 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-67/1/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

Let

R = X (l)

and

C

be its complement, with Hilbert-space dimension

d_{C}

A_{R} (t) = \frac{Tr _{C} A _{X} ( t )}{d _{C}} \otimes I_{C} = \int d U_{C} (I_{R} \otimes U_{C}) A_{X} (t) (I_{R} \otimes U_{C}^{†}) .

This is an operator supported on

R

. The normalization

1/ d_{C}

is essential: the partial trace alone would not fix an operator already supported on

R

Subtract the integrand from

A_{X} (t)

A - U A U^{†} = [A, U] U^{†}

and every unitary operator has norm one, the assumed Lieb-Robinson bound, applied to a unitary supported on the entire complement, gives

∥ A_{X} (t) - A_{R} (t) ∥ \leq 2∣ X ∣ ∥ A_{X} ∥ e^{- μ (v t + l)} (e^{2 k s t} - 1) .

The factor

min (∣ X ∣, ∣ C ∣)

was bounded by

∣ X ∣

; there is no sum over sites and hence no volume-dependent prefactor.

Choose

v = \frac{4 k s}{μ} .

For

t, l \geq 0

e^{2 k s t} - 1 \leq 2 k s t e^{2 k s t}

, and therefore

∥ A_{X} (t) - A_{R} (t) ∥ \leq μv t ∣ X ∣ ∥ A_{X} ∥ e^{- 2 k s t - μ l} \leq μv t ∣ X ∣ ∥ A_{X} ∥ e^{- μ l /2} .

Thus

A_{X (l)} (t) = A_{R} (t)

is the desired Lieb-Robinson localization by Haar twirling. The displayed choice assumes the usual positive constants

μ, s

; if the interaction bound has

s = 0

, the error is identically zero and any positive

v

works. For negative times the same argument uses

∣ t ∣

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