Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 5

Use $[a_i,a_j^{†}]={\delta }_{ij}$ and define $E_{ij}=a_i^{†}{a}_j$. A direct multiplication givesWith $S_{+}=\hslash a_{\uparrow }^{†}{a}_{\downarrow }$, $S_{-}=\hslash a_{\downarrow }^{†}{a}_{\uparrow }$, this implies $[S_{+},S_{-}]=2\hslash S_z$ and $[S_z,S_{\pm }]=\pm \hslash S_{\pm }$. Resolving the raising and lowering combinations givesThese operators realize the Schwinger boson representation of angular momentum. Each commutes with $\hat{N}=n_{\uparrow }+n_{\downarrow }$, so fixed total occupation is an invariant subspace.

The bosonic commutator is important in $S_{+}{S}_{-}={\hslash }^2{n}_{\uparrow }(n_{\downarrow }+1)$ and $S_{-}{S}_{+}={\hslash }^2{n}_{\downarrow }(n_{\uparrow }+1)$. ThereforeOn the $N$-particle sector,For example $N=0,1,2$ gives spin $0,\frac{1}{2},1$. The total number operator can be retained in the same identity before restricting to a sector.

Total symmetry makes a component depend only on its number $n_{\uparrow }$ of up indices. The possible values are $0,1,\dots ,N$, givingA normalized basis is $|n_{\uparrow },N-n_{\uparrow }⟩$, with $S_z$ eigenvalue $\hslash (n_{\uparrow }-N/2)$. It is precisely one spin-$N/2$ multiplet. These are independent complex amplitudes before normalization and removal of a common quantum phase, rather than $N+1$ independent real physical parameters.

The defining matrix conditions areThe group operation is matrix multiplication. Equivalently $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real $a_0,\mathbf{a}$ and $a_0^2+|\mathbf{a}{|}^2=1$, so it has three real parameters and is a compact connected simply connected Lie group. Its Hermitian spin generators obey the commutator from part (a); the corresponding Lie-algebra matrices themselves are traceless and anti-Hermitian.

The literal printed difference is zero, because the same spin-index product appears twice, and zero is invariant under every transformation. The intended SU(2)-invariant antisymmetric spinor contraction uses opposite indices in the second term. With $\epsilon =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, define $I={\varphi }^T\epsilon \chi$. For any two-by-two matrix,Thus for $U\in SU(2)$,Expanding the determinant identity proves the result without complex conjugating either spinor. This is the invariant alternating tensor behind the spin singlet state. If the two spinors are identical, the contraction vanishes. An antisymmetric internal two-boson state also needs an antisymmetric additional degree of freedom for overall bosonic symmetry; it does not lie in the fully symmetric fixed-two-mode sector discussed above.