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The rotation

\nabla v = (- u_{y}, u_{x})

from the Cauchy-Riemann equations is invertible. Hence

\nabla u = 0 ⟺ \nabla v = 0.

Thus their stationary points coincide, including degenerate ones.

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 7 / ii / Solution 2026-10-18

The Cauchy-Riemann equations give

\nabla v = (- u_{y}, u_{x})

. Therefore

\nabla u \cdot \nabla v = - u_{x} u_{y} + u_{y} u_{x} = 0.

The gradient is normal to a regular level set, so these curves intersect orthogonally wherever their gradients are nonzero. At a stationary point, where both gradients vanish, a regular curve tangent need not exist; the algebraic identity remains valid.

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 7 / iv / Solution 2026-10-18

For a harmonic function, put

a = u_{xx}

and

b = u_{x y} = u_{y x}

. Since

u_{yy} = - u_{xx}

, its Hessian matrix is

D^{2} u = (a b b - a), det D^{2} u = - a^{2} - b^{2} .

A nonzero determinant is therefore negative, and the two real Hessian eigenvalues are

\pm a^{2} + b^{2}

. At a stationary point the quadratic term of the Taylor expansion has both signs. Consequently every such nondegenerate stationary point is a saddle point. This is the nondegenerate stationary points of a planar harmonic function criterion. The same argument applies to

v

, whose stationary points coincide with those of

u

. A zero Hessian determinant is outside this second-order classification.

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 10 / c / Solution 2026-10-18

Differentiating the Gaussian variational bound for an attractive Gaussian well gives

E^{'} (a) = \frac{1}{2} - \frac{V _{0}}{2 a ( 1 + a ) ^{3/2}} .

Thus an interior stationary point obeys

a_{*} (1 + a_{*})^{3/2} = V_{0}, a_{*} (1 + a_{*})^{3} = V_{0}^{2} .

The left side of the first equation increases strictly from zero to infinity, so there is exactly one such point. More explicitly,

E^{''} (a) = \frac{V _{0} ( 1 + 4 a )}{4 a ^{3/2} ( 1 + a ) ^{5/2}} > 0,

E

is a strictly convex function on

a > 0

and

a_{*}

is its global minimum. At that minimum the stationarity equation gives

E (a_{*}) = \frac{a _{*}}{2} - a_{*} (1 + a_{*}) = - \frac{a _{*}}{2} - a_{*}^{2} < 0.

There is a continuous limiting zero at

a = 0

, although

y_{T} = 1

at exactly

a = 0

is not a square-integrable function and hence is not an admissible trial. For a positive zero, division by

a

and squaring gives

a (1 + a) = 4 V_{0}^{2}

, so

a_{z} = \frac{1 + 16 V _{0}^{2} - 1}{2}

(5)

is the unique positive zero. The other quadratic root is negative and inadmissible. Near zero,

E (a) = - V_{0} a + \frac{a}{2} + O (a^{3/2}),

(6)

so the graph leaves the origin with slope

- \infty

, falls to its unique negative minimum, rises through

a_{z}

, and grows as

a /2 - V_{0} + O (a^{- 1})

at infinity.

Figure 1.
The Gaussian variational bound for an attractive Gaussian well at $V_{0} = 1$ , showing its negative minimum and positive zero
. **Details**.

By the Rayleigh-Ritz variational principle,

E_{0} \leq E (a_{*}) = - \frac{a _{*}}{2} - a_{*}^{2} < 0.

(7)

Thus the attractive well supports a negative-energy bound state for every

V_{0} > 0

Past exam of the numerical linear algebra and continuous optimisation course of the University of Oxford / 2009 / B / 4 / c / Solution 2026-10-31

Choose

x_{0}

, fixed

\overset{α}{ˉ} > 0

σ, β \in (0, 1)

, and repeat for every

n \geq 0

: compute

g_{n} = \nabla f (x_{n})

, choose

α_{n}

by the backtracking line search in part (b) with

s_{n} = - g_{n}

, and set

x_{n + 1} = x_{n} - α_{n} g_{n}

. At a stationary point keep the sequence constant.

Assume

f \in C^{1}

has a globally Lipschitz gradient and is bounded below on its initial sublevel set

{x : f (x) \leq f (x_{0})}

. Then the steepest descent method satisfies

\nabla f (x_{n}) \to 0

. Indeed, the Armijo condition and the uniform lower step bound give

f (x_{n}) - f (x_{n + 1}) \geq σ α_{n} ∥ g_{n} ∥_{2}^{2} \geq σ c ∥ g_{n} ∥_{2}^{2} .

The objective values decrease and stay in the stated sublevel set. Summing the inequality and using its lower bound

f_{i n f}

yields

\sum_{n = 0}^{\infty} ∥ g_{n} ∥_{2}^{2} \leq \frac{f ( x _{0} ) - f _{i n f}}{σ c} < \infty,

which proves the limit. Every cluster point is a stationary point by continuity of the gradient. If the initial sublevel set is compact, cluster points exist. This global convergence result guarantees stationarity from arbitrary initial points under these assumptions; it does not, for a general nonconvex objective, guarantee convergence of the whole iterate sequence or a global minimum.

Past exam of the physics 1 paper of the University of Oxford / 2021 / 3 / Solution 2026-10-25

First take

a \neq = 0

, put

A = ∣ a ∣

, and suppose

k > 0

. The force is the negative derivative of the potential energy:

U (x) = - \frac{k x}{A ^{2} + x ^{2}}, F (x) = - U^{'} (x) = \frac{k ( A ^{2} - x ^{2} )}{( A ^{2} + x ^{2} ) ^{2}} .

The potential energy is an odd function, passes through

U (0) = 0

, and approaches zero from above as

x \to - \infty

and from below as

x \to + \infty

. Its stationary points are

x = \pm A

U (- A) = \frac{k}{2 A}, U (A) = - \frac{k}{2 A}, U^{''} (- A) = - \frac{k}{2 A ^{3}}, U^{''} (A) = \frac{k}{2 A ^{3}} .

Thus the sole stable equilibrium is the local minimum at

x = A

; the local maximum at

x = - A

is an unstable equilibrium. The requested values at the stable equilibrium are

x_{stable} = A, U_{stable} = - \frac{k}{2 A}, F_{stable} = 0.

The force is an even function, positive for

∣ x ∣ < A

, negative for

∣ x ∣ > A

, and zero at

x = \pm A

. Its central maximum is

F (0) = k / A^{2}

. Since

F^{'} (x) = \frac{2 k x ( x ^{2} - 3 A ^{2} )}{( A ^{2} + x ^{2} ) ^{3}},

its two minima occur at

x = \pm 3 A

, with

F = - k / (8 A^{2})

, and

F \to 0^{-}

at both ends. These values fix the shapes of both requested sketches.

Figure 1.
Dimensionless potential energy and force for positive k and nonzero a, with the stable and unstable equilibria marked
. **Details**.

In one dimension, attraction towards the origin means

x F (x) < 0

, while repulsion from the origin means

x F (x) > 0

. With that explicitly stated convention,

attraction: repulsion: - A < x < 0 or x > A, x < - A or 0 < x < A .

(5)

A positive force alone means rightward, not necessarily outward. At the origin

F (0) \neq = 0

, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls

a, k

merely constants. For

k < 0

both curves reverse sign, the stable equilibrium is

x = - A

, its potential energy is

- ∣ k ∣/ (2 A)

, and the attraction/repulsion intervals interchange. For

k = 0

the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If

a = 0

and

k \neq = 0

, then

U = - k / x

for

x \neq = 0

and

F = - k / x^{2}

, with no finite equilibrium.

Past exam of the physics 2 paper of the University of Oxford / 2015 / 11 / b / Solution 2026-10-25

Regard the refraction angle as a function of the angle of incidence,

β = arcsin (sin α / n)

. Implicit differentiation of Snell's law gives

β^{'} = cos α / (n cos β)

. Differentiating the deviation angle gives

δ^{'} = 2 - 4 β^{'}

. An interior stationary point therefore satisfies

cos α = \frac{n}{2} cos β, A = 2.

The stationary deviation of a primary rainbow ray is a minimum: differentiating

cos α = n cos β β^{'}

again and using Snell's law gives

β^{''} = tan β ((β^{'})^{2} - 1)

. At the stationary point,

β^{'} = 1/2

, and hence

δ^{''} = - 4 β^{''} = 3 tan β > 0

for a positive interior refraction angle. This minimum is responsible for the concentration of emerging light rays near the primary rainbow direction.

Past exam of the probability, statistics and options I course of the University of Oxford / 2007 / G / 1 / Solution 2026-10-31

Let

η

be a smooth variation with

η = η^{'} = 0

at both endpoints, and consider

y_{ϵ} = y + ϵη

. stationary point gives the first variation

0 = \frac{d}{d ϵ} J [y_{ϵ}]_{ϵ = 0} = \int_{x_{1}}^{x_{2}} (F_{y} η + F_{y^{'}} η^{'} + F_{y^{''}} η^{''}) d x .

Integrate the middle term once by parts and the last term twice. All boundary terms vanish because both

η

and

η^{'}

vanish there. Assuming the derivatives needed for the classical formula, the given fundamental lemma of the calculus of variations yields

F_{y} - \frac{d}{d x} F_{y^{'}} + \frac{d ^{2}}{d x ^{2}} F_{y^{''}} = 0.

Without assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.

For the specific integrand,

F_{y} = - 2 y

F_{y^{'}} = 0

F_{y^{''}} = 2 y^{''}

, so the equation is

y^{(4)} - y = 0

. Its weak solutions here are smooth, and its general classical solution is

y = A cos x + B sin x + C cosh x + D sinh x .

Put

ℓ = π /2

c = cosh ℓ

s = sinh ℓ

. The conditions at zero give

C = - A

and

D = 1 - B

. Those at

ℓ

give

B - A c + (1 - B) s = 0, - A - A s + (1 - B) c = 0.

Using

c^{2} - s^{2} = 1

, solve to obtain

A = c /2

B = (1 - s) /2

C = - c /2

D = (1 + s) /2

. Thus an extremal is

y (x) = \frac{1}{2} [c (cos x - cosh x) + (1 - s) sin x + (1 + s) sinh x] .

(5)

Substitution checks all four boundary values. The additive term

x^{2}

in the integrand contributes no variation in

y

, so it does not alter this equation.

Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / U / U2 / b / iii / Solution 2026-10-31

Here

f^{'} (x) = 4 x^{3} - 12 x^{2} + 4 x + 12

and

f^{''} (x) = 12 x^{2} - 24 x + 4

. This implementation uses the undamped Newton step, checks a stationary point's curvature, and reports failure on a two-cycle or exhausted iteration limit:

R
newton_min <- function(x0, tol = 1e-10, maxit = 100L) {
  fp <- function(x) 4*x^3 - 12*x^2 + 4*x + 12
  fpp <- function(x) 12*x^2 - 24*x + 4
  x <- x0
  trace <- x
  for (k in seq_len(maxit)) {
    g <- fp(x)
    h <- fpp(x)
    if (abs(g) <= tol) {
      return(list(x = x, converged = h > 0,
                  reason = if (h > 0) "minimum" else "not a strict minimum",
                  trace = trace))
    }
    if (!is.finite(h) || abs(h) <= .Machine$double.eps)
      stop("Newton step has zero or invalid curvature")
    next_x <- x - g/h
    if (!is.finite(next_x)) stop("Nonfinite Newton iterate")
    trace <- c(trace, next_x)
    n <- length(trace)
    if (n >= 3L && abs(trace[n] - trace[n - 2L]) <= tol*(1 + abs(next_x)))
      return(list(x = next_x, converged = FALSE,
                  reason = "two-cycle", trace = trace))
    x <- next_x
  }
  list(x = x, converged = FALSE, reason = "iteration limit", trace = trace)
}

A sufficiently close start near

- 0.7692923542

converges to the minimum. The derivative cubic has exactly one real root: after

x = 1 + t

it is proportional to

t^{3} - 2 t + 2

, whose local maximum and minimum are both positive, so it crosses zero only once, on the left. Since the quartic tends to

+ \infty

in both directions and its curvature at that stationary point is positive, this is the unique global minimum. The function reports convergence rather than assuming that every starting point must reach it.

Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / U / U2 / b / ii / Solution 2026-10-31

Let

e_{k} = x_{k} - x_{*}

, with

f^{'} (x_{*}) = 0

f^{''} (x_{*}) \neq = 0

. Taylor expansion gives

f^{'} (x_{*} + e) = f^{''} (x_{*}) e + \frac{1}{2} f^{(3)} (x_{*}) e^{2} + O (e^{3}), f^{''} (x_{*} + e) = f^{''} (x_{*}) + f^{(3)} (x_{*}) e + O (e^{2}) .

Division in the Newton update yields

e_{k + 1} = \frac{f ^{(3)} ( x _{*} )}{2 f ^{''} ( x _{*} )} e_{k}^{2} + O (e_{k}^{3}) .

Thus convergence is generically quadratic when that coefficient is nonzero. With a locally Lipschitz second derivative and nonzero

f^{''} (x_{*})

, the weaker estimate

∣ e_{k + 1} ∣ \leq C ∣ e_{k} ∣^{2}

also proves local quadratic convergence.

“Quadratic at best” is false as a universal upper bound. If the displayed coefficient vanishes, the order can be higher. For example

f (x) = x^{2} /2 + x^{4} /4

has a minimum at zero, and its Newton optimization update is

x_{k + 1} = 2 x_{k}^{3} / (1 + 3 x_{k}^{2})

, which has cubic convergence. For a nondegenerate quadratic objective the method reaches the stationary point in one step. The expansion proves the standard generic quadratic statement while retaining and correcting the source's stronger wording explicitly.

Past exam of the quantum mechanics quantum theory and quantum computers course of the University of Oxford / 2007 / B / 6 / Solution 2026-10-31

Use the normalized Rayleigh quotient

R (ψ) = ⟨ ψ, H ψ ⟩ / ⟨ ψ, ψ ⟩

. Its real directional derivative in an arbitrary complex direction

η

δ R = \frac{2 Re ⟨ η , ( H - R ) ψ ⟩}{⟨ ψ , ψ ⟩} .

It vanishes for every direction precisely when

(H - R) ψ = 0

, since testing

η = (H - R) ψ

gives its squared norm. Conversely an eigenvector makes every derivative zero. Stationarity over all states is stronger than stationarity in a restricted trial family.

Normalize the Gaussian as

ψ_{α} = (α / π)^{1/4} e^{- α x^{2} /2}

, with

α > 0

. Its moments are

⟨ X^{2} ⟩ = 1/ (2 α)

⟨ X^{4} ⟩ = 3/ (4 α^{2})

and

⟨ P^{2} ⟩ = ℏ^{2} α /2

. The printed potential energy consequently gives

E (α) = \frac{ℏ ^{2} α}{4 m} + \frac{m ω ^{2}}{4 α} + \frac{15 m ^{2} ω ^{3}}{256ℏ α ^{2}} .

Put

a = α ℏ/ (mω)

. Then

E / (ℏ ω) = a /4 + 1/ (4 a) + 15/ (256 a^{2})

, whose second derivative is

1/ (2 a^{3}) + 45/ (128 a^{4}) > 0

. It diverges at both positive endpoints, so its sole stationary point is the unique global minimum. Its equation is

a^{3} - a - \frac{15}{32} = 0, a_{*} ≃ 1.18179610.

The variational principle therefore yields the optimized upper bound

E_{0} \leq E (α_{*}) ≃ 0.548944745 ℏ ω, α_{*} ≃ 1.18179610 \frac{mω}{ℏ} .

The printed minimizer is inconsistent with the printed quartic coefficient. At

a = 3/2

the derivative is

5/48 > 0

, so this is not the stationary Gaussian for that potential energy. It is nevertheless a legitimate trial vector and gives the simple, weaker bound

E_{0} \leq 109ℏ ω /192

. If the quartic coefficient were instead

5 m^{2} ω^{3} / (16ℏ)

, four times the printed value, then

a = 3/2

would indeed minimize the family and its energy would be

31ℏ ω /48

. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is Gaussian trial for a quadratic-plus-quartic oscillator.

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S1 / 1 / 4 / Solution 2026-10-25

Along a level curve,

V_{x} d x + V_{y} d y = 0

, with

V_{x} = V_{0} θ / π

and

V_{y} = V_{0} (1 + lo g r) / π

. Therefore

\frac{d y}{d x} = - \frac{θ}{1 + lo g r} .

In the intended upper half-plane

0 < θ < π

, the slope is positive for

0 < r < e^{- 1}

, vertical at

r = e^{- 1}

, and negative for

r > e^{- 1}

. At the positive real boundary

θ = 0

the slope is zero except at the stationary point

(e^{- 1}, 0)

, where both derivatives vanish and the quotient is indeterminate. More generally, outside the chosen half-plane its sign is the sign of

- θ (1 + lo g r)

, with the argument branch held fixed. A convenient parametric description of the curves to sketch is

Vπ / V_{0} = r (sin θ lo g r + θ cos θ)

Figure 1.
Equipotentials for the upper-half-plane logarithm branch and illustrative square-well transmission
. **Details**.

Past exam of the soft condensed matter course of the University of Cambridge / 2022 / Section B / 4 / c / Solution 2026-10-18

Minimize the Helmholtz free energy at fixed

N, T

. Its derivative is

\frac{\partial F}{\partial ξ} = N k_{B} T (- \frac{2 b ^{2}}{ξ ^{3}} + \frac{ϵ b}{ξ ^{2}}) .

The physical stationary point for

ϵ > 0

ξ_{*} = \frac{2 b}{ϵ}, F_{ads, *} = - \frac{1}{4} N k_{B} T ϵ^{2} .

To check the minimum directly, put

x = b / ξ

. Then

F / (N k_{B} T) = (x - ϵ /2)^{2} - ϵ^{2} /4

, which is globally minimized at

x = ϵ /2

. The binding free energy gain per monomer is consequently

μ = k_{B} T ϵ^{2} /4 > 0

The weak-adsorption blob picture requires

b ≪ ξ_{*} ≪ b N

, or

2/ N ≪ ϵ ≪ 2

in this model. If

ϵ

is too large, the layer saturates at a monomer-scale thickness; if too small, a many-blob adsorbed layer has not formed.

Past exam of the soft condensed matter course of the University of Cambridge / 2022 / Section B / 4 / e / Solution 2026-10-18

During slow peeling in atomic force microscopy,

n

adjusts to minimize the Helmholtz free energy at each imposed height. Define the positive binding gain

μ = k_{B} T [ϵ b / ξ - (b / ξ)^{2}]

for the adsorbed segment. The stationary point satisfies

0 = (\frac{\partial F}{\partial n})_{z} = μ - \frac{3 k _{B} T z ^{2}}{2 b ^{2} n ^{2}}, n_{*} (z) = \frac{z}{b} \frac{3 k _{B} T}{2 μ} .

Since

\partial_{n}^{2} F = 3 k_{B} T z^{2} / (b^{2} n^{3}) > 0

, this is a minimum while

0 < n_{*} < N

. The upward pulling force is the height derivative of the minimized free energy. The implicit

n_{*} (z)

term vanishes because

\partial_{n} F = 0

, giving

f = \frac{d F ( n _{*} ( z ) , z )}{d z} = \frac{3 k _{B} T z}{b ^{2} n _{*}} = \frac{6 k _{B} T μ}{b} .

Using

ξ = 2 b / ϵ

f_{peel} = \frac{3}{2} \frac{ϵ k _{B} T}{b} .

It is independent of

z

: extra pulling height releases more monomers while leaving the fractional extension of the stretched part unchanged. Equivalently

F_{m i n} = - N μ + f_{peel} z

. This is the equilibrium peeling force of an adsorbed Gaussian chain.

The plateau ends when

n_{*} = N

, at

z_{det} = N b \frac{2 μ}{3 k _{B} T} = \frac{N b ϵ}{6} .

Thereafter all monomers are detached and the Gaussian chain force is

f = 3 k_{B} T z / (N b^{2})

, agreeing continuously with the plateau at

z_{det}

. Very near the start and final release, finite-chain and junction effects lie outside the blob approximation.

Past exam of the theoretical mechanics course of the University of Oxford / 2011 / B / 5 / a / i / Solution 2026-10-31

Take the nondegenerate stationary point in the interior of the integration interval. Choose a smooth cutoff equal to one near

k_{⋆}

. Away from its support,

ψ^{'}

is bounded away from zero. Integration by parts using

e^{i t ψ} = (i t ψ^{'})^{- 1} \partial_{k} e^{i t ψ}

makes that nonstationary contribution

O (t^{- 1})

, under the usual smooth-phase assumptions.

Near the stationary point use the signed coordinate

s = sgn (k - k_{⋆}) 2 (ψ (k) - ψ (k_{⋆})) .

Since

ψ^{''} (k_{⋆}) > 0

, this is locally invertible, with

d s / d k = ψ^{''} (k_{⋆})

at the point. The local integral becomes

e^{i t ψ (k_{⋆})} \int a (s) e^{i t s^{2} /2} d s, a (0) = \frac{f ( k _{⋆} )}{ψ ^{''} ( k _{⋆} )} .

Scale

s = 2/ t r

. The supplied Fresnel integral gives

\int_{- \infty}^{\infty} e^{i r^{2}} d r = e^{iπ /4} π

. Approximating the smooth amplitude by its value at zero, with a cutoff and a nonstationary estimate for the tails, therefore gives

I (t) = f (k_{⋆}) e^{i t ψ (k_{⋆}) + iπ /4} \frac{2 π}{t ψ ^{''} ( k _{⋆} )} + o (t^{- 1/2}) .

f (k_{⋆}) \neq = 0

, this is the requested asymptotic equivalence. The phase shift

π /4

comes from the positive quadratic curvature. A stationary point at an integration endpoint instead needs a one-sided formula.

The printed auxiliary estimate asserting an

O (γ^{- 1})

Fourier integral for every continuous amplitude is too strong without extra regularity. The argument here uses integration by parts only for the actual smooth, cutoff nonstationary amplitudes, where that estimate is justified; continuity alone generally guarantees convergence to zero, not this rate.

Past exam of the theoretical physics I course of the University of Cambridge / 2013 / Paper 1 / 6 / d / Solution 2026-10-20

Assume

b > 0

and temperature-independent

d

for the local model. The zero-field stationary points obey

m (a + d m + b m^{2}) = 0, m_{\pm} = \frac{- d \pm d ^{2} - 4 ab}{2 b} .

For

d \neq = 0

, the favoured ordered sign is opposite to

d

. The nonzero extrema first appear at

a = d^{2} / (4 b)

; their existence is not yet a coexistence condition. To find the equilibrium transition, require both

F^{'} (m_{*}) = 0

and

F (m_{*}) = F (0)

. Substituting

a = - d m_{*} - b m_{*}^{2}

into the free energy gives

F (m_{*}) - F_{0} = - \frac{d}{6} m_{*}^{3} - \frac{b}{4} m_{*}^{4} .

Hence

m_{*} = - \frac{2 d}{3 b}, a_{*} = \frac{2 d ^{2}}{9 b} > 0.

At coexistence the entire potential factorizes:

F - F_{0} = \frac{b}{4} m^{2} (m + \frac{2 d}{3 b})^{2} .

It has equal minima at zero and

m_{*}

separated by a barrier. Every fixed nonzero cubic coefficient produces a first-order transition in this quartic model, with a finite magnetization jump before the quadratic coefficient reaches zero.

For

d^{2} / (4 b) > a > a_{*}

, the ordered minimum is metastable and zero is globally stable. For

a_{*} > a > 0

, the ordered state is globally stable and zero is metastable. At

a = 0

the zero minimum loses local stability. Thus the spinodals are

a = d^{2} / (4 b)

and

a = 0

, distinct from

a_{*}

. For

a < 0

, the opposite-sign nonzero state can also be a local minimum, but the cubic term favours the state of sign

- d

Figure 1.
First-order coexistence, metastability and the magnetization jump in a cubic-quartic Landau model
. **Details**.

a = a_{0} (T - T_{c}) / T_{c}

, the transition is shifted to

T_{*} = T_{c} [1 + 2 d^{2} / (9 b a_{0})]

. With

b, d

constant and

a^{'} (T) > 0

, the entropy jump from the disordered to ordered phase is

- a^{'} (T_{*}) m_{*}^{2} /2

, giving a latent heat of magnitude

T_{*} a^{'} (T_{*}) m_{*}^{2} /2

. As

d \to 0

, the jump and shift vanish and the original continuous transition is recovered.

A genuine zero-field uniaxial ferromagnet with exact spin-reversal Z2 symmetry forbids this cubic term. The modified model therefore presumes that symmetry has been changed; it must not be presented as a symmetry-allowed correction for the original ferromagnet.

Past exam of the theoretical physics I course of the University of Cambridge / 2016 / Paper 1 / 4 / e / Solution 2026-10-20

Choose a basis whose first axis is

n

. The tensor with uniaxial nematic order then has eigenvalues

2 Q, - Q, - Q

. Hence

Tr S^{2} = 6 Q^{2}

Tr S^{3} = 6 Q^{3}

and

Tr S^{4} = 18 Q^{4}

. Substitution gives

f (Q) = 6 a Q^{2} + 6 b Q^{3} + 18 c Q^{4} .

For

a, c > 0

Q = 0

remains a local minimum with curvature

12 a

. Its nonzero stationary points satisfy

2 a + 3 b Q + 12 c Q^{2} = 0

, so

Q_{\pm} = \frac{- 3 b \pm 9 b ^{2} - 96 a c}{24 c} .

They first appear at

b_{s} = - 32 a c /3

as a stationary inflection point; for more negative

b

Q_{-}

is a local maximum and

Q_{+}

a local minimum. The latter is initially metastable, becomes degenerate with

Q = 0

b_{t} = - 12 a c

, and then becomes the global minimum as

b

decreases further. This is a first-order transition in a cubic-quartic Landau potential: the order changes discontinuously. The saddle-node bifurcation creating a local minimum is not the coexistence condition. The isotropic minimum never loses local stability in this fixed positive-

a

scan.

Figure 1.
Cubic nematic invariant creates a metastable minimum followed by first-order coexistence
. **Details**.

The illustrative plots set

a = c = 1

and show

f /6

. A cubic invariant is permitted because head-tail reversal leaves

S

unchanged; it does not implement

S \mapsto - S

. The physical range of a uniaxial nematic order parameter imposes

- 1/2 \leq Q \leq 1

; the usual small-order-parameter Landau result assumes the displayed minimum lies in that interval. The unrestricted quartic graphs express the intended Landau model, with the effect of enforcing the bound stated in part (f).^[ref]

Past exam of the theoretical physics I course of the University of Cambridge / 2022 / 3 / c / Solution 2026-10-18

The self-consistency equation requires that the mean spin computed using the single-site Boltzmann distribution equals the spin

S

used to form the effective field. With

h = 4 J S + B

⟨ s ⟩_{h} = \frac{\sum _{s = - 1, 0, 1} s e ^{β h s}}{\sum _{s = - 1, 0, 1} e ^{β h s}} = \frac{e ^{β h} - e ^{- β h}}{1 + e ^{β h} + e ^{- β h}} .

Thus

S = \frac{2 sinh [ β ( 4 J S + B )]}{1 + 2 cosh [ β ( 4 J S + B )]} .

Equivalently, differentiating part (b) gives

N^{- 1} \partial F_{MF} / \partial S = 4 J [S - ⟨ s ⟩_{h}]

, so stationarity of the Helmholtz free energy in the mean-field approximation is exactly this condition. Among multiple stationary points one must select the stable minimum.

Past exam of the theoretical physics major option course of the University of Oxford / 2018 / 2 / b / Solution 2026-10-25

Both the canonical kinetic term and

V (ϕ^{T} ϕ)

are invariant under the orthogonal group

O (n)

. The stated nonzero vacuum expectation value has length

∣ v ∣

. Its stabilizer is the orthogonal group

O (n - 1)

acting on the perpendicular subspace. Hence

N_{GB} = \frac{n ( n - 1 )}{2} - \frac{( n - 1 ) ( n - 2 )}{2} = n - 1.