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

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.