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Approximate quantum counting estimates the number $k$ of satisfying assignments of a Boolean function, using a quantum oracle and quantum measurement. Controlled repetitions of a Grover search algorithm iteration let quantum interference reveal the count. For one input bit, a Hadamard test can give Pauli Z gate expectation value $1-k$: with the complemented phase oracle $U_f=\mathrm{diag}(2f(0)-1,2f(1)-1)$ and $G=-X{U}_f$, use $⟨+|X=⟨+|$ to obtain $⟨+|G|+⟩=-[(2f(0)-1)+(2f(1)-1)]/2=1-k$. Repeated preparations estimate this expectation value; one binary measurement result does not distinguish all three counts $0,1,2$.

A distribution of quantum state populations over atomic magnetic sublevels with a nonzero expectation value of the angular momentum operator vector.

A Bell test compares experimental probabilities or expectation values with a Bell inequality. In a two-setting photon polarization test, each station chooses a polarizer orientation and records an outcome. A coincidence circuit identifies candidate pairs using an agreed timing rule. The choice of which trials enter the statistic is part of the experiment: correlations calculated only from detected pairs can differ from correlations of all emitted pairs. The Polarization CHSH count model separates ideal expectation values from this experimental selection problem.

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.

In the infinite square well on $(-a,a)$, extend $\sqrt{2/a}\sin (\pi x/a)$ from $(0,a)$ by zero. It is normalized and continuous, but its derivative jumps at zero. Its energy expectation value is the quadratic form of a positive quantum Hamiltonian ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$, giving the displayed value. It is not in the operator domain of the Dirichlet Hamiltonian operator: its distributional second derivative contains a Dirac delta. Its sine-basis coefficients are $a_2=-1/\sqrt{2}$, all other even coefficients zero, and $a_{2p+1}=4\sqrt{2}(-1{)}^p/[\pi (4-(2p+1{)}^2)]$. These make ${\sum }_n{E}_n|a_n{|}^2$ finite but ${\sum }_n{E}_n^2|a_n{|}^2$ divergent, independently distinguishing the form domain from the operator domain.

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.

Equality in the Heisenberg uncertainty relation requires the centred vectors $x^{\prime }\psi$ and $p^{\prime }\psi$ to be imaginary scalar multiples, where $x^{\prime }=x-x_{0}I$, $p^{\prime }=p-p_{0}I$ and $x_0,p_0$ are the expectation values. Thus, for some $s>0$,In the position representation, using the momentum operator turns this into a first-order differential equation, whose normalizable solutions areHence the quantum states saturating the Heisenberg uncertainty relation are these Gaussian wave packets, with arbitrary position and momentum means and global phase. Their width is ${\sigma }^2=\hslash /(2s)$. A Gaussian wave packet with an additional quadratic quantum phase need not saturate this product bound.

For an infinite square well on $[0,L]$, choose ${\varphi }_n=\sqrt{2/L}\sin (n\pi x/L)$ and $E_n=n^2{E}_1$, $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$. The state $\psi =({\varphi }_1{e}^{-i{E}_{1}t/\hslash }+{\varphi }_2{e}^{-i{E}_{2}t/\hslash })/\sqrt{2}$ has constant expectation value $⟨E⟩=5E_1/2$, butIndeed, the diagonal position matrix elements are $L/2$, whereas the off-diagonal element isA different initial relative phase shifts the oscillation in time. The oscillation comes from quantum interference, while the constant mean energy follows from the fixed populations of the two energy eigenstates.

Prepare a fresh copy of the same quantum state for each trial and measure the same observable. If preparations and readouts are independent, the outcomes $X_i$ are independent and identically distributed random variables with the Born rule probability distribution. Their sample mean is an unbiased estimator of the expectation value:The law of large numbers explains convergence of the sample mean, while the standard error describes its finite-sample spread. It takes four times as many independent trials to halve that standard error. This reduces estimation error; it does not reduce the quantum uncertainty of a single prepared copy.

For the spin expectation value, let $n_{+}$ and $n_{-}$ be the two outcome counts, with $N=n_{+}+n_{-}$. ThenThe binomial distribution gives the exact sampling model. For $\theta =\pi /3$, the standard error is $\hslash \sqrt{3}/(4\sqrt{N})$: approximately $0.1083\hslash$ for $N=16$ and $0.03827\hslash$ for $N=128$. A run can fluctuate away from the predicted mean even as its typical spread decreases.

Repeatedly measuring one surviving system is a different protocol. Under the ideal Lüders rule, outcome $a_r$ leaves $P_r|\psi ⟩/\sqrt{p_r}$ when $p_r>0$. With no intervening change of observable or dynamics moving the state out of that eigenspace, the next projective measurement returns $a_r$ with certainty. Such a record is correlated and does not sample the original Born rule distribution independently. The quantum nondemolition measurement discussion describes this repeatability; fresh quantum state preparations supply the ensemble needed for the estimator above.

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.

Exclusion constrains complete one-particle states, not a collection of little spatial boxes. In particular, two opposite-spin electrons in the same spatial orbital haveThe spatial factor is symmetric; the spin-one-half singlet state supplies the required antisymmetry. At ${\mathbf{r}}_1={\mathbf{r}}_2$, this wavefunction need not vanish for opposite spin labels. For equal spin labels it does vanish. In a continuous position description, an exact coincident point has zero measure anyway; the useful comparison concerns joint probability densities or detection in finite regions. Equal positions do not imply equal complete quantum states.

The fermionic occupation numbers $n_{\alpha }=0,1$ refer to a chosen orthonormal basis of one-particle states. A quantum superposition need not have definite occupation numbers in that basis, and an expectation value $⟨n_{\alpha }⟩$ can be fractional. For example, one electron in $(u+v)/\sqrt{2}$ has mean occupation number $1/2$ in each of two orthogonal spin orbitals, without splitting into two particles. The canonical anticommutation relations give $(a_{\alpha }^{†}{)}^2=0$ and make $a_{\alpha }^{†}{a}_{\alpha }$ a projector, so its expectation value stays between zero and one in every normalized quantum state.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

Under a quantum harmonic oscillator Hamiltonian, the Heisenberg equations of motion giveSubtract the expectation values, square the first equation, and use the symmetrized covariance $C_0=⟨\{x-⟨x⟩,p-⟨p⟩\}⟩/2$. This yieldsThese moment equations hold for any state with finite second moments; a quadratic Hamiltonian additionally preserves Gaussian wave packets. A coherent state has ${\sigma }_x(0{)}^2=\hslash /(2m\omega )$, ${\sigma }_p(0{)}^2=m\hslash \omega /2$, and $C_0=0$, so its width stays fixed while its center follows the classical orbit. A squeezed coherent state generally has a width that oscillates at $2\omega$.

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.

The Heisenberg uncertainty relation constrains the standard deviations of position and momentum in one preparation:On the real line, the position operator $x$ acts by multiplication, and the momentum operator is $p=-i\hslash d/dx$, with $\hslash$ the reduced Planck constant. Their canonical commutation relation $[x,p]=i\hslash I$ and the Robertson uncertainty principle give the bound on normalized quantum states satisfying the moment and product-domain assumptions above; Schwartz functions are a convenient sufficient class. Thus making a preparation more sharply localized in position forces a wider momentum distribution, even for ideal measurement in quantum mechanics.

An explicit saturating example is the normalized Gaussian wave packet, for $\sigma >0$,The wavefunction is a probability amplitude; its squared magnitude is the probability density function, by the Born rule. That density is a normal distribution of mean zero and standard deviation $\Delta x=\sigma$, rather than a density whose width is read directly from the amplitude's exponent. With the unitary, negative-exponent convention for the Fourier transform, its momentum amplitude isThe Gaussian integral evaluates this Fourier transform. Consequently $|{\tilde{\psi }}_{\sigma }(p){|}^2$ is another normal distribution, with mean zero and standard deviationFor example, in dimensionless units with $\hslash =1$, $\sigma =1/2,1,2$ give $(\Delta x,\Delta p)=(1/2,1),(1,1/2),(2,1/4)$. The Gaussian wave packets all saturate the same bound: narrowing one density broadens the other. More precisely, choose a reference length $L$ and set $X=x/L$, $P=pL/\hslash$; then $[X,P]=iI$, and the plotted widths are $\sigma /L$ and $L\Delta p/\hslash$.

The Heisenberg uncertainty relation connects localization to the cost in kinetic energy. In a quantum harmonic oscillator with mass $m>0$ and angular frequency $\omega >0$, the variance contribution to the expectation value of the energy is bounded using the arithmetic-geometric mean inequality:The mean-dependent contributions are nonnegative. The Gaussian eigenstate for a quadratic potential with ${\sigma }^2=\hslash /(2m\omega )$ and zero means reaches this ground state energy. This is why the tradeoff is more than a limitation on preparation: it predicts an unavoidable positive minimum energy.

Heisenberg's original article appeared in March 1927 as "Über den anschaulichen Inhalt der quantentheoretischen Kinematik und Mechanik", Zeitschrift für Physik 43, pp. 172–198. Its abstract discusses simultaneous imprecision of canonically conjugate quantities and explanatory thought experiments. The precise standard deviation formulation here distinguishes preparation spreads from a slogan about measurement disturbance. ^[ref].

A matrix element $A_{ij}$ is the entry in row $i$ and column $j$ of a matrix $A$.

For a linear operator $T$ on an inner product space with an orthonormal basis $(e_i)$, its matrix elements are $T_{ij}=⟨e_i,T{e}_j⟩$, using the convention that the inner product is linear in its second argument. For a finite-dimensional space, the expansion $T{e}_j={\sum }_i{T}_{ij}{e}_i$ explains why these numbers represent the linear operator in that basis. In quantum mechanics, the expression $⟨\varphi |T|\psi ⟩$ is also called a matrix element between the two states; it need not be diagonal or an expectation value.

Departure of an intrinsic nuclear shape from spherical symmetry. Here intrinsic means referred to axes attached to a model nuclear charge distribution. A nuclear quantum state with definite spin need not have the same laboratory charge density as that intrinsic distribution. In particular, the ground state of a deformed even-even nucleus can have $I=0$, a spherical laboratory expectation value, and a nonzero intrinsic nuclear quadrupole moment.

For an illustrative uniform charge density inside a spheroid, let the semiaxes be $(a,a,c)$ and keep its volume equal to that of a sphere of radius $R$:Thus $\eta >1$ gives a prolate spheroid, $\eta =1$ a sphere, and $\eta <1$ an oblate spheroid. These are three-dimensional equal-volume shapes: equal areas of their drawn cross-sections would be a different constraint. Real nuclear charge distributions have diffuse surfaces and need not be uniform, axially symmetric or confined to one shape.

The correlation function with a single field insertion is its expectation value. In a translation-invariant vacuum it is independent of position. A tadpole subtraction chooses a linear counterterm, or shifts the expansion point of the field, so that this expectation is zero. This removes attached tadpole diagrams from higher connected correlation functions without discarding the physical need to renormalize the chosen vacuum.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.