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A chirped Gaussian wave packet adds a quadratic quantum phase to the zero-mean Gaussian wave packet already evaluated above. Here $\sigma >0$ and $\kappa$ is a real dimensionless parameter. Its probability density function $|{\psi }_{\sigma ,\kappa }(x){|}^2$ is unchanged for every $\kappa$, so $⟨x⟩=0$ and $(\Delta x{)}^2={\sigma }^2$. Nevertheless, applying the momentum operator givesThe sign of the quantum covariance follows directly, without assigning a classical trajectory to the wavefunction:Here $r$ is a normalized symmetrized quantum covariance, analogous to a correlation coefficient. Positive $\kappa$ gives positive $C_{xp}$, and reversing $\kappa$ reverses $C_{xp}$ while leaving both variances unchanged. In the negative-exponent Fourier transform convention used above, the momentum density isConsequently the quantum uncertainty product and the covariance matrix determinant tell different stories:Every member of this pure state family saturates the Robertson-Schrödinger uncertainty relation. Only $\kappa =0$ saturates the Heisenberg uncertainty relation for this fixed pair of position and momentum axes. Thus being a Gaussian wave packet alone does not imply a minimum $\Delta x\Delta p$.

For a concrete comparison, choose a reference length $L$, put $X=x/L$, $P=pL/\hslash$, and set $\sigma =L$. The choices $\kappa =0,+2,-2$ all have $(\Delta X{)}^2=1$, but give $(\Delta P{)}^2=1/4,5/4,5/4$ and $C_{XP}=0,+1,-1$, respectively. The two nonzero cases have $r=\pm 2/\sqrt{5}$ and product $\sqrt{5}/2$; all three covariance matrices have determinant $1/4$.

The phase space curves in the figure satisfy $z^T{\Sigma }^{-1}z=1$ in the dimensionless coordinates, where $z=(X,P{)}^T$. They display second moments, not observed simultaneous position and momentum pairs. Their common area is $\pi \sqrt{\det \Sigma }=\pi /2$, although their coordinate projections differ.

This distinction is useful for predicting focusing. Under free particle evolution with mass $m$, $x(t)=x(0)+tp(0)/m$, soA negative chirp initially narrows the position distribution; positive chirp initially broadens it. For $\kappa <0$, the minimum occurs at $t_{\ast }=-2m{\sigma }^2\kappa /[\hslash (1+{\kappa }^2)]>0$ and has variance ${\sigma }^2/(1+{\kappa }^2)$. At that time the quantum covariance vanishes, and the unchanged momentum spread gives $\Delta x(t_{\ast })\Delta p=\hslash /2$. The extra initial product was a usable position-momentum correlation, not extra measurement backaction.

Suppose each Cartesian coordinate of a particle of mass $m$ is supported in an interval of length $L$, in $d$ dimensions. Popoviciu's inequality on variances gives $\Delta x_i\le L/2$, and the Heisenberg uncertainty relation gives $\Delta p_i\ge \hslash /L$. Since $⟨p_i^2⟩=(\Delta p_i{)}^2+⟨p_i{⟩}^2$,This is a rigorous but non-sharp confinement estimate. For a cubical infinite square well, separation of variables gives the stronger exact ground-state value $d{\pi }^2{\hslash }^2/(2m{L}^2)$. One must not identify a support length with a uniquely prescribed standard deviation or claim that the uncertainty estimate is the exact ground-state energy.

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.

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.

For a number operator eigenstate $|n⟩$, the annihilation operator and creation operator satisfy $⟨a⟩=⟨a^{†}⟩=⟨a^2⟩=⟨(a^{†}{)}^2⟩=0$, $⟨a^{†}a⟩=n$ and $⟨a{a}^{†}⟩=n+1$. Inserting $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$ and $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$ gives the displayed second moments and zero means.

Thus the kinetic energy and potential energy each have mean $\hslash \omega (n+1/2)/2$, and the uncertainty product is $\Delta x\Delta p=\hslash (n+1/2)$. Among number states only the ground state saturates the Heisenberg uncertainty relation. A quantum superposition can have nonzero position and momentum means despite each constituent number state's zero means.

Heisenberg's microscope is a thought experiment, not a report of a performed measurement. In Werner Heisenberg's Nobel lecture, The development of quantum mechanics, delivered on 11 December 1933, the imagined electron is illuminated with very short-wavelength light and observed through a high-resolution microscope. A scattered photon entering the observer's eye would, through Compton scattering, eject the electron from its original path. The passage challenges a classical trajectory picture; its microscope is hypothetical. The lecture date is 1933 even though the award is the Nobel Prize in Physics in 1932.^[ref]

The following idealized geometry makes the tradeoff visible. Let the collected light have wavelength $\lambda$ and let the microscope objective accept outgoing directions within a half-angle $\alpha$ about the vertical axis. Take $x$ transverse to that axis and hold the incoming photon momentum fixed. A diffraction-limited estimate of transverse resolution has scale $\delta x\sim \lambda /\sin \alpha$. The unobserved outgoing direction admits transverse photon momentum components between approximately $-(h/\lambda )\sin \alpha$ and $+(h/\lambda )\sin \alpha$, neglecting the angular variation of the scattered wavelength. By conservation of momentum, uncertainty in that component leaves uncertainty in the transverse recoil of the electron:Here $h=2\pi \hslash$ is the Planck constant; the two $\delta$ symbols denote heuristic resolution and recoil scales, not the standard deviations in the Heisenberg uncertainty relation. Increasing the acceptance angle improves the estimated resolution while broadening the allowed recoil range. Numerical factors depend on the resolution criterion and angular distribution; this scaling argument cannot supply the sharp constant $\hslash /2$.

The thought experiment makes the interaction needed to obtain an image memorable. The preparation uncertainty theorem has broader use: it applies to the prepared quantum state even when the two observables are measured on separate copies. No claim that Werner Heisenberg built this gamma ray apparatus is needed to understand either argument.

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].

The standard deviation of the momentum measurement distribution of a quantum state. A zero mean momentum need not imply zero momentum uncertainty. For a normalized state with finite moments, position and momentum uncertainties obey the Heisenberg uncertainty relation.

Awarded to Werner Heisenberg for creating quantum mechanics, including its application to forms of hydrogen. The Heisenberg uncertainty relation is one route into the resulting theory.^[ref]

The initial quantum state is a coherent state. Since $a|\alpha ⟩=\alpha |\alpha ⟩$ and the state is normalized, its mean occupation number is $N=⟨a^{†}a⟩=|\alpha {|}^2$. Each quantum harmonic oscillator energy eigenstate acquires the phase $e^{-i\omega (n+1/2)t}$, soThe prefactor is a global phase and does not change an expectation value. Inverting the definitions of the annihilation operator and creation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. If $\alpha =\sqrt{N}{e}^{i\phi }$, thenFor a classical harmonic oscillator with amplitude $A$, the conserved energy is $E=m{\omega }^2{A}^2/2$. Setting $E=\hslash \omega N$ gives exactly the same amplitude, $A=\sqrt{2\hslash N/(m\omega )}$. Thus these coherent-state oscillator expectation values follow the classical orbit.

The full quantum mean energy is nevertheless $⟨H⟩=\hslash \omega (N+1/2)$: the extra zero-point energy belongs to the nonzero position and momentum uncertainties required by the Heisenberg uncertainty relation, rather than to the motion of the packet centre. In particular, $N=0$ gives zero mean displacement but a finite ground-state width. A coherent state follows a classical mean trajectory while retaining finite position and momentum uncertainties.

For a normalized state and an observable $A$, define $⟨A⟩=⟨\psi ,A\psi ⟩$ and ${\Delta }_{\psi }A=\Vert (A-⟨A⟩)\psi \Vert$. Suppose first that the position and momentum variances are finite and the state has the regularity needed for the usual position-momentum quadratic forms. Put $a=⟨x⟩$, $b=⟨p⟩$, $u=(x-a)\psi$ and $v=(-i\hslash {\partial }_x-b)\psi$. The Cauchy-Schwarz inequality givesThe term involving $b$ has zero imaginary part. Integration by parts givesConsequently $\Im ⟨u,v⟩=\hslash /2$, andThis proves the Heisenberg uncertainty relation. For smooth rapidly decaying states the boundary terms vanish directly. The same proof extends to $x\psi \in L^2$ and weak derivative ${\psi }^{\prime }\in L^2$ using smooth cutoffs and taking their radius to infinity; the necessary mixed products are integrable by Cauchy-Schwarz. Normalization alone need not give finite variances or an operator-domain momentum expectation. A state with infinite spread is interpreted through the corresponding extended bound, rather than assigning a finite uncertainty to an undefined expression.

For the specified Gaussian wave packet, write $s=\hslash t$ and $a_t=1+is$. Its probability density isIt is a normalized centered normal distribution with variance $1+s^2$. Symmetry gives $⟨x⟩=0$, and henceAlso ${\partial }_x\psi =-x\psi /(2a_t)$, so $p\psi =i\hslash x\psi /(2a_t)$ and $⟨p⟩=0$ by oddness. Integration by parts, or directly taking the squared norm, givesThusThe packet saturates the bound at $t=0$ and has a larger position-momentum uncertainty product at every nonzero time. Its spatial spreading does not change the momentum distribution.

The commutator is $\boxed{[A,B]=AB-BA}$. To evaluate it for differential operators, apply it to a smooth test wavefunction $f$:Thus the canonical commutation relation is $\boxed{[x,p_x]=i\hslash I}$. The associated Heisenberg uncertainty relation is $\Delta x\Delta p_x\ge \hslash /2$. Position and momentum cannot both have arbitrarily sharp outcome distributions in the same prepared state, even with ideal measuring apparatus. This concerns quantum-state preparation, not merely a calibration error in a particular instrument.

The postulates of quantum mechanics give a framework for predicting observations from a prepared state. The experiments below motivate different features of that framework; no one experiment proves the whole set of postulates or specifies every Hamiltonian operator.

A pure quantum state is a normalized vector in a complex Hilbert space, with vectors differing by an overall quantum phase representing the same physical preparation. In position representation, $\int |\psi (\mathbf{r}){|}^2{d}^{3}r=1$. The linear state space permits quantum superposition, while the Born rule assigns probability ${\int }_{\Omega }|\psi {|}^2{d}^{3}r$ to a position outcome in a region $\Omega$. The wavefunction is a probability amplitude; it is not a classical distribution of definite particle positions and momenta.

An observable is represented by a self-adjoint operator. If $Q$ has spectral projectors $P_q$, its possible measurement outcomes are its spectral values and $\mathrm{Pr}(q)=⟨\psi |P_q|\psi ⟩$. The mean is $⟨Q⟩=⟨\psi |Q|\psi ⟩$. For an ideal repeatable projective measurement, obtaining $q$ leaves the normalized conditional state $P_q\psi /\sqrt{\mathrm{Pr}(q)}$. This projects onto an entire eigenspace if the outcome is degenerate. General detectors need more general measurement in quantum mechanics models; the projective update is an idealization. Noncommuting observables need not have simultaneous sharp values, as the canonical commutation relation and Heisenberg uncertainty relation illustrate.

Between measurements, an isolated system has unitary time evolution generated by its Hamiltonian operator:This preserves normalization and superpositions. Composite systems use a tensor product of state spaces, allowing entangled states as well as product states; identical particles additionally require the bosonic or fermionic exchange symmetry appropriate to their species. Statistical mixtures are described by a density matrix, with $\mathrm{Pr}(q)=\mathrm{tr}(\rho P_q)$. These rules separate preparation, coherent evolution and measurement.

The photoelectric effect exposes discrete energy transfer by light. Illuminate a clean metal cathode in an evacuated tube and measure the collected electron current while varying the light frequency, intensity and a retarding voltage. The stopping potential $V_s$ probes the largest emitted electron kinetic energy, whereas the photocurrent measures the rate at which charge is collected. In the ordinary single-photon regime there is a threshold frequency; above it the maximum energy grows with frequency, rather than with intensity. Changing intensity at fixed frequency mainly changes the number of emitted electrons. These observations are summarized by the Einstein photoelectric equationHere $\Phi$ is the surface work function, and the stopping voltage is understood after any photoelectric contact-potential correction. A photon delivers $h\nu$ in an absorption event; emission is allowed only when that energy can pay the escape cost. This accounts for the frequency threshold and the lack of the classical waiting time associated with slowly accumulating energy in a single electron. The experiment supports energy quantization of light; it does not imply that every incident photon emits an electron or that every emitted electron has the maximum energy. Surface losses and collection geometry affect the full current-voltage curve.

The Davisson–Germer experiment demonstrates a complementary wave property of matter. An electron gun uses a heated emitter, accelerating voltage and apertures to send a narrow beam onto a nickel single crystal in vacuum. A movable collector measures scattered electric current versus direction; rotating the crystal also tests its azimuthal symmetry. The crystal gives sharp angular beams whose positions depend on voltage and crystal orientation. A prominent feature occurs near $54\mathrm{V}$ accelerating voltage and roughly $50^{\circ }$ detector colatitude. The nonrelativistic de Broglie wavelength predicted at that voltage isDiffraction wavelengths inferred using the known crystal structure agree with this voltage dependence. These angular peaks arise from coherent addition of scattering amplitudes from the crystal, not from classical specular reflection of independent point projectiles. Bragg's law is $2d\sin {\theta }_B=r\lambda$ for lattice-plane spacing $d$ and angle to those planes; a detector colatitude is a different angle and must not be substituted without its geometry. The historical readout was current, not a photograph of individual electron trajectories.^[ref]

The double-slit experiment makes the role of amplitudes especially direct. A coherent source illuminates two narrow slits; an observation screen a distance $D$ away records the pattern. For slit spacing $d$ and small screen displacement $y$, the path difference is approximately $dy/D$. The two amplitudes add before probabilities are formed:For equal coherent illumination the interference term varies as $\cos (2\pi dy/(\lambda D))$, giving fringe spacing $\lambda D/d$ beneath the single-slit diffraction envelope. Blocking one slit removes that cross term. Young's optical experiment established wave interference; later particle-resolved realizations with photons or electrons show localized detection events whose accumulated distribution has the interference pattern. Low arrival rates demonstrate that this pattern does not require different incident particles to collide with one another.

If the paths become correlated with distinguishable detector states $|D_1⟩$ and $|D_2⟩$, the cross term acquires their overlap $⟨D_2|D_1⟩$. Perfectly distinguishable path records make that overlap zero, suppressing interference even if the spatial beams still overlap. This connects quantum entanglement, measurement and loss of observable coherence. Without a path measurement, the two alternatives are amplitudes for one preparation, rather than classical mutually exclusive events whose probabilities may simply be added.

Together these experiments motivate a description in which energy transfer and detector outcomes can be discrete, yet propagation retains complex quantum phases and interference. The Born rule connects those amplitudes to repeatable frequencies of outcomes. The Schrödinger equation supplies the dynamics once a Hamiltonian operator is specified, while observable operators, state preparation and the measurement model specify what the apparatus can read.

Write $L=10^{-9}\mathrm{m}$ and use the Heisenberg uncertainty relation separately in the three Cartesian directions. Since a position coordinate is confined to an interval of length $L$, Popoviciu's inequality on variances gives $\Delta x_i\le L/2$. Consequently $\Delta p_i\ge \hslash /L$, and the expectation value of the kinetic energy obeysThus an uncertainty-based lower estimate is $\boxed{K\gtrsim 0.11\mathrm{eV}}$. This confinement energy bound from position support is not sharp: the exact ground-state energy for a cubical infinite square well is $3{\pi }^2{\hslash }^2/(2m_e{L}^2)=1.13\mathrm{eV}$. The Heisenberg uncertainty principle establishes the confinement scale ${\hslash }^2/(m_e{L}^2)$; identifying a box size with an approximate position standard deviation changes its numerical coefficient.

A localized quantum particle is described by a wave packet: a coherent quantum superposition of momentum components, rather than a single delocalized plane wave. The Born rule assigns the position probability density $|\psi (x,t){|}^2$. A useful normalized initial Gaussian wave packet isIts mean position and momentum are $x_0$ and $p_0$, and its standard deviations satisfy ${\sigma }_x(0)={\sigma }_0$, ${\sigma }_p=\hslash /(2{\sigma }_0)$. This unchirped initial state saturates the Heisenberg uncertainty relation.

For a free particle, the momentum-space Schrödinger equation multiplies each amplitude by $e^{-i{p}^{2}t/(2m\hslash )}$. The inverse Fourier transform is a Gaussian integral, giving the free Gaussian wave packet evolutionOne way to obtain this expression explicitly is to write the initial momentum amplitude as proportional to $\exp [-{\sigma }_0^2(p-p_0{)}^2/{\hslash }^2-ip{x}_0/\hslash ]$, multiply by the free evolution phase and complete the square in $p$. Taking the modulus squared removes the position-dependent phase and leavesThe expectation value follows uniform classical motion while the density spreads. A narrower initial packet needs a wider momentum distribution and therefore spreads more rapidly. At long times its width grows linearly with $|t|$; its variance grows quadratically, unlike the linear variance growth of ordinary diffusion. The probability continuity equation preserves total probability despite the decreasing peak height. The mean momentum and the mean energy $(p_0^2+{\sigma }_p^2)/(2m)$ remain constant, since the momentum populations do not change.

The same result has a wave interpretation. For a free nonrelativistic particle $\omega (k)=\hslash k^2/(2m)$, so the group velocity is $d\omega /dk=\hslash k/m$, equal to $p_0/m$ at the packet center. The curvature $d^2\omega /d{k}^2=\hslash /m$ makes different momentum components travel with different group velocities, producing dispersion. The phase velocity is $\omega /k=\hslash k/(2m)$ for this energy convention and does not describe the motion of the envelope. The evolving quadratic spatial phase is essential to the spreading; a density snapshot alone does not specify the full quantum state.

A freely spreading Gaussian wave packet remains a pure Gaussian state but develops position-momentum correlations. Its symmetrized covariance is $C_{xp}=⟨\{x-⟨x⟩,p-⟨p⟩\}⟩/2=t{\sigma }_p^2/m$. ConsequentlyEquality in the Robertson-Schrödinger uncertainty relation survives at every time, whereas the simpler product ${\sigma }_x{\sigma }_p$ equals $\hslash /2$ only initially. A chirped Gaussian wave packet can already have $C_{xp}(0)\ne 0$. More generally, $x(t)=x(0)+p(0)t/m$ gives ${\sigma }_x(t{)}^2={\sigma }_x(0{)}^2+2t{C}_{xp}(0)/m+t^2{\sigma }_p(0{)}^2/m^2$. A sufficiently negative initial covariance permits initial contraction before subsequent spreading.

A potential energy changes this behavior. A quadratic potential preserves the Gaussian form because the evolution of position and momentum is linear. For a quantum harmonic oscillator, the Gaussian covariance in a harmonic potential isA coherent state has ${\sigma }_x^2=\hslash /(2m\omega )$, ${\sigma }_p^2=m\hslash \omega /2$ and zero centered covariance. Its width is constant, and its center follows $x_c(t)=x_0\cos \omega t+p_0\sin \omega t/(m\omega )$. In oscillator notation its coherent state label rotates as $\alpha (t)=\alpha (0)e^{-i\omega t}$, apart from the common ground-energy phase of the state vector. A squeezed coherent state generally has a width that breathes at $2\omega$. Thus being Gaussian does not imply either free spreading or constant width; the Hamiltonian operator and initial covariance determine the result.

For a general nonquadratic potential, different parts of the packet experience different nonlinear forces, and an initially Gaussian profile need not remain Gaussian. Ehrenfest theorem gives the exact relations $d⟨x⟩/dt=⟨p⟩/m$ and $d⟨p⟩/dt=-⟨V^{\prime }(x)⟩$. Replacing the last mean by $-V^{\prime }(⟨x⟩)$ is exact for a quadratic potential and approximate for a packet narrow compared with the scale on which the force varies. This connects packet motion to the correspondence principle without treating a quantum state as a definite classical path. Scattering can split a packet into reflected and transmitted components, while confinement can produce oscillations and interference between discrete energy eigenstates. Their relative phases evolve even when their populations remain constant. Localization, spreading and coherent motion all follow from the same linear unitary time evolution.

The standard deviations defining the position and momentum uncertainties satisfy $(\Delta x{)}^2=⟨x^2⟩-⟨x{⟩}^2$ and $(\Delta p{)}^2=⟨p^2⟩-⟨p{⟩}^2$. Part (a) makes both means zero. Using the second moments from part (b),Multiplication givesThe Heisenberg uncertainty relation requires only $\Delta x\Delta p\ge \hslash /2$. Among the oscillator energy eigenstates, equality occurs solely for the ground state $n=0$; the excited states have larger uncertainty products.

For Coulomb binding, the virial theorem gives a typical kinetic energy of the order of the binding energy. Taking $p_{\mathrm{typ}}\simeq \sqrt{2m_e|E_1|}$ and using the Heisenberg uncertainty relation estimate $r_{\mathrm{typ}}\sim \hslash /p_{\mathrm{typ}}$ givesThe Bohr model gives $r_1=a_0/Z=a_0/2=2.65\times 10^{-11}\mathrm{m}$ for a heavy helium nucleus, agreeing with the characteristic scale. Equivalently, minimizing the localization estimate $E(r)\sim {\hslash }^2/(2m_e{r}^2)-Z{e}^2/(4\pi {ϵ}_{0}r)$ gives that same scale.

This is an order-of-magnitude radius estimate, not an exact deduction of the cloud's expected value from the inequality alone. Numerical coefficients depend on which spread is used; $\Delta x\Delta p\ge \hslash /2$ does not assert $⟨r⟩=\hslash /p$. For the exact hydrogenic $1s$ state, $⟨r⟩=3a_0/(2Z)=3.97\times 10^{-11}\mathrm{m}$ at $Z=2$, one and a half times the Bohr orbital radius. An electron cloud has a radial distribution, whereas the Bohr model assigns a definite radius.

Both first moments vanish, so the standard deviations follow directly from the second moments:The Heisenberg uncertainty relation is satisfied and is not saturated. A Gaussian wave packet can saturate the position-momentum bound; this compactly supported polynomial is not Gaussian.

Preparation uncertainty describes the outcome spread intrinsic to a specified quantum state preparation. Operationally, prepare many copies of the same quantum state, measure position on one group and momentum on another, and estimate the two variances. Ideal measurements still have the spreads predicted by the Born rule. No earlier position measurement need occur in the momentum group. The Heisenberg uncertainty relation constrains these two distributions belonging to the same preparation.

Measurement backaction concerns how a particular measurement in quantum mechanics changes a quantum state. Instrument resolution concerns how its reported outcomes relate to the intended observable. Those questions require a model of the instrument and its interaction with the system. Substituting an instrument error and an induced change for $\Delta x$ and $\Delta p$ in the Heisenberg uncertainty relation does not follow from its proof. The practical value of preparation uncertainty is that it bounds achievable input spreads before any such instrument model is chosen.