Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 10C (Quantum Information and Computation)

Table of contents
- i 10C
  - Solution i
- ii 10C
  - Solution ii
- iii 10C
  - Solution iii
- iv 10C
  - Solution iv
- v 10C
  - Solution v

i

Solution

Using the Pauli matrices,

A = n_{x} σ_{x} + n_{y} σ_{y} + n_{z} σ_{z} = (n_{z} n_{x} + i n_{y} n_{x} - i n_{y} - n_{z}) .

ii

Solution

In a spherical coordinate system,

n = (sin θ cos ϕ, sin θ sin ϕ, cos θ) .

Since

n_{x} \mp i n_{y} = sin θ e^{\mp i ϕ}

,

A = (cos θ e^{i ϕ} sin θ e^{- i ϕ} sin θ - cos θ) .

iii

Solution

The Pauli multiplication relations and

∣ n ∣ = 1

give

A^{2} = (n \cdot σ)^{2} = I .

Therefore the possible measurement outcomes, the eigenvalues of

A

, are

+ 1 and - 1 .

iv

Solution

The spectral projector for outcome

+ 1

is

Π_{+} = \frac{I + A}{2} .

For

∣1 ⟩ = (0, 1)^{T}

, the Born rule gives

P (+ 1) = ⟨ 1∣ Π_{+} ∣1 ⟩ = \frac{1 - n _{z}}{2} = sin^{2} \frac{θ}{2} .

v

Solution

The Born rule says that the probability of a projective-measurement outcome is the squared norm of the state projected onto the corresponding eigenspace. The basis vectors in the state whose second qubit is one are

∣010 ⟩, ∣110 ⟩, ∣111 ⟩ .

They are mutually orthogonal, so

P (second qubit = 1) = ∣ b ∣^{2} + ∣ c ∣^{2} + ∣ d ∣^{2} .

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