Past exam of the groups and representations course of the University of Oxford / 2020 / 4

Differentiate a curve $U(t)\in SU(4)$ through $U(0)=I$, writing $X=U^{\prime }(0)$. Differentiating $U(t{)}^{†}U(t)=I$ gives $X^{†}+X=0$; differentiating $\det U(t)=1$ gives $\mathrm{tr}X=0$. Conversely, for every anti-Hermitian matrix $X$ with zero trace, the matrix exponential $e^{tX}$ is unitary and has determinant $e^{t\mathrm{tr}X}=1$. Thus the real Lie algebra isThe commutator again has zero trace and is anti-Hermitian, so this set is closed under the Lie bracket. Let $E_{jk}$ be the matrix units. A convenient real basis isThere are six independent $A_{jk}$, six independent $B_{jk}$, and three independent $H_a$. Every permitted off-diagonal complex entry is supplied by one real combination of $A_{jk},B_{jk}$; the imaginary diagonal entries have one trace constraint and are spanned by the $H_a$. ConsequentlyFor the rank of a semisimple Lie algebra, the diagonal Cartan subalgebra consists of $i\mathrm{diag}(t_1,t_2,t_3,t_4)$ with real $t_i$ and ${\sum }_i{t}_i=0$, of dimension three. Commuting anti-Hermitian matrices are simultaneously unitarily diagonalizable, so no larger commuting toral Lie subalgebra exists. In the common physics convention one uses the Hermitian generators $-i{A}_{jk},-i{B}_{jk},-i{H}_a$ and writes group elements as $e^{itT}$; these are the same generators with the factor of $i$ moved into the exponential.

The complexification of a Lie algebra gives $\mathfrak{su}(4{)}_{\mathbb{C}}={\mathfrak{sl}}_4(\mathbb{C})$, of type A3 root system. Choose simple roots ${\alpha }_1=e_1-e_2$, ${\alpha }_2=e_2-e_3$, ${\alpha }_3=e_3-e_4$, with squared lengths two. The Cartan matrix and Dynkin diagram areEach line represents a single bond. In the Dynkin label convention $w_a=⟨w,{\alpha }_a^{\vee }⟩$, the simple roots have coordinates equal to the rows of $A$.

In the defining representation, apply the simple coroots $E_{aa}-E_{a+1,a+1}$ to the four coordinate basis vectors. Their weights areEquivalently these are obtained by subtracting ${\alpha }_1,{\alpha }_2,{\alpha }_3$ successively from the highest weight. The dual representation has the negative weights, henceAll listed weight multiplicities are one, so both dimensions are four.

The highest weight $(1,0,1)$ is ${\alpha }_1+{\alpha }_2+{\alpha }_3$, the highest root. Its irreducible representation is therefore the Adjoint representation. Under the action on traceless matrices by conjugation, the off-diagonal matrix unit $E_{ij}$ has weight $v_i-v_j$, while the diagonal Cartan subalgebra has zero weight. ThusThe superscript $[3]$ indicates weight multiplicity three. The twelve nonzero weights each have weight multiplicity one, giving $12+3=15$ in total. This also follows from the tensor product representation $4\otimes \overline{4}=15\oplus 1$: the sixteen differences $v_i-v_j$ contain four zero weights, and removing the identity matrix removes one of them.

The projection deletes the third Dynkin label, so the four weights $v_i$ of solution restrict to $(1,0),(-1,1),(0,-1),(0,0)$. The first three are the defining SU(3) weight system, and the fourth is a singlet. Negating them gives the dual representation. HenceFor the Adjoint representation, the projected nonzero weights consist of the six A2 root system roots, the three defining weights and their negatives. There are also three zero weights. Two zero weights belong to the SU(3) Adjoint representation, leaving one singlet. Therefore the SU4 adjoint branching to SU3 isOne can see all four invariant subspaces directly by writing a traceless matrix asThe traceless part of $A$ is $8$, the column $b$ is $3$, the row $c$ is $\overline{3}$, and the remaining diagonal matrix is proportional to $\mathrm{diag}(1,1,1,-3)$ and is fixed by SU(3). These have dimensions $8+3+3+1=15$. Equivalently restrict $4\otimes \overline{4}-1$ and use $3\otimes \overline{3}=8\oplus 1$.

Take the quarks $q=(u,d,s,c{)}^T$ in the defining $4$ of flavour SU(4) and the antiquarks in $\overline{4}$. The flavour part of a meson is a quark-antiquark pair $M^i{}_j=q^i{\bar{q}}_j$. It transforms as $M\mapsto UM{U}^{-1}$, so the complete four-flavour meson multiplet isIts traceless part $M-\frac{1}{4}(\mathrm{tr}M)I_4$ is the Adjoint representation $15$, the direct four-flavour analogue of the three-flavour meson octet. The trace supplies an additional trivial representation. Thus “the mesons” can refer to the traceless multiplet or to the complete quark model tensor product; keeping both explicit avoids dropping a state.

Using solution, the SU(3) representation contents areDirectly, the upper-left block $q^i{\bar{q}}_j$ with $i,j\le 3$ is the old $3\otimes \overline{3}=8\oplus 1$; the three entries $q^i\bar{c}$ form $3$, the three entries $c{\bar{q}}_j$ form $\overline{3}$, and $c\bar{c}$ is another singlet. The dimensions are $9+3+3+1=16$. This is a classification by flavour symmetry, rather than colour or spin. The much larger charm quark mass breaks the flavour SU(4) approximation strongly; it does not change this representation counting.

Compared with the complete three-flavour meson tensor product, the additional flavour states are exactlyThus there are seven new flavour states. For the pseudoscalar mesons, the six open-charm states are ${\bar{D}}^0,D^{-},D_s^{-}$ and $D^0,D^{+},D_s^{+}$, respectively, using the D meson and Ds meson conventions. A pure $c\bar{c}$ pseudoscalar meson is the idealized flavour state associated with the eta c meson. The quark model flavour calculation alone does not specify spin or spatial excitations, so the same flavour labels also organize other meson species.

If comparing the traceless SU(4) $15$ with the traceless SU(3) meson octet, the extra seven states are instead the same two triplets and the normalized SU(3) singletIts coefficient matrix is traceless and has squared norm one. In particular, $c\bar{c}$ by itself is not a vector in the traceless $15$; its coefficient matrix has trace one.

To relate these two descriptions, let $S_4=(u\bar{u}+d\bar{d}+s\bar{s}+c\bar{c})/2$ be the normalized SU(4) singlet and $S_3=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ the old SU(3) singlet. ThenThis orthonormal change of basis shows explicitly why the complete tensor product contains two light-flavour singlets but adds only one state to the old light-flavour singlet sector.