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For the usual three-dimensional Coulomb potential Hamiltonian operator, a separated radial wavefunction $R(r)$ uses the reduced function $u=rR$, whose radial norm has measure $dr$. The regular physical operator domain imposes $u(0)=0$. Near the origin the leading radial equation gives the indicial equation $c(c-1)-\ell (\ell +1)=0$ for $u\sim r^c$, with roots $c=\ell +1$ and $c=-\ell$.

The regular root gives $R\sim r^{\ell }$. The other root gives $R\sim r^{-\ell -1}$, and its radial probability integral ${\int }_0^{\epsilon }{r}^2|R{|}^{2}dr\sim {\int }_0^{\epsilon }{r}^{-2\ell }dr$ diverges for $\ell \ge 1$. At $\ell =0$, $R\sim 1/r$ is locally square-integrable in three dimensions, but violates $u(0)=0$ and has divergent kinetic energy integral ${\int }_0^{\epsilon }{r}^2|R^{\prime }{|}^{2}dr\sim {\int }_0^{\epsilon }{r}^{-2}dr$. Thus local wavefunction normalization alone is insufficient to select the physical branch in this exceptional case; the standard Coulomb energy domain supplies the missing condition.

The energy eigenspace at $E$ is the vector subspacewhere $H$ is the Hamiltonian operator and $I$ is the identity operator. It is nonzero exactly when $E$ is an energy eigenvalue. Its dimension is the multiplicity of that energy eigenvalue; degenerate energy levels have multiplicity greater than one. The ground-state subspace is ${\mathcal{E}}_{E_0}$, and its dimension is the ground-state degeneracy. A vector in an energy eigenspace need not be one of a specially chosen orthonormal basis of energy eigenstates.

Linearity makes the useful point immediate: if $H{\varphi }_1=E{\varphi }_1$ and $H{\varphi }_2=E{\varphi }_2$, then $H(a{\varphi }_1+b{\varphi }_2)=E(a{\varphi }_1+b{\varphi }_2)$. Every normalized nonzero combination stays in the same energy eigenspace. For a time-independent Hamiltonian operator, its evolution is multiplication by the common global phase $e^{-iEt/\hslash }$, so it is a stationary state. A quantum superposition within a degenerate energy eigenspace can change spatial shape when its coefficients are chosen differently, but its density does not subsequently beat under this unchanged Hamiltonian operator.

Across distinct energy eigenspaces, relative quantum phases evolve. For orthonormal eigenfunctions ${\varphi }_1,{\varphi }_2$ of the Hamiltonian operator with energy eigenvalues $E_1\ne E_2$, an equal-weight quantum superposition hasand position probability densityThe interference term can change local probabilities even though the Born rule assigns the same time-independent probability $1/2$ to each energy measurement outcome. Orthogonality makes its integral zero, preserving wavefunction normalization. This pure state is not a stationary state; some particular measurement distributions may nevertheless remain constant. An incoherent mixed state with the same energy populations has no such interference term.

For the pictured infinite square well, $\theta =3E_{1}t/\hslash$ and the interference repeats after $T=2\pi \hslash /(3E_1)$. At $t=0$, direct integration gives $⟨x⟩=L/2-16L/(9{\pi }^2)$; halfway through the period it is $L/2+16L/(9{\pi }^2)$. This is a moving expectation value, rather than a trajectory of an individual particle. The distinction between an energy eigenspace and a combination of different energy eigenspaces is what makes the different behavior possible.

An energy eigenvalue is an eigenvalue of the Hamiltonian operator $H$: there is a nonzero vector $\psi$ in its operator domain $D(H)$ withFor a self-adjoint operator, $E$ is real. Normalizing $\psi$ produces an energy eigenstate; an ideal energy measurement then returns $E$ with probability one. The associated energy eigenspace includes all vectors with this same energy eigenvalue, including the zero vector, which represents no quantum state. Different energy eigenvalues have orthogonal energy eigenspaces. ^[ref].

The energy eigenvalues form the point spectrum, which can be only part of the spectrum. The term discrete spectrum more specifically means isolated eigenvalues of finite multiplicity. The usual fixed-nucleus hydrogen atom model has discrete negative bound state energy eigenvalues together with a continuum at nonnegative energy. ^[ref]. A quantum system need not have a ladder of discrete energy eigenvalues: a whole-line free particle has continuous spectrum. A plane wave used for a free particle is a generalized eigenfunction of the Hamiltonian operator, rather than a normalized vector in the physical Hilbert space. ^[ref].

For the infinite square well $0<x<L$, the Dirichlet boundary conditions giveSubstitution into $H{\varphi }_n=E_n{\varphi }_n$ and wavefunction normalization verify these statements directly. Here $E_1$ is the ground-state energy, even though its index is one; the labeling convention does not determine which quantum state is lowest. Equal populations of ${\varphi }_1$ and ${\varphi }_2$ have energy expectation value $(E_1+E_2)/2=5E_1/2$, but an energy measurement returns $E_1$ or $4E_1$, each with probability $1/2$. An average of allowed energy eigenvalues need not itself be an energy eigenvalue.

Consider an original one-dimensional effective mass model with $V(x)=0$ for $|x|<a$ and $V(x)=V_0>0$ outside. Take the same positive effective mass $m^{\ast }$ everywhere, an abrupt interface, and no interactions. The wavefunction represents a transverse envelope; free motion parallel to the layer is outside this calculation. A double heterostructure laser need not be thin enough for this quantization to dominate its operation.

Set $s=x/a$, $ϵ={\hslash }^2/(2m^{\ast }{a}^2)$, $z_0^2=V_0/ϵ$, $z^2=E/ϵ$ and $w=\sqrt{z_0^2-z^2}$. The time-independent Schrödinger equation becomesHere $u(s)$ denotes the dimensionless potential; a bound state has $0<z<z_0$. Matching the wavefunction and its derivative at $s=1$ givesFor the even bound state, the interior solution is $A\cos (zs)$ and the exterior solution is $A\cos z{e}^{-w(|s|-1)}$. For the odd bound state, replace $\cos (zs)$ by $\sin (zs)$ and $\cos z$ by $\mathrm{sgn}(s)\sin z$. The physical wavefunction is $\psi (x)=\varphi (x/a)/\sqrt{a}$.

Wavefunction normalization includes both exterior tails, not just the active layer. Direct integration givesThe first two terms are the interior integral; the last is the total exterior integral. Thus the interior probability is $P_{\mathrm{in}}=A^2[1\pm \sin (2z)/(2z)]$, with the upper sign for even parity.

Choose $z_0=2$. There is one even root in $(0,\pi /2)$ and one odd root in $(\pi /2,2)$; higher branches start beyond $z_0$. Each left-hand side increases on its stated interval while $\sqrt{4-z^2}$ decreases, proving uniqueness there. Numerically,These are calculated values of the chosen finite square well. The excited bound state lies closer to the barrier and has a longer exterior tail: being bound does not mean having unit probability inside the well.

This model demonstrates electronic confinement, not optical refractive index confinement or a complete laser. Unequal effective masses, band bending, interface disorder and interactions require a more detailed model. In particular, continuity of the ordinary derivative above uses the chosen constant effective mass assumption.

For a free particle of mass $m>0$ in nonrelativistic quantum mechanics, choose the potential energy zero and writeApplying the momentum operator $-i\hslash \nabla$ returns $\mathbf{p}\psi$; this fixes the sign convention. The angular frequency is measured relative to this chosen energy zero. Adding a constant energy changes only a time-dependent global phase, leaving the spatial de Broglie wavelength unchanged. No rest-energy term is included in this nonrelativistic quantum mechanics convention.

The plane wave has $|\psi {|}^2=|A{|}^2$ and probability current $\mathbf{j}=(\mathbf{p}/m)|A{|}^2$. Its real part can be drawn as a sine or cosine, but that graph is a component of a probability amplitude, not a transverse displacement. On infinite space this is a generalized momentum eigenstate, not a state satisfying wavefunction normalization. A narrow wave packet instead moves with group velocity ${\nabla }_{\mathbf{k}}\omega =\mathbf{p}/m$; the phase velocity $\omega /|\mathbf{k}|=|\mathbf{p}|/(2m)$ is not its transport speed.

The equality of canonical momentum and kinetic momentum here assumes a field-free region with the magnetic vector potential chosen zero. For a charged particle in a magnetic field, the kinetic momentum is $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$. An arbitrary canonical phase gradient must not then be substituted into $m\mathbf{v}$; the free particle formula needs its stated assumptions.

Use the quantum harmonic oscillator in units with reduced Planck constant $\hslash =1$, mass $m=1$, and angular frequency $\omega =1$. Its Hamiltonian operator and trial Gaussian wave packet areThis is the same Gaussian wave packet used for the Heisenberg uncertainty relation: $|{\psi }_{\sigma }{|}^2$ is a normal distribution with mean zero and variance ${\sigma }^2$. The Gaussian integral gives wavefunction normalization and $⟨x^2⟩={\sigma }^2$. These smooth, rapidly decaying trials lie in the one-dimensional harmonic oscillator form domain and even in the operator domain.

Since ${\psi }_{\sigma }^{\prime }=-x{\psi }_{\sigma }/(2{\sigma }^2)$, integration by parts with vanishing boundary terms gives the kinetic energy and potential energy:Narrow trials cost kinetic energy, while broad trials cost potential energy. The exact differenceshows that the unique optimal width is ${\sigma }^2=1/2$, with $E=1/2$. At $\sigma =1/2$ and $\sigma =1$, the trial energies are both $5/8=0.625$; the kinetic energy and potential energy contributions exchange places.

Minimizing within this trial family alone establishes $E_0\le 1/2$. To establish exactness, use the Heisenberg uncertainty relation to obtain $E_{\psi }\ge 1/2$ for every admissible quantum state, as in its existing oscillator example. Together the bounds give $E_0=1/2$. Direct substitution also yields $H{\psi }_{1/\sqrt{2}}={\psi }_{1/\sqrt{2}}/2$, agreeing with the Gaussian eigenstate for a quadratic potential. This is a useful distinction: the variational method supplies the upper bound, while the Heisenberg uncertainty relation supplies the complementary lower bound.

Consider hard walls at $x=0,3L$, zero potential energy on $0<x<L$ and $2L<x<3L$, and a central rectangular barrier whose height equals the ground state energy $E$. The time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=0$ on the central interval. Reflection symmetry and the nodeless theorem for a one-dimensional ground state require an even, positive state about $3L/2$, hence a constant central segment. Wavefunction matching at a finite potential step makes the derivatives vanish at $L,2L$. The side segments are proportional to $\sin (\pi x/(2L))$ and $\sin (\pi (3L-x)/(2L))$. Their common plateau value is $1/\sqrt{2L}$ after wavefunction normalization, and $E={\pi }^2{\hslash }^2/(8m{L}^2)$. Equality of barrier height and energy is essential: a higher barrier gives positive curvature of a positive wavefunction in the middle, while a lower barrier gives negative curvature.

The infinite square well is an ideal confinement model with potential energy $V(x)=0$ for $0<x<L$ and infinite walls outside. The walls impose Dirichlet boundary conditions $\varphi (0)=\varphi (L)=0$ and a vanishing exterior wavefunction. Inside, the time-independent Schrödinger equation isThe first wall removes the cosine term; the second requires $\sin (kL)=0$, hence $k=n\pi /L$. The $n=0$ solution is identically zero and cannot satisfy wavefunction normalization. Each energy eigenstate has $n-1$ interior nodes of a standing wave, at $x=jL/n$ for $j=1,\dots ,n-1$. More curvature costs more kinetic energy, explaining the $n^2$ growth. The positive ground-state energy is a confinement effect: a constant zero-energy wavefunction cannot satisfy both walls and remain normalized.

For an electron with mass $m_e=9.1093837139\times 10^{-31}\mathrm{kg}$ in a box of length $L=1\mathrm{nm}$, use $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$. These constants are from the 2022 CODATA adjustment. ^[ref]. Substitution givesDoubling $L$ quarters every energy eigenvalue; doubling $m$ halves them. This scaling makes the particle in a box useful for estimating confinement energies before attempting a more realistic calculation. A finite square well permits exterior tails for bound states and has only finitely many bound energy eigenstates. For an electron, the scalar box calculation assumes fixed spin and no spin-dependent interactions.

For the ground state, the probability of finding the particle in the middle half of the box is fully computable:Thus half the available length contains about $81.8\%$ of the probability. The figure contrasts the signed wavefunctions with their nonnegative probability densities: a negative lobe is an amplitude with opposite quantum phase, not negative probability.

An equal quantum superposition of the first two energy eigenstates gives a simple example of moving density:Its density period is $T=h/(E_2-E_1)\approx 3.66608\mathrm{fs}$ for the electron box above. At $t=0$ the cross term enhances the left side; at $T/4$ it vanishes; at $T/2$ it enhances the right side. More quantitatively,which takes the values $0.924413$, $0.5$, and $0.075587$ at those times. Orthogonality keeps the total probability one throughout. An energy measurement still returns $E_1$ or $E_2$ with equal probabilities: changing position density arises from relative quantum phase, not changing populations. A mixed state describing a statistical mixture with those populations has the constant probability density $({\varphi }_1^2+{\varphi }_2^2)/2$; it lacks this matter-wave interference term.

Robert Laughlin's 1983 variational method uses pairwise zeros to suppress close electron encounters within the lowest Landau level. For aligned spin, odd $m$ gives fermionic exchange symmetry.^[ref]

Here $\mathcal{N}$ enforces wavefunction normalization, ${\ell }_B$ is the magnetic length, and complex coordinates $z_j=x_j-i{y}_j$ describe electrons in a magnetic field $B\hat{\mathbf{z}}$, $B>0$, with the symmetric gauge. Reversing the orientation conjugates the polynomial without changing probability densities. The Gaussian function confines the wavefunction.

The exchange check is especially simple for two electrons. Writing $w=z_1-z_2$, exchange sends $w\mapsto -w$, so $w^m\mapsto (-1{)}^m{w}^m$. An aligned two-electron spin state is symmetric, requiring the spatial wavefunction to change sign. Hence $m=1,3,5,\dots$ works, whereas $m=2$ would give the wrong sign. At a close encounter, the probability density vanishes as $|w{|}^{2m}$: $m=3$ imposes a sixth-order zero, stronger than the quadratic zero at $m=1$. This is the local geometric mechanism for avoiding a large repulsive Coulomb potential energy.

A useful large-$N$ counting outline follows directly from the polynomial. Each complex coordinate has maximum degree $m(N-1)$. A lowest Landau level orbital proportional to $z^k{e}^{-|z{|}^2/(4{\ell }_B^2)}$ has its areal probability density maximum at $r^2=2k{\ell }_B^2$, by differentiating its logarithm. This suggests droplet area $A\simeq 2\pi mN{\ell }_B^2$, neglecting the boundary correction. Dividing $N/A$ by the degeneracy of a Landau level per area, $1/(2\pi {\ell }_B^2)$, gives filling factor $\nu \simeq 1/m$. This counting is a motivation for the fraction, not an independent proof that the wavefunction minimizes the interacting Hamiltonian.

The LCAO approximation builds a molecular orbital as a linear combination of single-centre atomic orbitals. For two equivalent normalized real $1s$ functions ${\varphi }_A$ and ${\varphi }_B$, choose their signs positive and define their orbital overlap $S$. At nonzero separation, $0<S<1$, and the normalized combinations areHere ${\sigma }_{+}={\sigma }_g$ and ${\sigma }_{-}={\sigma }_u^{\ast }$. Equal coefficients use the equivalence of the two centres; a general LCAO need not have equal coefficients. See Field, MIT 5.61, lecture 24, pp. 3–6: LCAO coefficients and overlap integrals.

For the Gaussian function toy's $a=1$, the two denominators are $\sqrt{2(1+S)}\approx 1.654013$ and $\sqrt{2(1-S)}\approx 1.124385$. To check wavefunction normalization, expand the square: $\int ({\varphi }_A\pm {\varphi }_B{)}^2=1+1\pm 2S$. Dividing by $2(1\pm S)$ gives one. The local probability density isThe cross term is the interference; adding the separate densities first would lose it. These formulas normalize the chosen linear combination exactly, while the restricted choice of atomic orbitals remains an approximation to an actual molecular orbital.

A coherent state is a normalized eigenstate of the annihilation operator, $a|\alpha ⟩=\alpha |\alpha ⟩$. Expanding in the Fock state eigenbasis gives a recurrence $c_{n+1}=\alpha c_n/\sqrt{n+1}$, and wavefunction normalization fixesThe displacement operator therefore shifts the oscillator ground state in phase space. Distinct coherent states are not orthogonal:

The occupation probabilities form a Poisson distribution,For the quantum harmonic oscillator, the expectation values of position and momentum areThe fluctuations equal those of the ground state,Thus the position wavefunction is a displaced Gaussian wave packet that saturates the Heisenberg uncertainty principle.

Time evolution under $H=\hslash \omega (a^{†}a+1/2)$ preserves this form:The center follows the classical oscillator trajectory while the packet width stays constant. Its mean energy is $\hslash \omega (N+1/2)$; it is generally a superposition of energy eigenstates, rather than a state of definite energy. For $N\gg 1$, the relative occupation fluctuation is $1/\sqrt{N}$, explaining the classical appearance of a highly excited coherent state.

Use $|m_1,m_2⟩$ for a product of the two individual spin-one spin eigenstates, with magnetic quantum numbers $m_i=-1,0,1$. Total $S_z=0$ requires $m_1+m_2=0$, leaving preciselyHere these are product labels, not labels of the total angular momentum. Write their coefficients in this order as $\alpha ,\beta ,\gamma$. With physical spin ladder operators $S_{\pm }=S_{1\pm }+S_{2\pm }$, each nonzero individual step has amplitude $\sqrt{2}\hslash$, soBoth must vanish for a singlet state. The independent product states therefore require $\alpha =\gamma =-\beta$. Wavefunction normalization gives the spin-one singlet stateup to an overall quantum phase. Conversely, $S_z|\psi ⟩=S_{+}|\psi ⟩=0$ implies $S^2|\psi ⟩=0$ through $S^2=S_{-}{S}_{+}+S_z^2+\hslash S_z$, so these conditions really select total spin zero. Distinguishability imposes no additional exchange restriction. The ladder hint uses dimensionless angular momentum units; restoring its common factor $\hslash$ does not change the coefficient ratios.

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.

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.

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 $\{{\varphi }_j\}$ be a complete orthonormal eigenbasis of the Hamiltonian, with $H{\varphi }_j=E_j{\varphi }_j$ and $E_j\ge E_0$. Expand a nonzero admissible trial wavefunction as $\tilde{\Psi }={\sum }_j{c}_j{\varphi }_j$. ThenConsequentlyEvery numerator term is nonnegative and the denominator is positive. Equality requires the trial to lie wholly in the ground-state subspace; it need not be an individually chosen eigenvector when the ground energy is degenerate. This is the quantum variational principle, and wavefunction normalization before calculation is optional because the quotient removes the overall scale.

Wavefunction normalization requires $\alpha >0$. For the normalized probability weight proportional to $e^{-\alpha x^2}$, the Gaussian integral and its derivatives giveSince ${\tilde{\Psi }}^{\prime \prime }=({\alpha }^2{x}^2-\alpha )\tilde{\Psi }$, the kinetic energy and potential energy expectations areThusThe second derivative is positive, and the energy diverges at both ends of $\alpha >0$, so this is the global minimum within the Gaussian function family. The best Gaussian function upper bound isThis is an upper bound, not the exact quartic ground energy.

Wavefunction normalization requires unit total probability:Choosing an irrelevant overall phase to be zero givesThe amplitude has units of inverse square root length, so $|{\Psi }_n{|}^2\mathrm{d}x$ is dimensionless.

Inside the infinite square well, the stationary wavefunctions can be chosen real:They vanish outside the well and at its impenetrable endpoints. A sine meets the left boundary condition; the right one requires $\sin (n\pi )=0$. The putative $n=0$ function is identically zero, so it is not a quantum state. The constants $C_n$ are fixed by wavefunction normalization.

Multiply the eigenfunction expansion by ${\psi }_m^{\ast }$ and integrate over the common domain. Orthogonality and wavefunction normalization giveThus $\boxed{c_m=⟨{\psi }_m,f⟩}$. For a normalized $f$, the Born rule identifies $|c_m{|}^2$ as the probability of this basis outcome.