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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-51/3/d/solution

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Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 51 / 3 / d / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Absorb

β

into the nonconstant couplings and write the corresponding dimensionless action as

H_{u}

. Choose its additive-constant convention so that one blocking step has

\sum_{σ} K (σ^{'} ∣ σ) e^{- H_{u} (σ)} = e^{- N g (u)} e^{- H_{R u} (σ^{'})} .

g (u)

is the identity-operator contribution to renormalization-group free energy: it records the extensive constant generated by eliminating short-distance degrees of freedom. Summing over

σ^{'}

gives

Z (u, 0, N) = e^{- N g (u)} Z (R u, 0, N^{'}), N^{'} = b^{- D} N .

For the dimensionless free-energy density

Φ (u) = - N^{- 1} ln Z (u, 0, N)

, this is the inhomogeneous recursion

Φ (u) = g (u) + b^{- D} Φ (R u) .

For the explicit constant

C

in the original partition function, the corresponding new constant obeys

β C^{'} = b^{D} [βC + g (u)]

;

C

and

g

are not additional interactions between spins.

Iterating the recursion

n

times yields

Φ (u) = \sum_{j = 0}^{n - 1} b^{- j D} g (R^{j} u) + b^{- n D} Φ (R^{n} u) .

An analytic background

a (u)

satisfying

a (u) = g (u) + b^{- D} a (R u)

absorbs the regular contribution. The remaining singular part

f_{s} = Φ - a

obeys

f_{s} (u) = b^{- D} f_{s} (R u) .

This separation explains both the scaling of the singular part and the role of the inhomogeneous term; simply dropping the generated constant would incorrectly change the full free energy.

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