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Time dependence in quantum mechanics describes how a prepared quantum state changes and thereby changes the probabilities of later measurements. For an isolated system, the Schrödinger picture assigns this evolution to a state vector $|\psi (t)⟩$, governed by the Time-dependent Schrödinger equationThe Hamiltonian is a self-adjoint operator; it has units of energy, while $\hslash$ has units of action, so $\hat{H}/\hslash$ has units of inverse time. For a particle with potential energy $V(\mathbf{r},t)$, this becomesGiven a suitable initial wavefunction and the physical boundary conditions, this first-order equation determines its subsequent propagation. Its linearity means that the evolution of a quantum superposition is the corresponding quantum superposition of evolved states.

For a real potential energy, this equation and its complex conjugate imply the local continuity equationHere $\rho$ is the probability density and $\mathbf{j}$ the probability current. Vanishing net boundary flux preserves wavefunction normalization, as required by the Born rule.

The time-evolution operator packages the solution for every initial state:Because ${\hat{H}}^{†}=\hat{H}$, differentiation givesTherefore $U^{†}U=I$: this is unitary time evolution, preserving inner products and all state norms. It is reversible, with $U(t,t_0{)}^{-1}=U(t_0,t)=U(t,t_0{)}^{†}$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.

For a time-independent Hamiltonian, the solution is the operator exponentialIf the Hamiltonian varies with time, the analogous simple exponential of its integral is valid when $[\hat{H}(t),\hat{H}(t^{\prime })]=0$ for all relevant times. Otherwise time ordering is essential:The meaning is explicit in the first terms of the ordered series,the later-time operator stands on the left. In the position representation, the same propagation is expressed by the quantum-mechanical propagatorIts initial value is the Dirac delta function ${\delta }^3(\mathbf{r}-{\mathbf{r}}^{\prime })$.

A time-independent Hamiltonian has energy eigenstates satisfying $\hat{H}|n⟩=E_n|n⟩$. A single such initial state evolves asIt is a stationary state: its global phase changes, but its probability density and the distributions of time-independent observables do not. Its energy is sharp. Any quantum superposition entirely within one degenerate energy eigenspace also acquires one common global phase and is stationary. Stationarity does not require zero probability current; for instance, $u_n(\varphi )=e^{in\varphi }/\sqrt{2\pi }$ on the ring has a constant density and persistent circulation when $n\ne 0$.

An arbitrary normalized state can instead contain different energy eigenvalues:A continuous spectrum requires corresponding integrals. For the discrete case, an energy measurement gives $E_n$ with fixed probability $|c_n{|}^2$ (summed over degeneracies), but relative quantum phases change with time. ConsequentlyThe interference terms can oscillate at angular frequencies $(E_m-E_n)/\hslash$. For two populated energy levels, this supplies a characteristic period $2\pi \hslash /|E_m-E_n|$ when the interference term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.

For a normalized state and an observable $\hat{A}(t)$, its expectation value isThis is the statistical mean over repeated preparations and measurements. In an eigenbasis of energy eigenstates for a time-independent Hamiltonian and a time-independent $\hat{A}$,Only matrix elements connecting different energies can generate these oscillations. If $[\hat{A},\hat{H}]=0$, its expectation value is constant for every initial state, provided it has no explicit time dependence.

Differentiating the expectation value and substituting the Schrödinger equation for both the bra and ket proves the general Ehrenfest theorem:Equivalently,These manipulations require the state to lie in the relevant operator domains. Taking $\hat{A}=\hat{H}$ gives $d⟨H⟩/dt=⟨{\partial }_t\hat{H}⟩$, so energy is conserved under a time-independent Hamiltonian.

For $\hat{H}={\hat{p}}^2/(2m)+V(\hat{x})$, the canonical commutation relation givesHence the particularly useful position-and-momentum form of Ehrenfest theorem isIt resembles Newton's second law, but the mean force generally differs from the force at the mean position. If a narrow wave packet samples a region where the force varies little, $⟨V^{\prime }(x)⟩\simeq V^{\prime }(⟨x⟩)$ gives approximate classical motion. For a quadratic potential energy this replacement is exact. In particular the quantum harmonic oscillator obeysThe centroid follows the classical harmonic oscillator, although this does not specify the full evolving wavefunction or its width.

The Heisenberg picture expresses the same physics by assigning evolution to observables instead:The expectation value is unchanged, and differentiation yields $d{\hat{A}}_H/dt=(i/\hslash )[{\hat{H}}_H,{\hat{A}}_H]+U^{†}({\partial }_t{\hat{A}}_S)U$. Thus state propagation, evolving expectation values, and the classical correspondence in Ehrenfest theorem are mutually consistent descriptions of the same unitary time evolution.

The expression is the canonical partition function for a system in thermal equilibrium with a large heat bath at fixed temperature $T$. The system exchanges energy with the bath, while its particle number and external mechanical parameters, such as volume, are fixed. The bath is large enough that its temperature is not appreciably changed by this exchange.

The system has allowed energy levels ${ϵ}_j$. The index $j$ labels distinct energy values, and the integer degeneracy factor $g_j$ counts distinct quantum states with that same energy. $k_B$ is the Boltzmann constant, and $\exp [-{ϵ}_j/(k_{B}T)]$ is a state's Boltzmann factor. The dimensionless sum $Z$ normalizes the probabilities:Thus the degeneracy multiplies the probability of occupying the level, not the probability assigned to each individual state within it.

The canonical model is not restricted to a classical ideal gas. It applies to a quantum rotor, for example, provided the sum counts all of that system's allowed states with their proper degeneracies. For a many-particle system, ${ϵ}_j$ would be the energy of a many-particle state, with any quantum-statistical restrictions already built into the state count.

The internal energy is the canonical mean of the energy:Holding the external mechanical parameters fixed, so that the spectrum and degeneracies do not change when differentiating with respect to $\beta$, givesComparison yieldsAn explicitly temperature-dependent effective Hamiltonian would require care: differentiating its partition sum would also differentiate the assigned energy levels. The stated result uses the fixed physical spectrum of the canonical system.

The one-dimensional time-independent Schrödinger equation isFor the infinite square well on $[0,a]$, $V=0$ inside and the wave function is zero outside. The boundary conditions are $\psi (0)=\psi (a)=0$. For positive integer $n$, ${\psi }_n=A\sin (n\pi x/a)$ satisfies them, and ${\psi }_n^{\prime \prime }=-(n\pi /a{)}^2{\psi }_n$. Substitution gives the energy levelsNormalization requires $1={\int }_0^a|{\psi }_n{|}^{2}dx=A^{2}a/2$. The requested real positive constant is therefore

A self-assembled InAs quantum-dot single-photon source uses small indium arsenide islands inside a larger-gap gallium arsenide host. Strain during epitaxial growth favors islands after a thin wetting layer has formed. Their characteristic diameter is of order $10$–$30\mathrm{nm}$ and their height a few nanometres, for example $2$–$5\mathrm{nm}$. Growth conditions determine the areal density: low-density samples suitable for isolating dots can be around $10^8$–$10^9\mathrm{c}{\mathrm{m}}^{-2}$, while ordinary dense layers can be much denser. The particular electrically driven device underlying the supplied image had about $5\times 10^8\mathrm{c}{\mathrm{m}}^{-2}$. These are characteristic scales, not dimensions deducible from the emission image.^[ref]

Quantum confinement in the dot traps both electrons and electron holes in three directions, giving discrete energy levels. In a junction device, injected carriers relax into these levels. Electron-hole recombination then emits a photon, and the dot must be repopulated before emitting again on that same transition. Spatial selection of one dot, spectral selection of one line, and well-separated injection pulses make this a single-photon source. Photon antibunching, rather than low average brightness alone, establishes suppression of simultaneous photons. In particular, collecting both photons of a cascade without selecting a line would not produce exactly one collected photon per excitation.

The lower-current feature $X$ is the transition from one confined exciton to the empty dot. The higher-current feature $X_2$ is the first step of the biexciton cascade $XX\to X\to 0$: it becomes strong when the dot can contain two electron-hole pairs. In the weak-pumping limit the probabilities of one and two pairs scale approximately as $I$ and $I^2$, respectively. Coulomb and confinement energies shift the two transitions relative to one another; $X_2$ is not a photon of twice the $X$ energy. In this particular image the $X_2$ photon is about $4$–$5\mathrm{meV}$ higher in energy. With $B_{XX}=2E_X-E_{XX}$, this is a negative biexciton binding energy, since $E_{XX}-E_X=E_X-B_{XX}$.

For the mass estimate, the original plot places the $X$ line at approximately $E_X=1.3945\mathrm{eV}$, with graph-reading precision of order $0.5\mathrm{meV}$. Under the prescribed identification of photon energy with the full lowest exciton excitation energy,The three-dimensional screened Wannier-Mott exciton has the hydrogenic exciton binding energyThereforeReading the line as $1.395\mathrm{eV}$ instead gives $0.331m_0$; about $0.33$–$0.34m_0$ is an appropriate precision. This inference uses the supplied pair-creation energy and a three-dimensional screened hydrogenic model. In a real dot, strain and confinement also alter transition energies, so it is an effective model estimate and does not independently determine either individual carrier mass.

The two-mode approximation for a Bose gas retains the real localized orbitals:It is valid when the lowest doublet is well separated from higher energy levels, and the temperature and interaction-induced changes of the orbitals are small on that separation. More explicitly, if $\Delta$ is the energy gap to discarded orbitals, sufficient low-energy conditions are $k_{B}T,t,N{U}_{0}I\ll \Delta$, where $I=\int d^{3}r|{\phi }_L{|}^4$. A deep double-well potential suppresses interwell interaction overlaps. Large $N$ alone does not justify the truncation: increasing $N$ at fixed coupling can violate the interaction condition.

Let $ϵ=(E_0+E_1)/2$ and use the even and odd eigenfunctions to evaluate the projected one-particle Hamiltonian operator:ThusThe same quantum tunnelling matrix element is $t=-\int d^{3}r{\phi }_L{h}_0{\phi }_R$. Reflection symmetry makes $\int |{\phi }_L{|}^4=\int |{\phi }_R{|}^4=I$. Neglecting mixed contact interaction integrals leavesThe last step is the normal-ordered boson pair count, following from the canonical commutation relation. Consequently the two-site Bose-Hubbard model has precisely the requested coefficientsandHere $\mu$ is the coefficient obtained by rewriting the fixed-particle number Hamiltonian operator, rather than an independently prescribed thermodynamic chemical potential. Adding a reservoir term $-{\mu }_{\mathrm{res}}\hat{N}$ would replace it by $\mu +{\mu }_{\mathrm{res}}$.

The negligible-overlap instruction applies to the mixed contact interaction integrals while the small quantum tunnelling matrix element is retained. Exactly disjoint smooth orbitals for a local $h_0$ would also give $t=0$; setting every overlap-related matrix element identically to zero would discard the physical splitting one is trying to describe.

For each spin let ${\mathcal{F}}_{\sigma }$ contain the $N_{\sigma }$ lowest tight binding energy levels, and set $n_{\sigma }=N_{\sigma }/M$ and $C_{\sigma }=M^{-1}{\sum }_{k\in {\mathcal{F}}_{\sigma }}{e}^{ik}$. The normalized trial Slater determinant isIts nearest-neighbour one-body density matrix is $⟨c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}⟩=C_{\sigma }$. By Wick's theorem, fermionic exchange symmetry subtracts the squared off-diagonal one-body density matrix from the direct contributionThis formula also specifies the finite-$M$ answer exactly, including partially filled terminal energy levels through the chosen occupied set. For $N_{\sigma }$ consecutive allowed wavevectors, $|C_{\sigma }|=|\sin (\pi N_{\sigma }/M)|/[M\sin (\pi /M)]$; its phase locates the centre of that occupied interval.

Write $\tau =|t|$. In the large-$M$ limit the Fermi sea is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with Fermi wavevector $k_{F,\sigma }=\pi n_{\sigma }$ measured from $k_0$. ConsequentlyThe subtraction due to fermionic exchange symmetry is essential: replacing $⟨n_j{n}_{j+1}⟩$ by $n^2$ would discard the suppression of neighbouring equal-spin occupation. A completely filled spin band has zero kinetic energy and interaction energy $MV$, as the formula also shows.

Write $G(r)=⟨{\psi }^{†}(0)\psi (\mathbf{r})⟩$, $r=|\mathbf{r}|$, and ${\ell }^2=\hslash /(m\omega )$. At zero temperature all $N$ particles occupy ${\varphi }_0(\mathbf{r})=(\ell \sqrt{\pi }{)}^{-1}{e}^{-r^2/(2{\ell }^2)}$. Expanding the bosonic annihilation operator in single-particle orbitals givesThis one-body density matrix is a Gaussian function of width $\ell$; it contains one wavefunction evaluated at $r$, so its exponent is half that of the spatial number density $N|{\varphi }_0(r){|}^2$. In particular,the total number density at the trap centre, including condensed and noncondensed particles.

At positive temperature, with occupations ${\overline{n}}_a$ of the trap energy eigenstates, the exact ideal-gas decomposition isFor $\hslash \omega \ll k_{B}T<k_B{T}_c$, the sketch has a broad Bose-Einstein condensate component on scale $\ell$ and a narrower thermal peak near the origin. On approaching $T_c$ the broad component loses its weight $N_0$, while the thermal part remains. Above $T_c$ there is no macroscopic broad Bose-Einstein condensate component; close to $T_c$, low energy levels can still give a shoulder, so a single narrow Gaussian function is only a high-temperature description. At $k_{B}T\lesssim \hslash \omega$ individual excited energy levels matter and the two-scale description need not apply.

For $T\gg T_c$ the Maxwell-Boltzmann distribution applies. The leading off-diagonal one-body density matrix isHere ${\sigma }_r$ is the Gaussian function width defined by $e^{-r^2/(2{\sigma }_r^2)}$, and ${\lambda }_T$ is the thermal de Broglie wavelength. The thermal cloud itself extends over the much larger spatial scale $\sqrt{k_{B}T/\kappa }$; this is a correlation width, not the cloud radius.

Since $P{\psi }_n=n\hslash {\psi }_n$, applying the Hamiltonian operator givesThus the energy eigenvalues and corresponding full stationary states areThe ground state is ${\psi }_0$ with energy zero and quantum degeneracy one. For each $n=1,2,\dots$, the positive energy level $n^2{\hslash }^2/(2m)$ has quantum degeneracy two: its energy eigenspace is $\mathrm{span}\{{\psi }_n,{\psi }_{-n}\}$. This is the free particle spectrum on a circle.

A stationary state has $\Psi (t,x)=e^{-iEt/\hslash }\psi (x)$. Substitution into the Schrödinger equation gives the energy eigenvalue equationThe trial wavefunction $C{e}^{ikx}$ satisfies it with $E={\hslash }^2{k}^2/(2m)$. The periodic boundary condition requires $e^{2\pi ikR}=1$, hence $k=n/R$, $n\in \mathbb{Z}$. The wavefunction normalization is $1=2\pi R|C{|}^2$, so a convenient common global phase gives the orthonormal basisIndeed, ${\int }_0^{2\pi R}{\varphi }_n^{\ast }{\varphi }_{q}dx={\delta }_{nq}$. This is the free particle spectrum on a circle. The ground state ${\varphi }_0$ has energy zero and is nondegenerate. Each positive energy level ${\hslash }^2{n}^2/(2m{R}^2)$, $n=1,2,\dots$, has energy eigenspace $\mathrm{span}\{{\varphi }_n,{\varphi }_{-n}\}$ and quantum degeneracy two. Real sine and cosine combinations give another basis of the same energy eigenspace.

The Hamiltonian operator separates into two commuting one-dimensional quantum harmonic oscillators. WriteProducts of harmonic oscillator eigenfunctions giveHere $H_n$ is the physicists' Hermite polynomial, so $H_0=1$, $H_1(u)=2u$ and $H_2(u)=4u^2-2$. The first six energy eigenstates, ordered with multiplicity, areFor this two-dimensional oscillator level ordering, the excitation energies obeyEvery state with $n_1+n_2\ge 3$ costs at least $3\hslash {\omega }_1$, so none has been missed. Equalities within the list occur when ${\omega }_2={\omega }_1=\omega$. Then the two-dimensional isotropic harmonic oscillator hasThe quantum degeneracy counts $n_1=0,\dots ,N$ with $n_2=N-n_1$. The six listed states occupy three distinct energy levels, of quantum degeneracies one, two and three.

Applying the Hamiltonian operator to ${\psi }_n$ gives ${\psi }_n^{\prime \prime }=-n^2{\psi }_n$, and thereforeThe ground state $n=0$ has zero kinetic energy. Every positive energy level is of quantum degeneracy two, because $n$ and $-n$ correspond to opposite angular momenta with equal squared magnitude.

The confinement scale suggests a tangential momentum scale $\hslash /r$ and thus an excitation energy scale ${\hslash }^2/(m{r}^2)$, consistent with the Heisenberg uncertainty principle. More precisely the de Broglie wavelength must fit an integer number of times around the circumference: $2\pi r=|n|\lambda$, so $|p|=h/\lambda =|n|\hslash /r$. This explains both the $r^{-2}$ dependence and the $n^2$ sequence.

One must not infer a nonzero ground-state energy by applying $\Delta \varphi \Delta L\ge \hslash /2$ naively. Angle is periodic and is not a globally canonical position operator on the circle. The uniform $n=0$ wavefunction, with $\Delta L=0$, is a concrete counterexample to that naive inequality. The ordinary Cartesian Heisenberg uncertainty principle is not violated by the zero-angular momentum state.

The emitted photon energy is the difference of the two energy levels:Using $E_{\gamma }=hc/\lambda$ givesNo numerical mass or radius is supplied, so this is the determinate result. It is the wavelength conditional on the transition occurring. An isolated particle on a circle with no coupling would remain in its energy eigenstate; a charged particle coupled to electromagnetic radiation can emit subject to its transition selection rules. The $3\to 2$ change has $\Delta n=-1$, compatible with an electric-dipole angular factor for a charged ring particle.

Separate center-of-mass motion and replace $m$ by the reduced mass $\mu =mM/(m+M)$ in the relative-coordinate formulas. Then $r_n$ is the proton-electron separation, energy levels and the Rydberg constant scale by $\mu /m$, and transition wavelengths increase by $m/\mu$. The electron's own orbital radius about the center of mass is $M{r}_n/(M+m)$; the proton radius is $m{r}_n/(M+m)$. The relative speed retains $\alpha c/n$, while the electron's individual speed acquires $M/(M+m)$. Replacing the mass in the radius alone but leaving energies unchanged would be inconsistent.

Use the ordered orthonormal basis $(|1s⟩,|2p_x⟩)$. The projected Hamiltonian between pulses is $H_0=-\Delta {\sigma }_z$, so its free time-evolution operator through a period isAcross one instantaneous pulse, integrating the Schrödinger equation givesusing the exponential of a Pauli vector. The finite free Hamiltonian contributes vanishing evolution during the zero-duration kick. A period measured immediately before successive pulses therefore consists of the kick followed by free evolution. Composition givesThis is the Floquet operator at the specified phase of the pulse train. Actual hydrogenic energy levels are not opposite numbers: the displayed convention subtracts their midpoint, with $\Delta =(E_{2p}-E_{1s})/2$. Restoring that midpoint multiplies the result by an unobservable global phase. Likewise $E$ in the given Hamiltonian denotes the electric coupling per unit length; if it is instead a physical electric-field strength, its charge factor must be included in $E{x}_0$.

Photoelectron spectroscopy measures the kinetic energies of photoelectrons produced by illuminating matter with photons of known energy. The photoelectric effect supplies the emission mechanism; spectroscopy turns the distribution of emitted electron energies into information about occupied energy levels and chemical bonds. A broad survey helps identify chemical elements, while small shifts within one element's lines can distinguish different chemical surroundings. These are different questions: finding carbon does not establish which chemical bonds its chemical atoms form.

The useful inversion is an energy balance. A known incoming photon energy pays for removal of an electron and for its outgoing kinetic energy. To apply that balance to a solid, specify the zero of energy and the work function of the instrument. Photoelectron binding energy reference derives the conversion for a conducting sample in electrical contact with its analyzer. A free molecule instead has an ionization energy referenced to an electron at rest in vacuum; it need not share a Fermi level with the analyzer.

Pieter Zeeman observed changes in emitted light when its source was placed in a magnetic field. Resolving the components and their wave polarization connected the spectral pattern with the direction of observation. The magnetic field changes atomic energy levels, while selection rules determine which polarized components are visible. Comparing observations along and across the magnetic field therefore tests more than a simple increase in line width: it connects the Zeeman effect to both atomic structure and experimental geometry.^[ref][ref]

A physicist.^[ref]

The Zeeman effect connects a laboratory magnetic field to the frequencies and wave polarization of light emitted by chemical atoms. Its experimental history makes the viewing direction and spectral resolving power essential to the inference.

The Planck constant $h$ connects the frequency $\nu$ of an electromagnetic field mode with a single photon's energy, $E=h\nu$. Its physical dimension is action, or energy multiplied by time. It sets the scale of quantum mechanics: the de Broglie wavelength of a particle is $\lambda =h/p$, and discrete energy levels often have spacings involving $h$.

In the International System of Units, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ exactly. Fixing this value defines the kilogram through already defined units of length and time; it does not remove experimental uncertainty from a practical mass measurement.^[ref] The reduced Planck constant $\hslash =h/(2\pi )$ is convenient when angular frequency replaces ordinary frequency.

The root mean square radius of the proton charge distribution. Finite nuclear size changes hydrogen atom energy levels and enters precision predictions for the Lamb shift.^[ref]

The quadratic Stark effect is an energy level shift whose leading term is quadratic in an applied electric field. For a nondegenerate energy eigenstate of definite parity, the diagonal matrix element of an odd electric-dipole perturbation vanishes. Second-order nondegenerate perturbation theory givesEvery term is negative for the ground state. If field reversal is implemented by the parity operator, an analytic nondegenerate energy branch is even in the field. This symmetry explains the absence of odd powers; it does not apply without qualification to a degeneracy split linearly by the field.