Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 52 / 2

A highest-weight vector is a nonzero vector $v_0$ satisfying $J_3{v}_0=j{v}_0$ and $J_{+}{v}_0=0$; its eigenvalue $j$ is the highest weight. Existence does not require the Lie algebra representation to be antihermitian. Since $V$ is a nonzero finite-dimensional complex vector space, $J_3$ has an eigenvector $v$ with some eigenvalue $\lambda$. The commutator with $J_{+}$ shows that each nonzero $J_{+}^{k}v$ has eigenvalue $\lambda +k$. Infinitely many such vectors would have distinct eigenvalues and be linearly independent, which is impossible. The last nonzero vector in this raising chain is therefore a highest-weight vector $v_0$.

Put $v_{\ell }=J_{-}^{\ell }{v}_0$. The ladder operator relations give $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. The raising coefficient follows by expanding the commutator of $J_{+}$ with a power of $J_{-}$:The factor $1/2$ is essential: these ladder operators have $[J_{+},J_{-}]=J_3$, rather than the convention with $2J_3$. The same finite-dimensionality argument gives a last nonzero lowered vector $v_r$, with $J_{-}{v}_r=v_{r+1}=0$. Applying $J_{+}$ to $v_{r+1}=0$ yieldsso $2j=r\in {\mathbb{Z}}_{\ge 0}$. All $v_0,\dots ,v_r$ are nonzero and have distinct eigenvalues. Their span is invariant under $J_3,J_{+},J_{-}$ and therefore under the original three generators. Irreducibility of the Lie algebra representation makes this span equal to $V$. HenceIn particular, the initial construction produces the largest eigenvalue of $J_3$, since its full spectrum is now $j,j-1,\dots ,-j$. This establishes the highest weight without assuming beforehand that all eigenvalues are real or that $J_3$ is diagonalizable.

For the additional antihermitian Lie algebra representation, the given definitions imply $J_3^{†}=J_3$ and $J_{+}^{†}=J_{-}$. Thus distinct $J_3$ eigenvalues give orthogonal weight vectors. With $\Vert v_0\Vert =1$, write $N_{\ell }=\Vert v_{\ell }{\Vert }^2$. The adjoint relation and the raising coefficient giveMultiplying this recursion proves the normalized highest-weight lowering formula in the present normalization:Every factorial argument is an integer, even for half-integral $j$. The displayed states have unit norm, mutually distinct eigenvalues, and span $V$, so they form an orthonormal basis. Their phase convention also fixes the positive ladder amplitudeswith the endpoint actions interpreted as zero. These amplitudes will determine the Clebsch-Gordan coefficients.

The Clebsch-Gordan decomposition for SU2 iswhere $j$ increases by one and every summand occurs once. The tensor product of Lie algebra representations is antihermitian for the product inner product, so an invariant subspace has an invariant orthogonal complement; this gives complete reducibility. To see the particular summands, suppose $j_1\ge j_2$. The product weight diagram has one state of weight $j_1+j_2$, two of weight $j_1+j_2-1$, and increasing multiplicities down to the plateau determined by $j_1-j_2$. Subtracting the weight string of $V_{j_1+j_2}$ leaves a single highest state of weight $j_1+j_2-1$. Repeating highest-weight character subtraction yields exactly the displayed sequence, down to $j_1-j_2$. Its dimensions exhaust the tensor product becauseThe case $j_2\ge j_1$ is identical with the factors exchanged.

Let $C_m^j(m_1,m_2)=⟨m_1:m_2|j,m⟩$ be the Clebsch-Gordan coefficients. Since the total $J_3$ is $J_3^{(1)}\otimes 1+1\otimes J_3^{(2)}$, orthogonality of its eigenspaces gives the selection rule $C_m^j(m_1,m_2)=0$ unless $m_1+m_2=m$. Normalize the coupled states by lowering from a unit highest-weight vector, as in part (i); a common phase for an entire summand is immaterial. ThenTake the inner product with $⟨m_1:m_2|$. On the other hand, the total ladder operator is $J_{\mp }^{(1)}\otimes 1+1\otimes J_{\mp }^{(2)}$, and $J_{\mp }^{†}=J_{\pm }$. Its action on the product bra is thereforeEquating the two evaluations and multiplying by $\sqrt{2}$ proves both signs of the ladder recurrence for Clebsch-Gordan coefficients:Terms with an index outside its allowed weight range are zero. Merely specifying eigenvectors of $J_3$ would allow unrelated phases for different $m$; the coherent positive-ladder convention above is what makes these recurrences hold with the displayed positive square roots.