Past exam of the theoretical physics II course of the University of Cambridge / 2016 / Paper 1 / 2

Assume the composite pure state has unit norm. Put its amplitudes into the matrixThe partial trace contracts the same $B$ index between ket and bra:ThusIt is Hermitian, positive semidefinite and has trace one. Its determinant is particularly useful:This follows from multiplicativity of the determinant, or by expanding the two-by-two expression. If the supplied amplitudes were not normalized, the physical reduced state would be $C{C}^{†}/\mathrm{tr}(C{C}^{†})$ instead.

The specified Bell-state phases giveThereforeThe sum contains complex squares, not squared moduli. Define $q={\sum }_j{b}_j^2$ and $\mathcal{C}=|q|$. Normalization gives ${\sum }_j|b_j{|}^2=1$, so the triangle inequality ensures $0\le \mathcal{C}\le 1$.

The characteristic polynomial is ${\lambda }^2-\lambda +{\mathcal{C}}^2/4$. HenceHere logarithms are natural and $0\ln 0$ means its limiting value zero. Using logarithms to base two gives entropy in bits instead. Product states have $\mathcal{C}=0$, spectrum $(1,0)$ and entropy zero; maximally entangled states have $\mathcal{C}=1$, spectrum $(1/2,1/2)$ and entropy $\ln 2$.

This phased Bell basis is a magic basis, and $\mathcal{C}$ is the pure-state concurrence. The paper's ordering is a permutation of the usual magic-basis ordering, which leaves $|{\sum }_j{b}_j^2|$ unchanged.^[ref]

A matrix $R$ belongs to the special orthogonal group in four dimensions precisely whenDifferentiating the orthogonality relation along a path $R(ϵ)=I+ϵX+O({ϵ}^2)$ gives $X^{\mathsf{T}}+X=0$. Thus its Lie algebra consists of real antisymmetric four-by-four matrices. Such a matrix has one free entry for each unordered coordinate pair, giving dimension $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$.

Each displayed generator has a single two-coordinate antisymmetric block. The $K$ matrices cover planes $(1,2),(1,3),(1,4)$; the $L$ matrices cover $(3,4),(2,4),(2,3)$, with their printed orientations. These are all six distinct planes. Therefore the generators are linearly independent and span the full antisymmetric space: they form a basis of $\mathfrak{so}(4)$.

Conversely, exponentiating any real antisymmetric $X$ gives $(e^X{)}^{\mathsf{T}}{e}^X=I$ and $\det (e^X)=e^{\mathrm{tr}X}=1$. This verifies that the infinitesimal matrices generate proper rotations rather than reflections.

Use ${ϵ}_{xyz}=+1$ and the supplied commutators. Antisymmetry also gives $[K_a,L_b]={ϵ}_{abc}{K}_c$. Expanding the combinations,The mixed commutator cancels:ThusThere are two mutually commuting three-generator rotation algebras. As real compact Lie algebras,For example, the anti-Hermitian basis $-i{\sigma }_a/2$ obeys the same commutation convention in each $\mathfrak{su}(2)$ factor. This is an algebra statement. Globally, the simply connected group is $\mathrm{Spin}(4)\cong SU(2)\times SU(2)$, andDropping that central quotient would incorrectly identify the global groups.

In the printed phased Bell basis, the actions on its four kets areEach matrix is purely imaginary and antisymmetric. Let $T$ have these four basis kets as columns in the computational basis. The infinitesimal generators of the standard rotation $U_A=\exp [-i\theta \mathbf{n}\cdot {\mathbf{\sigma }}_A/2]$ areUsing the displayed actions gives the phase-convention-specific identitiesThey are real antisymmetric matrices. A finite $SU(2)$ rotation on $A$ therefore acts on the magic-basis coefficient vector as $b\mapsto Ob$, with $O\in SO(4)$. ConsequentlyIt preserves concurrence and hence preserves the reduced eigenvalues and entanglement entropy. A general $U(2)$ operation adds a common phase to $b$; that changes $q$ by a phase of twice the angle but leaves $|q|$ unchanged. The real-rotation description of local transformations is a characteristic property of the magic basis.^[ref]

As an independent direct check, a local unitary sends ${\rho }_A$ to $U_A{\rho }_A{U}_A^{†}$, which is a unitary similarity transformation and cannot change its spectrum. This local-unitary invariance of entanglement entropy also explains why merely rotating one subsystem cannot create entanglement.