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In the Parity basis of a one-dimensional S-matrix, parity acts as $\mathcal{P}=\mathrm{diag}(1,-1)$. Parity invariance requiresand thereforeIndependently, unitarity givesCombining parity and unitarity leaves two phases,

For the symmetric double-delta potential, consider the odd channel on $x\ge 0$. Choose the asymptotic conventionand write $psi(x)=C\sin (kx)$ for $0<x<a$; the sine enforces odd parity at the origin. Continuity and the delta potential derivative jump at $x=a$ giveEliminating $C$ yields the Odd-parity S-matrix of a symmetric double-delta potentialUsing $2i\sin (ka)=e^{ika}-e^{-ika}$ rewrites this as the requested expressionFor real $k$, the two factors in the ratio are complex conjugates, directly confirming $|S_{--}|=1$.

A bound state occurs at a pole $k=i\kappa$, $kappa>0$. Setting the denominator to zero givesorFor $U_0<0$, write $g=-U_0>0$. The right-hand side $g(1-e^{-2\kappa a})$ is increasing and concave, starts at zero with slope $2ga$, and approaches $g$, whereas the left-hand side is the straight line $2\kappa$. A positive intersection therefore exists exactly whenThis is the odd bound state of a symmetric double-delta potential; equality is its zero-energy threshold.

For $U_0>0$, analytically continue the same pole equation into the lower half of the complex $k$-plane. PutIt becomesWhen $G\gg 1$, the lowest odd mode lies near the hard-wall value $z=\pi$. Writing $z=\pi +\eta$ and expanding givessoIn particular, the odd resonance of a strongly repulsive symmetric double-delta potential hasThe negative imaginary part identifies a decaying scattering resonance. Ifthen, to leading nonzero order,Its time-dependent amplitude is proportional toso its survival probability decays as $e^{-\Gamma t/\hslash }$ and its lifetime isPhysically, this is the lowest odd standing wave trapped for a long time between two strong repulsive barriers and leaking out by quantum tunnelling.

Using the zero test $s$ from part (i), define by primitive recursionSince $s(0)=1$ and $s(1)=0$, the values alternate:Thus $r$ is the required parity function and, like $s$, is primitive recursive without minimization.

Since $X=L(a+a^{†})$, the perturbation isEvery diagonal matrix element vanishes by parity, so all first-order shifts are zero. For the ground state,The second-order shift is thereforeThusFor every nonzero real $\lambda$, the cubic potential is unbounded below on one side, so there are no genuine bound-state eigenvalues continuously connected to the harmonic levels. The formal perturbation series does not converge; it is at best asymptotic to resonance energies.

The parity of a wavefunction is its eigenvalue under spatial reflection: $\psi (-x)=+\psi (x)$ for even parity and $\psi (-x)=-\psi (x)$ for odd parity. If $V(-x)=V(x)$, the Hamiltonian commutes with reflection, so its energy eigenspaces admit a parity eigenbasis. In particular, every nondegenerate bound-state energy eigenfunction has definite parity; within a degenerate scattering eigenspace one must choose the even and odd linear combinations.

To first order in $F$, time-dependent perturbation theory replaces the state on the right-hand side by its unperturbed value $e^{-i{E}_{0}t/\hslash }|0⟩$. The bound state is even, while the position operator $x$ is odd. The parity selection rule therefore givesso ${\dot{b}}_k=0$ and, with no initial continuum component,

Projection onto the odd state givesWriting ${\Omega }_k=(E_k-E_0-\hslash \omega )/\hslash$ and integrating from zero to $t$ yields

Write $m=|m|e^{i\theta }$. Parity reverses the spatial coordinates and must exchange the two chiral Weyl spinors. One convenient choice isThe identities ${\sigma }^0={\bar{\sigma }}^0$ and ${\sigma }^i=-{\bar{\sigma }}^i$ exchange the two kinetic terms after $\mathbf{x}\mapsto -\mathbf{x}$. The mass bilinear $m{\bar{\psi }}_R{\psi }_L$ maps to its complex conjugate $m^{\ast }{\bar{\psi }}_L{\psi }_R$, so the two mass terms are exchanged and the action is invariant. More generally, independent unit phases $a,b$ may multiply the two transformations provided $b^{\ast }a=m^{\ast }/m$.

Apply parity to the mixed anticommutator. The transformed left-hand side isbecause $|{\eta }_P|=1$. The Pauli-matrix identity represented by this product leaves the temporal matrix unchanged and reverses the three spatial matrices, so it equalsBut a parity transformation acts on momentum as $P^0\mapsto P^0$ and $P^i\mapsto -P^i$. This is exactly the parity transform ofso $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$ is consistent with the stated transformation law. The arbitrary intrinsic phase cancels.

With matrix-valued fields, charge conjugation acts asUnder parity, $A_0$ is even and $A_i$ is odd:Consequently $F_{0i}$ is odd and $F_{ij}$ is even. Under time-reversal symmetry, which is antiunitary, $A_0$ is even and $A_i$ is odd after $t\mapsto -t$; hence $F_{0i}$ is even and $F_{ij}$ is odd. Equivalently, the non-Abelian electric and magnetic fields transform as

There are three even faces, so $\mathbb{P}(A)=1/2$. The sum is even precisely when both faces have the same parity, giving $9+9=18$ outcomes and $\mathbb{P}(B)=1/2$. Both $A$ and $B$ occur precisely when both faces are even, giving nine outcomes. ThereforeThus $A$ and $B$ are independent events.

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.

The observed dominant two-body transition is $\boxed{{\Sigma }^0\to {\Lambda }^0\gamma }$. Both baryons have valence quarks $uds$ and the same strangeness; an electromagnetic interaction transition is therefore possible. Their mass difference is only about $77\mathrm{MeV}/c^2$, below the lightest pion mass, so no additional on-shell strongly interacting particle fits. A photon is neutral and massless, and can carry the required angular momentum through a magnetic dipole transition between the two positive-parity spin-$1/2$ states. A single neutrino would violate lepton number and would not give the required fermion parity.

At rest, four-momentum conservation gives $p_{\Lambda }=p_{\Sigma }-k$ and $k^2=0$. ThusThe result is slightly below $(M_{\Sigma }-M_{\Lambda })c^2$ because the Lambda baryon recoils; rounded masses give about $74.5\mathrm{MeV}$.

The channels $n{\pi }^0$ and $p{\pi }^{-}$ are actually kinematically open: their total rest energies are about $1075$ and $1078\mathrm{MeV}$, below $M_{\Sigma }{c}^2\simeq 1193\mathrm{MeV}$. Their suppression is therefore not a decay threshold prohibition. Each changes strangeness from $-1$ to zero and so requires a weak interaction. The much faster allowed electromagnetic transition makes such weak branching fractions extremely small. “Not observed” should be understood as overwhelming electromagnetic competition, not an exact ban in the Standard Model. The photon identification likewise uses the known low-energy spectrum and dominant transition mechanism, not neutrality alone.

Set $a=\sqrt{m\omega /(2\hslash )}x+\sqrt{\hslash /(2m\omega )}d/dx$ and let $a^{†}$ have the opposite derivative sign. On the bound-state domain, $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Thus every energy eigenvalue has $E/\hslash \omega -1/2=\nu \ge 0$.

The lowering operator maps an eigenstate of $a^{†}a$ of eigenvalue $\nu$ to one of eigenvalue $\nu -1$, with squared norm multiplied by $\nu$. Repeated lowering cannot pass below zero, so it must terminate at zero. Hence $\nu$ is a nonnegative integer. The equation $a{\psi }_0=0$ gives a Gaussian ground state, and repeated raising gives a nonzero square-integrable state for every integer $n\ge 0$. ThereforeThese are also the Gaussian-times-Hermite-polynomial solutions obtained by the suggested series method.

Put $\alpha =m\omega /\hslash$. Normalization gives ${\psi }_0=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, soThe expectation of $x$ itself is zero by parity; that cancellation does not apply to its absolute value.

The stationary Schrödinger equation isA nonzero bound-state wave function has ${\int }_{\mathbb{R}}|\psi {|}^{2}dx<\infty$, so it can be rescaled to have integral one. Odd parity means $\psi (-x)=-\psi (x)$.

For $-V_0<E<0$, set $k=\sqrt{2m(V_0+E)}/\hslash >0$ and $\kappa =\sqrt{-2mE}/\hslash >0$. Oddness removes the interior cosine solution, and normalizability removes the growing exponentials outside. A continuous odd candidate can therefore be writtenContinuity of the derivative at $a$ gives $k\cos (ka)=-\kappa \sin (ka)$; oddness gives the other boundary condition. A nonzero solution cannot have $\sin (ka)=0$, since the derivative condition would then force its interior coefficient to vanish. Thus, with $\xi =ka$ and $\eta =\kappa a$,Physical bound states require $\xi ,\eta >0$. On $0<\xi <\pi /2$, the cotangent condition gives negative $\eta$, so no physical intersection occurs. On $(\pi /2,\pi )$, the curve $-\xi \cot \xi$ rises from zero to positive infinity, while the positive quadrant of the circle decreases. Their first intersection exists exactly when the circle radius $R=\sqrt{2m{V}_0{a}^2/{\hslash }^2}$ exceeds $\pi /2$.

In particular,There is also no normalizable odd state at equality: the putative intersection has $\eta =0$, hence $E=0$, and its exterior tail does not decay. Higher positive branches begin beyond this first interval and require still larger $R$.

A nonzero polynomial solution of degree $n$ has $a_n\ne 0$ but $a_{n+2}=0$. The recurrence therefore forces $2n+1-ϵ=0$. Since $ϵ=2E/(\hslash \omega )$,Conversely, for this value of $ϵ$, choose only the parity sequence containing degree $n$ and set the other initial coefficient to zero. The recurrence terminates at $n$, producing a nonzero polynomial times the decaying Gaussian. Thus every displayed energy occurs under the supplied normalizability criterion.

Differentiate the power series twice and align the coefficient of ${\xi }^k$. The equation becomesThusEven coefficients are generated from $a_0$, and odd coefficients from $a_1$; the two parity sequences are independent.

The nuclear quadrupole vibration has five one-phonon magnetic substates, $m=-2,-1,0,1,2$. Bose-Einstein statistics restrict three identical phonons to the symmetric power $S^3(V_2)$, of dimension $\left(\genfrac{}{}{0ex}{}{5+3-1}{3}\right)=35$. Count occupation states $(n_{-2},\dots ,n_2)$ with $\sum n_m=3$ and $\sum m{n}_m=M$. For nonnegative $M$ the counts areEach spin-$J$ irreducible representation contributes one state at every projection from $-J$ to $J$. magnetic-substate subtraction therefore gives multiplicity $d_J-d_{J+1}$: one each at $J=0,2,3,4,6$, and none at $1,5$. Thus the three-quadrupole-phonon multiplet isThe dimension check is $1+5+7+9+13=35$. The parity is positive because each constituent phonon has positive parity. Pairwise angular-momentum triangle rules alone would include states incompatible with complete boson symmetrization.

beta decay is a weak charged current process. At quark level $d\to u{W}^{-\ast }\to u{e}^{-}{\bar{\nu }}_e$, with amplitude proportional to $G_F{V}_{ud}$ at low momentum transfer. A neutron in a nucleus can thereby become a proton:The second reaction denotes an energetically allowed nuclear conversion, not a decay of an isolated free proton. In a nuclear beta transition the mass number stays fixed and the atomic number changes by one. baryon number and electric charge are conserved; the neutrino or antineutrino supplies the missing energy, momentum and angular momentum. The three-body final state produces a continuous electron spectrum rather than a single electron energy. Neglecting neutrino mass and recoil, an allowed spectrum in electron total energy isThe Fermi function accounts for the electron's Coulomb interaction with the daughter; $E_0$ is the electron total-energy endpoint. forbidden beta decay adds momentum-dependent nuclear shape factors. The V-A interaction violates parity: it couples left-chiral fermion fields, not an equal mixture of left and right currents.

For allowed beta decay the leading leptons have zero orbital angular momentum. Fermi transitions have $\Delta J=0$; Gamow-Teller transitions have $\Delta J=0,1$ with $0\to 0$ excluded. Both preserve nuclear parity. Higher lepton partial waves permit forbidden transitions with extra suppression. Nuclear matrix elements and phase space, as well as these selection rules, determine lifetimes. Particle-physics examples include free-neutron decay and pion beta decay, ${\pi }^{+}\to {\pi }^0{e}^{+}{\nu }_e$. semileptonic decays test quark mixing and the universality of the weak charged current. beta decay changes quark flavour through virtual W exchange; its continuous spectrum reflects an additional lepton in the final state.

For the neutral quark-antiquark mesons here, $C=(-1{)}^{L+S}$. Thus $C({\eta }^{\prime })=C({\pi }^0)=+1$, while $C({\rho }^0)=C(\gamma )=-1$. The radiative ${\rho }^0\gamma$ mode conserves charge conjugation and admits a quark-line photon-emission diagram. Both components of the supplied ${\rho }^0$ wavefunction contribute: the charge-weighted flavour overlap is proportional to $Q_u-Q_d=1$, so it does not cancel. Its photon energy is $(m_{{\eta }^{\prime }}^2-m_{\rho }^2)/(2m_{{\eta }^{\prime }})\simeq 165\mathrm{MeV}$.

The ${\pi }^0\gamma$ mode changes charge conjugation and, more decisively, a spin-zero particle cannot decay into another spin-zero particle and one real photon. In the parent rest frame the photon has helicity $\pm 1$, while the opposite-moving pion has helicity zero; the spin-zero two-body helicity selection rule cannot be satisfied. Its on-shell two-body amplitude vanishes even if a weak interaction violates C. A virtual photon subsequently making another pair is a different final state.

For ${\pi }^0{\rho }^0$, angular-momentum conservation requires relative orbital angular momentum $L=1$. Its parity is then negative, matching the parent, but its charge-conjugation eigenvalue is $C_{\pi }{C}_{\rho }=-1$: the two distinct self-conjugate particles are not interchanged by C, so there is no additional $(-1{)}^L$ factor in this C eigenvalue. The decay is forbidden by strong and electromagnetic charge-conjugation conservation despite being above threshold. A weak contribution would require CP violation as well. At the strong/electromagnetic accuracy of the diagram classification,If extremely small weak CP-violating amplitudes are retained, ${\pi }^0{\rho }^0$ need not be exactly zero; the one-real-photon spin prohibition remains exact for ${\pi }^0\gamma$.

The gamma width is $0.10\mathrm{eV}=10^{-7}\mathrm{MeV}$, so the remaining proton width is ${\Gamma }_p=0.20-0.15-10^{-7}=0.0499999\mathrm{MeV}$. At the peak of the same resonance the total-width denominator and gamma exit width cancel. alpha particles and carbon-12 have spin zero; the proton and the stated nitrogen level each have spin $1/2$. Hence $g_{\alpha }/g_p=4$, independently of the unspecified resonance spin. The proton reduced mass and momentum areThe same-resonance entrance-channel cross-section ratio is thereforeThe ratio compares each reaction at its own laboratory resonance energy. Factors of ${\hslash }^2$ restoring the cross section's area units cancel. Only the supplied nitrogen spin magnitude is used: its printed positive parity is immaterial to this spin average, so no orbital-parity inference is made from that assignment.

A nuclear excited state has internal energy above its ground state, with levels labelled by spin and parity $J^{\pi }$ and, where useful, approximate isospin. In the shell model, nucleons occupy orbitals split by spin-orbit coupling; nuclear particle-hole excitation and broken pairs produce excited configurations. An orbital has parity $(-1{)}^l$, and an even-even paired ground state is commonly $0^{+}$. Residual interactions mix configurations, so shell spacings are not a complete spectrum.

Collective motion gives distinct patterns. A deformed nucleus can have a rotational band with $E_J-E_0={\hslash }^{2}J(J+1)/(2\mathcal{I})$, so a simple even-even $0^{+},2^{+},4^{+}$ band has $E_4/E_2=10/3$. Near spherical shape, a quadrupole vibration gives a one-phonon $2^{+}$ excitation; two such phonons can couple to $0^{+},2^{+},4^{+}$ near twice that energy in the harmonic limit. Anharmonicity and coupling to single-particle motion modify these ideal patterns.

De-excitation occurs by gamma emission, internal conversion, or particle emission if energetically open. photon multipolarity $\lambda \ge 1$ must satisfy the angular-momentum triangle rule. Electric multipoles change parity by $(-1{)}^{\lambda }$ and magnetic multipoles by $(-1{)}^{\lambda +1}$; a single photon cannot connect $0\to 0$. Large multipolarity or a small transition matrix element can make a long-lived nuclear isomer. Gamma energies, coincidence cascades, angular correlations and lifetimes constrain level schemes; inelastic scattering and transfer reactions supply complementary spin/configuration information. One observed line alone need not identify a unique level assignment.