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

Write the product state as $|\Psi ⟩=|a⟩\otimes |b⟩$. The two partial inner products areBoth vectors therefore lie in the same one-dimensional vector subspace, possibly with one or both equal to zero. If $(c_1,c_2)\ne (0,0)$, choose ${\lambda }_1=c_2$ and ${\lambda }_2=c_1$; then ${\lambda }_1|{\psi }_1^{\prime }⟩={\lambda }_2|{\psi }_2^{\prime }⟩$, and the multipliers are not both zero. If both $c_j$ vanish, choose $({\lambda }_1,{\lambda }_2)=(1,0)$.

Thus the conditional vectors are linearly dependent, with a nontrivial pair of multipliers in every case. The printed existential condition does not explicitly exclude $(0,0)$, which would make it vacuous for arbitrary vectors; the argument establishes the stronger intended statement. These unnormalized vectors need not themselves be valid unit state vectors, and a zero vector corresponds to a measurement outcome of probability zero. This is the forward direction of the conditional-vector product-state criterion.

Factor the coefficients:Therefore the first state is a product state, not entangled. Both factors have unit norm, so this factorization also checks normalization.

Taking partial inner products with $⟨0|$ and $⟨1|$ on the first wire gives $(|0⟩+|1⟩)/2$ and $(-|0⟩+|1⟩)/2$. They are nonzero and orthogonal, hence linearly independent. This contradicts the necessary product-state property proved in part (a), so the second state is entangled.

Equivalently, its tensor-product coefficient matrix is $C=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$, with $\det C=1/2\ne 0$. The two-qubit product-state determinant criterion gives the same conclusion. In fact $C{C}^{†}=I/2$, so its Schmidt coefficients are both $1/\sqrt{2}$: it is maximally entangled.

The corresponding conditional vectors are $|0⟩/\sqrt{2}$ and $|+⟩/\sqrt{2}$. The second has a nonzero $|1⟩$ component while the first does not, so they are linearly independent. Therefore the third state is also entangled. Orthogonality of these two conditional vectors is not required for this conclusion.

For an independent matrix check, expanding the second factor givesThe original expression is normalized because its two first-wire factors are orthogonal: its squared norm is $(1+1)/2=1$, despite the overlap between $|0⟩$ and $|+⟩$ on the second wire.

Measure the first qutrit in the computational basis and the second in the orthonormal basis $\{|{\varphi }_0⟩,|{\varphi }_1⟩,|{\varphi }_2⟩\}$ from Question 1(a), with those labels as outcomes. For first-party outcome 0, the unnormalized conditional vector of the second party is $|0⟩/\sqrt{3}$, so this outcome occurs with probability $1/3$ and leaves the normalized state $|0⟩$.

The second-party Born rule probabilities are then $|⟨{\varphi }_0|0⟩{|}^2=1/2$, $|⟨{\varphi }_1|0⟩{|}^2=0$, and $|⟨{\varphi }_2|0⟩{|}^2=1/2$. HenceThe relevant joint probabilities are $1/6,0,1/6$, each divided by $\mathrm{Pr}(A=0)=1/3$. This distinguishes the conditional statement from an unconditional outcome probability. More generally, conjugate-state steering from a maximally entangled pair explains how the first local measurement determines the second party's conditional vector; the basis choice here makes that vector especially simple.

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.