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

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.