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The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

The puzzle is the coexistence of reproducible spectral lines and stable chemical atoms. A classical circular orbit in a Coulomb potential can have any negative energy: Coulomb's law alone selects no special radius or emitted frequency. Moreover, its electron has centripetal acceleration. The Larmor formula then predicts positive radiated power, so conservation of energy makes the orbit shrink. Simply drawing concentric circles explains neither stability nor a discrete spectrum.

Niels Bohr postulated nonradiating stationary states and emission with $h\nu =E_i-E_f$. His circular-orbit condition is $L=n\hslash$. These assumptions override classical radiation; the emitted frequency need not equal an orbital frequency.^[ref]

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 eigenstate is a quantum state represented by an eigenvector of the Quantum Hamiltonian. For a time-independent Quantum Hamiltonian, a normalized eigenfunction $\varphi$ with energy eigenvalue $E$ evolves according to the Schrödinger equation asHere $\hslash =h/(2\pi )$ is the reduced Planck constant. The changing factor is a global phase, so this is a stationary state: its position probability density is constant. For example, the second infinite square well mode has two density lobes separated by a node of a standing wave at $x=L/2$; both lobes stay fixed while its quantum phase rotates. An ideal energy measurement returns $E$ with probability one, but a position measurement still has a distribution. Stationary density does not describe a particle sitting at the peak of a lobe. Distinct energy eigenvalues accumulate different quantum phases, making their quantum superposition behave differently.

Suppose a constant Hamiltonian operator has a complete discrete orthonormal basis of energy eigenstates, $H|n⟩=E_n|n⟩$. Expanding an initial pure state and applying the time-evolution operator term by term givesA projective measurement resolving these basis vectors therefore assigns the same Born rule probability $|c_n{|}^2$ to each label $n$ at every time. An energy measurement groups labels sharing the same energy eigenvalue: $P(E)={\sum }_{n:E_n=E}|c_n{|}^2$ is also constant. If $c_m,c_n\ne 0$, their probability amplitude ratio instead evolves asOnly energy differences enter this relative quantum phase. Adding a constant $\epsilon I$ to the Hamiltonian operator multiplies the entire pure state by $e^{-i\epsilon t/\hslash }$, a global phase, and leaves the density operator unchanged.

For a fixed observable $A$, the expectation value can nevertheless vary:Terms joining distinct energy eigenvalues can oscillate; diagonal terms cannot. If $[A,H]=0$, these terms between different energies vanish, so the expectation value is constant. This distinguishes conservation of energy from the stronger claim that every observable is stationary.

An energy eigenstate has only a global phase evolution and is a stationary state. A quantum superposition entirely within one degenerate energy eigenspace is also an energy eigenstate and is stationary. For a general density operator in finite dimension, stationarity under a constant Hamiltonian operator means $[H,\rho ]=0$: differentiating $U\rho U^{†}$ proves necessity, and commuting $\rho$ through the exponential proves sufficiency. Merely fixing the diagonal energy populations does not fix the off-diagonal quantum coherence.

A nonzero probability density inside a barrier does not by itself imply transport. In one dimension, the probability current of a wavefunction is $J=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$. A single evanescent wave $C{e}^{-\kappa x}$ has ${\psi }^{\ast }{\psi }^{\prime }=-\kappa |C{|}^2{e}^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a potential barrier extending to infinity with $E<V_0$.^[ref]

For the finite-barrier wavefunction $A{e}^{\kappa x}+B{e}^{-\kappa x}$, the cross terms instead giveThe relative quantum phase of the two probability amplitudes permits a nonzero, spatially constant probability current, even though neither exponential function separately carries it. The probability continuity equation gives ${\partial }_{x}J=0$ for this stationary state. Matching the wavefunction and its derivative also makes $J$ continuous at both faces, soHere $R=|r{|}^2$ is the quantum reflection probability, and $T=J_{\mathrm{tr}}/J_{\mathrm{inc}}$ is the transmission probability. The net exterior probability current equals the interior probability current; it is not lost as the wavefunction decreases. A physical incident wave packet supplies finite total probability, while the stationary state describes the energy component used to compute its transmitted flux.

In the stationary-state terminology used here, an excited state is a normalized energy eigenstate with energy eigenvalue $E>E_0$, where $E_0$ is an attained ground-state energy. Its excitation cost above the ground state is $\Delta E=E-E_0>0$. Several energy eigenstates can share one excited energy eigenvalue through degenerate energy levels. An arbitrary quantum superposition with energy expectation value above $E_0$ need not be a single stationary excited state; its energy content is described by its Born rule probabilities or, for a continuum, its scalar spectral measure.

An excited state can be a stationary state: the second infinite square well mode is an immediate example. Spontaneous photon emission is a transition caused by coupling matter to the electromagnetic field; it is absent from an isolated matter-only Hamiltonian operator. The relevant energy eigenstate of that isolated model is therefore not automatically an energy eigenstate of the full interacting system. This distinction explains how an ideal stationary excited state can approximate a state with a finite experimental lifetime. See electric-dipole interaction for the coupling used in the simplest optical model.

For a one-photon transition between stationary internal states, neglecting recoil, conservation of energy requiresHere $h$ is the Planck constant and $\nu$ is the photon frequency. Energy matching alone does not guarantee a transition: in the electric-dipole interaction approximation its strength also involves the matrix element $⟨f|\mathbf{d}\cdot \mathbf{ϵ}|i⟩$, where $\mathbf{d}$ is the electric dipole moment operator and $\mathbf{ϵ}$ specifies photon polarization. A zero matrix element gives an electric-dipole selection rule. Thus an absent optical line does not prove the corresponding energy eigenvalue is absent; a different interaction can couple different states. ^[ref]. The Franck-Hertz experiment gives a particularly instructive comparison between collision excitation and photon emission.

The first excited state terminology assumes a least energy eigenvalue above $E_0$. Such a next level is not guaranteed by quantum mechanics: a continuous spectrum can start immediately above the spectral bottom, or energy eigenvalues can accumulate there. For example, a Hamiltonian operator with energy eigenvalues $0,\epsilon ,\epsilon /2,\epsilon /3,\dots$, where $\epsilon >0$, has a ground-state energy zero but no least positive energy eigenvalue: every candidate $\epsilon /n$ is exceeded downward by $\epsilon /(n+1)$. A free particle on the whole line has neither a ground state nor a first stationary excited state. Each use of a discrete excitation ladder therefore needs a specified Hamiltonian operator and its boundary conditions.

Take a nonrelativistic particle of constant mass $m$ with energy $E>0$, and choose the same zero of potential energy on both sides of a rectangular potential barrier:The equality of the exterior potential energies matters: it gives the same exterior wavenumber $k=\sqrt{2mE}/\hslash$ and makes $T=|t{|}^2$. Here $\hslash =h/(2\pi )$ is the reduced Planck constant. This model has a finite jump in potential energy, no delta potential, and no absorbing term.

For $0<E<V_0$, the time-independent Schrödinger equation gives the spatial wavefunctionThe full stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. We set the incident probability amplitude to one and impose no incoming wave from the right; this scattering wavefunction is not a normalizable wavefunction. The tunnelling decay constant $\kappa$ describes exponential decay, rather than an oscillatory wavenumber in the barrier. Both exponential functions are needed on the finite interval: the growing term is not excluded by a condition at infinity.

The wavefunction and its derivative are continuous at each face, giving four linear matching conditions:Their solution supplies the transmission amplitudeA short derivation is to use the two right-face equations to express $A{e}^{\kappa a}$ and $B{e}^{-\kappa a}$ in terms of $t$, then substitute into the left-face equations. Taking the squared modulus gives the rectangular barrier transmission probability.

A ground state is a normalized energy eigenstate that attains the lowest possible energy of a quantum system. Precisely, let the Hamiltonian operator $H$ be a self-adjoint operator bounded below on a Hilbert space, and write $E_{\ast }=\inf \sigma (H)$ for the infimum of its spectrum. A ground state is a unit vector ${\psi }_0$ in the operator domain $D(H)$ satisfyingThus a ground state exists exactly when the spectral bottom $E_{\ast }$ is an energy eigenvalue. The lowest energy eigenvalue among some listed bound states is insufficient if other parts of the spectrum lie below it. The ground-state subspace collects all minimizing energy eigenstates; it can have ground-state degeneracy. The quantum variational principle characterizes $E_{\ast }$ as the infimum of normalized energy expectation values. ^[ref].

For the one-dimensional quantum harmonic oscillator, the ground state has energy $\hslash \omega /2$ above the minimum of the potential energy. This zero-point energy is compatible with being the lowest allowed energy: a localized wavefunction cannot simultaneously have zero kinetic energy. For a free particle on the whole line, the spectral bottom is zero but is unattained by any normalizable wavefunction; see ground-state energy. A lower bound therefore need not describe a physically preparable energy eigenstate.

The ground state supplies a reference for excitation costs and for the quantum variational principle. A ground state is also a stationary state under a time-independent Hamiltonian operator: its quantum phase changes, while its position probability density stays fixed. A nonzero ground-state energy does not imply motion along a classical trajectory, and ground-state degeneracy means there need not be a unique lowest quantum state.

When a ground state exists, its ground-state degeneracy is the dimension of the ground-state subspace:Here $I$ is the identity operator, and the kernel of a linear map is taken within the operator domain of the Hamiltonian operator $H$. The count concerns linearly independent energy eigenstates, rather than every normalized quantum superposition or every choice of global phase. Even a single energy eigenspace contains infinitely many representing vectors; its dimension measures independent freedom. Ground-state degeneracy is the lowest-energy instance of degenerate energy levels.

For an exact finite model with energy unit $\epsilon >0$, let the Hamiltonian operator be the diagonal matrixIts ground-state energy is zero and $g_0=2$: the first two coordinate unit vectors $e_1,e_2$ span the ground-state subspace. Every normalized $a{e}_1+b{e}_2$, with $|a{|}^2+|b{|}^2=1$, is a ground state, including $(e_1+e_2)/\sqrt{2}$. Its energy measurement is certain to return zero, even though a measurement distinguishing $e_1$ from $e_2$ need not have a definite outcome. Under the unchanged Hamiltonian operator, this quantum superposition is a stationary state because both components acquire the same global phase.

Ground-state degeneracy matters when an interaction selects or splits previously equal energy eigenvalues. In the finite model, adding $\epsilon \mathrm{diag}(\delta ,-\delta ,0,0)$ with $0<\delta <2$ gives the formerly lowest pair the distinct energies $\pm \epsilon \delta$, leaving a unique ground state. This direct calculation illustrates why degenerate perturbation theory first examines the perturbation inside the whole ground-state subspace.

A classical continuous-symmetry model can instead have a manifold of minimizing configurations, such as the sphere of orientations of a Néel state in an isotropic Heisenberg antiferromagnet. That geometric description does not by itself count the dimension of a finite quantum ground-state subspace.

Matrix mechanics represents an observable by an array of matrix elements between chosen quantum states. In an orthonormal basis of energy eigenstates, each a stationary state, the diagonal entries of the Hamiltonian operator are the energy levels; off-diagonal entries of other observables describe couplings between those states. The same array can therefore answer two different questions: which values can a measurement return, and which pairs of energy eigenstates can a specified interaction connect? The eigenvalues answer the first question; the relevant matrix elements, together with the energy differences, answer the second.

This makes linear algebra a practical language for atomic physics. One can calculate energy levels, electric-dipole selection rules and probability amplitudes for transitions without assigning an electron an unobserved classical orbit. The three-state example below separates a missing optical coupling from a missing energy level; the finite oscillator commutator boundary defect explains why a finite array requires care when approximating a quantum harmonic oscillator.

For a real potential energy and a stationary state, the probability current isThe time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$, so $j^{\prime }=\hslash \mathrm{Im}({\psi }^{\ast }{\psi }^{\prime \prime })/m=0$. In an exterior plane wave superposition, interference terms contribute no current andConsequently $|A{|}^2+|D{|}^2=|B{|}^2+|C{|}^2$ for every choice of incoming amplitudes. If $u=(A,D{)}^T$, this says $u^{†}(I-S^{†}S)u=0$ for every complex vector $u$. Since $I-S^{†}S$ is Hermitian, vanishing of its quadratic form implies that the matrix itself vanishes. ThereforeThe second identity follows by choosing $u=(1,0{)}^T$. Since $S$ is a square matrix, the first also implies $S{S}^{†}=I$, proving unitarity. A real delta potential preserves this proof: continuity of $\psi$ and the real derivative jump make the current continuous across the interaction point.

For the quantum harmonic oscillator, split the two expectation values into their variances and squared means:The last two terms are nonnegative, soApply the arithmetic-geometric mean inequality to the two remaining nonnegative terms, then the Heisenberg uncertainty principle:The argument applies to any normalized state with finite oscillator energy; the stated stationary state is a particular case. Equality requires zero position and momentum means, minimum quantum uncertainty, and balanced kinetic and potential contributions, as in the oscillator ground state.

For a time-independent Quantum Hamiltonian, a stationary state is an energy eigenstate. The Schrödinger equation separates its time dependence asIts probability density $|\psi (x,t){|}^2=|\chi (x){|}^2$ is time independent. Linearity of the Schrödinger equation givesPut $A_{jk}={\int }_{-\infty }^{\infty }{\chi }_j^{\ast }\hat{A}{\chi }_{k}dx$. The operator for a time-independent observable is a Hermitian operator, so $A_{21}=\overline{A_{12}}$. Substituting the evolved wavefunction into the expectation value givesand therefore, for a normalized state,The normalization is automatic for orthogonal ${\chi }_1,{\chi }_2$, in particular when $E_1\ne E_2$: self-adjointness of $\hat{H}$ gives $(E_1-E_2)⟨{\chi }_1,{\chi }_2⟩=0$. If the energies coincide, normalization of the individual eigenstates alone does not ensure normalization of their displayed sum; its squared norm is $1+\mathrm{Re}⟨{\chi }_1,{\chi }_2⟩$. For a nonzero sum not having unit norm, the physical expectation value is the displayed numerator divided by this squared norm. With orthonormal energy states, the stated formula holds in both the distinct and degenerate cases.

Put $D=1+i\hslash t/m$ and write the Gaussian wave packet as $\psi ={\pi }^{-1/4}{D}^{-1/2}{e}^{-x^2/(2D)}$. Differentiation givesHence $\boxed{i\hslash {\psi }_t=-({\hslash }^2/2m){\psi }_{xx}}$, the free Time-dependent Schrödinger equation.

With $\tau =\hslash t/m$, its probability density isThe stated Gaussian integral shows its integral is one. Symmetry gives $⟨x⟩=0$; differentiating that Gaussian integral with respect to its coefficient gives $⟨x^2⟩=(1+{\tau }^2)/2$. Thus the position standard deviation isIt is not a stationary state: its probability density changes with time as the packet spreads. A stationary state would have only an overall time-dependent phase and a fixed position density.

For the separated stationary state, restore its time factor $e^{-iEt/\hslash }$. Away from radial nodes, the quantum phase is $S=k\varphi -Et/\hslash$, up to a constant $0$ or $\pi$ where the real radial function has fixed sign. The guidance equation in plane polar coordinates therefore gives the Bohmian circulation of an angular-momentum eigenstateThe direction is $\hat{\mathbf{\varphi }}$ for $k>0$ and $-\hat{\mathbf{\varphi }}$ for $k<0$; the velocity is zero for $k=0$. Each admissible trajectory is a circle:The orbital angular momentum along the trajectory is $m{r}_0{v}_{\varphi }=\hslash k$. The origin or any zero-amplitude circle is excluded from this local formula. For the real degenerate superposition constructed above, the spatial quantum phase is constant on each nodal sector, so its Bohmian mechanics velocity is instead zero. Thus the circular motion is a conclusion about the separated angular-momentum eigenstate, not an arbitrary energy eigenstate.

The rotational invariance of a central-potential Hamiltonian allows a simultaneous eigenstate of energy and the two-dimensional orbital angular momentum $L_z=-i\hslash {\partial }_{\varphi }$. For a separated wavefunction $F(r)G(\varphi )$, the Laplacian in polar coordinates givesThe $\varphi$ term must be a constant; write $G^{\prime \prime }/G=-k^2$. The angular eigenfunctions can be chosen as $e^{ik\varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ imposes $e^{2\pi ik}=1$, hence $k\in \mathbb{Z}$. The radial equation in the separation of a two-dimensional central-potential eigenstate isFor a real central potential and the usual real self-adjoint radial boundary conditions, the radial equation admits a basis of real solutions: real and imaginary parts of a complex solution obey the same equation and boundary conditions. Choose a real normalized radial eigenfunction. Since the plane area element in plane polar coordinates is $rdrd\varphi$, the normalization isThis establishes the intended separated simultaneous eigenstate form. It is not the form of every stationary state. The radial equation depends on $k^2$, so the $k$ and $-k$ sectors have the same energy. For $k\ge 1$, their normalized superpositionis a single-valued stationary state with that energy but is not a single angular exponential. Quantum degeneracy is precisely why separation of variables selects a convenient eigenstate basis rather than all vectors in an energy eigenspace. The printed assertion needs this qualification. The printed polar-coordinate aid also labels a gradient component tuple as a divergence; the gradient used below is the two-dimensional vector $\hat{\mathbf{r}}{\partial }_r+\hat{\mathbf{\varphi }}{r}^{-1}{\partial }_{\varphi }$.

The squared sine has integral $a$ over the interval of length $2a$, since it contains an integer number of half-wavelengths. A real positive phase convention therefore givesInside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }_n^{\prime \prime }/(2m)=E_n{\psi }_n$. The wavenumber is $\pi n/(2a)$, soThe stationary states vanish at both walls and form an orthonormal basis of the square-integrable state space. A common arbitrary phase on each basis function does not alter any energy or measurement probability.

The Parseval identity for the orthonormal stationary-state expansion is ${\sum }_{n\ge 1}|c_n{|}^2=\Vert \psi {\Vert }^2=1$. Splitting the amplitudes into even and odd indices givesMultiply by ${\pi }^2/3$ to obtain the requested identity. Its integer constants areThe series converges absolutely, since both parity contributions are $O(p^{-2})$. This derivation uses the normalization of the prepared state and completeness of the well stationary states, rather than a separate summation trick.

For a finite potential energy discontinuity, the stationary state $\psi$ and its first derivative are continuous at $a$. Integrating the time-independent Schrödinger equation across a shrinking interval about $a$ gives continuity of ${\psi }^{\prime }$; a jump in $\psi$ would produce a distributional derivative incompatible with a finite potential energy.

For $E>V_0$, put $k=\sqrt{2mE}/\hslash$, $q=\sqrt{2m(E-V_0)}/\hslash$ and use unit incident amplitude:There is no wave incident from the right. The matching equations $1+r=t$, $k(1-r)=qt$ give $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. Comparing reflected and incident probability currents gives the reflection coefficientThe transmitted fraction is $T=(q/k)|t{|}^2=4kq/(k+q{)}^2$, so $R+T=1$.

For $0<E<V_0$, write $\kappa =\sqrt{2m(V_0-E)}/\hslash$ and retain only the decaying right-hand solution $t{e}^{-\kappa (x-a)}$. Matching gives $r=(k-i\kappa )/(k+i\kappa )$, which has modulus one. The evanescent wave carries no transmitted probability current. At $E=V_0$, the bounded right-hand solution is constant and matching again gives $r=1$. ThusClassically, a particle is wholly transmitted for $E>V_0$ and wholly reflected for $E<V_0$; there is no classical partial reflection above the step. At the threshold it has zero right-hand speed and no transmitted flux, a marginal case requiring a convention about motion exactly on the discontinuity. The quantum result approaches $R=0$ as $E/V_0\to \infty$.

The Time-dependent Schrödinger equation for the one-dimensional Quantum harmonic oscillator isA stationary state has $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, so cancellation of the common time factor gives

SetThe equation becomesWith $\psi =f(y)e^{-y^2/2}$,and hence $f$ obeys the Hermite differential equation

If $f$ is monic of degree $N$, the coefficient of $y^N$ in this equation is $ϵ-1-2N$. It must vanish, soWriting $f={\sum }_{k=0}^N{a}_k{y}^k$ gives the recurrenceThe coefficient of $y^{N-1}$ first gives $a_{N-1}=0$, and the recurrence preserves parity. If $N$ is even, every odd coefficient vanishes, so $f$ is an even function. The Gaussian factor is also even; therefore the stationary state $\psi$ has even parity.