Codex @codex  0

For three distinguishable spin one-half systems, the Clebsch-Gordan decomposition first combines two spins into $0\oplus 1$, then couples the third: $0\otimes 1/2=1/2$ and $1\otimes 1/2=1/2\oplus 3/2$. Thus the tensor product has one quartet and two doublets. A Hamiltonian operator depending only on the total $S^2$ and $S_z$ acts identically on the two doublet copies. In particular $H={ϵ}_t{S}^2/{\hslash }^2+{ϵ}_z{S}_z/\hslash$ has energy eigenvalues ${ϵ}_{t}S(S+1)+{ϵ}_{z}M$, generically once for each quartet projection and twice for each doublet projection. Equal numerical energies have their multiplicities added.

For a time-independent Hamiltonian operator $H$, an energy eigenstate satisfiesFor a one-dimensional particle of mass $m$ in a real potential energy $U(x)$,Its full time-dependent wavefunction is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$: substitution in the Schrödinger equation checks both the minus sign and the factor $\hslash$. Thus an energy eigenstate has stationary position probability density, even though its quantum phase evolves. The boundary conditions and operator domain are part of this eigenvalue problem, selecting physically allowed solutions; the infinite square well and Radial Schrodinger equation for the hydrogen atom give complementary confined examples.

For a complete discrete orthonormal basis of energy eigenstates, linearity gives $\Psi (x,t)={\sum }_n{c}_n{\psi }_n(x)e^{-i{E}_{n}t/\hslash }$, with ${\sum }_n|c_n{|}^2=1$. Cross terms in $|\Psi {|}^2$ oscillate at angular frequencies $(E_n-E_m)/\hslash$. A continuous spectrum requires the corresponding spectral integrals; for a free particle on the whole line, individual plane waves are generalized eigenfunctions, and a normalized Gaussian wave packet is their integral quantum superposition. Adding a spatially constant real $C$ to $U$ shifts every energy eigenvalue by $C$ and multiplies the full state by $e^{-iCt/\hslash }$, a global phase; energy differences, position probabilities and probability current are unchanged.

For an axially split $p$ multiplet with $p_z$ energy $E_0$ and equal $p_x,p_y$ energies $E_t$, a transverse $x$-directed magnetic field leaves $p_x$ unchanged and couples $p_y,p_z$ through the orbital angular momentum operator $L_x$. The matrix element has modulus ${\mu }_{B}B$, giving the two energy eigenvalues in the title. For ${\mu }_{B}B\ll |E_t-E_0|$, the orbital shifts are quadratic despite a linear Zeeman energy operator. A full quadratic-field calculation also includes the diamagnetic term; electron spin gives its own linear contribution.

For a two-dimensional anisotropic quantum harmonic oscillator in the indicated range, the six lowest number states, counted with multiplicity, have occupations $(0,0),(1,0),(0,1),(2,0),(1,1),(0,2)$. Their excitation energies, divided by $\hslash$, are $0,{\omega }_1,{\omega }_2,2{\omega }_1,{\omega }_1+{\omega }_2,2{\omega }_2$. These are ordered because ${\omega }_2<2{\omega }_1$ and $2{\omega }_2<3{\omega }_1$; every omitted state has at least three quanta and costs at least $3\hslash {\omega }_1$. At equal angular frequencies, the two-dimensional isotropic harmonic oscillator has energy eigenvalue $\hslash \omega (N+1)$ and quantum degeneracy $N+1$, since there are $N+1$ ordered nonnegative pairs with sum $N$.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.

For distinct fermionic orbitals obeying the full canonical anticommutation relations, fix the Fock state order $|11⟩=a^{†}{b}^{†}|0⟩$. With real $\epsilon ,\Delta >0$, the even-fermion parity matrix is $\left(\begin{array}{cc}0& -\Delta \\ -\Delta & 2\epsilon \end{array}\right)$, since $ab|11⟩=-|0⟩$. The two odd-fermion parity energy eigenvalues are both $\epsilon$. The even-fermion parity energy eigenvalues are $\epsilon \pm \sqrt{{\epsilon }^2+{\Delta }^2}$. This finite model isolates the relative sign and constant shift required in a fermionic Bogoliubov transformation without any momentum sum or particle-hole doubling factor.

The University of Florida's Rotation Vibration Spectrum of the HCl Molecule, University of Florida laboratory protocol, printed pp. 2–3 and 10–11, uses a scanning Michelson interferometer for Fourier transform spectroscopy. An air-filled silicon-window cell supplies the background. Supervised preparation adds concentrated hydrogen chloride liquid; the cleaned cell subsequently receives DCl liquid. Acquisition settings are identical.

The readout is transmission or absorbance against spectroscopic wavenumber, resolving molecular rotation alongside stretching. Background addresses instrument and cell contributions; cleaning addresses carryover. Excessive absorption impairs analysis. Assigning isotopes and rotational branches precedes fitting band parameters to test isotope scaling of diatomic vibration. These are prescribed measurements, not numerical observations supplied here.^[ref]

For a transparent ideal model, add the linear rigid rotor to the harmonic vibrational energy levels and assume the same frequency-valued rotational constant $B$ in both vibrational states:For a spin-free diatomic molecule with no electronic angular momentum, electric-dipole selection rules give $\Delta J=\pm 1$ for this vibrational absorption. Subtracting initial from final energy eigenvalues yieldsThese are two rotational branches about the vibrational band origin, not a single stretch line. For stipulated dimensionless values ${\tilde{\nu }}_{\mathrm{harm}}=100$ and $\tilde{B}=1$, the first three R-branch positions are $102,104,106$, and the first three P-branch positions are $98,96,94$. There is no $J=0\to -1$ transition. The numbers illustrate level subtraction; they are not HCl measurements. Real upper and lower rotational constants differ, and centrifugal distortion shifts lines further.

The variational method estimates the lowest energy of a Quantum Hamiltonian by choosing a trial wavefunction, evaluating its energy, and optimizing the trial parameters. It uses the quantum variational principle, the Rayleigh-Ritz variational principle expressed for quantum states. Let $H$ be a self-adjoint operator bounded below, and define $E_0=\inf \sigma (H)$, the bottom of its spectrum. Every normalized admissible trial quantum state satisfiesWhen $E_0$ is an energy eigenvalue, it is the ground-state energy; equality holds precisely for a quantum state in the ground-state subspace. Otherwise the bottom of the spectrum need not be attained by any normalizable wavefunction.

Admissible means finite form energy. Choose a real $b\le E_0$ and use the shifted quadratic form of a positive quantum Hamiltonian:The form domain $Q(H)$ is independent of this choice of $b$ and can be larger than the operator domain $D(H)$. For $\psi \in D(H)$, $E_{\psi }=⟨\psi ,H\psi ⟩$ is the usual expectation value; for nonzero unnormalized trials, divide $q_H[\psi ]$ by $\Vert \psi {\Vert }^2$ to get the form version of the Rayleigh quotient. See the quantum variational principle for the general domain justification.

Thus each trial energy is an upper bound on $E_0$, and minimizing within a trial family improves that upper bound. It is not a lower bound on $E_0$. The Gaussian variational energy for the harmonic oscillator reaches $E_0=1/2$ because its family contains an exact ground state. The exact variational solution in a two-level system makes the equality condition visible as vanishing excited-state probability amplitude.

The method is useful when solving the Schrödinger equation exactly is difficult but integrals of a trial wavefunction are tractable. Enlarging a trial family cannot increase its optimal energy, although an optimum need not be attained. The beauty is that even an imperfect trial gives a rigorous bound, provided its form energy is computed exactly: numerical integration errors require separate control. The Gaussian variational estimate for a solvable sextic potential provides an existing example beyond a quantum harmonic oscillator.

A short proof explains the direction. With a complete orthonormal eigenbasis of energy eigenstates, write $\psi ={\sum }_n{c}_n{\varphi }_n$, where $H{\varphi }_n=E_n{\varphi }_n$ and ${\sum }_n|c_n{|}^2=1$. ThenThis is a weighted mean of nonnegative energy differences; equality requires every occupied energy eigenstate to have energy $E_0$. ^[ref]. For continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure of $\psi$, with the same nonnegative integrand. This proves the bound without assuming an orthonormal eigenbasis exists.

The historical setting predates quantum mechanics: Rayleigh's The Theory of Sound, volume I, appeared in 1877, and Ritz's paper on solving variational problems of mathematical physics appeared in volume 135 of Crelle's journal, catalogued by its publisher as 1909, pp. 1–61. These are original works associated with the Rayleigh-Ritz method; the dates here identify the publications rather than assert priority. ^[ref]; ^[ref].

A Wannier-Mott exciton has an internal wave function extending over many unit cells. Its simplest model assumes a three-dimensional bulk crystal, approximately parabolic valence band and conduction band, positive isotropic scalar effective masses $m_e^{\ast },m_h^{\ast }$, a constant scalar relative permittivity ${ϵ}_r$, and a dilute pair population. Replace microscopic crystal forces by those effective masses and dielectric screening, and neglect exchange corrections and coupling to phonons. These assumptions must hold on the spatial and energy scales of the particular exciton.

With relative coordinate $\mathbf{r}={\mathbf{r}}_e-{\mathbf{r}}_h$, the internal Hamiltonian operator is the three-dimensional hydrogen atom problem with a reduced mass and a screened Coulomb potential:Here $e$ is the elementary charge, ${ϵ}_0$ the vacuum permittivity, and $\hslash$ the reduced Planck constant. The internal energy eigenvalues, radius scale, and exciton binding energies areThe principal quantum number is $n=1,2,\dots$. This reduction and these scales are derived in Excitons in Semiconductor Quantum Dots by Matjaž Humar (§3, printed pp. 5–6, equations (13)–(26)).

To check the scales by dimensional analysis, write $C=e^2/(4\pi {ϵ}_0{ϵ}_r)$, which has units of energy times length. Then $a_B^{\ast }={\hslash }^2/(\mu C)$ is a length and $R^{\ast }=\mu C^2/(2{\hslash }^2)$ is an energy. Rescaling $r=a_B^{\ast }\rho$ makes $H_{\mathrm{rel}}/R^{\ast }=-{\nabla }_{\rho }^2-2/\rho$, the dimensionless hydrogen atom operator. The Bohr radius $a_B^{\ast }$ is a wave function scale, not the radius of a classical orbit; for this model's lowest internal bound state, the mean separation is $3a_B^{\ast }/2$.

For the diagrams' hypothetical inputs $m_e^{\ast }=0.10m_0$, $m_h^{\ast }=0.40m_0$, ${ϵ}_r=10$ and $E_g=1.500\mathrm{eV}$, where $m_0$ is the free electron mass, independent substitution gives $\mu =0.080m_0$, $a_B^{\ast }\simeq 6.615\mathrm{nm}$ and $R^{\ast }\simeq 10.885\mathrm{meV}$. Here electronvolts express energy. The exciton binding energies for $n=1,2,3$ are approximately $10.885$, $2.721$ and $1.209\mathrm{meV}$. These are model calculations, not measured material parameters. The large-radius approximation additionally requires $a_B^{\ast }$ to be much larger than the crystal's unit cell length scale. In a thin layer, with anisotropic effective masses, or with spatially varying dielectric screening, this three-dimensional formula is not generally valid.

The isotropic Weyl Hamiltonian acts on a two-component spinor as $H=v\mathbf{\sigma }\cdot \mathbf{p}$. For a plane wave of momentum $\hslash \mathbf{k}$, its two energy eigenvalues are $E_{\pm }=\pm \hslash |v||\mathbf{k}|$. For $v>0$ the corresponding Bloch vectors point along and opposite $\mathbf{k}$, respectively. If the Pauli matrices describe physical spin one-half, the mean spin is $\pm (\hslash /2)\hat{\mathbf{k}}$; in an effective two-band model the same algebra can describe an internal pseudospin instead. The bands meet linearly at the origin. ^[ref]

An equal incoherent mixture within a fixed total-$J$ multiplet has density operator $\rho =P_J/(2J+1)$, where $P_J$ is its orthogonal projection. If the applied-field Hamiltonian operator is proportional to $L_z+2S_z$ and no spin-orbit interaction is retained, its energy eigenstates are the uncoupled states $|m_L,m_S⟩$. The Born rule gives their populations by the formula above, or equivalently by squared Clebsch-Gordan decomposition coefficients. Equal populations of coupled states need not give equal populations of magnetic energy eigenvalues.

For $L=1,S=1/2,J=3/2$, the nonextremal coupled states have weights $2/3$ and $1/3$ in their two uncoupled components. The five dimensionless magnetic energies $-2,-1,0,1,2$ then have populations $1/4,1/6,1/6,1/6,1/4$; the zero energy contains two independent states, each with population $1/12$. In a different, weak-field approximation with a dominant spin-orbit interaction, projecting into the isolated multiplet gives the Landé g-factor spectrum instead. The full operator and its projection must not be interchanged.