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In the position representation, $\hat{p}=-i\hslash \nabla$. The orbital angular momentum operator and its squared magnitude areFor example, with $D=x\cdot \nabla$, the latter is $-{\hslash }^2[r^2\Delta -D^2-D]$. Operator products here act on smooth wavefunctions in a common invariant domain.

To calculate the angular momentum commutation relations, use $[x_a,p_b]=i\hslash {\delta }_{ab}$ and the vanishing position-position and momentum-momentum commutators. Expanding the products givesMultiplication by ${ϵ}_{iab}{ϵ}_{jcd}$ then givesFinally, the product commutator rule yieldsThe final cancellation pairs the antisymmetric Levi-Civita symbol with an expression symmetric in $j,k$. Thus

The orbital angular momentum operators are $\mathbf{L}=\mathbf{x}\times \mathbf{p}$, soThe canonical commutation relations are $[x_i,x_j]=[p_i,p_j]=0$ and $[x_i,p_j]=i\hslash {\delta }_{ij}$. On a common domain of smooth wavefunctions, the product commutator identity givesContract this with ${ϵ}_{iab}{ϵ}_{jcd}$. The first term contributes $i\hslash ({\delta }_{ij}\mathbf{x}\cdot \mathbf{p}-x_j{p}_i)$ and the second contributes $i\hslash (x_i{p}_j-{\delta }_{ij}\mathbf{x}\cdot \mathbf{p})$. Their sum isThis verifies the angular momentum commutation relations explicitly. Similarly,

In the position representation $\mathbf{p}=-i\hslash \nabla$. For a radial wavefunction, $\nabla {\Psi }_0={\Psi }_0^{\prime }(r)\mathbf{x}/r$, parallel to $\mathbf{x}$, soWrite ${\Phi }_i=h(r)x_i$ with $h(r)={\Psi }_0^{\prime }(r)/r$. The operator $L_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$ annihilates the radial factor. It gives $L_3{x}_1=i\hslash x_2$, $L_3{x}_2=-i\hslash x_1$, and $L_3{x}_3=0$, henceThus these components, when nonzero, are the specified eigenfunctions with eigenvalues $0,+\hslash ,-\hslash$. This is an example of angular momentum of a radial function times a linear polynomial.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

In the position representation, the orbital angular momentum operator isOn a common smooth domain use $[x_a,p_b]=i\hslash {\delta }_{ab}$, $[x_a,x_b]=[p_a,p_b]=0$. Expanding the product commutator givesContracting with the two alternating tensors gives the orbital angular momentum commutation relations:For $L_{\pm }=L_1\pm i{L}_2$, these imply $[L_3,L_{\pm }]=\pm \hslash L_{\pm }$. Therefore $L_3\psi =0$ yieldsThusThe nonzero qualification is necessary: a rotationally invariant wavefunction is annihilated by all three components, and the zero vector is not an eigenvector.

The rotation unitary operator is $U(\mathbf{n},\theta )=\exp (-i\theta \mathbf{n}\cdot \mathbf{J}/\hslash )$. A state with zero orbital angular momentum has a rotationally invariant orbital factor; a spinless $j=0$ state is wholly unchanged. Zero orbital angular momentum alone does not rule out a nontrivial rotation of intrinsic spin angular momentum.

For spin one-half, $\mathbf{J}=\hslash \mathbf{\sigma }/2$. Direct multiplication of the Pauli matrices gives ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$. The antisymmetric part vanishes when contracted with $a_i{a}_j$, henceFor $\mathbf{a}\ne 0$, the operator has trace zero, so its eigenvalues are $\pm \hslash |\mathbf{a}|/2$; for $\mathbf{a}=0$ both are zero. With unit $\mathbf{n}$, the even and odd powers in the exponential give

Take ${\mathbf{n}}^{\prime }=(\sin \theta ,0,\cos \theta )$. The rotated state $|m^{\prime }{⟩}_{\theta }=U(\mathbf{y},\theta )|m^{\prime }⟩$ satisfies $({\mathbf{n}}^{\prime }\cdot \mathbf{J})|m^{\prime }{⟩}_{\theta }=m^{\prime }\hslash |m^{\prime }{⟩}_{\theta }$, because $U{J}_3{U}^{†}={\mathbf{n}}^{\prime }\cdot \mathbf{J}$. By the Born rule, the probability of this outcome in the initial state $|m⟩$ is $|{}_{\theta }⟨m^{\prime }|m⟩{|}^2=|⟨m|U(\mathbf{y},\theta )|m^{\prime }⟩{|}^2$. In the $J_3$ eigenbasis,The transition from $m=1/2$ to $m^{\prime }=-1/2$ thus has probability ${\sin }^2(\theta /2)=1-{\cos }^2(\theta /2)$. $\boxed{A=1,B=-1}$.

Applying the angular momentum operator directly givesFor an eigenvalue $\lambda$, comparison with $\lambda (f(x)+iy)z$ gives $\hslash x=\lambda f(x)$ and $\hslash f^{\prime }(x)=\lambda$. The first identity excludes $\lambda =0$; differentiating it and using the second gives ${\lambda }^2={\hslash }^2$. No additive constant in $f$ is permitted. The possibilities are $f(x)=x$, $\lambda =\hslash$, and $f(x)=-x$, $\lambda =-\hslash$.

For $f=x$, the polynomial $\psi =(x+iy)z$ is homogeneous of degree two and satisfies $\Delta \psi =0$. Expanding the three squared orbital angular momentum operators gives, with $D=x{\partial }_x+y{\partial }_y+z{\partial }_z$,Here $D\psi =2\psi$ and $D^2\psi =4\psi$. Therefore ${\hat{L}}^2\psi =6{\hslash }^2\psi$. This is the degree-two case of angular momentum of a homogeneous harmonic polynomial; its angular labels are $l=2$, $m=1$.

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

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

With $\hslash =1$, angular momentum eigenstates have $j=0,\frac{1}{2},1,\dots$ and $m=-j,-j+1,\dots ,j$. Their $J^2$ and $J_3$ eigenvalues are $j(j+1)$ and $m$. Adding spins $1/2$ and $j$ gives $J=j+1/2$ and, if $j\ge 1/2$, $J=j-1/2$. For the Clebsch-Gordan coefficients to be nonzero one requires $\boxed{M=r+m}$.

For $j=2$, write $|\pm ,m⟩$ for $|1/2,\pm 1/2⟩|2,m⟩$. The highest state and one application of the angular momentum lowering operator giveThe lowering coefficient of the highest $J=5/2$ state is $\sqrt{5}$, while the two product-state coefficients before division are $1$ and $2$. The last line is its normalized orthogonal complement. Its coefficient with maximal $m=2$ is positive, fixing the requested phase convention. All other coefficients for the allowed $M\ge 3/2$ vanish.

For the decay, the final total intrinsic spin is $1/2$. Combining it with integer orbital angular momentum $\ell$ to obtain $J=3/2$ permits only $\ell =1,2$. Parity conservation gives ${\eta }_X={\eta }_A{\eta }_B(-1{)}^{\ell }=(-1{)}^{\ell }$. Thus $\boxed{{\eta }_X=+1:\ell =2;{\eta }_X=-1:\ell =1}$.

For $\ell =2$ and initial $M=3/2$, the last displayed coupled state shows that the spin-up probability, after ignoring the orthogonal orbital states, is $1/5$. For $\ell =1$, $J=3/2$ is the maximal coupled angular momentum and its highest state is $|1/2,1/2⟩|1,1⟩$. Therefore $\boxed{P=1/5for{\eta }_X=+1,P=1for{\eta }_X=-1}$.

The agreement is in the energy levels, not in electron trajectories. The Bohr model assumes a particle on a circular orbit, with orbital angular momentum quantized as $N\hslash$. The Coulomb bound states here are spatially distributed wavefunctions with no definite classical orbit. Their probability density is spherically symmetric, and their orbital angular momentum is zero because $\ell =0$. Their radial distributions also depend on $N$ and can have nodes. A real radial wavefunction times a global stationary phase has zero probability current, unlike the circulating point charge in the Bohr picture.

The orbital angular momentum linear operator is the cross product of the position operator and the momentum operator:In position space, $\hat{\mathbf{p}}=-i\hslash \nabla$, so $\hat{\mathbf{L}}=-i\hslash \mathbf{r}\times \nabla$. The Levi-Civita symbol gives ${\hat{L}}_1={\hat{x}}_2{\hat{p}}_3-{\hat{x}}_3{\hat{p}}_2$ and its cyclic analogues.

In atomic units the time-independent Schrödinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

The intrinsic parity $\eta$ of a particle is the parity eigenvalue of its one-particle state at rest:For the particles in this question, $\eta =\pm 1$. A two-particle state has multiparticle parity ${\eta }_1{\eta }_2(-1{)}^{\ell }$, where $\ell$ is its relative orbital angular momentum.

The pionic-deuterium ground state has atomic $\ell =0$. Since the pion has spin zero, its total angular momentum is therefore the deuteron spin, $J=1$, and its parity is ${\eta }_d{\eta }_{\pi }={\eta }_{\pi }$. The two neutrons are identical fermions. Their antisymmetric spin-one-half singlet state would require even orbital $\ell$, but $S=0$ and $J=1$ would instead require $\ell =1$. Hence they must occupy the symmetric spin-one-half triplet state, so the orbital state is antisymmetric and $\ell$ is odd. The addition of angular momentum with $S=J=1$ permits $\ell =0,1,2$, leaving $\ell =1$. Thus parity conservation givesso the pion has intrinsic parity $\boxed{{\eta }_{\pi }=-1}$.

In ${\Delta }^{-}\to {\pi }^{-}+n$, the final total spin is $S=1/2$. Coupling it to $\ell$ to obtain $J=3/2$ givesso the allowed orbital angular momenta are $\boxed{\ell =1,2}$. The observed angular probability is ${\cos }^2\theta =|P_1(\cos \theta ){|}^2$, where $P_1$ is a Legendre polynomial; it therefore selects the $\ell =1$ partial wave rather than $\ell =2$. The parity selection rule for a two-body decay now givesTherefore the delta particle has intrinsic parity $\boxed{{\eta }_{{\Delta }^{-}}=+1}$.

The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

The Hamiltonian is the sum of three commuting one-dimensional harmonic-oscillator Hamiltonians. Its product eigenstates arewith energiesEquivalently, the level with $N=n_x+n_y+n_z$ has energy $E_N=\hslash \omega (N+3/2)$, as in the three-dimensional isotropic harmonic oscillator.

Up to normalization, the ground-state wavefunction isIt is radial, so $\nabla {\Psi }_{000}$ is parallel to the position vector. Since the orbital angular momentum is $L=-i\hslash r\times \nabla$, every component of $L$ annihilates this state. Thereforeand the ground state has $\ell =m=0$.

The first excited level has $N=1$ and is spanned byThe statesatisfies $L_z{\Phi }_0=0$. Applying the ladder operators givesHence convenient $m=\pm 1$ eigenstates areup to normalization and irrelevant overall phases.

Finally, the isotropic Hamiltonian is rotationally invariant. The rotational invariance of a central-potential Hamiltonian givesThese commuting self-adjoint operators can be simultaneously diagonalized within each energy eigenspace, which is why joint eigenstates of $H,L^2,L_z$ must exist.

Parity conservation and the negative parity of $X$ require odd orbital angular momentum $L$. Since the two $Y$ particles are identical bosons, their spin state must then be antisymmetric, so their total spin is $S=1$. Total angular momentum zero forces $L=S=1$.

Conditioning on spin projection $m_S=1$ leaves orbital projection $m_L=-1$. The angular density is therefore proportional to $|Y_{1,-1}(\theta ,\phi ){|}^2\propto {\sin }^2\theta$. Including the solid-angle measure, the requested probability is

The planar orbital angular momentum isThe canonical commutation relations giveConsequentlyThus angular momentum preserves each energy eigenspace. More explicitly,soWithin a degenerate energy eigenspace, this argument places no further restriction on the matrix of $L$.

A constant-width Breit-Wigner resonance cross section is the simplest nonrelativistic curve; a P-wave scattering width and relativistic kinematics explain important expected distortions. A comparison to specific observed deviations cannot be made without the missing plot.

Write $E$ for the centre-of-mass frame kinetic energy, $\mu =m_p{m}_{\pi }/(m_p+m_{\pi })$ for the reduced mass, and $k$ for the relative wave number. The nonrelativistic relations areFor an isolated particle resonance of spin $J$, averaging over the initial proton spin and using the zero pion spin givesFor the nearly elastic ${\Delta }^{++}$ state, $J=3/2$ and ${\Gamma }_{\pi p}\simeq \Gamma$, soFreezing $k$ at $k_R$ gives a symmetric Lorentzian centered at $E_R$, with full width at half maximum $\Gamma$ and height $8\pi /k_R^2$. This is the local curve to sketch. If the displayed variable is the beam kinetic energy $T$, its width is $\Delta T\simeq (m_p+m_{\pi })\Gamma /m_p$ within this nonrelativistic approximation. Keeping $k(E)$ changes the prefactor and hence the symmetry. Extending a constant-width approximation all the way to $E=0$ produces a spurious divergence, so it is not a valid threshold model.

The pion has intrinsic parity $-1$ and the proton has $+1$. The positive parity and $J=3/2$ of the Delta baryon require odd orbital angular momentum, with the lowest choice $l=1$. The finite-range partial-wave threshold law gives ${\Gamma }_{\pi p}(k)\propto k^{2l+1}=k^3$ near threshold. Inserting this in the elastic numerator makes ${\sigma }_{\mathrm{el}}\propto k^4$ away from the pole as $k\to 0$, rather than diverging. Equivalently the elastic scattering phase shift has ${\delta }_1=O(k^3)$, and ${\sin }^2{\delta }_1/k^2=O(k^4)$. Away from threshold, form factors and other channels modify the simple cubic width.

At the illustrative peak of part (d), the incident pion kinetic energy is comparable to its rest energy, so a nonrelativistic pion treatment is already inaccurate. The exact relative momentum iswith $\sqrt{s}=\sqrt{(m_p+m_{\pi }{)}^2+2m_{p}T}$. Relative to the symmetric local estimate, this changes the horizontal conversion, peak normalization and tails. The illustrative $1232\mathrm{MeV}$ peak has $p_{\ast }\simeq 227\mathrm{MeV}/c$, giving $8\pi (\hslash c{)}^2/(p_{\ast }c{)}^2\simeq 190\mathrm{mb}$ before small inelastic effects.

Finally, nonresonant partial waves can interfere with the resonant scattering amplitude, higher particle resonances and inelastic scattering channels can enter, and instrumental resolution can broaden a measured peak. These are physically distinct possible deviations; the missing data do not justify claiming any particular one was actually observed. Both the missing graphical overlay and its data-specific explanation remain source gaps.

A kaon is a pseudoscalar meson, with intrinsic parity $-1$. At rest, $P|\pm ⟩=-|\pm ⟩$. Combining this with the specified CP eigenvalues givesEquivalently, in the exchange-positive CP convention for neutral kaon mixing fixed by the question, $C|K^0⟩=-|{\bar{K}}^0⟩$ and $C|{\bar{K}}^0⟩=-|K^0⟩$. This sign is a convention for the flavor states; the specified CP eigenvalues and physical predictions must be kept consistent with it.

For two spin-zero pions, conservation of total angular momentum fixes the relative orbital angular momentum to zero. The resulting ${\pi }^{+}{\pi }^{-}$ or ${\pi }^0{\pi }^0$ state has CP $+1$. The lowest partial waves of ${\pi }^{+}{\pi }^{-}{\pi }^0$ and $3{\pi }^0$ have CP $-1$. Thus the dominant nonleptonic assignments in the CP symmetry approximation areThe two-pion mode has substantially more available kinetic energy and larger Lorentz-invariant phase space; the three-pion mode has an additional final particle and lies closer to its decay threshold. This helps explain why the CP-odd state has the longer mean lifetime. The actual mean lifetimes also depend on the decay amplitudes and on semileptonic channels, so counting final particles alone is not a quantitative prediction.

The middle panel of the figure in part (a) shows $\bar{s}\to \bar{u}{W}^{+}$, $W^{+}\to u\bar{d}$, with the $d$ as a spectator quark. The outgoing $d\bar{u}$ and $u\bar{d}$ form ${\pi }^{-}$ and ${\pi }^{+}$. In the bottom panel, an additional gluon produces $d\bar{d}$, supplying the neutral pion component for ${\pi }^{+}{\pi }^{-}{\pi }^0$. The charge-conjugate diagrams supply the other component of each of the kaon CP eigenstates. The drawn gluon splitting represents nonperturbative hadronization, rather than a quantitatively reliable perturbative prediction at the kaon scale.

These multiplicities describe the dominant low-orbital angular momentum modes, not an absolute rule for every three-pion configuration. In ${\pi }^{+}{\pi }^{-}{\pi }^0$, odd orbital angular momentum values of the charged pair and of its motion relative to ${\pi }^0$ can couple to total zero and give CP $+1$, but are strongly suppressed near decay threshold. Physical $K_L\to 2\pi$ also occurs through CP violation.

Model alpha decay as a preformed alpha particle making repeated escape attempts from a short-range nuclear potential well, then undergoing quantum tunnelling through the external Coulomb potential. Write the daughter atomic number as $Z_D$, the nuclear radius as $R$, and the relative reduced mass as $\mu$. For zero orbital angular momentum, the WKB approximation givesSet $v=\sqrt{2Q/\mu }$ and $r=r_2{\sin }^2\theta$. ThenWhen $R/r_2\ll 1$, the leading term isHere $P_{\alpha }$ is the alpha preformation probability and ${\nu }_{\mathrm{attempt}}$ the attempt frequency. The stated proportionality retains only the dominant barrier exponent. It assumes a thick classically forbidden barrier, nonrelativistic relative motion, a spherical unscreened external Coulomb potential, and slowly varying prefactors. Finite $R$, changes in alpha preformation probability, and a centrifugal barrier for nonzero orbital angular momentum produce significant corrections. The precise charge in the barrier is $Z_D=Z_P-2$; using the parent charge is an approximation for large $Z_P$. The measured alpha particle speed approximates $v$ only when daughter recoil is negligible.