 Solution

ID: past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2015/7/b/solution

H = [\dot{ϕ}^{2} + (\partial_{x} ϕ)^{2} + m^{2} ϕ^{2}] /2

. In the vacuum only

⟨ a_{n} a_{m}^{†} ⟩ = δ_{nm}

contributes. Hence

⟨ \dot{ϕ}^{2} ⟩ = \sum_{n} \frac{ω _{n}}{L} sin^{2} (k_{n} x), ⟨(\partial_{x} ϕ)^{2} ⟩ = \sum_{n} \frac{k _{n}^{2}}{ω _{n} L} cos^{2} (k_{n} x), ⟨ ϕ^{2} ⟩ = \sum_{n} \frac{s i n ^{2} ( k _{n} x )}{ω _{n} L} .

Substitution, using

ω_{n}^{2} = k_{n}^{2} + m^{2}

, yields

⟨ H (x)⟩ = n = 1 \sum \infty [\frac{ω _{n}}{2 L} - \frac{m ^{2}}{2 ω _{n} L} cos (2 k_{n} x)] .

This Dirichlet scalar vacuum energy density illustrates why a massive field's local unregularized energy density need not be uniform. The cosine integrates to zero over the interval, so

E_{0}^{BC} (L) = \int_{0}^{L} d x ⟨ H (x)⟩ = \frac{1}{2} n = 1 \sum \infty ω_{n} .

These divergent expressions are understood with a common regulator before exchanging sums, integrals or derivatives. Normal ordering the Hamiltonian operator separately in each geometry would erase the vacuum terms one is trying to compare; a common regulated subtraction is used instead in c and d.

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