 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2014/paper-1/4/d/solution

Write

T_{p} = p^{2} / (2 m)

and

V = m ω^{2} x^{2} /2

. Dropping commutators of order

β^{2}

permits

e^{- β H} ≃ e^{- β V} e^{- β T_{p}}

. With

⟨ x ∣ p ⟩ = e^{i p x /ℏ} / 2 π ℏ

, the momentum Gaussian gives

⟨ x ∣ e^{- β H} ∣ x^{'} ⟩ ≃ \frac{m}{2 π β ℏ ^{2}} exp [- \frac{m ( x - x ^{'} ) ^{2}}{2 β ℏ ^{2}} - \frac{β m ω ^{2} x ^{2}}{2}] .

Integrating the diagonal yields

Z ≃ 1/ (β ℏ ω)

⟨ x ∣ ρ ∣ x^{'} ⟩ ≃ ω \frac{β m}{2 π} exp [- \frac{m ( x - x ^{'} ) ^{2}}{2 β ℏ ^{2}} - \frac{β m ω ^{2} x ^{2}}{2}] .

This unsymmetrized product has an apparent off-diagonal asymmetry at the omitted order; reversing the product puts

x^{'}

in the potential factor. A manifestly Hermitian approximation uses

e^{- β V /2} e^{- β T_{p}} e^{- β V /2}

, replacing

β m ω^{2} x^{2} /2

β m ω^{2} (x^{2} + x^{'2}) /4

with the same leading normalization. The condition is

β ℏ ω ≪ 1

; the Gaussian off-diagonal width is of order

ℏ β / m

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