Let $S_N={\sum }_{n=0}^N{\sum }_{k=0}^{2^n-1}{a}_{n,k}{H}_{n,k}$. The assumed orthonormality of the Haar wavelets givesEnumerate the coefficients level by level as $A_1,A_2,\dots$. They are independent and identically distributed standard normal random variables, so the squares have mean one. The strong law of large numbers impliesConsequently,A convergent sequence in a Hilbert space must be bounded in norm. Hence this random Haar wavelet series diverges in $\mathcal{H}$ almost surely.

In the Lévy construction of Brownian motion, start with $B_0=0$ and $B_1=Z_0$, where $Z_0$ has the standard normal distribution. Suppose values are defined at the endpoints $u=k{2}^{-n}$ and $v=(k+1)2^{-n}$. Setusing fresh independent standard normal random variables $Z_{n,k}$, also independent of everything previously used. Join the values by straight segments to obtain a continuous piecewise linear process $B^{(n)}$ on the dyadic grid.

To verify the Brownian increments, write $D=B_v-B_u$ and $\ell =v-u$. The two new increments are $D/2+\sqrt{\ell }{Z}_{n,k}/2$ and $D/2-\sqrt{\ell }{Z}_{n,k}/2$. If $D\sim N(0,\ell )$, their variances are $\ell /2$ and their covariance is zero. They are Jointly Gaussian, so they are independent. Induction proves the required Brownian increments on every dyadic grid.

For convergence, successive interpolants differ by tents of disjoint interiors, soThe Gaussian tail bound and the union bound giveThese probabilities are summable. The First Borel-Cantelli lemma and ${\sum }_{n}n{2}^{-n/2}<\infty$ therefore prove uniform convergence of the interpolants almost surely. Their limit has continuous sample paths. Approximating any finite set of times by dyadic times passes the Jointly Gaussian increment laws to the limit, giving a standard Brownian motion on $[0,1]$.

Equivalently, with ${\psi }_{n,k}(t)={\int }_0^t{H}_{n,k}(s)ds$, these tents give the Lévy-Ciesielski expansion

On a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, a standard one-dimensional Brownian motion is a real-valued stochastic process $B=(B_t{)}_{t\ge 0}$ with $B_0=0$ almost surely, continuous sample paths almost surely, and independent increments, such thatHere the second parameter is the variance. Thus its Brownian increments have the normal distribution with mean zero and variance equal to elapsed time. Equivalently, relative to a specified filtration $({\mathcal{F}}_t)$, require $B$ to be an adapted process and $B_t-B_s$ to be independent of ${\mathcal{F}}_s$ with this normal distribution. Taking the natural Brownian filtration recovers the first definition.

Complete the square:The minimum is stable because $d^{2}E/d{ϵ}^2=\beta >0$. HenceHere the Jahn-Teller stabilization energy is the positive magnitude of the energy reduction.

The term $-\alpha ϵ$ is the leading electronic energy gain from splitting the initially degenerate orbitals and occupying the lower levels. Its negative sign expresses the lowering of the occupied-electron energy for the chosen elongation branch of the Jahn-Teller effect. The positive coefficient $\alpha$ is a linear electron-lattice coupling in this reduced model.

The term $\beta {ϵ}^2/2$ is the harmonic elastic cost of moving the ligands away from the undistorted reference structure. Mechanical stability requires $\beta >0$. The competition between linear gain and quadratic cost produces a finite equilibrium distortion rather than indefinite elongation.

For axial compression along $z$, the ordering is reversed: $d_{xy}$ lies below the $d_{xz},d_{yz}$ doublet, and $d_{x^2-y^2}$ lies below $d_{z^2}$ within $e_g$. One electron occupies $d_{xy}$ and the other occupies either $d_{xz}$ or $d_{yz}$, with parallel spins in the weak-distortion regime. The orbital degeneracy is therefore two, with $S=1$.

The retained orbital degeneracy allows further symmetry lowering and explains why this compressed arrangement does not provide the fully quenched equilibrium of the elongated case in the specified simple model.

Take the elongation axis as $z$. The axial oxygen ligands move farther away while the equatorial ones move closer, at fixed mean bond length. Orbitals containing $z$ are stabilized relative to their in-plane partners. Thus $d_{xz},d_{yz}$ move below $d_{xy}$ within $t_{2g}$, while $d_{z^2}$ moves below $d_{x^2-y^2}$ within $e_g$. The schematic level ordering appears in the figure below; the octahedral gap remains open for a small distortion.

The two parallel-spin electrons occupy $d_{xz}$ and $d_{yz}$, one each. Filling the entire lower doublet gives a unique orbital configuration, so its orbital degeneracy is one and the spin is still $S=1$. This is the orbitally quenched configuration relevant to the later spin Hamiltonian.

The energetic preference can be made explicit in a traceless linear splitting model. For an elongation coordinate $Q>0$, write the $t_{2g}$ levels relative to their barycentre as $(-aQ,-aQ,2aQ)$ with $a>0$. Their two-electron ground energy decreases by $2aQ$. For an equal-magnitude compression, the levels are $(-2aQ,aQ,aQ)$ and the two-electron energy decreases only by $aQ$. With the same quadratic elastic cost, minimizing $-2aQ+k{Q}^2/2$ gives four times the stabilization of minimizing $-aQ+k{Q}^2/2$. Thus the small-distortion Jahn-Teller effect favors elongation for this high-spin $d^2$ filling. The argument assumes the linear coupling and harmonic elastic model, rather than claiming that symmetry alone fixes all real-material distortions.

For the regular octahedron, repulsion from the six oxygen ligands raises the two $e_g$ orbitals, $d_{z^2}$ and $d_{x^2-y^2}$, above the three $t_{2g}$ orbitals, $d_{xy},d_{xz},d_{yz}$. The latter point between the metal-ligand axes and experience less direct repulsion. The crystal field splitting is therefore a lower triplet and an upper doublet.

For the $3d^2$ V${}^{3+}$ ion, Hund's rules put the two electrons into different $t_{2g}$ orbitals with parallel spins, giving $S=1$. There are three choices for the unoccupied $t_{2g}$ orbital, so the orbital ground state is triply degenerate. Its spin degeneracy is separate. This orbital degeneracy makes a symmetry-lowering Jahn-Teller effect possible.

For $\mathbf{B}\parallel [111]$, its squared projection onto the long axis of every valley is $1/3$. All six valleys therefore have the same cyclotron effective mass:At the fixed photon energy,Thus the two-peak spectrum is replaced by one peak near $1.4\mathrm{T}$. The source does not determine the linewidth or absolute oscillator strength for the new direction; the sketch below shows only a normalized illustrative line at the computed position.

The photon energy fixes the resonant cyclotron frequency by $\hslash \omega =575\mu \mathrm{eV}$. ConsequentlyRead the two absorption maxima as approximately $B_L=0.95$--$0.97\mathrm{T}$ and $B_H=2.10\mathrm{T}$. For $\mathbf{B}\parallel [001]$, the two valleys whose long axis is along $z$ have $m_{\mathrm{cr}}=m_1$, whereas the other four have $m_{\mathrm{cr}}=\sqrt{m_1{m}_2}$. HenceFor the representative reading $B_L=0.96\mathrm{T}$, the values are $m_1=0.193m_e$ and $m_2=0.925m_e$. The graph does not warrant additional precision. The larger longitudinal effective mass means flatter dispersion and a longer constant-energy ellipsoid along the valley axis.

$P_1$ has primitive coordinates $(1/2,1/2)$ and therefore fails the first-order optical momentum condition; its local $V_1$ pattern also has zero dipole derivative. $P_2$ is a zone-centre mode with the dipole-active $V_2$ pattern, so its molecular dipole changes add coherently.The allowed absorption for $P_2$ uses an polarization of an electromagnetic wave with an out-of-plane component. At normal incidence on an ideal free-standing layer the electric field is in the plane, so that specific measurement geometry would suppress it despite the mode's intrinsic vibrational infrared activity.

For ordinary one-phonon infrared optical absorption, both an energy and a momentum selection rule must hold. The optical photon has negligible crystal momentum on the scale of the Brillouin zone, so crystal momentum conservation selects $\mathbf{q}\simeq 0$ modulo a genuine reciprocal lattice vector. In addition, the mode must change the macroscopic electric polarization, with a component along the incident polarization of an electromagnetic wave.

In terms of atomic displacements ${\mathbf{u}}_j$ and Born effective charge tensors $Z_j^{\ast }$, the relevant derivative is proportional to ${\sum }_j{Z}_j^{\ast }{\mathbf{u}}_j$ per primitive unit cell. Thus an optical phonon at $\mathbf{q}=0$ is not automatically infrared active: its dipole contributions may cancel. Disorder-assisted and multiphonon processes relax these simple first-order selection rules.

Apply the same change of reciprocal lattice coordinates to the phonon wave vectors:Although $(1,0{)}_r$ is an integer coordinate of the rectangular reciprocal grid, it is not a reciprocal lattice vector of the actual primitive crystal. Thus $P_1$ is not equivalent to the Brillouin zone centre.

Comparing the two coefficients in the reciprocal-coordinate transformation givesThe inverse transformation is $h_r=h_p+k_p$, $k_r=k_p-h_p$. This provides an independent check on the signs associated with the two primitive arrows in the source.

Let the reciprocal lattice vector be $\mathbf{G}=h_r{\mathbf{a}}_r^{\ast }+k_r{\mathbf{b}}_r^{\ast }=h_p{\mathbf{a}}_p^{\ast }+k_p{\mathbf{b}}_p^{\ast }$, with each dual pair obeying $\mathbf{a}\cdot {\mathbf{a}}^{\ast }=\mathbf{b}\cdot {\mathbf{b}}^{\ast }=2\pi$ and zero cross products. Its coefficients are obtained by dotting with the corresponding direct-space primitive lattice translation vectors:HenceFor actual integer-indexed reciprocal lattice vectors, this also explains the centred-rectangular condition that $h_r$ and $k_r$ have the same parity. General phonon wave vectors need not have integer coefficients.

The primitive translations connect the two centred-rectangular sublattices. In the orientations marked in the source, ${\mathbf{a}}_p$ points right and down, whereas ${\mathbf{b}}_p$ points right and up. ThusTheir parallelogram has half the area of the rectangular cell, confirming that the latter contains two primitive unit cells.

There are six atoms in the nonlinear oxalate ion, hence $3\cdot 6=18$ positional degrees of freedom. Subtract three rigid translations and three rigid rotations to obtain $3N-6=12$ vibrational normal modes. Planarity does not remove out-of-plane motions. Since three modes have been displayed,

Use the indicated out-of-plane displacements and the equivalence of the four oxygen sites and of the two carbon sites in the isolated planar oxalate ion. In $V_1$, the two oxygen atoms on one side move oppositely to those on the other side, and the two carbon atoms also move oppositely. Their dipole changes cancel. The mode is even under molecular inversion symmetry, which independently forbids an electric-dipole transition.

In $V_2$, all four oxygen atoms move together out of the plane while both carbon atoms move in the opposite direction. The positive and negative charge distributions acquire a relative out-of-plane displacement, producing a nonzero derivative of the molecular electric dipole moment.

In $V_3$, two oxygen atoms move out and two move in, in a checkerboard arrangement, while the carbon atoms are stationary. The equivalent oxygen contributions cancel. This mode illustrates that being odd under inversion symmetry is necessary but not sufficient for vibrational infrared activity: its full symmetry does not transform like any component of an electric dipole moment.These conclusions concern the isolated-molecule displacements drawn in the source; a different crystal environment can alter selection rules through additional symmetry breaking.

Infrared spectroscopy measures absorption as the incident photon frequency is swept through molecular normal mode frequencies. In the electric-dipole approximation the interaction is $-\mathbf{p}\cdot \mathbf{E}$. Expanding the molecular electric dipole moment in a normal coordinate $Q$ givesThe fundamental one-quantum vibrational transition dipole moment is proportional to the derivative multiplying $Q$. Thusis the criterion for vibrational infrared activity; the derivative must also have a component along the incident polarization of an electromagnetic wave. A permanent molecular electric dipole moment is not necessary. Absorption occurs near $\hslash \omega =E_{v+1}-E_v$, with broadened lines reflecting finite lifetimes and the molecular environment.

At $B_z=8\mathrm{T}$,The cyclotron spacing is $\hslash e(8\mathrm{T})/m^{\ast }$, smaller than at 10 T, and each resolved level's degeneracy per area is also reduced by $4/5$. Thus the same carrier density occupies more levels. This ratio is determined without the missing graph, but it does not specify whether the chemical potential lies in a localized tail or an extended-state centre.

TODO — missing density/plateau assignment: a numerical index and chemical-potential marker at 8 T require the unavailable density information. For example the first illustrative density in part (i) gives ${\nu }_8\simeq 2.5$, while $N_S=3.87\times 10^{15}{\mathrm{m}}^{-2}$ instead gives ${\nu }_8\simeq 2$. These produce different schematic placements even though both satisfy every surviving sample parameter. Unresolved spin degeneracy and disorder add further qualifications; no unique 8 T label should be invented.