Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 45 (Symmetry and Particle Physics)

For the addition of angular momentum, the uncoupled orthonormal basis is $|j_1{m}_1⟩\otimes |j_2{m}_2⟩$, whereas the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2$ and $J_3$ for $\mathbf{J}={\mathbf{J}}_1+{\mathbf{J}}_2$. Define the Clebsch-Gordan coefficients byThe Clebsch-Gordan decomposition allows $|j_1-j_2|\le J\le j_1+j_2$ in unit steps, and a coefficient vanishes unless $m_1+m_2=M$. Taking the inner product of the expansion with itself provesWith the customary real phase convention the modulus signs can be omitted, giving the sum of squares used in the question. In an arbitrary phase convention the modulus-square identity is the invariant statement.

Treat the strong interaction as exactly isospin-invariant for this calculation. Its transition operator $T$ commutes with the three isospin generators and hence with their quadratic Casimir operator. Acting on the initial state $|IM⟩$, it can therefore produce only final states with the same $I$ and $M$. The tensor product representation of final multiplets $I_1,I_2$ contains one copy of each permitted total isospin, so $T|IM⟩=a_M|(I_1{I}_2)IM⟩$. Commuting $T$ with the lowering operator givesEvery connecting ladder factor is nonzero, so all $a_M$ coincide with one reduced coefficient $a$. Expanding the final coupled state now gives the isospin-invariant two-body decay amplitudeThe reduced coefficient may depend on the particle species, momenta and spin channel, but not on the three isospin projections. Summing the squared amplitude over the orthogonal charge channels givesA physical decay rate also includes phase space, angular integration and any spin sums. In the isospin-degenerate limit these factors are common to the charge channels; absorbing their square root into $a$ gives ${\Gamma }_{IM}=|a{|}^2$ for this specified two-body multiplet channel. If several distinct species or independent spin channels are included in “anything”, their reduced rates must additionally be summed. There is no extra factor $2I+1$ when the initial magnetic component is fixed.

The Delta baryon states have $I=3/2$ and projections $3/2,1/2,-1/2,-3/2$ for ${\Delta }^{++},{\Delta }^{+},{\Delta }^0,{\Delta }^{-}$. The proton and neutron form $I=1/2$, with projections $+1/2,-1/2$, and the pion triplet has $I=1$ with projections $+1,0,-1$. The highest coupled state isApply the total lowering operator $I_{-}=I_{-}^{(N)}+I_{-}^{(\pi )}$. Its coefficient on the left is $\sqrt{3}$, while on the right the nucleon and pion coefficients are $1$ and $\sqrt{2}$:The squared Clebsch-Gordan coefficients are consequently $1/3$ and $2/3$, compared with coefficient one for the ${\Delta }^{++}$ channel. Thus the Delta baryon pion branching ratios areAt the opposite end of the same isospin multiplet, $|3/2,-3/2⟩=|n{\pi }^{-}⟩$, so the strong decay is ${\Delta }^{-}\to n{\pi }^{-}$, with the same reduced rate in this symmetry limit. Physical mass splittings and electromagnetic effects perturb these idealized equal-kinematics ratios.

Let $V={\mathbb{C}}^3$ be the defining flavor SU(3) representation. Under $U\in SU(3)$, $q^{\alpha }\mapsto U^{\alpha }{}_{\beta }{q}^{\beta }$. The antiquark transforms in the dual representation:Since $U$ is unitary, its lower-index column of components transforms by $U^{\ast }$, the complex conjugate numerical matrix. Thus $q^{\alpha }{\bar{q}}_{\alpha }$ is invariant. The defining triplet and its conjugate are inequivalent complex representations; conjugation reverses their weights. Write $3=V$ and $\overline{3}=V^{\ast }$, and distinguish a direct sum $\oplus$ from a tensor product $\otimes$.

First split $V\otimes V$ into symmetric tensors and totally antisymmetric tensors:The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant Levi-Civita symbol gives $A_{\gamma }=\frac{1}{2}{ϵ}_{\gamma \alpha \beta }{A}^{\alpha \beta }$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha \beta }={ϵ}^{\alpha \beta \gamma }{A}_{\gamma }$. HenceThe symmetric part is the irreducible highest-weight representation with Dynkin labels $(2,0)$; its six monomials give the standard symmetric-square representation.

For three triplets, the antisymmetric first-pair sector becomes $V^{\ast }\otimes V$. Its tensors $M^{\alpha }{}_{\beta }$ split into their scalar trace and traceless part:The traceless part is the adjoint representation of SU(3); conjugation acts as $M\mapsto UM{U}^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex special linear Lie algebra ${\mathfrak{sl}}_3$.

The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot 3=18$. Complete symmetrization maps it onto ${\mathrm{Sym}}^{3}V$, whose dimension is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e={ϵ}_{ebc}{T}^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $10\oplus 8$. Combining the sectors gives the tensor cube of the defining SU(3) representationHere the decuplet has Dynkin labels $(3,0)$, and the singlet is the fully antisymmetric contraction ${ϵ}_{abc}{T}^{abc}$. The two octets are distinct copies, not two different irreducible types.

For two conjugate SU(3) triplets with a triplet, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by ${ϵ}^{bcd}$ with an upper index, giving $V\otimes V=6\oplus \overline{3}$. The symmetric lower-pair sector has dimension $3\cdot 6=18$, and its contraction is the equivariant mapIt is onto $V^{\ast }$, since the invariant embeddingsatisfies $C\iota (v)=v$. Every such tensor therefore splits uniquely as $\iota (C(T))+(T-\iota (C(T)))$. The latter tensor is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the highest-weight vector $e_1\otimes (e_3^{\ast }{)}^2$ of Dynkin labels $(1,2)$; contraction vanishes because its upper and lower indices differ. The Weyl dimension formula for an SU(3) module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These tensor representations are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-weight summand containing this vector exhausts the fifteen-dimensional kernel. ThusThe question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different tensors. Conjugating the triplet-cube decomposition givesThe octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.

Now consider the flavor diquark $D_{\alpha }\propto {ϵ}_{\alpha \beta \gamma }{q}^{\beta }{q}^{\gamma }$. This notation describes an antisymmetric two-quark flavor state; color and spin factors are separate. In particular $D_3=(ud-du)/\sqrt{2}$ has isospin zero, baryon number $2/3$, strangeness zero and flavor hypercharge $2/3$. The strange antiquark has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3{D}_3\bar{s}$ hasThe charge follows from the Gell-Mann--Nishijima formula. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric tensor component $A_{333}$. It lies entirely in ${\mathrm{Sym}}^3{V}^{\ast }=\overline{10}$, the pentaquark antidecuplet. Equivalently, the symmetric diquark pair belongs to $\overline{6}$, and $\overline{6}\otimes \overline{3}=\overline{10}\oplus 8$, with this highest-hypercharge state in the antidecuplet.

It cannot be in an octet. The octet has the weights of a traceless $q^{\alpha }{\bar{q}}_{\beta }$ tensor, whose hypercharges are $Y_{\alpha }-Y_{\beta }$. For $(u,d,s)$ the quark values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ weight. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-representation assignment for a possible pentaquark, not a proof that the configuration binds or that its full color-spin-orbital wavefunction satisfies all Fermi statistics constraints.

In the two-body centre-of-mass frame, spatial reflection reverses the relative coordinate of a quark-antiquark pair. Its orbital state of angular momentum $L$ has orbital parity $(-1{)}^L$, as follows from $Y_{Lm}(-\hat{\mathbf{r}})=(-1{)}^L{Y}_{Lm}(\hat{\mathbf{r}})$. Opposite constituent intrinsic parities multiply to minus one. Thus the meson parity and charge conjugation rules begin withTo obtain charge conjugation, consider a self-conjugate flavor state and write its spin-orbital wavefunction using ordered creation operators $b_{\sigma }^{†}(\mathbf{p})d_{\tau }^{†}(-\mathbf{p})$. Charge conjugation interchanges the quark and antiquark operators. Reordering them back to quark-first order produces a fermionic sign minus one. Interchanging the two spin-$1/2$ labels has eigenvalue $-1$ in the spin singlet state ($S=0$) and $+1$ in the spin triplet ($S=1$), namely $(-1{)}^{S+1}$. Reversing the relative momentum contributes $(-1{)}^L$. ThereforeReciprocal phases for the two conjugate creation operators cancel, so the result is independent of that convention. This applies to a state that is mapped to itself by charge conjugation, not to every electrically neutral flavor state. For example a neutral meson with unequal quark and antiquark flavors can be exchanged with a different antimeson. Charged states have no individual $C$ eigenvalue.

Restrict flavor to the $u,d$ isospin doublet. Its product with the conjugate doublet decomposes as $2\otimes \overline{2}=3\oplus 1$. Convenient normalized flavor states, with irrelevant overall phase choices, areOrbital excitation normally costs energy, so the simplest low-lying quark model states have $L=0$. Coupling the two constituent spins gives $S=0$ or $S=1$, and then $J=S$. The isotriplet with $S=0$ is the pion triplet, while the isotriplet with $S=1$ is the rho meson triplet. The isosinglet with $S=1$ is the nonstrange component of the omega meson. Their quantum numbers areThe charged pions have $J^P=0^{-}$ and the charged rho mesons have $J^P=1^{-}$; charge conjugation exchanges the positive and negative members. The same tensor product also allows an $S=0$, $I=0$ pseudoscalar state, so symmetry alone does not assert that only the listed mesons exist or predict their masses. The physical isosinglet spectrum involves further dynamics and, when strange flavor is included, mixing.

For a vector-meson decay to two spinless pions, conservation of angular momentum forces their relative orbital value to be $L=1$. Their combined parity is $(-1{)}^2(-1{)}^L=-1$, and the neutral charge combination has $C=(-1{)}^L=-1$, matching both ${\rho }^0$ and $\omega$. Thus $C$ and $P$ alone do not distinguish these two decays. The decisive restriction is isospin together with Bose statistics.

Two isotriplet pions have $3\otimes 3=1\oplus 3\oplus 5$, or total $I=0,1,2$. Under exchange the isospin-$I$ state has sign $(-1{)}^{2-I}$: the $I=2$ highest state and its lowered partners are symmetric, the $I=1$ highest combination $(|+0⟩-|0+⟩)/\sqrt{2}$ is antisymmetric, and the singlet $(|+-⟩+|-+⟩-|00⟩)/\sqrt{3}$ is symmetric. The $L=1$ orbital state is antisymmetric. Since pions are spin-zero bosons, their total state must be symmetric, requiring the antisymmetric isospin sector $I=1$. This allows ${\rho }^0\to {\pi }^{+}{\pi }^{-}$ but excludes an isospin-conserving $\omega \to {\pi }^{+}{\pi }^{-}$ amplitude.

There is no analogous obstruction for three pions. In a Cartesian isospin basis $a,b,c$, their totally antisymmetric combination ${ϵ}_{abc}|{\pi }^a{\pi }^b{\pi }^c⟩$ is an $I=0$ state. It can multiply a totally antisymmetric momentum amplitude proportional in the rest frame toMomentum conservation ${\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3=0$ makes this amplitude change sign under every exchange of two pion labels. The product of the flavor and momentum factors therefore obeys Bose statistics. The cross product supplies $J=1$ and an even orbital parity; multiplying by the three negative pion intrinsic parities gives total $P=-1$. In the charged component, charge conjugation exchanges ${\pi }^{+}$ and ${\pi }^{-}$ and changes the momentum factor's sign, giving $C=-1$. Thus $\omega \to {\pi }^{+}{\pi }^{-}{\pi }^0$ is compatible with $I=0$ and $J^{PC}=1^{--}$.

The same distinction is encoded in G parity: $G=C(-1{)}^I$ for the neutral member gives $G_{\pi }=-1$, $G_{\rho }=+1$, $G_{\omega }=-1$. An $n$-pion state has $G=(-1{)}^n$, so the G-parity selection rule for pion multiplicities allows two pions for the rho and three for the omega. The printed prohibition of the omega two-pion mode is an exact-isospin strong-interaction selection rule. In the physical theory, isospin-breaking effects allow a small two-pion mode; it is not an absolute prohibition. The scope of C parity is also stated in the Particle Data Group quark-model review, and the physical two-pion mode is recorded in its omega listing.

In the finite-dimensional complex semisimple Lie algebra setting, the rank of a semisimple Lie algebra is the dimension of a Cartan subalgebra $\mathfrak{h}$, equivalently the number of independent commuting Cartan generators. For a compact real algebra, use its complexification and the corresponding maximal torus. Simultaneously diagonalizing the commuting adjoint actions of $\mathfrak{h}$ gives the root-space decompositionThe nonzero linear functionals $\alpha$ for which this space is nonzero are the roots of a root system. An invariant inner product identifies them with vectors in a real Euclidean root space. These are the usual semisimple assumptions behind the rank and root language here.

The displayed generators $H,E^{+},E^{-}$ form an sl2 triple. Start with a nonzero lowest-weight element $X_0$ and define $X_{n+1}=[E^{+},X_n]$. The Jacobi identity givesInduction starting with $[H,X_0]=-\lambda X_0$ consequently gives the finite sl2 lowest-weight ladder weightsNow commute the lowering equation with $E^{+}$ and again use the Jacobi identity:The left side is $q_n{X}_n$. Whenever $X_n\ne 0$, coefficient comparison givesThe initial condition is $q_0=0$, meaning $[E^{-},X_0]=0$; directly $[E^{-},X_1]=[-H,X_0]=\lambda X_0$, so $q_1=\lambda$. Summing the recurrence from $0$ to $n-1$ yieldsIn fact the vector identity $[E^{-},X_n]=n(\lambda -n+1)X_{n-1}$ follows inductively from the same Jacobi identity without needing a separately assumed lowering coefficient.

If $X_{n_0}$ is the last nonzero element and $[E^{+},X_{n_0}]=0$, apply $E^{-}$ to this zero next element. The lowering identity givesso $\lambda =n_0$, a nonnegative integer. The nonzero endpoint is necessary. Literally allowing $X_{n_0}=0$ would make the printed conclusion false: take $X_0=E^{-}$ in the adjoint representation of ${\mathfrak{sl}}_2$. Then $\lambda =2$, $X_1=H$, $X_2=-2E^{+}$, $X_3=0$. Choosing $n_0=3$ gives a vanishing next bracket but not $\lambda =3$. The intended endpoint is the first terminating bracket after the nonzero ladder.

For the plus-sign nested commutators, use $H=H_1$, $E^{\pm }=E_1^{\pm }$ and $X_0=E_2^{+}$. The cross relation gives $[E_1^{-},E_2^{+}]=0$, while $[H_1,E_2^{+}]=K_{21}{E}_2^{+}$. Thus the lowest-weight label is $\lambda =-K_{21}$. If the $n$-fold raised element is nonzero and the $(n+1)$-fold element vanishes, the preceding endpoint argument gives $\lambda =n$ and henceFor the minus-sign ladder, use the sl2 triple $H^{\prime }=-H_1$, $E^{\prime +}=E_1^{-}$, $E^{\prime -}=E_1^{+}$ and start at $X_0=E_2^{-}$. Its lowest-weight equation is again $[H^{\prime },X_0]=K_{21}{X}_0$, and its lowering bracket vanishes by the cross relation. The same argument proves exactly the same result, including the endpoint assumption that is explicit in the nested-commutator question.

For arbitrary roots of a root system $\alpha ,\beta$, choose the normalized sl2 subalgebra associated with a root $\beta$, whose generators obey $[H_{\beta },E_{\beta }^{\pm }]=\pm 2E_{\beta }^{\pm }$ and $[E_{\beta }^{+},E_{\beta }^{-}]=H_{\beta }$. The coroot normalization givesThe normalization has a concrete construction. Use the nondegenerate Killing form $B$, and define $t_{\beta }\in \mathfrak{h}$ by $B(t_{\beta },H)=\beta (H)$. Invariance makes root spaces orthogonal unless their roots sum to zero, so there exist $e\in {\mathcal{L}}_{\beta }$ and $f\in {\mathcal{L}}_{-\beta }$ with $B(e,f)=1$. Invariance then gives $B([e,f],H)=B(e,[f,H])=\beta (H)$, hence $[e,f]=t_{\beta }$. With $(\beta ,\beta )=\beta (t_{\beta })$, take $E_{\beta }^{+}=e$, $E_{\beta }^{-}=2f/(\beta ,\beta )$ and $H_{\beta }=2t_{\beta }/(\beta ,\beta )$. Their brackets have the required normalization and $\alpha (H_{\beta })=2(\alpha ,\beta )/(\beta ,\beta )$. Let $v$ be a nonzero $\alpha$ root vector with $H_{\beta }$ weight $w=\alpha (H_{\beta })$. Repeatedly apply $\mathrm{ad}{E}_{\beta }^{-}$ until reaching a last nonzero vector $v_0=(\mathrm{ad}{E}_{\beta }^{-}{)}^{p}v$. Its weight is $w-2p$. This process terminates because different weights are linearly independent and the Lie algebra is finite-dimensional. Starting at $v_0$, repeatedly raise; this also terminates for the same reason. The proven lowest-weight endpoint formula says $w-2p=-N$ for a nonnegative integer $N$. ThereforeThis proves integrality of Cartan integers directly from the finite ladder, rather than assuming the crystallographic axiom.

Finally use the indices actually printed in the PDF: $E_1^{+}=R^1{}_2$, $E_2^{+}=R^2{}_3$, $E_1^{-}=R^2{}_1$, $E_2^{-}=R^3{}_2$. The converted TeX has corrupted these indices. The prescribed bracket is that of trace-free matrix units. The sum of the diagonal generators is zero, leaving two independent diagonal elements, and their pair brackets giveFor example $[R^1{}_2,R^2{}_1]=R^1{}_1-R^2{}_2$, and the analogous second pair gives $H_2$. The cross raising-lowering brackets vanish. A diagonal $H$ with entries $h_r$ has $[H,R^r{}_s]=(h_r-h_s)R^r{}_s$. Thus $H_1$ acts on $E_1^{+},E_2^{+}$ with weights $2,-1$, and $H_2$ acts on them with weights $-1,2$. The negative generators have the opposite weights. In the question's index ordering,This is the Cartan matrix of the A2 root system, and the Lie algebra generated by these matrix units has rank two. As an additional bracket check, $[E_1^{+},E_2^{+}]=R^1{}_3\ne 0$ whereas $[E_1^{+},R^1{}_3]=0$, so the nested ladder has $n=1$ and indeed $K_{21}=-1$. The negative ladder likewise terminates after one nonzero bracket.