Past exam of the computer science course of the University of Oxford / 2020 / Quantum information / 2 / d

Fix the two subsystem dimensions. For reflexivity, $|\Psi ⟩=(I_d\otimes I_{d^{\prime }})|\Psi ⟩$, and both identity operators are unitary.

For symmetry, if $|\Psi ⟩=(U\otimes V)|\Phi ⟩$, apply the inverse local unitary transformation:The conjugate transposes of unitary operators are unitary, so the reverse relation holds.

For transitivity, if additionally $|\Phi ⟩=(R\otimes S)|\Theta ⟩$, the tensor product multiplication rule givesBoth products are unitary, for example $(UR{)}^{†}(UR)=R^{†}{U}^{†}UR=I$. Thus local-unitary equivalence is reflexive, symmetric and transitive, and hence is an equivalence relation.

Use $S=\mathrm{diag}(1,i)=T^2$ and the Pauli X gate. Starting with the Bell state $|{\Phi }^{+}⟩$, the exact constructions areFor the first, $I\otimes X$ changes the two terms to $|01⟩$ and $|10⟩$, and $S\otimes I$ multiplies just the second term by $i$. For the second,All factors are single-party unitary operators, proving both claimed local-unitary equivalences. The equations hold as vectors, without even an additional global phase.

Extend the normalized vector $|\varphi ⟩$ to an orthonormal basis $\{|f_j⟩{\}}_{j=1}^d$ with $|f_1⟩=|\varphi ⟩$, and extend $|\psi ⟩$ to an orthonormal basis $\{|e_j⟩{\}}_{j=1}^d$ with $|e_1⟩=|\psi ⟩$. The Gram-Schmidt process provides such extensions. DefineThen $U^{†}U={\sum }_j|f_j⟩⟨f_j|=I$ and $U{U}^{†}=I$, so $U$ is unitary and $U|\varphi ⟩=|\psi ⟩$. Construct $V$ on the second subsystem in the same way, sending $|{\varphi }^{\prime }⟩$ to $|{\psi }^{\prime }⟩$. ConsequentlyThus any two normalized product states on the same pair of subsystem spaces are local-unitary equivalent. The dimension and normalization requirements matter because a unitary preserves both its domain and vector norms.

A local unitary transformation preserves a product factorization:The output is therefore another product state. If an entangled state could be locally mapped to a product state, applying the inverse $U^{†}\otimes V^{†}$ would make the original entangled state a product as well, a contradiction. Hence a product state and an entangled state are never local-unitary equivalent.

In terms of a tensor-product coefficient matrix, $C$ changes to $UC{V}^T$, with an ordinary transpose on the second factor. Both multiplying matrices are invertible, so the matrix rank, equivalently the Schmidt rank, is unchanged. This supplies an algebraic form of the same argument: a nonzero pure state is a product state exactly when that rank is one.