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Feynman diagrams organize the perturbative calculation of the matrix element in Fermi's golden rule. Each external line labels an incoming or outgoing particle state; each internal line contributes a propagator for a generally off-shell intermediate field; and each interaction vertex contributes a coupling and the appropriate spin, momentum and internal-symmetry structure. All indistinguishable allowed diagrams must be summed as probability amplitudes before taking the absolute square, so their interference contributes to the rate.

In the radiative pion decay diagram in part (b), solid arrowed lines are fermions; the wavy line joining the hadronic and leptonic vertices is a W boson propagator; and the wavy outgoing line is an external photon. Arrows track fermion number, so an outgoing antimuon has an arrow towards its production vertex. Each vertex conserves four-momentum; an internal propagator need not satisfy the on-shell energy–momentum relation. A closed loop Feynman diagram additionally requires an integration over an internal four-momentum.

The final Lorentz-invariant phase space, conservation Dirac delta functions, and the initial-state normalization or incident flux turn the squared scattering amplitude into a decay rate or cross section. A diagram supplies the dynamics; it is not itself the final density of states.

Beta decay converts a neutron to a proton or vice versa through the weak interaction, leaving the mass number fixed:Inside a nucleus the full initial and final nuclear masses determine whether either process is possible. Electron capture, $p+e^{-}\to n+{\nu }_e$, is the related bound-electron process. The emitted charged-lepton spectrum is continuous because energy is shared with the neutrino and nuclear recoil.

For allowed beta decay, the lepton wavefunctions' leading spatial terms are constant over the nucleus. A Fermi transition has $\Delta J=0$ and no parity change. A Gamow-Teller transition has $\Delta J=0,1$ and no parity change, excluding $0\to 0$. Thus a same-parity transition with nonzero initial and final spins and $\Delta J=0$ can contain both contributions. First-forbidden transitions change parity and permit $|\Delta J|=0,1,2$; the $|\Delta J|=2$ case is unique forbidden beta decay. Higher forbiddenness requires higher powers of the small lepton momentum times nuclear size. “Forbidden” denotes suppression of the leading allowed operator, not an impossible decay.^[ref]

Fermi's golden rule expresses a rate as a squared nuclear matrix element times leptonic phase space. Neglecting charged-lepton mass and Coulomb distortion, a constant allowed matrix element givesThis is the approximate Sargent rule, $\lambda \propto Q^5$ for comparable allowed strengths; it is not an exact law across arbitrary transitions. Finite electron mass, the Fermi function, forbidden shape factors and nuclear structure change the comparison. A microscopic first-forbidden unique operator introduces additional momentum factors.

Using bare nuclear masses $M_i,M_f$, the kinetic-energy release isThe daughter is different in the two cases: its charge is $Z+1$ for beta minus and $Z-1$ for beta plus. The relevant mass difference must exceed one electron mass. For neutral atomic masses, to leading order in small electronic binding corrections, the corresponding expressions are $Q_{{\beta }^{-}}=(M_i^{\mathrm{at}}-M_f^{\mathrm{at}})c^2$ and $Q_{{\beta }^{+}}=(M_i^{\mathrm{at}}-M_f^{\mathrm{at}}-2m_e)c^2$. Electron capture is energetically possible approximately when $M_i^{\mathrm{at}}>M_f^{\mathrm{at}}$, with available shell binding energy and daughter excitation included. Nuclear masses and atomic masses must not be mixed.

The bracketed diagram energies are nuclear, so subtract one electron rest energy from each parent-to-daughter energy difference. The three kinetic-energy endpoints areThe observed partial-rate ratios are ${\lambda }_N/{\lambda }_{Ne}=0.19672$ and ${\lambda }_N/{\lambda }_F=0.24918$.

To obtain the simple Sargent-law estimates, define an effective strength by $\lambda =C|M_{\mathrm{eff}}{|}^2{Q}^5$, using the same normalization $C$. ThenThus the two forbidden branches have effective strengths of the same broad order, whereas the N ground-state branch is highly suppressed relative to the allowed F branch after this approximate fifth-power kinematic scaling. The first comparison involves mirror transitions with related nuclear configurations; order-of-magnitude similarity is reasonable, without expecting exact equality from this estimate.

These effective strengths must not be confused with precisely extracted microscopic reduced nuclear matrix elements. Both N and Ne ground-state branches are first-forbidden unique, for which the first-forbidden unique beta-decay phase space contains an additional factor proportional to $p_e^2+E_{\nu }^2$. In the massless, no-Coulomb limit this gives $Q^7$, so using the same microscopic unique-forbidden normalization for those two branches instead yieldsComparing a unique-forbidden reduced operator directly with an allowed reduced operator also requires a convention for its powers of length and momentum. The number $8.05\times 10^{-5}$ is consequently an effective Sargent-normalized suppression, not a universal dimensionless ratio between different operator ranks. Electron-mass and opposite-sign Coulomb corrections matter, particularly for the low-energy F branch. The data justify approximate strength comparisons; they do not fix precision nuclear matrix elements without these extra ingredients.

The dominant interaction is weak. In panel (iv),The quark vertex carries the Cabibbo-Kobayashi-Maskawa matrix factor $V_{ud}$. The down quark and anti-up quark form a color singlet pion. The pion bracket represents its bound-state matrix element, conventionally specified by the pion decay constant; it is not an additional weak vertex. The neutrino carries the initial tau lepton number.

Assume allowed beta decay, so the leading matrix element has negligible lepton-momentum dependence. Neglect nuclear recoil and neutrino mass. Let $Q$ be the kinetic energy available to the electron and antineutrino, $E_0=m_e+Q$, and $E=\sqrt{p^2+m_e^2}$. The neutrino energy and momentum magnitude are $k=E_0-E$.

Fermi's golden rule weights accessible final states. After angular integration, the momentum-space factors are $p^{2}dp$ and $k^{2}dk$. The energy-conserving Dirac delta function removes the neutrino integral:The constant includes the allowed nuclear matrix element, weak-coupling and angular factors. Including the usual Fermi function for electron-nucleus Coulomb distortion givesFor the kinetic-energy spectrum, $dp/dT=E/p$ instead yieldsThe two distributions differ by a Jacobian; forgetting it changes the shape.

A pure fifth-power law additionally neglects electron mass and Coulomb distortion and keeps the nuclear matrix element fixed. Then $E=p$, $E_0\simeq Q$, andTwo momentum powers come from each final particle and one from the remaining integration. This is the origin of the Sargent rule. Exact $Q^5$ dependence requires the relativistic-electron, constant-matrix-element limit. For $Q\ll m_e$, the same approximation without Coulomb distortion has $pE\simeq m_e\sqrt{2m_{e}T}$ and instead scales as $m_e^{3/2}{Q}^{7/2}$. Forbidden-transition factors, nuclear-structure changes and Coulomb distortion also modify the simple law.

For gamma decay, the nuclear gamma-ray multipole selection rules require the radiation's angular momentum $L$ to satisfy the triangle inequalityHere $J_i=J_f=5/2$, so $0\le L\le 5$. A real emitted photon has no monopole radiation mode, leaving $L=1,2,3,4,5$.

The two nuclear quantum states have the same parity. An electric multipole has parity factor $(-1{)}^L$, while a magnetic multipole has factor $(-1{)}^{L+1}$. Thus electric radiation requires even $L$ and magnetic radiation odd $L$:In particular, an $E0$ nuclear transition cannot emit the single real photon requested here; processes such as internal conversion are different decay channels.

For usual nuclear gamma ray energies, $kR\ll 1$, where $k=E_{\gamma }/(\hslash c)$ and $R$ is the nuclear size. Higher multipole radiation orders bring extra powers of this small parameter, so $M1$ and $E2$ are the likely important contributions, with $M3$, $E4$ and $M5$ generally much smaller. Often $M1$ is dominant, but an allowed $M1$ matrix element can be hindered and an $E2$ transition can be collectively enhanced. The quantum numbers alone do not fix a numerical relative contribution or an unconditional ordering of the $M1$ and $E2$ rates.

The nuclear determination of $|V_{ud}|$. A $0^{+}\to 0^{+}$ beta decay transition has no nuclear parity change and can proceed with zero lepton orbital angular momentum. The allowed scalar operator gives a Fermi transition; the rank-one Gamow-Teller transition operator cannot connect total nuclear spins zero to zero. The strong isobaric-analog transitions used for precision measurements are superallowed beta decays. A generic $0^{+}\to 0^{+}$ assignment alone would not guarantee superallowed strength if the states were nonanalog.

At the quark level the weak charged current converts $d\leftrightarrow u$ and carries $V_{ud}$ in the amplitude. The nuclear Fermi operator is an isospin ladder operator, givingHere $f$ denotes the known phase space dependence, with the dimensional factors included in the proportionality constant. In the isospin-symmetric limit, $M_F$ between analog states is fixed by isospin algebra, for example $|M_F{|}^2=2$ for the usual $T=1$ analog transitions. A measured partial half-life and the computed phase space factor give the comparative half-life; after radiative corrections and isospin breaking corrections, its inverse determines $|V_{ud}{|}^2$. The decay rate measures a magnitude squared, not the complex phase of $V_{ud}$.

Ds decay diagrams and CKM comparisons. The original PDF has $K^{+}{\bar{K}}^0$, with ${\bar{K}}^0=s\bar{d}$. The local TeX drops that overbar; using $K^0=d\bar{s}$ instead would give a different flavour channel. Use the PDF's neutral antikaon throughout.

In the representative Ds nonleptonic tree amplitudes, the Ds meson spectator is $\bar{s}$. The elementary weak transitions are

In the first two channels, the produced $u$ joins the spectator $\bar{s}$ to form $K^{+}$; the other quark joins $\bar{d}$. For the first channel this uses the $-d\bar{d}/\sqrt{2}$ component of the neutral pion. In the third, $s\bar{s}$ forms the phi meson and $u\bar{d}$ forms ${\pi }^{+}$. The two possible final pairings have different colour contractions and hadronic matrix elements even when their weak CKM factors coincide.

The purely CKM rate suppression of $c\to d$ relative to $c\to s$ isThus $K^{+}{\pi }^0$ is singly Cabibbo suppressed, while $K^{+}{\bar{K}}^0$ and $\varphi {\pi }^{+}$ are Cabibbo-favored at this level. If reduced hadronic strengths and phase space effects were equal, including the neutral-pion component would giveThe supplied branching fractions instead giveThe first observed ratio is $0.814$ times the neutral-pion-corrected simple estimate, requiring only an order-one remaining hadronic and phase space factor. Relative to $\varphi {\pi }^{+}$, that remaining factor is $0.525$. The two Cabibbo-favored channels differ by a factor $0.644$ despite equal CKM magnitudes. The CKM hierarchy therefore explains why one channel is much rarer, but it does not calculate the three exclusive branching fractions; spin, phase space, color charge and strong interaction final-state effects also matter.

What spin and parity establish. In either pseudoscalar-pseudoscalar final state, the total intrinsic spin is zero. Conservation of angular momentum for a spin-zero parent forces $L=0$, so the final parity isopposite to the parent $0^{-}$ Ds meson. ConsequentlyFor a vector meson plus a pion, the intrinsic final spin is one. Coupling to total $J=0$ requires $L=1$, giving $P_f=(-1)(-1)(-1{)}^1=-1$, equal to the parent parity. Thus spin and parity alone permit a parity-conserving $\varphi {\pi }^{+}$ transition. That is the limited sense in which it is not necessarily weak from this information alone. In reality it is also weak: the initial charm flavour disappears, which the strong interaction and electromagnetic interaction conserve.

W branching fractions. In the massless-daughter, tree-level estimate, each leptonic channel contributes one common unit of decay width. Each accessible quark pair contributes $3|V_{ij}{|}^2$ because of the three final colours. The charged-current W branching-fraction counting denominator isOnly $u$ and $c$ are accessible up-type quarks; the large $V_{tb}$ is irrelevant because a W boson cannot decay into the heavier top quark. HenceA final bottom quark arises in $W^{-}\to b\bar{u}$ or $b\bar{c}$, soThis counts the primary decay channels; finite daughter masses and QCD corrections refine the approximation.

The requested top-lifetime estimate and its limitation. Under the instructed equal-coefficient Sargent's rule scaling, $\Gamma \propto m^5$ and the mean lifetime is inversely proportional to that fifth power. Converting both masses to GeV givesThis is the requested order-of-magnitude estimate, assuming the same effective coefficient for the two total decay rates. It corresponds to $c\tau \simeq 54\mathrm{fm}$ and is longer than the hadronization time, roughly a few $10^{-24}\mathrm{s}$. Taken literally, it would allow top-flavored hadrons to form; it does not establish the physical statement that the top decays before hadronizing.

The extrapolation fails because Sargent's rule uses a low-energy Fermi interaction approximation, whereas $m_t>m_W+m_b$ permits $t\to bW$ with an on-shell W boson. The physical leading decay width instead scales asUsing $m_W\simeq 80.4\mathrm{GeV}$, the Fermi constant $G_F=1.16638\times 10^{-5}\mathrm{Ge}{\mathrm{V}}^{-2}$ and $|V_{tb}|=0.999$ gives about $1.55\mathrm{GeV}$ at the supplied top mass. Then ${\tau }_t=\hslash /{\Gamma }_t$, with $\hslash =6.58212\times 10^{-25}\mathrm{GeV}\mathrm{s}$, gives ${\tau }_t\simeq 4.2\times 10^{-25}\mathrm{s}$. This corrected regime explains top-quark decay before hadronization.^[ref]

Count strangeness before drawing a weak interaction diagram. An Omega baryon has $S=-3$, a Lambda baryon has $S=-1$, a negatively charged kaon has $S=-1$, and a pion has $S=0$. Thus the two nonleptonic baryon decays requireThe strong interaction cannot supply either net change of strangeness.

In the upper Feynman diagram, one initial strange quark undergoes the weak charged current transitionThe remaining two strange quarks are spectator quarks. Combine the outgoing $u,d$ and one spectator $s$ into the Lambda baryon, and the remaining $s$ with $\bar{u}$ into the kaon. This uses one W boson propagator and two weak interaction vertices. Its Cabibbo-Kobayashi-Maskawa matrix factor has magnitude $|V_{us}{V}_{ud}|\simeq \sin {\theta }_C\cos {\theta }_C$.

For the lower Feynman diagram, let the upper initial $s$ first emit a $W^{-}$ and become a virtual $u$, then emit a $W^{+}$ and become a $d$. The $W^{+}$ is absorbed by a second initial $s$, turning it into a $u$; the third initial $s$ remains a spectator quark. The emitted $W^{-}$ produces $d\bar{u}$. The complete quark flavour flow is thereforeThe final $d,u,s$ form the Lambda baryon and the other $d\bar{u}$ form the pion. Electric charge conservation holds at every vertex. Internal lines are virtual; this is a contribution to one decay amplitude, not a sequence of freely propagating on-shell constituent-quark decays.

This is a possible lowest-weak-order diagram for a second-order strangeness-changing weak decay: it has two W boson propagators and four weak interaction vertices. Its Cabibbo-Kobayashi-Maskawa matrix factor has magnitude $|V_{us}{|}^2|V_{ud}{|}^2$ in the displayed light-quark topology. Other contributions can exist, but a single charged-current insertion cannot turn two net strange quarks into nonstrange ones. No extra loop Feynman diagram is compulsory: the displayed two-W tree already realizes the required quark flavour change.

At low energies, one-W exchange produces one power of the Fermi constant in an effective decay amplitude; the second process requires an additional weak insertion. Hadronic matrix elements and propagators supply the accompanying dimensions, so a numerical ratio cannot be obtained by simply squaring a dimensionful constant. The robust comparison isThe dominant reason is the extra weak order, with additional Cabibbo suppression in this representative topology. It is not a comparison between a strong decay and a weak decay.

The local Fermi interaction gives a three-body leptonic decay width of the format tree level with negligible neutrino masses and universal weak coupling constant values. The fifth power follows from the dimension of $G_F$ and the three-body phase space; $f$ results from integrating the V-A interaction matrix element over the reduced phase space when the charged daughter has nonzero mass. It is more than a purely geometric volume correction because the spin-dependent matrix element is included.

As $z\to 0$, $f(z)\to 1$; increasing daughter mass reduces the available energy and suppresses the rate. The largest ratio is $m_{\mu }/m_{\tau }$, so the $\tau \to \mu$ correction is visibly below one. The electron is extremely light relative to either parent. Using the lepton masses givesThese reproduce the supplied factors to their stated accuracy. Radiative and finite-W boson corrections are separate small corrections when a precision lepton universality test is performed.

The Cabibbo-Kobayashi-Maskawa matrix is the unitary change of basis between down-type quark mass eigenstates and the combinations appearing in the weak charged current. In particular, $u\bar{d}$ and $u\bar{s}$ production carry $V_{ud}$ and $V_{us}$. Exclusive tau lepton decays into a charged pseudoscalar meson giveup to small radiative corrections. The pseudoscalar decay constants specify the vacuum-to-meson axial-current matrix elements, $⟨0|\bar{q}{\gamma }^{\mu }{\gamma }^{5}u|P(p)⟩=i{f}_P{p}^{\mu }$.

The observed branching fraction ratio is $0.70/10.83=0.06464$, and the kinematic factor is approximately $0.8622$. Hence the directly determined combination isWith an independent, representative lattice input $f_K/f_{\pi }\simeq 1.193$, this becomes $|V_{us}/V_{ud}|\simeq 0.2295$, close to the source's rounded expectation $0.224/0.975=0.2297$. Conversely that expectation and the hadronic input predict a branching fraction ratio near $0.0648$. Without the decay-constant input the table does not isolate a pure Cabibbo-Kobayashi-Maskawa matrix ratio. Setting $f_K=f_{\pi }$ would give approximately $0.0455$, demonstrating the importance of flavor symmetry breaking.^[ref]

Use the explicit charged channels $D^{+}\to {\pi }^{+}{\pi }^{+}{\pi }^{-}$ and $K^{+}\to {\pi }^{+}{\pi }^{+}{\pi }^{-}$, labeling the two positive pions as 1,2 and the negative one as 3. With $m_{\pi }=0.139\mathrm{GeV}$, the opposite-charge pair mass squares have the rangesTheir exact correlated boundaries follow by taking the pair rest-frame angle to its endpoints. For general daughter masses, define $\Lambda (a,b,c)=a^2+b^2+c^2-2ab-2ac-2bc$. At fixed $s=s_{13}$,The charged-pion thresholds give the following numerical comparison:

The area is about $3150$ times smaller for the kaon. Including the $M_0^{-2}$ phase-space prefactor, its integrated labelled three-body phase space is about $220$ times smaller; neither comparison predicts a ratio of weak decay widths without the matrix elements.

A ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ intermediate state produces horizontal or vertical bands near $s_{13}$ or $s_{23}=m_{\rho }^2\simeq (0.775\mathrm{GeV}{)}^2$ in the charm decay. Scalar pion-pair amplitudes, including broad low-mass $S$-wave contributions and accessible $f_0(980)$, can also populate the plot. The kaon pair mass cannot exceed $0.355\mathrm{GeV}$, so it cannot contain an on-shell ${\rho }^0$ or $f_0(980)$ band; smooth nonresonant and rescattering effects remain possible. Channels containing neutral pions use their supplied $0.135\mathrm{GeV}$ mass and shift these boundaries slightly without changing the conclusion. The figure shows kinematic domains and representative accessible bands, not an assumed measured intensity.

In a relativistic two-body decay, four-momentum conservation in the parent's rest frame requires ${\mathbf{p}}_1^{\ast }=-{\mathbf{p}}_2^{\ast }$ andFor fixed daughter masses, this fixes a unique $p^{\ast }$, independent of direction. In the Lorentz-invariant phase space integral, the Dirac delta distribution enforcing conservation of energy therefore performs the radial momentum integration. Only the solid angle remains to be integrated; the resulting $p^{\ast }$ can be taken outside that integral, even if the matrix element depends strongly on direction.

For three or more final particles, four-momentum conservation generally leaves continuously variable daughter energies and momenta, so the decay rate includes integrations over them, or equivalently over independent invariant masses in a Dalitz plot. Integrating over a variable daughter invariant mass is also necessary when an unstable daughter is treated with a finite decay width rather than a fixed mass.

Define the Mandelstam variables for the scattering labels by $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2=q^2$ and $u=(p_1-p_4{)}^2$. Dropping the electron and muon masses givesThese identities follow by expanding the corresponding squared four-momenta, using $p_i^2=0$ and four-momentum conservation. Substitution into the supplied spin-averaged matrix element yieldsThe $1/4$ initial spin average is already included in the given expression; no further averaging factor belongs here.

For normalized amplitudes $c_n,c_{n^{\prime }}$ and a time-independent observable $A$, the Schrödinger equation givesThe interference term therefore hasThis is the period of two-state quantum interference. If the off-diagonal matrix element or either amplitude vanishes, the expectation value is constant, so this is not its unique fundamental period. Equal energies likewise give no beat.

For the Quantum harmonic oscillator defineThe Canonical commutation relation gives $[\hat{a},{\hat{a}}^{†}]=1$, andThus the creation operator and annihilation operator raise or lower the energy by one quantum. Positivity of ${\hat{a}}^{†}\hat{a}$ and termination of downward stepping give $E_n=\hslash \omega (n+1/2)$, $n=0,1,\dots$. Normalized actions are $\hat{a}|n⟩=\sqrt{n}|n-1⟩$ and ${\hat{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The annihilated ground state obeysThe nonzero Ground state energy is the zero-point energy, consistent with position-momentum quantum uncertainty. Since $\hat{x}=\sqrt{\hslash /(2m\omega )}(\hat{a}+{\hat{a}}^{†})$, a perturbation linear in position connects only neighbouring number states. The algebra is especially useful for matrix elements and selection rules.^[ref]

The infinite square well has energy levels $E_n=n^2{h}^2/(8m_e{L}^2)$, with $n=1,2,\dots$. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely $n=1\to 2$. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap givesHere the photon gap is approximately $1.55\mathrm{eV}$, not the ground-state energy itself.

Choose the two real energy eigenstates in part (a) and take their coefficients to be in phase initially. A different initial relative phase only shifts the oscillation in time. The normalized quantum superposition isOrthogonality removes the cross terms in the energy expectation value, soFor position, the diagonal matrix elements equal $L/2$ by reflection symmetry, but the off-diagonal element is nonzero. Using the product-to-sum identity and ${\int }_0^{L}x\cos (j\pi x/L)dx=L^2[(-1{)}^j-1]/(j^2{\pi }^2)$ givesTherefore the equal superposition of the first two box states hasThe amplitude is $\boxed{16L/(9{\pi }^2)\simeq 0.1801L}$, and the ordinary frequency is $\boxed{f=\Omega /(2\pi )=3h/(8m{L}^2)}$. The motion of the mean position is an interference effect between distinct energies, while their populations, and hence the mean energy, remain fixed.

Inverting the definitions of the creation operator and annihilation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. Their actions and the orthogonality of the oscillator energy eigenstates giveThus the electric dipole transition selection rule iswith the lower alternative absent for $n_1=0$. The field exchanges one oscillator quantum in first-order dipole coupling. For a long weak monochromatic drive, its frequency must also be resonant with the level separation, here $\omega$, to give a substantial transition probability. Higher-order multiphoton processes are beyond the stated single-position-matrix element approximation.

For the harmonic-oscillator number-state second moments, the number operator gives $⟨a^{†}a{⟩}_n=n$, while the canonical commutation relation gives $⟨a{a}^{†}{⟩}_n=n+1$. Terms with $a^2$ or $(a^{†}{)}^2$ have zero diagonal matrix elements, since they change the occupation number by two. ThereforeConsequently the potential energy and kinetic energy means areTheir sum is the oscillator energy eigenvalue $E_n=\hslash \omega (n+1/2)$. Their equality is also the virial theorem result for a quadratic potential, here derived directly from the ladder operators.

Each energy eigenstate picks up its own phase under unitary time evolution. The relative phase therefore changes at $(E_1-E_0)/\hslash =\omega$. After removing an irrelevant global phase, the state isThe diagonal position matrix elements vanish, but the annihilation operator and creation operator giveThe two off-diagonal cross terms in the expectation value yieldThus the angular frequency is $\omega$, and the magnitude of the oscillation, meaning its amplitude, isThe peak-to-peak displacement is $2A$. The phase $\varphi$ fixes where the oscillation starts but does not change $A$. A pure $|0⟩$ or $|1⟩$ preparation has no mean oscillation, whereas an equal-weight quantum superposition maximizes its amplitude within this two-state family. The motion is produced by interference between energy eigenstates, rather than by a changing occupation of either one.