Past exam of the computer science course of the University of Oxford / 2023 / Quantum information / 3 / c

Table of contents
- i c
  - Solution i
- ii c
  - Solution ii

i

Solution

The sign factor separates as

(- 1)^{x + y} = (- 1)^{x} (- 1)^{y}

. Consequently the quantum state factors into a tensor product of quantum systems:

\frac{1}{2} \sum_{x, y = 0}^{1} (- 1)^{x + y} ∣ x ⟩ ∣ y ⟩ = (\frac{∣0 ⟩ - ∣1 ⟩}{2}) \otimes (\frac{∣0 ⟩ - ∣1 ⟩}{2}) = ∣ - ⟩ \otimes ∣ - ⟩ .

It is a product state, so it is not entangled.

ii

Solution

Grouping by the first qubit gives

\frac{1}{2} (∣00 ⟩ + ∣01 ⟩ + ∣10 ⟩ - ∣11 ⟩) = \frac{∣0 ⟩ ∣ + ⟩ + ∣1 ⟩ ∣ - ⟩}{2} .

The vectors on both sides form orthonormal bases, so this is a Schmidt decomposition with two nonzero Schmidt coefficients

1/ 2

. Hence the state is maximally entangled. Independently, its coefficient matrix has determinant

det [\frac{1}{2} (11 1 - 1)] = - \frac{1}{2} \neq = 0,

which excludes a product state by the two-qubit product-state determinant criterion. Each reduced density matrix is

I /2

.

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