Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 35B (Principles of Quantum Mechanics)

Table of contents
- a 35B
  - Solution a
- b 35B
  - Solution b
- c 35B
  - Solution c

a

Solution

A density operator satisfies

ρ = ρ^{†}, ρ ⪰ 0, tr ρ = 1.

An

N \times N

Hermitian matrix has

N^{2}

real parameters, and the trace condition removes one, leaving

N^{2} - 1

real free parameters.

If

A_{H} (t) = U (t)^{†} A_{S} U (t)

, equality of expectation values requires

tr (ρ_{H} U^{†} A_{S} U) = tr (U ρ_{H} U^{†} A_{S}) .

Therefore

ρ_{S} (t) = U (t) ρ_{H} U (t)^{†} .

b

Solution

The spin commutators are

[S_{i}, S_{j}] = i ℏ ε_{ijk} S_{k}, [S^{2}, S_{i}] = 0.

r_{i} = \frac{2 ⟨ S _{i} ⟩ _{ψ}}{ℏ} .

Purity gives

r_{x}^{2} + r_{y}^{2} + r_{z}^{2} = 1

, so the supplied sign determines

r_{y} = sgn ⟨ S_{y} ⟩_{ψ} 1 - r_{x}^{2} - r_{z}^{2} .

For

r_{z} \neq = - 1

, choose the overall phase so that the upper component is nonnegative. The normalized state is

∣ ψ ⟩ = \frac{1 + r _{z}}{2} ∣ + ⟩_{z} + \frac{r _{x} + i r _{y}}{2 ( 1 + r _{z} )} ∣ - ⟩_{z} .

If

r_{z} = - 1

, it is simply

∣ - ⟩_{z}

up to phase.

c

Solution

For a mixed spin-half state, the Bloch vector formula gives

ρ = \frac{1}{2} (I + \frac{2}{ℏ} ⟨ S_{x} ⟩ σ_{x} + \frac{2}{ℏ} ⟨ S_{y} ⟩ σ_{y} + \frac{2}{ℏ} ⟨ S_{z} ⟩ σ_{z}) .

Physical consistency is equivalent to the resulting Bloch vector having length at most one.

Three linearly independent Hermitian operators do not always suffice. The operators

I, σ_{x}, σ_{z}

are linearly independent, but the first expectation is always one and none detects the

y

component. States with Bloch vectors

(0, c, 0)

and

(0, - c, 0)

have identical expectations for all three. The correct criterion for informationally complete three-observable qubit tomography is that the three traceless parts span the Pauli space.

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