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Expand the trial in a complete orthonormal eigenbasis of the Quantum Hamiltonian, $\tilde{\Psi }={\sum }_n{c}_n{\varphi }_n$, with $\hat{H}{\varphi }_n=E_n{\varphi }_n$. Orthonormality givesEvery summand in the numerator is nonnegative because $E_n\ge E_0$. This proves the quantum variational principle for the stated discrete expansion; the trial energy is a weighted mean of exact energy eigenvalues.

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.

The equal supplied masses give equal effective density of states. At the specified temperature,Using the same energy units in the exponential,For equal electron and hole mobilities, both carriers contribute to the intrinsic electrical conductivity:The factor two is essential. A dimensional check givesusing $\mathrm{A}=\mathrm{C}{\mathrm{s}}^{-1}$.

Both in-plane orientations minimize the one-site energy. Since every adsorption site is occupied, the sites label independent choices even though the molecules are chemically identical. The ground state of the whole system therefore hasThe ground-state degeneracy is extensive: independently choosing either in-plane orientation at each site costs no energy. There is no extra vacancy or free-translational contribution in this model.

For the same atoms in a three-dimensional volume $V$, the one-atom translational partition function is $q_{\mathrm{trans}}^{(3)}=V/{\Lambda }^3$. Comparing the same $N$ atoms at the same temperature cancels the common $N!$ in their canonical partition functions. Therefore the molar adsorption confinement free energy, with the sign defined as adsorbed minus free, isThe logarithm is dimensionless: $V/\sigma$ and $\Lambda$ are both lengths. For a macroscopic gas depth $V/\sigma \gg \Lambda$, confinement raises the translational Helmholtz free energy. Equivalently, the translational state count is reduced by a factor $\sigma \Lambda /V$ per atom. The question supplies neither a volume nor an adsorption binding energy, so this answer is symbolic and describes the confinement cost, not a numerical net adsorption free energy.

For example, a binding energy $D>0$ per atom would supply a separate $-N_{\mathrm{A}}D$ to $\Delta F_m$, together with any normal-motion internal contribution retained by a more detailed model. Adsorption can then be favourable despite the loss of free translational phase space. Merely calling the change a loss must not reverse the sign of the explicitly defined $F_{\mathrm{ads}}-F_{\mathrm{gas}}$.

A planar rigid rotor has angular eigenfunctions $e^{im\varphi }$ and single-valuedness imposes $m=0,\pm 1,\pm 2,\dots$. Its energy depends on $m^2$, so if the level is labelled by $J=|m|$, thenThe two states at $J>0$ represent opposite directions of angular momentum. If the signed quantum number itself is called $J$, each signed state is counted once instead; one must not both sum over both signs and multiply by two again.

For an unrestricted three-dimensional linear rigid rotor, $J=0,1,2,\dots$ has $M=-J,\dots ,J$ and state degeneracy $2J+1$, with $E_J={\hslash }^{2}J(J+1)/(2I)$. Planar confinement removes one rotational coordinate and changes both the spectrum and degeneracy. These are the spatial counts before nuclear spin statistics. For homonuclear hydrogen, the allowed even/odd spatial states must also be paired with the appropriate parahydrogen or orthohydrogen nuclear states, as considered explicitly in A8.

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

Let $\mathcal{A}$ be the total two-dimensional area. There are $\mathcal{A}/(2\pi {)}^2$ wavevector states per reciprocal-space area per spin. Including spin degeneracy two and filling a disk givesFor areal number density $n_e=N_e/\mathcal{A}$ this yieldsThe disk is centred at zero wavevector, and its boundary is a fermi circle. Omitting the two spin states would incorrectly double the energy inferred for a given number density.

For the free electron model, the Fermi energy is the highest occupied one-electron energy at zero temperature, measured relative to the bottom of the free-electron band. The occupied wavevectors fill a disk in two dimensions or a sphere in three dimensions. The radius is the Fermi wavevector $k_F$, withEach allowed wavevector has two spin states. At nonzero temperature, the Fermi-Dirac distribution rounds the occupation boundary; the zero-temperature Fermi energy remains a useful density-defined scale and should not be defined as the thermal energy $k_{\mathrm{B}}T$.

Multiply the first-order equation by ${\Psi }_i^{(0)\ast }$ and integrate. The first term vanishes because Hermitian conjugation moves the unperturbed Hamiltonian onto the known bra:Here $E_i^{(0)}$ is real. The remaining equation isUsing wavefunction normalization therefore yieldsThus the first-order energy coefficient is the perturbation's expectation value in the unperturbed quantum state.

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.

Project the first-order equation onto ${\Psi }_j^{(0)}$, with $j\ne i$. By orthonormality and the Hermitian operator property,ConsequentlyandThe numerator order is important: the question defines $H_{ij}$ with $i$ on the bra, whereas projecting the correction onto state $j$ gives $H_{ji}$. The two coincide only when the relevant matrix element is real.

Excluding an $i$ component chooses $⟨{\Psi }_i^{(0)}|{\Psi }_i^{(1)}⟩=0$, a permitted first-order phase and wavefunction normalization convention. A component parallel to ${\Psi }_i^{(0)}$ is not determined by the projected equation. Nondegeneracy ensures nonzero denominators; a small energy gap can still make the perturbative expansion inaccurate unless $|\lambda H_{ji}|\ll |E_i^{(0)}-E_j^{(0)}|$.

Let $P_{12}$ exchange the complete labels of the two electrons. The spatial factors obey $P_{12}{\Phi }_{\pm }=\pm {\Phi }_{\pm }$. The singlet state spin factor is antisymmetric, while each triplet state spin factor is symmetric. Thus in all four casesas required by the Pauli exclusion principle. In particular, the $M_S=0$ triplet state remains a triplet despite having opposite measured spin projections in each product term.

For a spin-independent Hamiltonian and the same fixed spatial atomic orbitals, direct evaluation giveswhere $J_{ab}$ is the direct Coulomb integral and $K_{ab}>0$ is the exchange integral for the helium $1s,2s$ pair. ThereforeThe antisymmetric spatial wavefunction vanishes when the two positions coincide and reduces their Coulomb repulsion. The lower energy of the maximum-spin member of this fixed electron configuration is the first of Hund's rules.

For the $p_z$ atomic orbital basis, only orbitals whose centres remain fixed contribute to a character. Each such orbital contributes $+1$ if its $z$ direction is preserved and $-1$ if it is reversed. In the standard D4h molecular point group class order,No nonidentity operation other than ${\sigma }_h$ leaves any carbon site fixed. Reflection in the molecular plane fixes all eight sites but reverses every $p_z$ function.

Molecular character reduction givesThe dimensions sum to $1+1+1+1+2\cdot 2=8$. The two copies of $E_g$ are separate equivalent irreducible representations which may mix with one another; they do not force all four functions to have a common energy.

At ${\beta }_L=0$, the preceding matrix separates into identical $({\varphi }_1,{\varphi }_2)$ and $({\varphi }_3,{\varphi }_4)$ blocks. For ${\beta }_S<0$, a normalized pair at the lower energy $\alpha +{\beta }_S$ isA pair at the upper energy $\alpha -{\beta }_S$ isThe four coefficient vectors are mutually orthonormal. Any orthogonal mixing within either equal-energy pair is an equally valid choice of molecular orbitals. At this limiting parameter value, additional coincidences with the one-dimensional levels occur: the full eight-orbital system is four isolated dimers, with four bonding and four antibonding levels.

Use the four supplied opposite-site combinations in the order $({\varphi }_1,{\varphi }_2,{\varphi }_3,{\varphi }_4)$. Their Hückel method matrix isThe negative corner is essential. For example, $⟨{\varphi }_4|H|{\varphi }_1⟩$ includes the long bonds $4$–$5$ and $8$–$1$, each with opposite coefficient signs, giving $-{\beta }_L$.

Because the basis is orthonormal, the secular determinant isIt factors as $[(\alpha -E{)}^2-{\beta }_S^2-{\beta }_L^2{]}^2=0$, giving two doubly degenerate energies, $\alpha \pm \sqrt{{\beta }_S^2+{\beta }_L^2}$.

Write $p_j=p_{zj}$ and use the orthonormal basis of the Hückel method. The four normalized one-dimensional symmetry-adapted linear combinations areFor example, the $xz$ mirror plane exchanges sites $1\leftrightarrow 6$, $2\leftrightarrow 5$, $3\leftrightarrow 4$, $8\leftrightarrow 7$, without changing the sign of an individual $p_z$ orbital. The all-positive combination is unchanged by every vertical mirror but changes sign under ${\sigma }_h$, giving $A_{2u}$. The alternating combination changes sign under both sorts of vertical mirror, as required for $A_{1u}$. The two remaining sign patterns distinguish $B_{1u}$ from $B_{2u}$ in the printed axes. The signs refer to the upper $p_z$ lobes; the lower lobes have opposite quantum phases, and neither colour denotes an electric charge.

For a normalized coefficient vector $\mathbf{c}$, the energy isEvery site has one short-bond and one long-bond neighbour. Equivalently, apply the Hückel method matrix to any coefficient and read off its two neighbouring signs. The results areWith negative resonance integrals and $|{\beta }_S|>|{\beta }_L|$, the $A_{2u}$ and $B_{2u}$ levels are occupied, while $B_{1u}$ and $A_{1u}$ are unoccupied in the closed-shell alternating-bond state.

Fill four spatial molecular orbitals with the eight $\pi$ electrons. For negative ${\beta }_S,{\beta }_L$ and $|{\beta }_S|>|{\beta }_L|$, the occupied energies are $\alpha +{\beta }_S+{\beta }_L$, $\alpha +{\beta }_S-{\beta }_L$, and the pair at $\alpha -\sqrt{{\beta }_S^2+{\beta }_L^2}$. ThereforeFor equal integrals, ${\beta }_S={\beta }_L=\beta <0$, this becomesAt equal bond lengths, the last two electrons occupy the frontier degeneracy at $E=\alpha$, so their distribution within that degenerate space does not change the bare Hückel method energy.

The uniform planar eight-electron ring has a partially filled frontier degeneracy. Bond-length alternation splits those levels, permitting the two frontier electrons to occupy the lowered orbital and opening a gap to the empty partner. This is the electronic mechanism for destabilizing the equal-bond planar geometry.

A controlled comparison keeps the mean resonance integral fixed. PutUsing part (i), the change from the uniform ring isThus a small alternation lowers the electronic energy, with a leading gain linear in $\delta$. An ordinary elastic cost starts quadratically, so the equal-bond planar state is unstable in this model. The equilibrium distortion requires those geometric and sigma-bond costs as well.

The eight $\pi$ electrons constitute a $4n$ cyclic count rather than the closed-shell $4n+2$ case associated with aromaticity. The result concerns the imposed planar geometry; it does not establish the unconstrained three-dimensional equilibrium shape of cyclooctatetraene.