Variational Ground-State Energy of Helium

Derivation

Variational Ground-State Energy of Helium

D-305 Home PU-306 Threads energy · matter Depends on variational-principle, Exchange Symmetry and the Ortho/Para Splitting

Statement

For the helium atom (nuclear charge \(Z=2\), two electrons) we estimate the non-relativistic ground-state energy by applying the variational principle to a trial wavefunction built from two identical hydrogenic \(1s\) orbitals of adjustable effective charge \(\zeta\). Working in atomic units (energies in Hartree \(E_h\), lengths in Bohr \(a_0\)), the energy functional evaluates to \(E(\zeta)=\zeta^2-2Z\zeta+\tfrac{5}{8}\zeta\); minimising over \(\zeta\) gives the optimal screened charge \(\zeta^\star=Z-\tfrac{5}{16}=\tfrac{27}{16}\) and the upper-bound energy \(E_{\min}=-(\zeta^\star)^2=-\left(\tfrac{27}{16}\right)^2 E_h\approx-2.848\,E_h\approx-77.5\ \text{eV}\).

Why it matters

Helium is the simplest atom whose Schrödinger equation cannot be solved in closed form: the electron–electron term \(1/r_{12}\) couples the two coordinates irremovably. It is therefore the canonical proving ground for approximate many-body methods, and the screened-charge variational estimate is the first calculation every student meets that turns a qualitative idea — each electron partially shields the nucleus from the other — into a quantitative number good to about two percent.

The single variational parameter \(\zeta\) carries real physics: its optimum \(\zeta^\star=27/16\) is smaller than the bare charge \(Z=2\) by exactly \(5/16\), a screening constant that emerges from the mathematics rather than being assumed. The method also demonstrates, concretely, that a variational energy is always an upper bound, and that relaxing an orbital (letting \(\zeta\) float away from the naive value \(Z\)) lowers the energy toward the truth.

Assumptions

Non-relativistic, fixed point nucleus.Dropping this — relativistic kinematics, spin–orbit coupling, or finite nuclear mass — shifts the true energy by tens of meV; the clamped-nucleus, Coulomb-only Hamiltonian below would no longer be the exact target the variational bound refers to. Product trial form \(\psi=\phi_\zeta(\mathbf r_1)\phi_\zeta(\mathbf r_2)\) with hydrogenic \(1s\) orbitals.If the trial space excluded such correlated-but-uncorrelated products, no analytic screened-charge optimum would exist; a richer basis (Hylleraas \(r_{12}\) terms) lowers the bound further but forfeits the closed form. Spatial part symmetric under \(\mathbf r_1\leftrightarrow\mathbf r_2\), spins in a singlet.The ground state has both electrons in the same spatial \(1s\) orbital, which is only Pauli-allowed if the spin state is the antisymmetric singlet (from two-electron-exchange-symmetry). Drop the singlet and the spatial function would have to be antisymmetric, forbidding double occupancy and changing the whole calculation. No explicit inter-electronic coordinate \(r_{12}\) in \(\psi\).The trial function contains zero radial/angular correlation, so it misses the Coulomb cusp at \(r_{12}\to0\); the estimate is systematically too high (too weakly bound) and cannot be pushed below \(-2.848\,E_h\) within this family.

Derivation

\[ \hat H=-\tfrac{1}{2}\nabla_1^2-\tfrac{1}{2}\nabla_2^2-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{r_{12}} \]

The clamped-nucleus helium Hamiltonian in atomic units: kinetic energy of each electron, Coulomb attraction to the charge-\(Z\) nucleus, and mutual repulsion \(1/r_{12}\), with \(r_{12}=|\mathbf r_1-\mathbf r_2|\). A

\[ \psi(\mathbf r_1,\mathbf r_2)=\phi_\zeta(\mathbf r_1)\,\phi_\zeta(\mathbf r_2),\qquad \phi_\zeta(\mathbf r)=\sqrt{\frac{\zeta^{3}}{\pi}}\;e^{-\zeta r} \]

Trial state: a product of two normalised hydrogenic \(1s\) orbitals with a shared variational exponent \(\zeta\) (an "effective charge"). The spatial function is symmetric, so the spins form a singlet and Pauli exclusion is satisfied (two-electron-exchange-symmetry). Each \(\phi_\zeta\) is already normalised, so \(\langle\psi|\psi\rangle=1\). B

\[ E(\zeta)=\langle\psi|\hat H|\psi\rangle=2\langle T\rangle+2\langle V_{\mathrm{ne}}\rangle+\langle V_{\mathrm{ee}}\rangle \]

By the variational principle, \(E(\zeta)\ge E_0\) for any \(\zeta\). Because \(\psi\) is a normalised product and the one-body operators act symmetrically on the two identical orbitals, the six terms collapse into twice a single kinetic expectation, twice a single nuclear-attraction expectation, plus one repulsion integral. A

\[ \langle T\rangle=\tfrac12\!\int|\nabla\phi_\zeta|^2\,d^3r=\frac{\zeta^{2}}{2} \]

The kinetic energy of a hydrogenic \(1s\) orbital scales as the square of its exponent. Concretely, \(\phi_\zeta\) is the exact ground state of a hydrogenic ion of charge \(\zeta\), whose kinetic energy is \(\zeta^2/2\) by the virial theorem; direct evaluation of the gradient integral confirms it. B

\[ \left\langle \frac{1}{r}\right\rangle=\int\frac{|\phi_\zeta|^2}{r}\,d^3r=\zeta \quad\Rightarrow\quad \langle V_{\mathrm{ne}}\rangle=-Z\left\langle\tfrac1r\right\rangle=-Z\zeta \]

Using \(\int_0^\infty 4\pi r^2\,(\zeta^3/\pi)e^{-2\zeta r}\,r^{-1}\,dr=\zeta\). Each electron's attraction to the true charge \(Z\) is \(-Z\zeta\); note the attraction is set by the true \(Z\), while the orbital's spread is set by \(\zeta\). B

\[ \langle V_{\mathrm{ee}}\rangle=\iint\frac{|\phi_\zeta(\mathbf r_1)|^2|\phi_\zeta(\mathbf r_2)|^2}{r_{12}}\,d^3r_1\,d^3r_2=\frac{5}{8}\,\zeta \]

Expand \(1/r_{12}\) in Legendre multipoles \(\sum_\ell (r_<^\ell/r_>^{\ell+1})P_\ell(\cos\theta_{12})\); the spherical \(|\phi_\zeta|^2\) kills every term except \(\ell=0\). The remaining radial double integral of two exponential charge clouds evaluates to \(\tfrac58\zeta\) — the classic Coulomb self-energy of overlapping \(1s\) densities. C

\[ E(\zeta)=2\cdot\frac{\zeta^2}{2}-2Z\zeta+\frac{5}{8}\zeta=\zeta^{2}-2Z\zeta+\frac{5}{8}\zeta \]

Assemble the three evaluated pieces from Steps 4–6. Grouping the linear terms, \(E(\zeta)=\zeta^2-\left(2Z-\tfrac58\right)\zeta\) — a simple downward parabola in \(\zeta\). A

\[ Z=2:\qquad E(\zeta)=\zeta^{2}-\left(4-\tfrac58\right)\zeta=\zeta^{2}-\frac{27}{8}\,\zeta \]

Insert the physical nuclear charge of helium, \(Z=2\), only now that the algebra is complete. A

\[ \frac{dE}{d\zeta}=2\zeta-\frac{27}{8}=0\quad\Longrightarrow\quad \zeta^\star=\frac{27}{16}=Z-\frac{5}{16} \]

Stationarity of the variational functional. The second derivative \(d^2E/d\zeta^2=2>0\) confirms a minimum. The optimum effective charge is the bare charge reduced by a screening constant \(\sigma=5/16=0.3125\). A

\[ E_{\min}=E(\zeta^\star)=(\zeta^\star)^2-2\zeta^\star\cdot\zeta^\star=-(\zeta^\star)^2=-\left(\frac{27}{16}\right)^{2}E_h \]

Substitute \(\zeta^\star\) back. Because the minimum of \(\zeta^2-b\zeta\) at \(\zeta=b/2\) is \(-b^2/4=-(\zeta^\star)^2\), the optimised energy is simply minus the square of the effective charge. A

Result

\[ \boxed{\;\zeta^\star=Z-\frac{5}{16}=\frac{27}{16},\qquad E_{\min}=-\left(Z-\frac{5}{16}\right)^{2}=-\left(\frac{27}{16}\right)^{2}E_h\approx-2.848\,E_h\approx-77.5\ \text{eV}\;} \]

Reading. Each helium electron does not feel the full \(+2\) nuclear charge; the other electron shields \(5/16\) of a unit, leaving an effective \(+27/16\approx1.69\). Using this self-consistently softened orbital gives a ground-state energy of \(-77.5\) eV, an upper bound lying \(1.5\) eV (about \(1.9\%\)) above the experimental \(-79.0\) eV. Letting \(\zeta\) relax from the naive value \(2\) recovers roughly two-thirds of that gap.

Units check. In atomic units \(\zeta\) is dimensionless and every term in \(E(\zeta)=\zeta^2-2Z\zeta+\tfrac58\zeta\) carries the unit \(E_h\) (Hartree). Restoring SI, \(E_h=\dfrac{m_e e^4}{(4\pi\varepsilon_0)^2\hbar^2}=27.211\ \text{eV}\), so \(-(27/16)^2 E_h=-2.8477\times27.211\ \text{eV}=-77.49\ \text{eV}\): an energy, as required.

Limiting cases

No repulsion (\(\langle V_{\mathrm{ee}}\rangle\to0\)): the \(5/8\) term vanishes, \(\zeta^\star\to Z\), and \(E\to-Z^2=-4\,E_h\) — two independent hydrogenic electrons, no screening.
Frozen orbital (fix \(\zeta=Z=2\), skip the optimisation): \(E(2)=4-\tfrac{27}{8}\cdot2=-2.75\,E_h=-74.8\) eV, worse by \(2.7\) eV — the price of not letting the electron relax.
Large \(Z\) isoelectronic ions (He-like C\(^{4+}\), etc.): the fractional screening \(5/(16Z)\) shrinks, so the bare-charge estimate becomes increasingly accurate and repulsion becomes a small perturbation on the hydrogenic \(-Z^2\).
\(Z=1\) (H\(^-\)): \(\zeta^\star=11/16\), \(E_{\min}=-(11/16)^2=-0.473\,E_h=-12.86\) eV, which lies above \(-13.6\) eV (H \(+\) free \(e^-\)) — this simple trial wrongly predicts H\(^-\) is unbound.

Breaks when

Weakly bound / diffuse two-electron systems (H\(^-\)). When the true binding relies almost entirely on angular and radial correlation, an uncorrelated product of identical orbitals cannot capture it: the method predicts H\(^-\) unbound, whereas it is in fact bound by \(\approx0.75\) eV. The variational bound is valid but too crude to see the physics.
Heavy nuclei where relativity matters. The Hamiltonian in Step 1 omits kinetic-relativistic, spin–orbit, and QED shifts. For high-\(Z\) He-like ions these grow like \(Z^4\) (or faster) and eventually exceed the correlation error, so the non-relativistic bound stops describing reality even though it remains a correct bound on the non-relativistic Hamiltonian.
Excited or open-shell configurations. The single-exponent \(1s^2\) ansatz encodes only the singlet ground state; for the \(1s2s\) triplet, near-degenerate states, or autoionising resonances the product form has the wrong structure and the optimisation is meaningless.

Failure modes

Using \(Z\) in the repulsion but forgetting to keep it in \(\zeta\). The nuclear-attraction coefficient is the true charge \(2Z=4\); the repulsion coefficient \(5/8\) multiplies \(\zeta\), not \(Z\). Mixing them gives the wrong linear term and a spurious \(\zeta^\star\).
Setting \(dE/d\zeta=0\) but plugging \(\zeta=2\) into \(E\). Optimising then evaluating at the un-optimised charge defeats the purpose and gives \(-2.75\,E_h\) while claiming it is the variational minimum.
Double counting the electrons. Writing \(4\langle T\rangle\) or forgetting the factor of \(2\) on the nuclear term. There are two electrons: \(2\langle T\rangle\), \(2\langle V_{\mathrm{ne}}\rangle\), and exactly one pair repulsion.
Believing \(-77.5\) eV can be below \(-79.0\) eV. A variational energy is an upper bound; a "result" beneath the experimental value signals an arithmetic error or a non-normalised trial function, not a better estimate.
Forgetting normalisation of \(\phi_\zeta\). The prefactor \(\sqrt{\zeta^3/\pi}\) depends on \(\zeta\); using an unnormalised orbital corrupts every expectation value and the location of the minimum.

Discussion

The single number \(\sigma=5/16\) is the whole story in miniature. It says each electron, sitting in a cloud that on average overlaps the other, sees the nucleus dressed by roughly a third of an electronic charge. This is the microscopic origin of Slater's empirical screening rules and of the "effective nuclear charge" concept that runs through the entire periodic table. That the number falls out of a first-principles minimisation, with no fitting, is what makes the calculation persuasive rather than merely descriptive.

Two theorems underwrite the result. The variational principle guarantees \(E(\zeta)\ge E_0\), turning "guess a wavefunction" into "obtain a rigorous ceiling." The virial theorem explains the tidy form \(E_{\min}=-(\zeta^\star)^2\): at the optimum the trial state behaves like the exact ground state of a fictitious hydrogenic ion of charge \(\zeta^\star\), for which \(E=-\zeta^2/2\) per electron and \(\langle T\rangle=-\tfrac12\langle V\rangle\). The optimisation is precisely the demand that the trial state satisfy the virial relation.

The residual \(1.5\) eV that the method misses is the correlation energy — the part of the electron–electron interaction that no independent-particle (product) wavefunction can represent, because it lives in the explicit dependence on \(r_{12}\) and in the Coulomb cusp as the electrons meet. Hylleraas famously added terms linear in \(r_{12}\) and drove the energy to within microhartrees of experiment; modern configuration-interaction and coupled-cluster methods are, in spirit, systematic ways of restoring exactly this correlation on top of a screened-orbital reference.

A subtler point: the optimised \(\zeta^\star=27/16\) is not the value that makes the orbital resemble the true one-electron density most closely, nor the one that satisfies the cusp condition at the nucleus (\(\zeta=Z=2\)). The variational optimum trades a distorted short-range shape for a better average energy, because the energy functional weights the whole density, not any single point. This is a recurring theme in electronic-structure theory — an energy-optimal orbital is a compromise, and different observables are extremised by different \(\zeta\); one should never read too much physical literalness into a variational parameter beyond the energy it was tuned to.

Common misconceptions. The effective charge \(\zeta^\star\) is not a measured or "real" charge and it is not the same as the charge felt near the nucleus; it is an energy-optimised orbital exponent. And "screening" here is a variational bookkeeping outcome, not a claim that one electron sits statically between the other and the nucleus.

Worked examples

Helium ground-state energy and percentage error. Evaluate the optimised bound in eV and compare with experiment (\(E_{\exp}=-79.01\) eV, the sum of the two ionisation energies \(24.59+54.42\) eV). A

\[ \zeta^\star=\frac{27}{16}=1.6875,\qquad E_{\min}=-(\zeta^\star)^2\,E_h=-2.84766\,E_h \]

\[ E_{\min}=-2.84766\times27.2114\ \text{eV}=-77.49\ \text{eV} \]

\[ \text{error}=\frac{|{-77.49}-({-79.01})|}{|-79.01|}=\frac{1.52}{79.01}=1.9\% \]

\[ E_{\min}=-77.5\ \text{eV},\quad 1.9\%\ \text{above experiment (a valid upper bound).} \]

Reading. A one-parameter trial function reaches within \(2\%\) of the true energy; the \(1.5\) eV shortfall is the correlation energy.

Units check. Multiplying a dimensionless Hartree count by \(E_h=27.211\) eV yields eV. The error is a pure ratio.

The isoelectronic ion Li\(^{+}\). Repeat for the two-electron ion with true charge \(Z=3\) and compare with \(E_{\exp}=-198.09\) eV (\(=-(I_2+I_3)=-(75.64+122.45)\) eV of lithium). B

\[ E(\zeta)=\zeta^2-2Z\zeta+\tfrac58\zeta=\zeta^2-\left(6-\tfrac58\right)\zeta=\zeta^2-\frac{43}{8}\zeta \]

\[ \zeta^\star=Z-\frac{5}{16}=\frac{43}{16}=2.6875,\qquad E_{\min}=-(\zeta^\star)^2=-\left(\frac{43}{16}\right)^2=-7.2227\,E_h \]

\[ E_{\min}=-7.2227\times27.2114\ \text{eV}=-196.55\ \text{eV},\qquad \text{error}=\frac{1.54}{198.09}=0.78\% \]

\[ \zeta^\star=\tfrac{43}{16},\quad E_{\min}=-196.6\ \text{eV}\ \ (0.78\%\ \text{high}). \]

Reading. The fractional screening \(5/(16Z)\) is smaller for larger \(Z\), so the method is more accurate for Li\(^+\) than for He: \(0.78\%\) versus \(1.9\%\).

Units check. Same Hartree-to-eV conversion; \(\zeta^\star\) dimensionless, energy in eV.

Problems

(A) Frozen-charge estimate. Compute the helium energy with the un-optimised choice \(\zeta=Z=2\) and quote it in eV. How much does optimisation to \(\zeta^\star\) gain?
Solution
\(E(2)=2^2-\tfrac{27}{8}(2)=4-6.75=-2.75\,E_h=-2.75\times27.211=-74.83\) eV. Optimising to \(\zeta^\star=27/16\) gives \(-77.49\) eV, a gain of \(-77.49-(-74.83)=-2.66\) eV, i.e. \(2.66\) eV lower. Relaxing the orbital recovers about \(2.66/(79.01-74.83)=64\%\) of the missing binding.
(A) Screening constant. From the general result \(\zeta^\star=Z-\tfrac{5}{16}\), state the screening constant \(\sigma\) for helium and interpret it physically.
Solution
\(\sigma=Z-\zeta^\star=5/16=0.3125\). Each electron sees an effective nuclear charge \(1.6875\) rather than \(2\); the partner electron screens \(0.3125\) of a unit charge on average. This is the variational, first-principles analogue of Slater's empirical screening constants.
(B) General minimum. For any downward-shifted parabola \(E(\zeta)=\zeta^2-b\zeta\) with \(b>0\), show the minimum lies at \(\zeta^\star=b/2\) with \(E_{\min}=-(\zeta^\star)^2\), and identify \(b\) for He-like ions.
Solution
\(dE/d\zeta=2\zeta-b=0\Rightarrow\zeta^\star=b/2\); \(d^2E/d\zeta^2=2>0\) (minimum). Then \(E_{\min}=(b/2)^2-b(b/2)=-b^2/4=-(\zeta^\star)^2\). For a two-electron ion \(b=2Z-\tfrac58\), so \(\zeta^\star=Z-\tfrac{5}{16}\) and \(E_{\min}=-\left(Z-\tfrac{5}{16}\right)^2 E_h\), reproducing He (\(Z=2\)) and Li\(^+\) (\(Z=3\)).
(B) The hydride ion. Apply the method to H\(^-\) (\(Z=1\)). Compute \(\zeta^\star\) and \(E_{\min}\) in eV, and explain why comparing to \(-13.61\) eV shows the trial function fails here.
Solution
\(\zeta^\star=1-\tfrac{5}{16}=\tfrac{11}{16}=0.6875\); \(E_{\min}=-(11/16)^2=-0.47266\,E_h=-12.86\) eV. A hydrogen atom plus a free electron has energy \(-13.61+0=-13.61\) eV. Since \(-12.86>-13.61\), this trial predicts H\(^-\) would spontaneously eject an electron — it is estimated unbound. In reality H\(^-\) is bound by \(\approx0.75\) eV, but that binding comes entirely from electron correlation, which an uncorrelated product orbital cannot represent. The bound is correct; the ansatz is simply too poor.
(C) Deriving the attraction integral. Show that \(\langle 1/r\rangle=\zeta\) for the normalised orbital \(\phi_\zeta=\sqrt{\zeta^3/\pi}\,e^{-\zeta r}\), and hence that \(\langle V_{\mathrm{ne}}\rangle=-Z\zeta\) per electron.
Solution
\(\displaystyle\left\langle\frac1r\right\rangle=\int|\phi_\zeta|^2\frac1r\,d^3r=\frac{\zeta^3}{\pi}\int_0^\infty\!\!\frac{e^{-2\zeta r}}{r}\,4\pi r^2\,dr=4\zeta^3\int_0^\infty r\,e^{-2\zeta r}\,dr.\) Using \(\int_0^\infty r\,e^{-2\zeta r}dr=1/(2\zeta)^2=1/(4\zeta^2)\), this is \(4\zeta^3\cdot\tfrac{1}{4\zeta^2}=\zeta\). Therefore the nuclear-attraction energy of one electron is \(\langle-Z/r\rangle=-Z\langle1/r\rangle=-Z\zeta\), confirming Step 5.