponential moments yields

$$\left\langle \frac{1}{r_{12}} \right\rangle = \frac{5}{8}\zeta.$$

The variational energy is therefore

$$E(\zeta) = \zeta^2 -2Z\zeta + \frac58\zeta.$$

For helium, $Z=2$:

$$E(\zeta) = \zeta^2 - \frac{27}{8}\zeta.$$

Optimized Estimate

Minimize:

$$\frac{dE}{d\zeta} = 2\zeta-\frac{27}{8} = 0.$$

Thus

$$\zeta_\star = \frac{27}{16} = 1.6875.$$

The optimized energy is

$$E_{\mathrm{var}} = - \frac{729}{256} = -2.84765625 \ \text{Hartree}.$$

This is an upper bound on the exact nonrelativistic ground-state energy. It is much better than using unscreened hydrogenic orbitals with $\zeta=2$, which gives

$$E(2)=-2.75 \ \text{Hartree}.$$

The improvement comes from allowing the electrons to screen each other on average.

What the Ansatz Captures

The screened-charge trial state captures:

• nuclear attraction and kinetic-energy balance;
• average screening of each electron by the other;
• the correct spherical symmetry of the ground state;
• the spin singlet structure through a symmetric spatial state.

It misses:

• explicit electron-electron correlation;
• angular correlation between electron positions;
• the electron-electron cusp condition;
• orbital relaxation beyond a single exponential and configuration mixing;
• the electron–nucleus cusp, because the exponent is $\zeta$ rather than $Z$.

Finite nuclear mass, relativity and QED modify the Hamiltonian itself; improving this ansatz alone cannot include them.

Explicit factors depending on $r_{12}$ describe correlation. The difference between this one-exponent estimate and the exact energy is not entirely the conventional correlation energy: that term means the difference between the exact and complete-basis Hartree–Fock energies. Our restricted orbital has additional mean-field error.

Exercises

A helium-like ion and its threshold

For nuclear charge $Z$, optimize $E(\zeta)=\zeta^2-2Z\zeta+5\zeta/8$ over $\zeta>0$. Does this family prove that $H^-$ ($Z=1$) is bound against losing one electron?

Solution

For $Z>5/16$, $\zeta_*=Z-5/16$ and $E_{\mathrm{var}}=-(Z-5/16)^2$. The one-electron ionization threshold is $-Z^2/2$. At $Z=1$, the estimate is $-121/256\simeq-0.4727$, above $-1/2$: it fails to prove binding, and does not prove nonbinding. A richer correlated trial state is needed. For $0<Z\le5/16$, this family has infimum zero as $\zeta\to0^+$, not a positive-width minimum.

Why screening does not describe correlation

For the product ansatz, compute the conditional probability density of the second electron’s position given the first. What changes when a factor $F(r_{12})$ is included?

Solution

The conditional density is simply $|\phi_\zeta(\mathbf r_2)|^2$, independent of $\mathbf r_1$. A common optimized exponent changes the average cloud but not this independence. A nonconstant $F(r_{12})$ generally makes the normalized conditional density depend on the first position and can suppress close electron encounters.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• A. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover, 1996.