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The four direct beta decay branching fractions exhaust the parent decays in the model, soHere the question's ${\Gamma }_i$ are dimensionless branching fractions, despite the conventional use of $\Gamma$ for a dimensionful decay width. Its “zeroth excited state” means the daughter ground state. Cascades through an excited level do not count as an additional direct beta branch to it.

As an estimate, model the confinement by an infinite square well of width $L$ and use the stipulated electron mass. Its energy eigenvalues are $E_j=j^2{\pi }^2{\hslash }^2/(2m{L}^2)$, $j=1,2,\dots$, so the gap controlling thermal occupation of a quantum well isThus $T\ll 130\mathrm{K}$ suppresses excitation: $p_2/p_1=e^{-\Delta E/(k_{B}T)}$. For example $T\lesssim 28\mathrm{K}$ makes this ratio below one percent in this model. There is no sharp cutoff. The ground state energy itself corresponds to about $44\mathrm{K}$, but it is the excitation gap that enters the Boltzmann factor. Finite barrier height and a different effective mass would change the numerical scale; neither is supplied.

The parity operator acts as $P\psi (x)=\psi (-x)$, with $P^2=I$. For an even potential energy $[H,P]=0$, so energy eigenstates can be chosen even or odd, with parity eigenvalues $+1$ and $-1$. One-dimensional bound states are nondegenerate; the ground state is nodeless and even.

For the finite square well, square integrability excludes the growing exterior exponential, while even parity excludes the interior sine. Thus the interior solution is proportional to $\cos kx$, with matching decaying tails andContinuity at $a$ gives $B\cos ka=A{e}^{-qa}$ and $-Bk\sin ka=-qA{e}^{-qa}$. Their ratio yieldsThe conditions at $-a$ follow from the same parity.

For a narrow weak well, $ka\ll 1$, hence $q\simeq k^{2}a\simeq 2m{V}_{0}a/{\hslash }^2$ andThis is self-consistent because $|E_0|/V_0\simeq 2m{V}_0{a}^2/{\hslash }^2\ll 1$. It is the shallow finite-square-well energy, equivalently the attractive delta potential limit with integrated strength $2a{V}_0$. An arbitrarily weak attractive well of this form still has its even bound ground state.

The first excited state must be odd. Replace the interior cosine by $\sin kx$ and choose exterior tails with opposite signs. The matching condition becomes $q=-k\cot ka$. Its first root requires $\pi /2<ka<\pi$. At the binding threshold $q\to 0$, sois necessary and sufficient for the first odd bound state. Equality gives a zero-energy nonnormalizable threshold solution, not an additional bound state.

For a normalizable state take $a>0$. inverse construction of a quantum potential gives $V=E+{\hslash }^2{\Psi }^{\prime \prime }/(2m\Psi )$. differentiation yields ${\Psi }^{\prime \prime }/\Psi =16a^2{x}^6-12a{x}^2$, so the chosen potential energy zero fixes $E=0$. ThusThis energy lies above the negative minima of the potential energy; its zero value does not mean zero kinetic energy. The nodeless eigenfunction is the ground state.

Put $\kappa =e^2/(4\pi {ϵ}_0)$. Substitution of $\psi =R(r)\Phi (\varphi )$ into the time-independent Schrödinger equation, followed by division by $R\Phi$ and multiplication by $r^2$, separates the angular dependence:Thus $\Phi =e^{im\varphi }/\sqrt{2\pi }$ andThe single-valued wavefunction condition, $\Phi (\varphi +2\pi )=\Phi (\varphi )$, requires $m\in \mathbb{Z}$. The regular boundary condition at the origin selects $R\sim r^{|m|}$. Accordingly use $s=|m|$ in the radial ansatz; the printed $x^m$ notation describes the $m\ge 0$ representatives.

For the two-dimensional Coulomb bound state, substitution of $R=x^s{e}^{-x/2}G$ with the stipulated radial and energy scales givesThe coefficient of $x^n$ in the power series isThe regular solution otherwise has an exponentially growing contribution at large $x$, incompatible with being normalizable. Termination at degree $n_r\ge 0$ therefore requiresThe resulting $G$ is proportional to a Generalized Laguerre polynomial.

Define $R_{\mathrm{y}}=\mu {\kappa }^2/(2{\hslash }^2)$, the hydrogenic Rydberg energy for the stipulated mass. The first three distinct two-dimensional energies are $-4R_{\mathrm{y}},-4R_{\mathrm{y}}/9,-4R_{\mathrm{y}}/25$, whereas in three dimensions they are $-R_{\mathrm{y}},-R_{\mathrm{y}}/4,-R_{\mathrm{y}}/9$. Both series approach zero from below, but the planar ground state is four times more tightly bound. Without spin the respective quantum degeneracies are $1,3,5$ and $1,4,9$; if individual degenerate states are counted, the second and third planar states already share the first excited energy. This comparison retains the supplied $1/r$ interaction; it does not replace it by the logarithmic electrostatic potential of a genuinely two-dimensional electromagnetic theory.

The canonical commutation relations are $[x,p_x]=[y,p_y]=i\hslash$, with $[x,y]=[p_x,p_y]=[x,p_y]=[y,p_x]=0$. Operators from different Cartesian modes commute, so $(p_x{p}_y{)}^{†}=p_y{p}_x=p_x{p}_y$ and $(xy{)}^{†}=yx=xy$. Thus $J_1$ is Hermitian and represents an observable on the usual oscillator domain.

Introduce annihilation operators $a_x=\sqrt{m\omega /(2\hslash )}x+i{p}_x/\sqrt{2m\hslash \omega }$ and similarly $a_y$. Set $N_x=a_x^{†}{a}_x$, $N_y=a_y^{†}{a}_y$ and $N=N_x+N_y$. The Schwinger boson representation givesFor $A=a_x^{†}{a}_y$ and $B=a_y^{†}{a}_x$, $AB=N_x(N_y+1)$ and $BA=N_y(N_x+1)$. ThereforeThe two-mode oscillator Casimir identity isSince $[N,A]=[N,B]=0$, $[H,J_1]=0$, and $J_1$ is conserved. However, $[J_1,J_2]=i\hslash J_3$ is not the zero operator, so there is no general joint eigenbasis. The exception is important: a common eigenstate of two angular-momentum components must satisfy $J_3\psi =0$, then the other commutators force $J_1\psi =J_2\psi =0$. Thus only the $j=0$ sector admits a common state; here that is the oscillator ground state.

Using $J^2={\hslash }^{2}j(j+1)$, with $j=0,1/2,1,\dots$, and selecting the positive energy branch givesEquivalently $N=2j=0,1,\dots$, so $E_N=\hslash \omega (N+1)$ and the degeneracy is $N+1$, corresponding to $N_x=0,\dots ,N$.

For observables with finite variances, the Heisenberg uncertainty principle gives $\Delta A\Delta B\ge |⟨[A,B]⟩|/2$; in particular $\Delta x\Delta p\ge \hslash /2$. It concerns the spreads of measurements in a prepared state, rather than only disturbance by a measuring device. Identical integer-spin particles have symmetric full states and Bose-Einstein statistics; half-integer-spin particles have antisymmetric full states and Fermi-Dirac statistics. The latter imply Pauli exclusion from one complete single-particle state, including spin.

Inside an infinite square well, solutions satisfying both zero boundary conditions arewith zero wavefunction outside. For the ground state, direct integration givesThe infinite-well ground-state uncertainty product follows:The ground state is not a minimum-uncertainty Gaussian. Here momentum variance is the usual kinetic-momentum second moment of the confined state; the hard-wall momentum operator has domain subtleties that do not alter this kinetic-energy result.

For spin degeneracy in filling a one-dimensional well, let $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$ and assume the spin levels are degenerate and available. Four spin-zero bosons all occupy the lowest orbital, giving $4E_1$. Four spin-$1/2$ fermions fill two spin states in each of $n=1,2$, giving $10E_1$. Spin-$3/2$ fermions have four spin projections, so all four can occupy $n=1$ in different complete states, giving $4E_1$. These fillings do not impose full spin polarization.

The anisotropic quantum harmonic oscillator is a sum of three commuting one-coordinate oscillator Hamiltonians. Substituting $\psi =X(x)Y(y)Z(z)$ and dividing by $XYZ$ gives a sum of three functions of independent coordinates, so each equals a separation constant. ConsequentlyFor a time-independent operator, conservation for arbitrary states follows from $[H,Q]=0$, since $d⟨Q⟩/dt=(i/\hslash )⟨[H,Q]⟩$. The first labelled operator is $i{L}_z/\hslash$; the second is $-L^2/{\hslash }^2$. The kinetic energy is rotationally invariant. The potential is invariant under rotations about $z$ precisely when (i) ${\omega }_x^2={\omega }_y^2$; full rotational invariance requires (ii) ${\omega }_x^2={\omega }_y^2={\omega }_z^2$. For positive oscillator frequencies these are equality of the frequencies themselves. Acting with the commutators on arbitrary test functions excludes anisotropic potentials in case (ii).

In the isotropic case let $b=m\omega /\hslash$. ThenThe polynomial $x$ is the first excited oscillator factor and the other two factors are ground-state Gaussians. The first excited spatial level has three independent choices, proportional to $x,y,z$. Odd parity gives $⟨x⟩=0$, and Gaussian integration yieldsFor oscillator uncertainty and allocation of classical energy, the $x$ mode has energy $3\hslash \omega /2$, whose classical amplitude is $A_x=\sqrt{3\hslash /(m\omega )}$, giving $\Delta x=A_x/\sqrt{2}$. If “same energy” instead means assigning the entire three-dimensional energy $5\hslash \omega /2$ to one classical coordinate, its amplitude is $\sqrt{5\hslash /(m\omega )}$ and the ratio is $\sqrt{3/10}$. A general three-dimensional classical orbit's $x$ amplitude is not fixed by total energy without specifying its allocation among modes.

The capacity of an isotropic oscillator shell, including the two electron spin states, is $(N+1)(N+2)$ electrons. Nine non-interacting electrons occupy shells as $2,6,1$ for $N=0,1,2$. ThusThe least positive excitation promotes an electron by one oscillator quantum. Rearrangements within the partly filled degenerate shell remain ground states and cost no energy.

For the quantum harmonic oscillator, put ${\ell }_0=\sqrt{\hslash /(m\omega )}$ and $\xi =x/{\ell }_0$. The normalized energy eigenstates arewhere $H_n$ is a Hermite polynomial. Successive energies are equally spaced; the ground state retains zero-point energy $\hslash \omega /2$. The $n$th wavefunction has $n$ nodes of a wavefunction and parity $(-1{)}^n$. In particular, the first three shapes are proportional to $e^{-{\xi }^2/2}$, $\xi e^{-{\xi }^2/2}$, and $(2{\xi }^2-1)e^{-{\xi }^2/2}$. The classical turning points satisfy $x=\pm {\ell }_0\sqrt{2n+1}$; the wavefunctions extend beyond them and decay, rather than ending abruptly.

The wavefunctions are plotted with arbitrary vertical scale about their energy levels. Their signed excursions are amplitudes, not additional energies or probability densities.

Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.

Evolution of one energy eigenstate. For a time-independent Hamiltonian operator, write $|\psi (t)⟩=f(t)|{\varphi }_n⟩$. Substitution into the Time-dependent Schrödinger equation gives $i\hslash \dot{f}=E_{n}f$, soOnly a global phase changes, leaving all measurement probabilities unchanged.

Delayed energy measurement. Expand the unchanged initial state in the complete new energy eigenbasis,For the fixed-left-wall geometry, direct overlap integration givesorthogonality isolates the new ground-state amplitude as $c_1{e}^{-i{E}_1^{\prime }t/\hslash }$. Its phase cancels on squaring, so the ground-state probability remains $32/(9{\pi }^2)$ for every delay, provided the expanded well remains fixed and no intervening measurement or environment changes the state. The position probability density generally evolves because different eigenstates acquire different relative phases.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

identical particles have the same intrinsic properties, including mass, charge and spin. Their labels are bookkeeping devices rather than measurable identities. When wave functions overlap, quantum mechanics supplies no observable trajectory that permanently identifies “particle 1” independently of “particle 2”. A many-particle description must therefore treat exchange consistently. The Hamiltonian and physical observables are invariant under relabelling identical particles.

For two particles, the exchange operator obeys ${\hat{P}}_{12}^2=\hat{I}$. The usual three-dimensional symmetrization postulate selects states symmetric under exchange for bosons and antisymmetric for fermions. In relativistic quantum theory, the Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions under its locality and positivity assumptions; the connection is not derived merely from ${\hat{P}}_{12}^2=\hat{I}$ in elementary nonrelativistic mechanics. For orthonormal one-particle states $u$ and $v$, the two-particle spatial combinations areThe sign concerns the complete state, including spin and other internal variables, not necessarily the spatial factor alone.

The antisymmetric state vanishes if both fermions occupy the same one-particle state: this is the Pauli exclusion principle. For $N$ fermions in orthonormal spin orbitals ${\chi }_j$, the normalized Slater determinant $\Psi =(N!{)}^{-1/2}\det [{\chi }_j(i)]$ supplies the required exchange sign. Bosonic occupation has no corresponding upper limit, and symmetrized products describe many bosons. Fermionic occupation is zero or one per complete one-particle state; two electrons can still occupy the same spatial atomic orbital with opposite spin.

Spin and spatial symmetry are linked. A two-electron spin singlet state is antisymmetric, so its spatial factor must be symmetric. A symmetric triplet state instead requires an antisymmetric spatial factor. Exchange then changes pair correlations and the energies available when interactions depend on particle separation, even without introducing a new classical “exchange force”. It contributes to atomic shell structure, bonding and magnetic behaviour.

In an occupation-number description, bosonic creation and annihilation operators satisfy $[{\hat{a}}_i,{\hat{a}}_j^{†}]={\delta }_{ij}$, whereas fermionic operators satisfy $\{{\hat{c}}_i,{\hat{c}}_j^{†}\}={\delta }_{ij}$ and anticommute among themselves. The latter imply $({\hat{c}}_i^{†}{)}^2=0$. This formalism implements the same exchange constraints without assigning permanent particle labels.

For noninteracting particles in thermal equilibrium, the mean occupation of a state of energy $ϵ$ isFermi-Dirac statistics produces a filled Fermi sea at zero temperature and degeneracy pressure; Bose-Einstein statistics permits macroscopic Ground state occupation when the density of available states and thermodynamic conditions allow condensation. The different statistics therefore have measurable collective consequences, not just different notation.^[ref]

When phase-space occupation is dilute, both distributions reduce to the Maxwell–Boltzmann statistics limit. Likewise, well-separated localized wave packets have negligible overlap and small exchange corrections, so effectively labelled classical trajectories can become a useful approximation. In two spatial dimensions more general exchange statistics are possible; the standard boson/fermion account here concerns ordinary three-dimensional particles and their total states.

For the Quantum harmonic oscillator defineThe Canonical commutation relation gives $[\hat{a},{\hat{a}}^{†}]=1$, andThus the creation operator and annihilation operator raise or lower the energy by one quantum. Positivity of ${\hat{a}}^{†}\hat{a}$ and termination of downward stepping give $E_n=\hslash \omega (n+1/2)$, $n=0,1,\dots$. Normalized actions are $\hat{a}|n⟩=\sqrt{n}|n-1⟩$ and ${\hat{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The annihilated ground state obeysThe nonzero Ground state energy is the zero-point energy, consistent with position-momentum quantum uncertainty. Since $\hat{x}=\sqrt{\hslash /(2m\omega )}(\hat{a}+{\hat{a}}^{†})$, a perturbation linear in position connects only neighbouring number states. The algebra is especially useful for matrix elements and selection rules.^[ref]

The original diagram places hard walls at $0$ and $3L$, and the central plateau on $L<x<2L$ is exactly at the indicated ground-state energy. This equality determines the middle of the requested sketch. In the time-independent Schrödinger equation,the central interval has $V=E$, hence ${\psi }^{\prime \prime }=0$. The ground state can be chosen real and positive inside the walls by the nodeless theorem for a one-dimensional ground state. Reflection symmetry makes it even about $x=3L/2$, so its linear central segment must be constant.

The hard-wall boundary conditions are $\psi (0)=\psi (3L)=0$. On both outer intervals $V=0<E$, so the positive wavefunction has negative curvature and is sinusoidal. Wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to remain continuous at $L,2L$; the derivative therefore vanishes where each sinusoidal flank meets the flat middle. A normalized representative isEach side contributes $L/2$ and the plateau contributes $L$ to the squared norm before multiplying by $1/(2L)$. The corresponding consistency condition is $E={\pi }^2{\hslash }^2/(8m{L}^2)$; a numerical energy is not required for the sketch.

Thus the ground state at the top of a central rectangular barrier has no internal nodes, continuous value and slope, and a flat middle, rather than an exponential dip through a classically forbidden region. Multiplication by an overall constant phase changes none of these physical features.

Hamiltonian and commutators. Use the canonical commutation relation $[\hat{x},\hat{p}]=i\hslash I$ and define $c_0=\sqrt{m\omega /(2\hslash )}$. Expanding the product of the given creation and annihilation operators,Therefore the quantum harmonic oscillator Hamiltonian operator isSimilarly,Using $[AB,C]=A[B,C]+[A,C]B$ gives

Lowering a normalized energy eigenstate. The commutator impliesFor $n>0$, nondegeneracy of this one-dimensional energy level means $\hat{a}|{\varphi }_n⟩=c_n|{\varphi }_{n-1}⟩$. Its norm is fixed by the number operator:Choosing successive eigenstate phases in the conventional way makes $c_n=\sqrt{n}$, soFor $n=0$, the same norm is zero, hence $\hat{a}|{\varphi }_0⟩=0$. A nonzero lowered ground state would have energy below the lowest allowed level, and positivity of ${\hat{a}}^{†}\hat{a}$ forbids that possibility.

Coherent-state eigenvalue. Applying the annihilation operator term by term, with the $n=0$ contribution zero, givesThusThe coherent state is normalized because $e^{-|\alpha {|}^2}{\sum }_n|\alpha {|}^{2n}/n!=1$. Its annihilation operator is not Hermitian, so the eigenvalue $\alpha$ may be complex.

Time evolution. Under the time-independent Hamiltonian operator, each energy component acquires its own phase:Since $|\alpha e^{-i\omega t}|=|\alpha |$, the normalization prefactor is unchanged. HenceThe state remains a coherent state, with its complex parameter rotating at the oscillator's angular frequency; the displayed extra factor is a global phase from the zero-point energy.

The Bohr model combines classical circular motion in the Coulomb potential with explicitly quantum postulates. A one-electron atom of nuclear charge $+Ze$ has electron–nucleus reduced mass $\mu$. With $C=Z{e}^2/(4\pi {\epsilon }_0)$, the relative circular orbit obeysBohr postulates permitted nonradiating stationary orbits with $L=n\hslash$, for $n=1,2,\dots$. These assumptions giveFor hydrogen with the nucleus taken infinitely heavy, the $n=1$ radius is the Bohr radius, $a_0\simeq 5.29\times 10^{-11}\mathrm{m}$. The finite reduced mass gives the small nuclear-motion correction, and increasing $Z$ contracts the one-electron orbit.

The circular-motion equation gives $T=C/(2r)$ while the potential energy is $-C/r$, so the total energy is $-C/(2r)$. Substituting the quantized radius yieldsFor the heavy-nucleus hydrogen approximation, $E_n\simeq -13.6\mathrm{eV}/n^2$. The ionization threshold is zero energy, so removing the ground-state electron requires about $13.6\mathrm{eV}$.

The model further postulates that radiation is emitted or absorbed only in transitions between stationary orbits, with photon energy $h\nu =|E_i-E_f|$. Emission from $n_i>n_f$ givesThus it reproduces the principal hydrogen spectral series, their limiting wavelengths, and the gross hydrogenic energy levels. For example, $n=3\to 2$ in hydrogen has $\Delta E\simeq 1.89\mathrm{eV}$ and $\lambda \simeq 656\mathrm{nm}$, a visible line. Absorption reverses the transition when the photon matches the energy gap.

The achievement is that quantized energies explain discrete spectra, but the model's stationary orbits are postulates rather than consequences of classical electrodynamics: an accelerating classical charge would radiate. It does not provide a general calculation of line intensities or transition selection rules and fails as a quantitative theory of many-electron atoms. Spin, fine structure and the full Zeeman effect require further physics. Modern quantum mechanics replaces definite electron orbits by wavefunctions and quantum probabilities. In particular, the true hydrogen ground state has $\ell =0$, so its orbital angular momentum vanishes; the model's rule $L=n\hslash$ should not be identified with the modern orbital eigenvalue rule $L^2=\ell (\ell +1){\hslash }^2$. Bohr's energy quantization remains a successful hydrogenic approximation, while its literal orbital picture does not.

The infinite square well has energy levels $E_n=n^2{h}^2/(8m_e{L}^2)$, with $n=1,2,\dots$. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely $n=1\to 2$. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap givesHere the photon gap is approximately $1.55\mathrm{eV}$, not the ground-state energy itself.

Assume $a>0$ and define $\omega =\sqrt{2a/m}$ so that the potential is that of a quantum harmonic oscillator. The two trial forms have derivativesSubstitution into the stationary Schrödinger equation cancels the $x^2$ terms precisely when $a=2{\hslash }^2{C}^2/m$. Normalizability selects the positive root,The remaining constants giveFor later overlaps choose positive normalization constants. The Gaussian integral gives $A=(2C/\pi {)}^{1/4}$, while its second-moment integral gives $B=2\sqrt{C}A$. Thus both wavefunctions have unit norm and are orthogonal by their opposite parity.

The real Gaussian is positive everywhere and has no nodes; the odd state has exactly one node at $x=0$. The Sturm oscillation theorem orders one-dimensional bound states by their node count. Consequently these are the ground state and first excited state, respectively, also agreeing with the oscillator spectrum $E_n=(n+1/2)\hslash \omega$.

The infinite wall imposes $\boxed{\Psi (0)=0}$, and the wavefunction vanishes for $x<0$. The even full-oscillator ground state fails this boundary condition; the lowest admissible state is the positive-half restriction of the odd first excited state. Multiplying by $\sqrt{2}$ restores normalization because half its probability originally lay on either side.

With $\ell =\sqrt{\hslash /(m\omega )}$, the half-harmonic oscillator ground state isIt rises from zero at the wall, reaches its wave-amplitude maximum at $x=\ell$, and decays to zero as $x\to \infty$. It is positive throughout the interior of the allowed region, with no interior node, as required for its ground state. The wall need not make the derivative continuous with the forbidden half-line.

For removal of a half-oscillator wall, the sudden approximation keeps the state unchanged during the change of potential. Its probability in the new full-oscillator ground state ${\Psi }_0={\pi }^{-1/4}{\ell }^{-1/2}{e}^{-x^2/(2{\ell }^2)}$ is the squared overlap:where ${\int }_0^{\infty }x{e}^{-x^2/{\ell }^2}dx={\ell }^2/2$. ThereforeThe integral runs only over the initially occupied half-line. The initial state is not the odd full-line eigenstate: it is zero, rather than negative, on the left. Indeed its overlap with that final first excited state is $1/\sqrt{2}$, giving probability $1/2$, and the remaining probability lies in higher states. Subsequent evolution under the new fixed Hamiltonian operator changes the quantum phases of its energy components without changing these energy-measurement probabilities.