Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 43 (Symmetry and Particle Physics)

The Adjoint representation of a Lie algebra acts on the algebra itself by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives $[{\mathrm{ad}}_X,{\mathrm{ad}}_Y]={\mathrm{ad}}_{[X,Y]}$, so this is a Lie algebra representation. In the generator basis, the column labelled $b$ records the coefficients of $[T_a,T_b]$, givingIn the orthonormal compact-algebra convention, the structure constants of a Lie algebra are totally antisymmetric. With the row index written first this is $(t_a^{\mathrm{Ad}}{)}_{bc}=-i{f}_{abc}$.

Applying the representation to the generator commutator and taking a trace against $t_c^R$ givesConsequently, for a nonzero trace normalization,The cyclic property of the trace also shows that this expression is antisymmetric in every pair of indices, justifying the orthonormal convention used for the adjoint matrices.

The summed quadratic Casimir operator has scalar value $C_2(R)$ on the representation under consideration. Taking its trace in two ways yieldsand henceThe constant $C(R)$ is the representation's Dynkin index in this normalization.

Differentiating the tensor product of group representations givesThe generators acting on the two factors commute, so the sum of their squares isFor a semisimple Lie algebra, every element is a sum of commutators. Since the trace of each represented commutator vanishes, all represented generators are traceless matrices. The trace of the cross term is therefore zero. Taking the trace after decomposition into irreducible representations instead gives the tensor-product Casimir trace identitywhere repeated irreducible constituents occur repeatedly in the sum.

This step requires tracelessness, as holds for the special unitary groups used below. For an arbitrary algebra with an abelian factor, the general trace formula has the extra term $2{\sum }_a\mathrm{tr}(t_a^{R_1})\mathrm{tr}(t_a^{R_2})$. For example, one-dimensional representations of an abelian generator with charges one and one have individual Casimirs one, while the product generator has charge two and Casimir four. Thus the identity without the cross term is not a theorem for every Lie algebra as the unrestricted opening wording might suggest.

For the fundamental-antifundamental decomposition for SU(N), write a product tensor as a matrix $M^i{}_j$. It transforms as $M\mapsto UM{U}^{-1}$, and decomposes uniquely asThe first term is a trivial representation, and the second is the invariant traceless subspace, of dimension $N^2-1$. Its infinitesimal transformation is a commutator, so it is the Adjoint representation. This representation is irreducible for $N\ge 2$: an invariant subspace is an ideal of ${\mathfrak{sl}}_N$; commuting a nonzero ideal element with diagonal and elementary matrices produces off-diagonal elementary matrices, whose further commutators generate all off-diagonal matrices and traceless diagonal matrices. A purely diagonal nonzero element also produces an off-diagonal element unless it is scalar, and a traceless scalar in characteristic zero is zero. ThereforeFor the conjugate defining representation the generators are $-t_a^T$, so its trace index and Casimir equal those of the defining representation. With $C(\mathbf{N})=1/2$, the trace formula gives $C_2(\mathbf{N})=(N^2-1)/(2N)$. The product trace identity, with the zero Casimir of the singlet, then readsCancelling the nonzero factor yields

A Lie algebra representation is a linear map $\rho :\mathfrak{g}\to \mathrm{End}(V)$ preserving brackets: $\rho ([X,Y])=[\rho (X),\rho (Y)]$. Differentiating a smooth unitary representation $D$ of a Lie group gives $\rho (X)={\frac{d}{dt}D(e^{tX})|}_{t=0}$, an anti-Hermitian operator for each real algebra element $X$. In the physics convention one uses Hermitian generators $t_X=i\rho (X)$ and writes $D(e^{tX})=e^{-it{t}_X}$. Conversely, integration of a Lie-algebra representation gives a representation of the connected simply connected group. A representation of another group with the same algebra additionally has to be trivial on the kernel of its covering map.

For the angular momentum commutation relations, take $J_3$ Hermitian and $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space. A highest-weight vector has $J_3|j,j⟩=j|j,j⟩$ and $J_{+}|j,j⟩=0$. Set $v_{\ell }=(J_{-}{)}^{\ell }|j,j⟩$. Commuting $J_3$ through each lowering operator gives $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. Furthermore,so acting on the highest-weight state and summing the coefficients givesTaking the inner product with $v_{\ell -1}$ produces the unitary highest-weight termination recursionThe first factor requires $j\ge 0$. If $2j$ is not an integer, the first integer $\ell >2j+1$ would make the norm negative, while every preceding norm is positive. ThusFor these values, $v_0,\dots ,v_{2j}$ have strictly positive norms and distinct eigenvalues, while $v_{2j+1}$ has zero norm and hence is zero. They therefore span a space of dimension $2j+1$, invariant under all three generators. Its weights are $j,j-1,\dots ,-j$. This proves integer or half-integer spin and finite dimension $2j+1$. The positive-definite unitary assumption matters: an unrestricted algebraic highest-weight module need not terminate. Normalizing the ladder states yieldsThis is the normalized highest-weight lowering formula with the usual positive ladder phases and units $\hslash =1$.

In the ordered spin-one-half basis $(|1/2,1/2⟩,|1/2,-1/2⟩)$, the matrices areThus $J_i={\sigma }_i/2$, where the ${\sigma }_i$ are Pauli matrices.

Choose the active-rotation convention. The SU(2) lift of an axis-angle rotation isThe Pauli matrix multiplication law shows that conjugating $\mathbf{x}\cdot \mathbf{\sigma }$ by this matrix produces $(R\mathbf{x})\cdot \mathbf{\sigma }$, withThis is the ordinary three-dimensional rotation. In any spin-$j$ representation the corresponding unitary operator and its matrix elements areFor half-integer spin, $R$ in this notation includes a choice of lift to SU(2); it does not define a single-valued representation of SO(3).

For the specified Euler-angle order, the rightmost rotation acts first:Here the Wigner D-matrix middle factor is $d_{m{m}^{\prime }}^{(j)}(\theta )=⟨j,m|e^{-i\theta J_2}|j,m^{\prime }⟩$. It can be calculated explicitly by realizing spin $j$ as the symmetric product of $2j$ spin-one-half factors. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The fundamental rotation sends $|+⟩$ to $c|+⟩+s|-⟩$ and $|-⟩$ to $-s|+⟩+c|-⟩$. Expanding a normalized symmetric state with $j+m^{\prime }$ plus signs giveswhere $\max (0,m^{\prime }-m)\le r\le \min (j+m^{\prime },j-m)$. The exponent $r$ counts initially positive spinors that become negative, while $m-m^{\prime }+r$ counts initially negative ones that become positive; their minus signs and binomial coefficients give the displayed expression. Every factorial argument is an integer even for half-integer spin.

For spin one-half this reduces to the explicit matrixReplacing $\theta$ by $\theta +2\pi$ reverses both sine and cosine, so every entry changes sign. In particular, a pure $2\pi$ rotation gives $D^{(1/2)}=-I$, while a $4\pi$ rotation gives $I$.

Finally, the Adjoint double cover from SU(2) to SO(3) is surjective by the axis-angle construction. Its kernel consists of matrices commuting with all Pauli matrices, hence scalar matrices; within SU(2) these are $\pm I$. ThusTheir Lie algebras are isomorphic, but their global topology and representations differ. In spin $j$, the central element $-I$ acts as $(-1{)}^{2j}I$, so exactly the integer-spin representations descend to ordinary SO(3) representations. This is the descent of an SU(2) representation to SO(3) criterion.

Let $V={\mathbb{C}}^3$ be the defining SU(3) representation. A mixed SU(3) tensor representation belongs to $V^{\otimes k}\otimes (V^{\ast }{)}^{\otimes l}$, with transformation lawThe unrestricted tensor has dimension $3^{k+l}$. The upper and lower counts are independent; setting them equal would lose part of the printed generality.

For an irreducible tensor, impose the appropriate Young symmetrizer on each set of indices and subtract all upper-lower contractions. Antisymmetric pairs can be converted using the invariant alternating tensor, since ${\bigwedge }^{2}3\cong \overline{3}$ and ${\bigwedge }^2\overline{3}\cong 3$. A full alternating triple is a singlet because $\det U=1$. Traces use the invariant Kronecker delta and yield lower-rank tensors. Repeating these operations gives invariant irreducible subspaces, with multiplicities retained.

An explicit algorithm for the complete decomposition starts at $(p,q)=(0,0)$, tensors with $3$ exactly $k$ times, and then with $\overline{3}$ exactly $l$ times. At each step applyomitting terms with negative labels and adding multiplicities of coincident terms. The first is the sl3 highest-weight tensor rule: a Young diagram of row lengths $(p+q,q,0)$ gains a box in one of its three rows. The third-row term loses a determinant column of height three, leaving labels $(p,q-1)$. The second rule follows by conjugation, which exchanges $p$ and $q$. Thus the procedure determines all constituents of the general tensor, not just one selected symmetry type.

The separately symmetric, traceless component has labels $(k,l)$. Before imposing traces its dimension is $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$. The contraction map has target the separately symmetric tensors of degrees $(k-1,l-1)$ and is onto: in polynomial coordinates it is ${\sum }_i{\partial }_{x_i}{\partial }_{y_i}$, the adjoint of multiplication by ${\sum }_i{x}_i{y}_i$ in the monomial-factorial inner product. Multiplication is injective, so the adjoint is surjective. Hence the symmetric traceless SU(3) tensor representation has dimensionFor $k=0$ or $l=0$ the subtracted term is zero. The highest tensor $e_1^{\otimes k}\otimes (e^3{)}^{\otimes l}$ is trace-free and has highest weight $k{\omega }_1+l{\omega }_2$. The three positive-root factors in the Weyl dimension formula are $k+1$, $l+1$ and $(k+l+2)/2$, giving exactly the kernel dimension above. Thus the whole symmetric traceless kernel is that irreducible constituent.

There is a genuine distinction between this highest-weight constituent and the literal largest-dimensional constituent. For example, $3^{\otimes 5}$ contains $(3,1)$, obtained by antisymmetrizing one pair, with dimension $24$, whereas the fully symmetric $(5,0)$ has dimension $21$. Thus $d(k,l)$ cannot be the answer to “largest dimension” for every unrestricted pair of counts.

The maximal dimension in mixed SU(3) tensor powers can nevertheless be determined exactly. Every constituent highest weight differs from $k{\omega }_1+l{\omega }_2$ by $a{\alpha }_1+b{\alpha }_2$ with nonnegative integers $a,b$, where ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. Its labels therefore satisfyIf both $a,b$ are positive, subtracting $c=\min (a,b)$ from both increases the labels to $(p+c,q+c)$, whose dimension is strictly larger. The resulting boundary labels occur in the tensor: use $a$ disjoint antisymmetric upper pairs or $b$ disjoint antisymmetric lower pairs, and then take the separately symmetric traceless component. Consequently the exact answer is the finite formulaThis supplies a practical closed finite computation for arbitrary $k,l$. To reduce it to at most four evaluations, put $K=k+1$, $L=l+1$ and $r=1+\sqrt{3}$. The first continuous maximum is at $a_{\ast }=(K-rL)/(2+r)$, clipped to $[0,k/2]$; test its two neighbouring allowed integers. The second is similarly at $b_{\ast }=(L-rK)/(2+r)$, clipped to $[0,l/2]$. Indeed differentiating the first cubic gives a derivative with the sign of $(K-2a{)}^2-2(K-2a)(L+a)-2(L+a{)}^2$, which changes sign once at that root. For equal upper and lower counts, both maxima are at zero and the familiar result is $(k+1{)}^3$.

For three quarks, the unconstrained flavour product isThe brackets indicate permutation symmetry: symmetric, mixed, and alternating. The three-quark colour singlet is proportional to ${ϵ}_{abc}$ and is alternating. The fermion wavefunction must be alternating overall. In the usual spatially symmetric ground-state S-wave, the spin-flavour part must therefore be symmetric.

The three-spin-one-half space has a symmetric spin-$3/2$ sector and mixed-symmetry spin-$1/2$ sectors; it has no fully alternating sector because ${\bigwedge }^3{\mathbb{C}}^2=0$. Symmetric flavour can pair with symmetric spin, giving the baryon decuplet of spin $3/2$. The flavour-octet multiplicity space and the spin-$1/2$ multiplicity space both transform as the two-dimensional standard representation of the symmetric group. Their product contains exactly one symmetric combination: its characters at the identity, a transposition and a three-cycle are $4,0,1$, so the symmetric multiplicity is $(4+3\cdot 0+2\cdot 1)/6=1$. This gives one baryon octet of spin $1/2$. Alternating singlet flavour would need alternating spin to make the spin-flavour part symmetric, and that sector does not exist. Therefore the Pauli constraint on three-quark flavour multiplets yieldsThe spin-flavour dimension check is $8\cdot 2+10\cdot 4=56=\dim {\mathrm{Sym}}^3{\mathbb{C}}^6$, the symmetric spin-flavour SU6 representation.

The spatial-symmetry qualification matters if the wording is interpreted as total $L=0$ alone. Rotationally scalar excited orbital functions can have mixed permutation symmetry. For example, the two independent differences among $r_{12}^2,r_{23}^2,r_{31}^2$, multiplied by a symmetric radial factor, span such a space while remaining rotational scalars. Combining this standard permutation representation with the spin-$1/2$ standard representation gives an alternating spin-orbital combination; multiplying by alternating singlet flavour makes it symmetric, and the colour factor restores overall fermionic antisymmetry. Thus total $L=0$ without a symmetric ground-state orbital assumption would not by itself exclude a singlet.

For a quark-antiquark pair, colour singletness selects the contraction ${\delta }^a{}_b$, and the independent flavour product givesQuark and antiquark are distinguishable, so there is no three-identical-quark exclusion of either flavour multiplet or either spin. Both the octet and singlet admit spin singlets and spin triplets. Their baryon number is zero. The flavour octet has rows $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$, while the singlet has $(Y,I)=(0,0)$. The octet's zero-isospin state is $(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$, and the singlet is $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. Each row contains all $I_3=-I,\dots ,I$, and electric charge is $Q=I_3+Y/2$.

For ground-state mesons with $L=0$, opposite intrinsic quark-antiquark parities give $P=-1$. The neutral self-conjugate states have charge conjugation $C=(-1{)}^{L+S}$, so both flavour multiplets occur as $0^{-+}$ pseudoscalar mesons and $1^{--}$ vector mesons, forming a flavour nonet in each channel. Charged and open-flavour states have no individual charge-conjugation eigenvalue. The meson request does not restrict orbital excitation: in general $J=|L-S|,\dots ,L+S$, $P=(-1{)}^{L+1}$, and the same formula for $C$ applies where defined. Thus the flavour representations stay $1\oplus 8$, with these additional orbital-spin quantum numbers.

A SU(2) matrix has the form $\left(\begin{array}{cc}a& b\\ -\overline{b}& \overline{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$. Comparing this with the Pauli expansion givesConsequently all four Pauli coefficients must be real, and their norm condition isConversely, the Pauli matrix multiplication law gives $A{A}^{†}=(u_0^2+|\mathbf{u}{|}^2)I$ and $\det A=u_0^2+|\mathbf{u}{|}^2$, so these conditions imply unitarity and determinant one. For complex coefficients the determinant equation alone would not imply unitarity.

On either open hemisphere,Both signs are necessary globally. The two hemispheres meet along $u_0=0$, and the full four-coefficient constraint is exactly the unit sphere in ${\mathbb{R}}^4$. The coefficient map and its inverse are smooth, establishing SU(2) as the three-sphere as a manifold, not merely a dimension count.

For two real coefficient quadruples, expanding the matrix product with the Pauli matrix multiplication law givesFor an infinitesimal second factor, $v_0=1+O(|\mathbf{v}{|}^2)$. Thereforewhile $d{u}_0=-u_i{v}_i$. These variations are tangent to the unit sphere because $u_{0}d{u}_0+u_{i}d{u}_i=0$.

On a hemisphere, differentiation of the norm constraint gives ${\partial }_{u_i}{u}_0=-u_i/u_0$, and henceContracting with the Pauli-coordinate left-invariant vector fields on SU(2) and using ${\mu }_{ji}{u}_i=u_0{u}_j$ givesThe Pauli matrix multiplication law gives exactly the same expression for $-A{\sigma }_j$, soIt follows thatThis uses the Pauli matrix commutator identity. The commutator of two first-order vector fields is again a vector field, and the matrix entries of $A$ span the four coordinate functions $u_0,u_1,u_2,u_3$. Equality of their actions on these entries determines the tangent field. ThereforeThe real tangent field underneath $T_j$ is generated by the flow $A\mapsto A{e}^{it{\sigma }_j}$; left multiplication transports that tangent consistently, explaining its left invariance. The rescaled generators $-T_j/2$ have the usual positive-sign angular-momentum commutator.

To determine the Haar measure, compute the metric induced from the unit three-sphere. Since $d{u}_0=-u_{i}d{u}_i/u_0$, its line element isThe rank-one matrix determinant lemma givesThus the Pauli-coordinate Haar measure on SU(2) isup to an overall normalization. To check its group invariance, fix one unit coefficient quadruple and multiply by it. The product formulas define a linear transformation of the other quadruple. The norm of a product is the product of norms, as follows either from $A{A}^{†}$ or directly from the dot and cross products. Fixed multiplication by a unit quadruple is therefore an orthogonal transformation of ${\mathbb{R}}^4$. It preserves the round sphere metric and volume, on both the left and the right, proving that this is the invariant measure.

The same density can be checked directly against the vector fields. On a fixed hemisphere, $u_0/|u_0|$ is constant, andThe second term vanishes because its contractions involve either ${\delta }_{ik}$ or $u_i{u}_k$, both symmetric in $i,k$. Thus the density has zero divergence along every invariant tangent direction.

The hemisphere chart becomes singular at the equator, but its volume is finite:Both hemispheres give total volume $2{\pi }^2$, so probability-normalized Haar measure is $d^{3}u/(2{\pi }^2|u_0|)$ on each hemisphere. Finally, the three-sphere is closed and bounded in ${\mathbb{R}}^4$, hence compact; its smooth identification with SU(2) proves that $SU(2)$ is a compact group. The finite integral reflects this compactness, and the equatorial divergence is only a coordinate artefact.