Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 33B (Principles of Quantum Mechanics)

On the tensor-product Hilbert space, addition of angular momentum givesSuppressing the identity operators, its components and ladder operators are

The Clebsch-Gordan decomposition permitsThus a total-angular-momentum singlet $j=0$ occurs exactly when $j_1=j_2$.

Put $J=j_1+j_2$. The only state with $m=J$ is the maximal highest-weight stateApplying $J_{-}=J_{1-}+J_{2-}$ and normalizing givesThe orthonormal vector in the same two-dimensional $m=J-1$ subspace isIt is annihilated by $J_{+}$, so it is the highest-weight state of the $j=J-1$ multiplet. These are all states with $m\ge J-1$. The last state exists when both subsystem spins are nonzero.

A product state factors as $|{\psi }_1⟩\otimes |{\psi }_2⟩$; an entangled state admits no such factorization. The state $|J,J⟩$ is a product state. When $j_1,j_2>0$, each state with $m=J-1$ displayed in part (iii) has two nonzero terms with orthogonal states in both subsystems, so its Schmidt rank is two and it is entangled. If one subsystem has spin zero, only the $j=J$ multiplet remains and its states factor through the unique state of that subsystem.

For $|J,J⟩$, tracing out subsystem 2 gives the reduced density matrixwhich is pure, so its entanglement entropy is zero.

For $|J,J-1⟩$, the two subsystem-2 factors in part (iii) are orthogonal. The partial trace therefore removes the cross terms and givesIts nonzero eigenvalues are $j_1/J$ and $j_2/J$, so the Von Neumann entropy isWhen $j_1=j_2$, both eigenvalues are $1/2$ and $S=\log 2$, the maximal entropy for a rank-two reduced state.

A $j=0$ state exists only when $j_1=j_2=:j$. Since its total magnetic quantum number is zero, every uncoupled basis term must have $m_1+m_2=0$, soApply the total raising operator $J_{+}=J_{1+}+J_{2+}$. The coefficient of
$|j,m+1{⟩}_1|j,-m{⟩}_2$ isThe angular momentum singlet state is annihilated by $J_{+}$, henceAll coefficients consequently have the same modulus and alternating signs. Normalization fixes that modulus to $(2j+1{)}^{-1/2}$, and an arbitrary overall phase may be chosen so thatThus