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Attosecond streaking uses a synchronized reference electric field to change a photoelectron's outgoing momentum. Scanning relative delay produces an energy-versus-delay distribution. A short emission interval samples only a small part of the reference cycle; a longer interval samples a range of kicks and can broaden or reshape the distribution.

For an explicit classical limit, use the electric dipole approximation and choose the magnetic vector potential $A(t)$ along the detection axis with $E(t)=-dA/dt$ and $A(\infty )=0$. Neglect the residual ion's Coulomb force. An electron released at $t_i$ with mechanical momentum $p_0$ obeysIndeed, integrating $\dot{p}=-eE=e\dot{A}$ from release to the end of the field gives the stated minus sign. The useful clock variable here is $A$, rather than the instantaneous $E$ itself. If $A$ is nearly linear across a narrow emission window with temporal standard deviation ${\sigma }_t$, and initial momentum is sharp, thenThis local mapping fails near a zero of $E$, where first-order timing sensitivity vanishes. A real inversion must include the incident bandwidth, initial photoelectron distribution, collection angles and target–field response.

For an original dimensionless example in atomic units, choose $p_0=2$, $m=e=1$ and $A(t_i)=0.10$. Then $p_f=1.9$ and $K_f=1.805$, compared with initial kinetic energy $2$. Reversing $A$ gives $K_f=2.205$: the shifts are asymmetric because of the $A^2/2$ term. A measurement that keeps only the term linear in $A$ would miss $0.005$ in both cases.

In atomic units, use the radial trial wavefunction $\psi (r)=\max (1-r/\beta ,0)$ with $\beta >0$. Its radial norm integral is $\int r^2{\psi }^{2}dr={\beta }^3/30$, its gradient contribution is $\int r^2{\psi }^{\prime 2}dr=\beta /3$, and its Coulomb contribution is $-2\int r{\psi }^{2}dr=-{\beta }^2/6$. In the hydrogen radial Rayleigh quotient, this gives the displayed energy, minimized at $\beta =4$. It is above the exact hydrogen ground-state energy $-1/2$, as required by the quantum variational principle. The derivative jumps at $r=\beta$, but the function has a square-integrable weak first derivative and is an admissible trial for the energy quadratic form. One must use that form rather than assume that a classical second derivative exists at the join.

In atomic units, the displayed integral is the direct electrostatic contribution between two normalized atomic orbitals $a,b$. It enters a two-electron energy expectation together with the exchange integral.

A product of two normalized $1s$ orbitals proportional to $e^{-\zeta r}$, combined with a singlet state, gives the fixed-nucleus helium atom trial energy $E(\zeta )={\zeta }^2-4\zeta +5\zeta /8$ in atomic units. Minimization gives $\zeta =27/16$ and $E=-729/256$ Hartree. Nuclear charge remains two in the Hamiltonian operator; varying $\zeta$ changes only the trial orbitals. The reduced orbital charge describes screening, while explicit dependence on electron separation can further lower the trial energy under the quantum variational principle.

A helium atom has two electrons bound to nuclear charge $+2e$. In the nonrelativistic fixed-nucleus approximation, using atomic units,The last term couples the electrons, so a product of independent hydrogenic orbitals is not an exact solution. The dominant ground configuration is $1s^2$. Its spatial wavefunction is symmetric, so Fermi antisymmetry requires the spin singletA triplet cannot put both electrons in the same $1s$ spatial orbital. Correlation changes the exact spatial function, including its dependence on $r_{12}$, while preserving this ground-state symmetry.

A useful helium variational effective charge takes normalized orbitals ${\varphi }_{\zeta }(\mathbf{r})=({\zeta }^3/\pi {)}^{1/2}{e}^{-\zeta r}$ for both electrons and multiplies their product by ${\chi }_{00}$. The energy in Hartree units isThe reduced effective charge expresses screening. Neglecting repulsion gives $-4E_h$, while its expectation in the unscreened $\zeta =2$ product gives $-2.75E_h$; optimizing the orbital improves that trial estimate. The exact ground energy is lower than the optimized product energy, since the product omits explicit electron correlation. This calculation keeps nuclear charge two in $H$; only the trial orbital parameter is varied.

For a singly excited helium state with distinct orthonormal orbitals $a=1s$ and $b=n\ell$, the spatial combinations areThe symmetric ${\Phi }_{+}$ pairs with the antisymmetric spin singlet, while antisymmetric ${\Phi }_{-}$ pairs with the symmetric spin triplet. With the same orbitals, their leading energies areHere $J_{ab}$ is the direct Coulomb integral and $K_{ab}$ is the exchange integral. The latter is positive for the Coulomb interaction, so the triplet is lower for this fixed-orbital comparison. singlet-triplet exchange splitting in helium arises from antisymmetry and spatial correlation in the Coulomb interaction, not from inserting a separate classical spin-spin force. Spin-singlet helium states are parahelium; spin-triplet helium states are orthohelium. The two spectra approximately form separate families because the leading electric dipole operator does not change total spin.^[ref]

The lowest $1s2s$ terms are ${}^3{S}_1$ and ${}^1{S}_0$, with the triplet lower. The $1s2p$ configuration gives ${}^3{P}_{0,1,2}^{\mathrm{o}}$ and ${}^1{P}_1^{\mathrm{o}}$, where the odd parity allows singlet $P$ to connect to the even-parity ground singlet by an electric dipole transition. Both $1s2s$ levels lack an allowed direct electric dipole decay to $1s^2$: the triplet is spin forbidden, and singlet $S\to S$ is forbidden by orbital angular momentum and parity. This explains helium metastability without implying an infinite lifetime once weaker radiative processes or collisions are allowed.

The observed first $1s2s$ excitation energies are approximately $19.82$ and $20.62\mathrm{eV}$ for the triplet and singlet, respectively. Higher singly excited Rydberg series approach the He II ground-state ionization threshold; doubly excited states above that threshold can autoionize and are not all stable bound excited levels.^[ref]

The spherical harmonic is also an eigenfunction of the $z$ component of orbital angular momentum:These are atomic units; restoring units gives $m\hslash$. The integer $m$ is the magnetic quantum number.

It represents the squared orbital angular momentum. In atomic units, $\hslash =1$, so its eigenvalues are $l(l+1)$. In ordinary units the corresponding observable is ${\hslash }^{2}l(l+1)$, with orbital angular momentum quantum number $l=0,1,2,\dots$.

For this quantum harmonic oscillator, ${\hat{H}}^{(1)}=x^4$. The Gaussian integral givesThe distinction between the coefficient and the actual first-order shift matters: the energy shift is $3\lambda /4$, so $E_0=1/2+3\lambda /4+O({\lambda }^2)$ in the stated atomic units.

In atomic units, $\hslash =m_e=e=4\pi {\epsilon }_0=1$, and the hydrogen atom hasThe allowed quantum numbers areHere the fixed electron spin quantum number is $s=1/2$. The energy eigenvalues are in hartrees; restoring units gives ${\hslash }^{2}l(l+1)$ and $\hslash m$ for the orbital operators and the corresponding factors for spin. The spin-independent hydrogen Hamiltonian does not split the two spin projections.

Keep one normalized spin function fixed and write $A=1/\sqrt{24}$ and $f_m(\theta ,\varphi )=\sin \theta e^{im\varphi }$, with $m=\pm 1$. Overall constants and the spin factor pass unchanged through the spatial operators.

First, in atomic units, ${\hat{l}}_z=-i{\partial }_{\varphi }$, soFor the squared orbital angular momentum, ${\partial }_{\varphi }^2{f}_m=-f_m$ andTherefore the two terms of the angular operator giveusing $1-{\cos }^2\theta ={\sin }^2\theta$. Thus ${\hat{l}}^2\Psi =2\Psi$.

Finally, the spatial hydrogen Hamiltonian isFor $R=Ar{e}^{-r/2}$,Substitute the already established angular eigenvalue $l(l+1)=2$:Hence $E=-1/8$ hartree, consistently with $n=2$. The angular calculation must precede the radial substitution if one wants to avoid differentiating the entire product repeatedly.

In atomic units, define the Coulomb integral and exchange integral byExpanding the four products in the expectation of $r_{12}^{-1}$ yieldsThe one-electron terms have the same value for these two states. Within this common-orbital approximation,so $\boxed{E_S-E_T=2K}$. The exchange integral is nonnegative: it is the positive Coulomb quadratic form of the transition density $a^{\ast }b$, or equivalently has a nonnegative Fourier representation. The triplet is lower. Actual orbital relaxation and correlation make the common-orbital $2K$ an approximate physical splitting.

At this separation the orbital overlap and interatomic two-electron exchange integrals are negligible. Each localized electron then contributes essentially the isolated hydrogen energy, while the interatomic electrostatic terms cancel to leading order for neutral fragments. Including nuclear repulsion in the molecular energy givesThe closed-shell RHF determinant instead doubly occupies $g\simeq (A+B)/\sqrt{2}$. It retains ionic configurations with both electrons on the same atom, even at dissociation. Using atomic units and the Hartree energy in the stated unrelaxed hydrogenic 1s basis, the same-atom two-electron Coulomb integral is $J_{AA}=J_{BB}=5E_h/8$. Consequently at infinite separationwhere energies and integrals inside the first expression are in atomic units. It is much too high compared with two neutral atoms. Retaining the leading intersite Coulomb term at $R=10\text{A˚}=18.90a_0$ gives $E_{\mathrm{RHF}}\simeq -11/16-1/(2R)\simeq -0.714E_h$, still about $0.286E_h$ above UHF. The numerical RHF estimates specify the supplied fixed 1s basis; a larger optimized orbital space changes them. If nuclear repulsion is excluded by an electronic-energy convention, subtract the same $1/R$ from both quoted total molecular energies. The restricted/unrestricted comparison and their dissociation behaviour are unchanged.^[ref]

Define $h_{pq}=⟨p|\hat{h}|q⟩$, the one-electron kinetic plus nuclear-attraction integral, and physicists' two-electron integralsin atomic units. With $E_{\mathrm{NN}}$ the zero-electron nuclear-repulsion term, the Slater determinant expectation isThe occupied sums include spin-orbitals, so the exchange integral vanishes for incompatible spins without needing a separate closed-shell factor.

Far from a neutral bound system, the slowest-decaying electron-removal amplitude is controlled by the first ionization energy $I=E_{N-1}-E_N>0$. Its asymptotic wavefunction satisfies a free-particle-like equation at negative energy $-I$, giving a decay exponent $\kappa =\sqrt{2m_{e}I}/\hslash$. Density is quadratic in the amplitude, soor $a=2\sqrt{2I}$ in atomic units. Power-law radial factors and angular dependence do not alter this leading exponential coefficient. It is the ionization energy, not the electron affinity, that controls this tail.

The Wigner-Seitz radius $r_s$ is the radius of a sphere containing one electron at the local uniform density:In atomic units, the unpolarized homogeneous electron gas has exchange energy per electron $-0.4582/r_s$. The local density approximation applies that uniform-gas result at each position using the local density, then integrates the result weighted by $\rho$. It assumes local uniform-gas behavior rather than resolving nonlocal exchange explicitly.

For a tail ${\rho }_{\sigma }\sim A_{\sigma }{e}^{-ar}$,At large $r$, the B88 correction integrand becomesSumming the spin densities and dividing by total density givesin atomic units. The exponentially decaying LDA term is negligible in this limit. The parameter $\beta$ cancels from the leading asymptotic term.

The electrostatic repulsion integral $J_{pq}=\iint |{\varphi }_p(1){|}^2|{\varphi }_q(2){|}^2{r}_{12}^{-1}d1d2$ in atomic units. For a normalized hydrogenic 1s orbital of exponent $\zeta$, the same-orbital value is $5\zeta /8$. It is a two-electron integral, unlike a diagonal Hückel one-electron parameter.

For real spatial orbitals, $K_{pq}=\iint {\varphi }_p(1){\varphi }_q(1)r_{12}^{-1}{\varphi }_q(2){\varphi }_p(2)d1d2$ in atomic units. It enters the energy of a Slater determinant through antisymmetry and becomes small for well-separated localized orbitals.