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For a regular self-adjoint Sturm-Liouville problem with positive weight $w$ and self-adjoint boundary conditions, eigenfunctions with distinct eigenvalues are orthogonal in the weighted inner product $\int w\overline{u}v$. Integration by parts proves this by equating $\int \overline{u}Lv$ and $\int \overline{Lu}v$.

For ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}$, separated normal modes decay at rates $B{\alpha }_n^4/({\zeta }_{\perp }{L}^4)$. Clamping at zero and vanishing moment and shear at $L$ give $\cos {\alpha }_n\cosh {\alpha }_n=-1$. A mode isThe fourth-order operator is self-adjoint: two integration by parts operations give $\int f{g}^{\prime \prime \prime \prime }=\int f^{\prime \prime }{g}^{\prime \prime }$ under these boundary conditions. Its zero eigenspace is trivial and its eigenfunctions form an orthogonal complete basis. Thus a nonzero projection on the first mode gives long-time decay proportional to ${\varphi }_1(x/L)\exp [-B{\alpha }_1^{4}t/({\zeta }_{\perp }{L}^4)]$, where ${\alpha }_1=1.8751040687\dots$. If this projection vanishes, the lowest excited mode replaces it.

For an infinite square well of width $a$, the normalized trial $\tilde{\psi }=\sqrt{30/a^5}x(a-x)$ obeys both Dirichlet boundary conditions and has no interior nodes of a wavefunction. Its kinetic energy isThe ratio to the exact ground-state energy is $10/{\pi }^2=1.01321\dots$. Its expansion in the normalized sine eigenfunctions has coefficients $4\sqrt{60}/({\pi }^3{n}^3)$ for odd $n$ and zero for even $n$. This supplies a concrete check of the quantum variational principle and shows that a small energy error can coexist with excited-state admixtures.

Choose the growth direction as $z$ and the well as $0<z<L$. In the envelope function approximation, write a subband state as $A^{-1/2}{e}^{i{\mathbf{k}}_{\parallel }\cdot {\mathbf{r}}_{\parallel }}{F}_n(z)u_{j0}(\mathbf{r})$, where $u_{j0}$ is the band-edge cell-periodic Bloch factor. The infinite square well hasThe envelope has the same shape for electron and hole states: changing their effective masses changes the eigenenergies, but not these infinite-well eigenfunctions. With in-plane polarization, the leading interband momentum matrix element acts on the Bloch factors, while the slow envelopes giveorthonormality therefore yields the infinite-well interband selection rule $n^{\prime }=n$. A nonzero band-level matrix element is still required; the overlap rule does not replace polarization selection rules. Finite barriers can make electron and hole envelopes different, and interface asymmetry or band mixing can relax this ideal same-index rule.

Use a three-dimensional electron gas with a uniform positive background, the jellium model. The Hartree-Fock method minimizes the expectation of the interacting Hamiltonian over normalized Slater determinants. Antisymmetry is exact within that trial class, but correlations requiring a superposition of determinants are omitted. For orthonormal occupied spin orbitals ${\varphi }_i$, the energy iswhere $J_{ij}$ is the direct Coulomb integral and $K_{ij}$ the exchange integral; the spin overlap makes exchange vanish between opposite-spin orbitals. The self-interaction terms $J_{ii}$ and $K_{ii}$ cancel. Variation gives a self-consistent Fock operator $\hat{f}=h_0+{\sum }_j({\hat{J}}_j-{\hat{K}}_j)$; exchange is nonlocal. Its eigenfunctions define occupied orbitals of a stationary determinant, and the determinant must be compared with competing self-consistent solutions to select a variational ground state.

The eigenvalues are orbital Lagrange multipliers. Their sum counts each interaction twice:Frozen-orbital removal energies may be estimated using Koopmans' theorem, but measured quasiparticle energies can contain relaxation and correlations absent from this approximation. Unoccupied orbital energies are not automatically an exact excitation spectrum. In particular, unscreened exchange in a uniform three-dimensional gas produces an unphysical logarithmic slope singularity at its Fermi surface; a one-determinant mean field does not reproduce the interacting Fermi liquid's screened response.

For a spatially uniform plane-wave determinant, electron-background and background-background terms cancel the uniform direct Coulomb contribution. Retain the background when making this cancellation: an isolated infinite electron charge distribution would have a divergent direct energy. The surviving contributions per particle are kinetic energy and negative same-spin exchange energy. Let $(3/(4\pi n){)}^{1/3}=r_s{a}_B^{\ast }$ define the Dimensionless Wigner-Seitz radius, with $a_B^{\ast }=4\pi \epsilon {\hslash }^2/(m^{\ast }{e}^2)$, $E_H^{\ast }={\hslash }^2/(m^{\ast }{a}_B^{\ast 2})$ and $\zeta =(n_{\uparrow }-n_{\downarrow })/n$. ThenFor the unpolarized state this is approximately $1.105/r_s^2-0.4582/r_s$. The kinetic contribution penalizes spin polarization because it enlarges an occupied Fermi sphere; exchange favors polarization because equal-spin occupation gains more exchange energy. As density falls, the $r_s^{-1}$ exchange scale overtakes the $r_s^{-2}$ kinetic scale. The uniform fully polarized and unpolarized energies cross atThis is a variational comparison within the homogeneous Hartree-Fock electron gas, not an exact physical ferromagnetic phase boundary. It demonstrates how Pauli kinetic energy and exchange compete and why the allowed trial states matter.

Allowing spatial symmetry breaking changes the result: the Overhauser instability lowers the paramagnetic determinant's Hartree-Fock energy through a spin density wave, even in the high-density limit. A calculation restricted to uniform Fermi spheres therefore does not give the unrestricted Hartree-Fock ground state.^[ref] At still lower density, localization and a Wigner crystal compete with fluid states. electron-gas correlation energy beyond a single determinant is important in assessing their competition. Quantum Monte Carlo provides a concrete route to comparing correlated fluid and crystal energies; the original stochastic calculation demonstrates why exchange-only energy differences are insufficient.^[ref]

Hartree-Fock is a useful variational baseline with exact exchange, not a controlled complete low-density phase diagram. At high density the kinetic scale dominates, but even a small symmetry-breaking energy gain can invalidate a restricted determinant as the exact Hartree-Fock minimizer. At low density neglected correlations, screening and localization can change both transition locations and the ordering of candidate phases. A numerical phase boundary should therefore state dimension, spin constraints, background model and whether correlation and spatially modulated states were included.

LetThe normalized orbital energy eigenstates vanish outside the well and have one-particle energies $n^2\epsilon$. Define the normalized symmetric and antisymmetric combinations for $r\ne s$ byThe normalized spin functions are the singlet state ${\chi }_{00}=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ and the triplet states ${\chi }_{11}=|\uparrow \uparrow ⟩$, ${\chi }_{10}=(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, ${\chi }_{1,-1}=|\downarrow \downarrow ⟩$. Their exchange energies are respectively $3J/4$ and $-J/4$.

For $0<J\ll \epsilon$, the three lowest distinct energy levels of the exchange-coupled spin-one-half particles in an infinite square well areThe spatially doubly occupied orbital can only occur with the singlet state. The next orbital sum is $1^2+2^2=5$, where either exchange symmetry is possible. The next unused sum is $2^2+2^2=8$, well separated under the small-coupling assumption. The paper's inequality is equivalent, up to a numerical factor, to $|J|\ll \epsilon$; the factor ${\hslash }^2$ in $J$ is essential because the spin operators carry units of angular momentum.

The sign of $\lambda$ is not explicitly specified. The table assumes the usual intended positive coupling in the displayed minus-sign convention. For $J<0$ with $|J|\ll \epsilon$, the same three sets of eigenfunctions occur, but the two rows at $5\epsilon$ exchange order. At $J=0$ those rows merge into a single energy level of multiplicity four; if three distinct energies are requested exactly at that point, the third is $8\epsilon$ with the single state ${\Phi }_{22}^{+}{\chi }_{00}$.

For distinguishable particles, the ordered spatial products are independent states, and neither singlet states nor triplet states are excluded from any orbital occupation. The unperturbed energy eigenvalues areFor $0<J\ll \epsilon$, the three lowest distinct energy levels arewhere $M=-1,0,1$. The first two levels do have zero first-order correction, since their spatial expectation value factorizes as $⟨1|x|1{⟩}^2=0$.

The last instruction in the original PDF is false for the third level. Although every diagonal product-state matrix element vanishes, the two orbital states have degenerate energy levels. Degenerate perturbation theory requires diagonalizing the perturbation within their span. For each $M$, its matrix in the ordered basis $(u_1{u}_2,u_2{u}_1)$ isThe eigenfunctions are ${\Phi }_{12}^{\pm }{\chi }_{1M}$, withThus the sixfold unperturbed level splits into two triplets. Its average shift is zero, and the diagonal expectation in an individual ordered product state is zero, but neither fact makes all its energy eigenvalue corrections zero. This is the degenerate orbital exchange splitting from a product perturbation; distinguishability allows both combinations rather than removing their coupling.

For negative $J$, the lowest two rows exchange order and the third unperturbed level is instead the twofold singlet orbital pair at $5\epsilon +3J/4$. It splits into ${\Phi }_{12}^{\pm }{\chi }_{00}$ with the same shifts $\pm \alpha D$, each nondegenerate. At $J=0$, the three lowest distinct unperturbed energies are $2\epsilon ,5\epsilon ,8\epsilon$, with multiplicities $4,8,4$. The first and third have zero first-order shift; the middle splits by $\pm \alpha D$, each fourfold. The printed zero-shift assertion therefore fails also in the zero-exchange limit.

For the multiplication operator, $\Vert Tf{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }\Vert f{\Vert }_{\infty }=\Vert f{\Vert }_{\infty }$. Taking $f=1$ gives equality, so $\Vert T\Vert =1$.

An eigenfunction obeys $(g(t)-\lambda )f(t)=0$ at every $t$. If $\lambda \in [0,1)$, its level set is the single point $t=\lambda$, and a continuous function supported at a single point is zero. If $\lambda \notin [0,1]$, that level set is empty. For $\lambda =1$, every continuous function vanishing on $[0,1]$ is an eigenfunction, and nonzero examples exist, such as $f(t)=\max \{t-1,0\}$. Thus ${\sigma }_p(T)=\{1\}$.

If $\lambda \notin [0,1]$, multiplication by $(g-\lambda {)}^{-1}$ is a bounded inverse of $T-\lambda I$. If $\lambda \in [0,1]$, choose $t_0$ with $g(t_0)=\lambda$. Every function in the range of $T-\lambda I$ vanishes at $t_0$, so the range omits the constant function one. Hence it is not onto. Therefore

The squared angular momentum operator and one selected component are commuting operators, so a complete set of common eigenfunctions can be chosen: for the $z$ component these are the usual $|j,m⟩$ angular momentum eigenstates. Distinct components do not have a common complete eigenbasis, because their commutator is another nonzero component. A $j=0$ state is an exception in which all components vanish; noncommutation does not forbid every individual common eigenfunction.

Separation of variables makes the angular factor a spherical harmonic. Applying the Quantum Hamiltonian givesConsequently $R{Y}_{lm}$ is an energy eigenfunction provided the radial factor satisfiesThis is the radial Schrödinger equation for a central potential, here with the attractive Coulomb potential. An arbitrary $R$ does not give an eigenfunction; regularity, decay and wavefunction normalization select admissible bound quantum states.

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.

Twice differentiation the sine givesBoth Dirichlet boundary conditions hold when $n$ is an integer. The zero function at $n=0$ is not an eigenfunction, and negative $n$ merely changes an overall sign. ThereforeThe corresponding normalized eigenfunction is $\sqrt{2/a}\sin (n\pi x/a)$.

Up to a global phase, $x(a-x)$ is real, positive throughout the interior, has no interior nodes of a wavefunction, and is symmetric about $x=a/2$. It is smooth inside the box and has one central maximum. These are the shape and reflection symmetry properties of the lowest particle in a box eigenfunction, $\sin (\pi x/a)$; they make it a plausible variational method trial without making it exact.

The trial is even under reflection about the midpoint. The normalized box eigenfunctions satisfy ${\varphi }_n(a-x)=(-1{)}^{n+1}{\varphi }_n(x)$, so orthogonality between even and odd functions gives $c_n=0$ for even $n$. If the higher odd contributions $n\ge 5$ are neglected, the weighted mean from the quantum variational principle becomesRearranging yields the two-level variational contamination estimateThe required assumption concerns the neglected contributions to the energy as well as their small probabilities. Numerically the two-level estimate is $0.0016542$. It is not an exact identity: direct expansion of this parabolic trial gives $c_n\propto n^{-3}$ for odd $n$, hence the exact ratio is $1/729=0.0013717$. The omitted higher odd states account for the difference.

A symmetry orbital is a molecular orbital when it is also an eigenfunction of the effective one-electron Hamiltonian operator. If an irreducible representation occurs once in a complete invariant model space, a symmetry-invariant Hamiltonian acts as a scalar on that copy by Schur's lemma. Its basis functions are then degenerate eigenfunctions. When the same symmetry occurs in several copies, those copies can mix; one must diagonalize the corresponding block. Merely having a one-dimensional symmetry label is insufficient if other orbitals have that same label.

A symmetry orbital is a linear combination of atomic orbitals transforming as a component of an irreducible representation of the molecular point group. A molecular orbital is a one-electron wavefunction for the molecule, obtained as an eigenfunction of the chosen effective one-electron Hamiltonian operator. Symmetry organizes the eigenproblem, but a symmetry label alone does not determine its coefficients.

The unperturbed Hamiltonian satisfiesThe supplied eigenfunctions can be chosen as an orthonormal eigenbasis, with $⟨{\Psi }_j^{(0)}|{\Psi }_i^{(0)}⟩={\delta }_{ji}$.

Linearity and the shared $l=1$ angular eigenvalue giveUsing the relative phases in part (c),Thus each real orbital remains an energy eigenstate with $E=-1/8$ hartree and has definite orbital angular momentum magnitude $\sqrt{2}\hslash$, but neither is an eigenfunction of ${\hat{l}}_z$. A measurement of $l_z$ gives $+\hslash$ or $-\hslash$, each with probability $1/2$. Its mean is zero and its variance is ${\hslash }^2$; a zero mean does not mean that a measurement gives zero.

For two eigenfunctions, apply the same Hermitian operator identity across the two states:Hence $(q_n-q_m)⟨{\psi }_m,{\psi }_n⟩=0$. Distinct eigenvalues therefore imply orthogonal eigenfunctions. Within a degenerate eigenspace, orthogonality is a possible basis choice, rather than a consequence for every pair of vectors.

The decisive observation is that $x$ is itself an eigenfunction:On the regular Legendre polynomial domain, integration by parts has endpoint factor $x^2-1$, so the boundary form vanishes for regular functions at $x=\pm 1$. The interval alone is not an endpoint condition; regularity is the usual condition selecting the Legendre polynomials. Their eigenvalues are $n(n+1)$, and the $n=1$ normalized function isThereforeEquivalently, the Hermitian operator identity gives $(q_n-2)c_n=0$, and nondegeneracy leaves only one possible nonzero coefficient. The unnormalized function $x$ has squared norm $2/3$, consistent with this coefficient.