Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 / 1 / c

Solution

At fixed uniform

ϕ

, the nonconserved variable

p

can relax freely, so mean-field approximation minimizes

f (ϕ, p) = \frac{a}{2} ϕ^{2} + \frac{b}{4} ϕ^{4} + \frac{A + c ϕ}{2} p^{2} + \frac{B}{4} p^{4}

with respect to

p

. The stationary solutions obey

p (A + c ϕ + B p^{2}) = 0.

Thus

p = 0

when

A + c ϕ \geq 0

, while

p^{2} = - (A + c ϕ) / B

when

A + c ϕ < 0

. Substitution gives

f (ϕ) = \frac{a}{2} ϕ^{2} + \frac{b}{4} ϕ^{4} - \frac{( A + c ϕ ) ^{2}}{4 B} h e t a (- A - c ϕ) .

Writing this in the requested form yields

ϕ_{o} = - \frac{A}{c}, C = \frac{c ^{2}}{2 B},

because

A + c ϕ = c (ϕ - ϕ_{o})

. The value

ϕ_{o}

is the composition at which the coefficient of

p^{2}

changes sign, so it is the mean-field threshold for polar ordering.

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