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For a hydrogen-like ion, the Coulomb potential is stronger by the nuclear atomic number $Z$, and the energy levels scale as $Z^2$:Here $Z$ is the number of protons in the atomic nucleus; the ion has exactly one electron, so its net electric charge is $(Z-1)e$. The zero of energy is a separated stationary electron and atomic nucleus at infinite distance. This is the nonrelativistic hydrogenic approximation using the supplied $R_H$; a more precise isotope-specific result also corrects the reduced mass.

For the outer $ns,np$ valence electrons, Slater's rules predict a modest initial rise down group 15, followed by approximately constant values. Write the effective nuclear charge as $Z_{\mathrm{eff}}=Z-S$, where $S$ is the shielding constant. Each atom has $n{s}^{2}n{p}^3$, so the other four electrons in the outer group contribute $4(0.35)=1.40$. Electrons in shell $n-1$ contribute $0.85$ each, and those deeper contribute $1.00$ each.^[ref]

Applying those weights gives the following original calculations:The increasing nuclear electric charge is largely offset by additional inner-electron shielding; once the preceding shell contains $18$ electrons, these simple weights give a plateau. The $n-1$ shell includes its d electrons. This trend concerns outer-shell values, not every orbital: a $1s$ electron is shielded only by its partner, so $Z_{\mathrm{eff}}(1s)=Z-0.30$ rises strongly down the group. The larger principal shell also means nearly constant outer $Z_{\mathrm{eff}}$ does not imply nearly constant atomic size or binding energy.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The cover establishes Part B, Trinity Term 2019, 13 June 2019, A13485W1; the annual bundle's 2018 filename does not determine the examination year. This is distinct from the Part C major option. All four questions are solved. Use SI magnetic susceptibility, $\chi =M/H$, and write $e>0$ for the elementary electric charge. In question 3, $n,p$ and $n_d$ are densities per area, and the band model has one parabolic extremum per carrier species with twofold spin degeneracy.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The cover identifies the Part C major option, Trinity Term 2019, 17 June 2019, A11046W1; the bundle's 2018 filename does not determine the examination year. All six questions are treated. Figure 2(A) has missing image tiles, including part of its lower-left plot and some path labels. The readable portions and the standard silicon valley model support the conclusions below; the missing image content is not reconstructed. Numerical values read from plotted curves are estimates. Write $e>0$ for the elementary electric charge.

TODO: Recover the missing tiles of original Figure 2(A), affecting questions 2(a) and 2(e). The solutions distinguish the surviving figure and standard silicon-band interpretation from unavailable path labels and precise missing valence-sheet contours.

Integrate the surface charge density using the surface integral in spherical coordinates:Since $r_0>a$, the elementary integral isSubstitution yields the total induced electric chargeThe equality with the image charge also follows by applying Gauss's law to the exterior electric field: the physical and fictitious descriptions give the same electric flux through a sphere enclosing the conductor but excluding the physical point charge.

Choose a spherical Gaussian surface of radius $s$ with $a<s<r_0$. Its physical enclosed electric charge is the total induced electric charge $Q_a$ on the inner sphere. The method of images produces the same electric field on that Gaussian surface, so Gauss's law identifies $Q_a$ with the sum of the enclosed image charges. These are the first family with $n\le -1$ and the second family with $n\le 0$:Summing these geometric series givesThe electric field in the empty central cavity vanishes by electrostatic shielding, so it contributes no additional inner-face surface charge density. The formula tends to $-q$ as $r_0\downarrow a$ and to zero as $r_0\uparrow b$, as expected when the point charge approaches one grounded boundary or the other.

Put $k=b/a>1$ and ${\mathbf{e}}_z=(0,0,1)$. The complete construction of image charges between concentric grounded spheres consists of two families, indexed by every integer:The term $(q_0,{\mathbf{r}}_0)$ is the physical point charge; every other term is an image charge. In the first family $n\le -1$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. In the second family $n\le 0$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. Thus no image charge lies in the physical gap.

To derive and verify these families, reflection of an axial point charge $(Q,z)$ in a grounded sphere of radius $R$ replaces it by the image charge $(-RQ/z,R^2/z)$. Reflection in $a$ sends $(q_n,{\mathbf{r}}_n)$ to $({\tilde{q}}_{-n},{\tilde{\mathbf{r}}}_{-n})$; reflection in $b$ sends it to $({\tilde{q}}_{1-n},{\tilde{\mathbf{r}}}_{1-n})$. Successive reflections therefore generate exactly the listed families. The method of images givesThe physical point charge itself is excluded when evaluating this formula. On $r=a$, pair the first-family term $n$ with the second-family term $-n$; their electric potentials cancel by the single-sphere distance identity. On $r=b$, pair $n$ with $1-n$ instead. These reindexings are justified by absolute convergence, as follows.

For $n\to -\infty$, both source positions tend to the origin while their electric charges have size $O(k^n)$. On the closed gap their distances are eventually at least $a/2$, so both electric potential terms are $O(k^n)$. For $n\to +\infty$, both distances are at least a positive constant times $k^{2n}$ and the electric charges are $O(k^n)$, giving $O(k^{-n})$ terms. Both tails are bounded by convergent geometric series. This proves absolute convergence and uniform convergence of the tails on the closed gap. On any compact set avoiding the physical point charge, the differentiated tails obey the same type of bounds: for example, the large positive tail of the electric field is $O(k^{-3n})$. Thus termwise differentiation is valid and the sum is a harmonic function except at the physical point charge, where it has the correct Coulomb's law singularity. The boundary cancellations and uniqueness of Poisson equation complete the construction.

Write the constant spherical harmonic $Y_{00}=C$. Its orthonormality gives $4\pi |C{|}^2=1$, independently of its phase convention. Orthogonality to $Y_{00}$ givesOnly the $\ell =m=0$ term contributes to the surface integral of the surface charge density. Multiplying by the surface-area factor $a^2$ givesThus the spherical harmonic calculation reproduces the total induced electric charge:

Translate the concentric spheres so that their centre is $(0,0,-a)$, take $b=a+d$, and keep the point charge fixed at $(0,0,h)$. Its distance from the centre is $r_0=a+h$. As $a\to \infty$, the upper portions of the inner and outer spheres approach the planes $z=0$ and $z=d$. The preceding total induced electric charge becomesThe outer sphere has total induced electric charge $Q_b=-q-Q_a$: a Gaussian surface within its conducting material has zero electric flux, and the grounded exterior is field-free by uniqueness of Poisson equation. Therefore $Q_b\to -qh/d$.

One can justify the total-charge limit without assuming that local convergence of electric fields preserves their surface integrals. Use induced charge as harmonic measure: if a harmonic function $w$ is one on the selected grounded boundary and zero on the other, Green's second identity with the electric potential givesIndeed, ${\nabla }^2\varphi =-q{\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)/{ϵ}_0$, $\varphi =0$ on the boundaries, and their unit normal outward from the gap satisfies $\sigma ={ϵ}_0{\partial }_n\varphi$. For the spheres the weighting harmonic function is $w_a(r)=a(b-r)/(r(b-a))$, exactly the factor already obtained. For the slab the two weighting harmonic functions are $w_0=1-z/d$ and $w_d=z/d$. Applying the same identity to the decaying grounded-slab electric potential, whose boundary term at lateral infinity vanishes, establishes directlyTheir sum is $-q$, and their relative sizes measure how close the point charge lies to each grounded plane.

Use the constitutive relations $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/\mu$, where $\mathbf{D}$ is the electric displacement field, $\mathbf{H}$ the magnetic field strength, $ϵ$ the permittivity, and $\mu$ the magnetic permeability. Let ${\rho }_f$ be the charge density of free charge and ${\mathbf{J}}_f$ its current density. In this homogeneous medium, Gauss's law and the magnetostatic Ampère law are, respectively,Here $W$ is a volume with closed boundary $\partial W$, and $S$ is an oriented spanning surface whose boundary orientation follows the right-hand rule. Thus the electric field form is $\nabla \cdot \mathbf{E}={\rho }_f/ϵ$ and ${\oint }_{\partial W}\mathbf{E}\cdot \mathbf{n}dS=Q_f/ϵ$; the magnetic field form is $\nabla \times \mathbf{B}=\mu {\mathbf{J}}_f$ and ${\oint }_{\partial S}\mathbf{B}\cdot d\mathbf{\ell }=\mu I_f$.

The remaining static Maxwell equations areThese express Gauss's law for magnetism and the static limit of Faraday's law. The volume and circulation forms follow from the divergence theorem and Stokes theorem. The electric charge and electric current on the right of the macroscopic laws are free sources; material electric polarization and magnetization are already included in $\mathbf{D}$ and $\mathbf{H}$.

Use the sign $s=\pm 1$ to label a point charge $sq$ at $s\mathbf{p}/(2q)$, with velocity $s{\mathbf{p}}^{\prime }/(2q)$. All quantities on the right below are evaluated at $t_0=t-r/c$. To first order,The denominator in each Liénard–Wiechert potential therefore expands assoIn the sum of the two electric potentials, the $q/r$ terms cancel and the finite terms add. This gives the retarded potentials of a point electric dipole:For the magnetic vector potential, the signed electric charge and signed velocity multiply to give the same sign for both contributions:Since ${\mu }_0{ϵ}_0{c}^2=1$,The ${\mathbf{p}}^{\prime }$ term in the electric potential arises from the moving-source denominator and remains finite in the point-electric dipole limit; discarding it would fail to give the full retarded electric field.

The gauge fields require their own Yang-Mills Lagrangian, $-W_{\mu \nu }^a{W}^{a\mu \nu }/4$. The non-Abelian field strength contains a term quadratic in the gauge potentials, so squaring it produces three-vector and four-vector self-interaction vertices, with derivative momentum factors in the cubic vertex and SU(2) structure constants in both. They exist in either scalar phase and must be added to the scalar-generated vertices. A gauge-fixed quantum calculation also has gauge-fixing and, where applicable, ghost terms; those are not extra physical boson species.

For ${\mu }^2>0$, the global phase transformation $\Phi \mapsto e^{i\alpha }\Phi$ leaves both the scalar potential and the covariant kinetic term unchanged while leaving $W_j$ fixed. Its Noether charge counts scalar particles minus antiparticles. Thus ${\varphi }_a{\varphi }_b\to {\varphi }_a{\varphi }_b$ and ${\varphi }_a{\varphi }_a^{\ast }\to W_3{W}_3$ are compatible with this charge, whereas ${\varphi }_a{\varphi }_b\to W_j{W}_k$ is forbidden: its initial charge is two and final charge zero. Charge conservation is necessary, not sufficient, for a particular amplitude to be nonzero; other quantum numbers and kinematics still apply.

After adding the hypercharge field, the neutral-vector mass term becomesIts mass matrix has zero determinant. With $G=\sqrt{g^2+g^{\prime 2}}$, the orthonormal combinations areThe unbroken generator is electric charge, $Q=T_3+Y/2$, because the lower doublet component has $T_3=-1/2$ and $Y=1$. The charged combinations $W^{\pm }=(W_1\mp i{W}_2)/\sqrt{2}$ keep mass $gv/2$.

The observed charged weak interaction couples left-chiral fermions within SU(2) doublets. In the Standard Model, $Q_L=(u_L,d_L)$ and $L_L=({\nu }_L,e_L)$ are doublets, while $u_R,d_R,e_R$ are singlets. A vectorlike assignment would also give right-handed charged weak currents, contrary to this observed chiral structure. Once the measured electric charges are specified, $Y=2(Q-T_3)$ fixesAn optional right-handed neutrino is a singlet with $Y=0$. Conjugate left-chiral fields have the opposite hypercharges to these right-chiral fields. These assignments permit gauge-invariant Yukawa interactions; they are not obtained by assigning one common hypercharge to every chirality.

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.

For the usual colour SU(3) fixing the fourth coordinate of the SU(4) defining representation,Taking the exterior square givesThe identification ${\Lambda }^{2}3\cong \overline{3}$ follows by contracting an antisymmetric tensor with the invariant three-index Levi-Civita symbol. Combining this with c gives all colour multiplets:A colour triplet has the color charge transformation law of a quark, an antitriplet that of an antiquark, and a colour singlet is colour-neutral, as a lepton is. Thus both parent representations permit quark-like and antiquark-like colour sectors; the vector has one neutral sector and the spinor two. These are colour assignments only. They do not determine particle spin, statistics, electric charge, weak quantum numbers or a full generation of the Standard Model. In a real parent field the conjugate colour sectors may describe one complex colour field and its conjugate, rather than independent particle species.

Global group qualification. The printed chain is correct at the Lie algebra level, $\mathfrak{su}(3)\subset \mathfrak{su}(4)\cong \mathfrak{so}(6)\subset \mathfrak{so}(7)$. Globally its simply connected form is $SU(3)\subset SU(4)\cong \mathrm{Spin}(6)\subset \mathrm{Spin}(7)$. The standard map of SU(4) into $SO(7)$ through $6\oplus 1$ has a nontrivial kernel: the concrete central element $-I_4$ acts as $+I_6$ on ${\Lambda }^{2}4$ and trivially on the singlet representation. Its image is $SO(6)\cong SU(4)/\{\pm I_4\}$, rather than an injective copy of SU(4) under this map.

The global form obstruction for the Spin7 spinor is explicit: the central element $-1$ of the Spin group $\mathrm{Spin}(7)$ maps to the identity in $SO(7)$ but acts as $-I_8$ on the eight-dimensional spin representation. Equivalently $-I_4$ acts as $-I$ on both $4$ and $\overline{4}$ in its restriction. Therefore the spinor does not descend to a genuine SO(7) group representation. A theory containing that spinor needs the Spin group global form, or must understand the source's notation as specifying only its Lie algebra. The displayed colour branchings remain valid with this correction.

For an A2 root system weight $w=(w_1,w_2{)}^T$, the specified electric charge and flavour hypercharge are the linear functionalsApply these to the meson octet weights in solution. The results, with electric charges measured in units of the positive elementary electric charge, areThe table illustrates the meson octet of pseudoscalar mesons; the same weight calculation applies to another meson species arranged in the same representation. Here flavour hypercharge is $Y=B+S$, with $B$ the baryon number and $S$ the strangeness, rather than weak hypercharge. If the full quark model tensor product $3\otimes \overline{3}$ is retained, its additional trivial representation is a ninth meson with $(q,y)=(0,0)$. This singlet is separate from the octet and need not share its mass.

A G2 multiplet containing the known meson octet must restrict to a representation containing $8$. Among the two fundamental representations just computed, $7$ fails this test, while the Adjoint representation $14$ contains the required octet. Thus the minimal such choice isLarger representations containing an octet would be other model choices; the octet alone does not force a unique arbitrarily large representation.

A dual vector must be pulled back by $P^T$, since $q_A^{T}Pw=(P^T{q}_A{)}^{T}w$. Applying this to the electric charge and flavour hypercharge vectors givesThus a G2 root system weight $(a,b)$ has $(q,y)=(a+2b/3,a+b/3)$. This pullback ensures that every known octet meson retains its values from solution. The extra $3\oplus \overline{3}$ gives six additional mesons, one at each ofFor instance the first three come from the G2 root system weights $(0,1)$, $(1,-2)$ and $(-1,1)$; the other three come from their negatives. The weight diagram displays the preserved octet and the new states.

The fractional electric charges make these additional states qualitatively different from ordinary quark-antiquark mesons, whose electric charges are integers. This is a consequence of the proposed symmetry extension, not evidence for such particles. If “known mesons” also includes a flavour singlet from $3\otimes \overline{3}$, that singlet can be accommodated by a separate G2 trivial representation: the minimal octet-plus-singlet construction is then $14\oplus 1$, still with six new states. A demand for one irreducible G2 multiplet containing the entire nonet is a different extra assumption; $14$ alone does not contain that singlet.

Use the electron electric charge $-e$, so $E=-{\partial }_x\phi$ gives acceleration $+(e/m_e){\partial }_x\phi$. The Vlasov equation, its average, and its linearized fluctuating part areThe quasilinear approximation retains the averaged quadratic term but drops products of fluctuations from the fluctuating equation. It freezes the slowly evolving $f_0$ during the fast response calculation. A real electrostatic potential has ${\phi }_{-k}={\phi }_k^{\ast }$, and the sum includes both signs of every nonzero wavenumber. Spatial averaging selects the opposite Fourier series harmonics; statistical homogeneity supplies the corresponding diagonal noise covariance.

Let $\hat{g}(p)={\int }_0^{\infty }{e}^{-pt}g(t)dt$. Zero initial $\delta f_k$ givesInsert this into the electrostatic self-consistency relation. Using ${\omega }_{pe}^2=4\pi e^2{n}_e/m_e$ and $p+ikv=ik(v-ip/k)$ givesThis derives the stated dielectric function with the Landau contour. A noise-free growing collective response is a zero of $ϵ$.

The Laplace transform of the white noise covariance is particularly simple. Writing $q=p^{\prime \ast }$,For clarity about contour signs, introduce the inverse-transform measuresThe first Bromwich inversion formula contour runs upward on $\mathrm{Re}p=\sigma$; its conjugate contour runs downward on $\mathrm{Re}q=\sigma$. Both measures therefore have the same positive orientation in their real frequency parameters. Combining the averaged equation with ${\widehat{\delta f}}_k$ yieldsAt a zero of $f_0^{\prime }$, the last expression means its continuous coefficient obtained before cancelling $f_0^{\prime }$. Substitution of the two inverse Laplace transforms and the transformed covariance givesThis is the paper's expression, with the orientation of $d{p}^{\prime \ast }$ made explicit. Reality follows by pairing $k$ and $-k$. The double integral retains switch-on transients; replacing it by a time-independent diffusion coefficient requires the additional long-time limit in b.

Use $Q=T_3+Y$ and $T_a={\tau }_a/2$. For each generation $r=e,\mu ,\tau$, the leptons areBoth are color singlets. The conjugate antilepton fields transform in the conjugate representations and have opposite hypercharge; they are already included by the conjugate fields in the Lagrangian. The two components of $L_r$ have electric charges $0,-1$, and $R_r$ has charge $-1$. There is no right-handed neutrino representation in this field content. These assignments use the electroweak representation and hypercharge table in the same hypercharge normalization as the scalar doublet.

For a doublet of hypercharge $Y$, defineThe $SU(2)$ term is absent for a singlet. Under the local transformation $\psi \mapsto e^{iY\omega }U\psi$, the gauge fields transform so that $D_{\mu }\psi \mapsto e^{iY\omega }U{D}_{\mu }\psi$. Consequently the required gauge-covariant kinetic terms areThe fields of definite chirality are defined with $P_{L,R}=(1\mp {\gamma }_5)/2$. Since $\bar{L}$ has hypercharge $+1/2$, the Yukawa interactionis invariant: its hypercharges add to zero, and $\bar{L}\varphi$ contracts the two doublet indices. A bare charged-lepton mass violates these assignments. No renormalizable neutrino Yukawa term is available without a right-handed neutrino. A gauge-invariant scalar potential depending on ${\varphi }^{†}\varphi$ can select the nonzero vacuum, without altering these kinetic and Yukawa constructions.

There is an explicit local unitary gauge transformation whenever $r_{\varphi }=(|{\varphi }_1{|}^2+|{\varphi }_2{|}^2{)}^{1/2}\ne 0$:Thus, with the real field $\rho =\sqrt{2}{r}_{\varphi }-v$,This proves the polar coordinates for the electroweak Higgs doublet constructively. The gauge choice is local about the nonzero vacuum; it need not be nonsingular at zeros of the doublet.

The three identical-generation kinetic terms permit unitary flavor changes. A singular value decomposition gives $U_L^{†}{Y}_{\ell }{U}_R=\mathrm{diag}(y_e,y_{\mu },y_{\tau })$ with $y_r\ge 0$. Rotate both components of each left doublet together, and the right singlets independently. In this basis,The left/right cross terms combine into this Dirac mass. Since the neutrinos have no mass matrix, their simultaneous left rotation preserves a diagonal charged current. This is charged-lepton Yukawa matrix diagonalization; the neutrinos remain massless in the stated renormalizable model.

The scalar kinetic term givesIntroduce $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The combinations $W^{+}=(A_1-i{A}_2)/\sqrt{2}$, $W^{-}=(W^{+}{)}^{\ast }$, $Z=c_W{A}_3-s_{W}B$ and $A_{\gamma }=s_W{A}_3+c_{W}B$ therefore diagonalize the quadratic terms:The factor $1/2$ applies to the real neutral field, not to the complex charged field. The massless photon corresponds to the unbroken generator $T_3+Y$, which annihilates the vacuum. This gives the electroweak doublet gauge-boson mass matrix and its diagonal spectrum.

Use units in which the proton electric charge is one. The up quark and down quark both have spin angular momentum $1/2$ and form a flavour isospin doublet:Their antiquarks have the opposite charges and third components of isospin, with the same total isospin and spin. This is strong-interaction flavour isospin, not the chiral weak isospin assignment.

The Clebsch-Gordan decomposition for SU2 says that spins $j_1,j_2$ combine into every $j$ from $|j_1-j_2|$ to $j_1+j_2$ in unit steps. It applies independently to spin and isospin. ThusBefore exchange symmetry is imposed, three light quarks have total spin $S=1/2$ or $3/2$, and total isospin $I=1/2$ or $3/2$; the doublet occurs twice in the uncoupled tensor cube. With zero orbital angular momentum, $J=S$.

Physical baryons must also be colour singlets. The three-quark colour singlet is completely antisymmetric, so the Pauli exclusion principle requires symmetric spatial-spin-flavour symmetry. For a symmetric spatial S-wave state, the spin-flavour factor is symmetric. The $I=3/2$ flavour quartet is symmetric, whereas the $I=1/2$ flavour doublets have mixed permutation symmetry; the spin factors have the analogous symmetries. Symmetric flavour therefore pairs with symmetric spin, while the two mixed factors pair in their unique symmetric combination. This is the Pauli constraint on three-quark flavour multiplets; it leavesEquivalently the symmetric cube of the four-dimensional two-flavour spin space has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20=2\cdot 2+4\cdot 4$. The unmatched spin-isospin pairings in the bare tensor product do not describe the usual symmetric-spatial ground-state baryons.

For three light quarks, charge addition gives $Q=I_3+1/2$. The observed nucleon doublet and Delta baryon quartet have the following assignments; the listed contents specify flavour counts, not the complete spin-flavour wavefunction:The proton and neutron have $J=1/2$; all four Delta baryons have $J=3/2$. These states have positive parity in the zero-orbital ground-state model.

For a light quark-antiquark meson, both spin and flavour doublets combine into singlet plus triplet. In a conventional choice of phases the flavour states areEach row can have $J=0$ or $1$, since the two constituent spins couple to zero or one. At zero orbital angular momentum their parity is negative, following the meson parity and charge conjugation rule $P=(-1{)}^{L+1}$. The spin-zero isotriplet is the pion multiplet, and the spin-one isotriplet the rho meson multiplet. The nonstrange spin-one isosinglet is the idealized omega meson flavour direction. Its spin-zero counterpart is the nonstrange direction contributing to the eta and eta prime mesons; it is not automatically a physical mass eigenstate once other flavours and mixing are included.

Now introduce an isospin-zero quark $h$ of spin $1/2$ and charge $q_h$. For baryons with one isospin-singlet quark, the two light quarks have either the antisymmetric flavour singlet $[ud]=(ud-du)/\sqrt{2}$ or the symmetric triplet $uu,\{ud\},dd$. Their colour pair is antisymmetric. Hence light-diquark spin-isospin correlation gives $S_{ud}=0$ for $I=0$ and $S_{ud}=1$ for $I=1$. Coupling the $h$ spin producesThe three triplet rows have $I_3=1,0,-1$. For a strange-like charge $q_h=-1/3$, these are the neutral spin-one-half Lambda baryon, the spin-one-half Sigma baryon triplet of charges $1,0,-1$, and its spin-three-half starred partners. For a charm-like charge $q_h=2/3$, they are the spin-one-half Lambda c baryon of charge $1$, and the Sigma c baryon triplets of charges $2,1,0$ with spins $1/2$ and $3/2$. Both possible new charge assignments have thus been covered; charge-conjugate antibaryons have opposite charges and unchanged $I,J$.

The meson doublets with one isospin-singlet quark are $h\overline{u},h\overline{d}$, withFor $q_h=-1/3$ their charges are $-1,0$, giving $K^{-},{\overline{K}}^0$-like kaons and their vector partners; the conjugate doublet has $u\overline{h},d\overline{h}$ and charges $1,0$. For $q_h=2/3$ the charges are $0,1$, giving $D^0,D^{+}$-like D mesons and their vector partners, with charge-conjugate partners as well. Finally a hidden-flavour pair $h\overline{h}$, if included, is neutral with $I=0$ and $J=0$ or $1$. These are quantum-number classifications; charge and isospin alone do not determine particle masses or mixing angles.