Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford

labeled bonds, choose either orientation of one Ising spin, then alternate across every other bond. This constructs exactly two spin configurations per position. Thus

E_{0} = - J (N - 2), g_{0} = 2 N (N odd) .

(3)

This is geometric frustration: every local interaction prefers opposite Ising spins, but the closed odd ring prevents all preferences being satisfied at once. No particular location is preferred for the single frustrated bond in an antiferromagnetic Ising ring.

Solution

 0  0

Take the thermodynamic limit

N \to \infty

at fixed separation

n

and fixed finite

K = β J > 0

. Since

∣ t ∣ = tanh K < 1

, the terms

t^{N - n}

and

t^{N}

in the finite-ring antiferromagnetic Ising correlation vanish, through either even or odd

N

. Therefore

C_{n} ⟶ t^{n} = (- 1)^{n} (tanh K)^{n} = (- 1)^{n} exp [n lo g (tanh K)] .

(1)

Matching the decaying exponential function gives the correlation length, in units of the site spacing:

⟨ S_{1} S_{n + 1} ⟩ = (- 1)^{n} e^{- n / ξ}, ξ = - \frac{1}{lo g tanh ( β J )} = \frac{1}{lo g coth ( β J )} .

(2)

If the physical spacing is

a

, the physical correlation length is

a ξ

. The alternating sign records the preferred opposite orientation of neighboring Ising spins; the finite correlation length shows that this preference does not maintain long-range Néel order at positive temperature.

For large

K

tanh K = 1 - 2 e^{- 2 K} + O (e^{- 4 K})

, so

ξ \sim e^{2 K} /2

: the correlation length becomes large on cooling, but diverges only at zero temperature. At infinite temperature,

K \to 0^{+}

and

ξ \to 0

; the separate equal-site two-point correlation function remains

C_{0} = 1

d

 0  0

Solution

 0  0

Keep

N

fixed and put

q = tanh K

, so

q \to 1^{-}

as temperature tends to zero. For even

N

, the finite-ring antiferromagnetic Ising correlation becomes

C_{n} = (- 1)^{n} \frac{q ^{n} + q ^{N - n}}{1 + q ^{N}} ⟶ (- 1)^{n} (N even) .

(1)

Both alternating ground states have

S_{1} S_{n + 1} = (- 1)^{n}

, so their equal-weight canonical ensemble has the same two-point correlation function despite its zero one-point correlation function.

For odd

N

, the signs instead give

C_{n} = (- 1)^{n} \frac{q ^{n} - q ^{N - n}}{1 - q ^{N}} .

(2)

Direct substitution gives an indeterminate ratio. Expanding at fixed

N, n

with

q = 1 - δ

q^{n} - q^{N - n} = (N - 2 n) δ + O (δ^{2}), 1 - q^{N} = N δ + O (δ^{2}) .

(3)

Consequently the zero-temperature correlation of an odd antiferromagnetic Ising ring is

β J \to \infty lim ⟨ S_{1} S_{n + 1} ⟩ = ⎩ ⎨ ⎧ (- 1)^{n}, (- 1)^{n} (1 - \frac{2 n}{N}), N even, N odd .

(4)

The odd-ring factor has a direct probability interpretation. At zero temperature, the canonical ensemble gives equal weight to its

2 N

ground states. The single frustrated bond in an antiferromagnetic Ising ring is uniformly located, with probability

n / N

of lying on the chosen

n

-bond path. If it is absent from that path,

S_{1} S_{n + 1} = (- 1)^{n}

; if it is present, this product is

- (- 1)^{n}

. Averaging gives

(- 1)^{n} [(1 - n / N) - n / N]

Thus the decay with separation here comes from averaging a movable, compulsory frustrated bond in an antiferromagnetic Ising ring, even after additional thermally excited frustrated bonds in an antiferromagnetic Ising ring disappear. In particular

C_{1} = - 1 + 2/ N

for odd

N

, and

J N C_{1} = - J (N - 2)

reproduces the ground-state energy. The formula also respects

C_{n} = C_{N - n}

, although its factor

1 - 2 n / N

changes sign halfway around the ring.

Figure 1.
Ring parity, unavoidable frustration, and the zero-temperature correlation
. Original Ising spin configurations show satisfied bonds in gray and the unavoidable odd-ring parallel bond in red. The plots show the exact finite-ring antiferromagnetic Ising correlation, with its alternating sign removed, and the zero-temperature limits for even and odd rings. **Details**.

The order and scale of limits matter. Taking the thermodynamic limit first at positive temperature gives the exponential function in part (c). Cooling an odd finite ring first leaves the linear factor

1 - 2 n / N

. Both orders subsequently give

(- 1)^{n}

at fixed

n

, but this agreement is not uniform in separation: when

n / N

stays nonzero, the odd-ring linear factor survives in the zero-temperature expression.

2

 0  0

a

 0  0

Solution

 0  0

Use the standard canonical anticommutation relations for the two distinct fermionic orbitals. In particular

{a, b^{†}} = {a^{†}, b} = 0

, in addition to the relations explicitly listed in the paper; these mixed relations are part of the usual two-orbital fermionic Fock space interpretation. Normalize the original vacuum state by

⟨ 0∣0 ⟩ = 1

and fix the ordering

∣ n_{a} n_{b} ⟩ = (a^{†})^{n_{a}} (b^{†})^{n_{b}} ∣0 ⟩ .

(1)

The canonical anticommutation relations give

a^{2} = (a^{†})^{2} = 0

and

a a^{†} = 1 - a^{†} a

. Consequently the number operator is an orthogonal projector:

N_{a}^{2} = a^{†} a a^{†} a = a^{†} (1 - a^{†} a) a = N_{a} .

(2)

The same holds for

N_{b}

. Hence each eigenvalue obeys

n^{2} = n

and is either

0

1

, expressing the Pauli exclusion principle. Also

[N_{a}, N_{b}] = 0

, so the two number operators have a simultaneous orthonormal basis of Fock states:

state ∣00 ⟩ = ∣0 ⟩ ∣10 ⟩ = a^{†} ∣0 ⟩ ∣01 ⟩ = b^{†} ∣0 ⟩ ∣11 ⟩ = a^{†} b^{†} ∣0 ⟩ N_{a} eigenvalue 0101 N_{b} eigenvalue 0011

(3)

Each number operator has eigenvalues $0, 1$ , each with state degeneracy two. For

N_{a}

, the zero eigenspace is spanned by

∣00 ⟩, ∣01 ⟩

and the unit eigenspace by

∣10 ⟩, ∣11 ⟩

; for

N_{b}

, these are respectively

∣00 ⟩, ∣10 ⟩

and

∣01 ⟩, ∣11 ⟩

. Any normalized linear combination within one such eigenspace is also an eigenstate of that number operator.

The ordering convention will matter below:

b^{†} a^{†} ∣0 ⟩ = - ∣11 ⟩

and

ab ∣11 ⟩ = ab a^{†} b^{†} ∣0 ⟩ = - a a^{†} b b^{†} ∣0 ⟩ = - ∣0 ⟩ .

(4)

b

 0  0

Solution

 0  0

Writing the Bogoliubov transformation componentwise, together with its Hermitian adjoint, gives

a = u c + v d^{†}, b = - v c^{†} + u d, a^{†} = u c^{†} + v d, b^{†} = - v c + u d^{†} .

(1)

The full canonical anticommutation relations for

c, d

imply

{a, a^{†}} = u^{2} {c, c^{†}} + v^{2} {d^{†}, d} = u^{2} + v^{2}, {b, b^{†}} = u^{2} + v^{2} .

(2)

Equal-type anticommutators vanish because

c^{2} = d^{2} = 0

and

{c, d^{†}} = {c^{†}, d} = 0

{a, a} = {b, b} = 0.

(3)

For the mixed anticommutators, the only potentially nonzero terms cancel:

{a, b} = - uv {c, c^{†}} + uv {d^{†}, d} = - uv + uv = 0,

(4)

while

{a, b^{†}} = 0

follows from

{c, c} = {d^{†}, d^{†}} = {c, d^{†}} = 0

. Taking the Hermitian adjoint supplies the remaining relations. Thus the necessary and sufficient condition is

u^{2} + v^{2} = 1.

(5)

Equivalently, the matrix

R = (u - v v u)

is an orthogonal matrix,

R^{T} R = I

, giving a canonical rotation of two fermionic modes. Its inverse is therefore

c = u a - v b^{†}, d = v a^{†} + u b .

(6)

c

 0  0

Solution

 0  0

Let

N_{c} = c^{†} c

N_{d} = d^{†} d

and

P = c d + d^{†} c^{†}

. Substitution of the Bogoliubov transformation, with normal ordering using

d d^{†} = 1 - d^{†} d

and

c c^{†} = 1 - c^{†} c

, gives

a^{†} a + b^{†} b ab + b^{†} a^{†} = (u^{2} - v^{2}) (N_{c} + N_{d}) + 2 v^{2} - 2 uv P, = 2 uv (N_{c} + N_{d}) - 2 uv + (u^{2} - v^{2}) P .

(1)

For example, the

v^{2} d d^{†}

term in

a^{†} a

contributes

v^{2} - v^{2} N_{d}

; these constants must be retained to obtain the ground-state energy. The two-orbital fermionic pairing Hamiltonian is consequently

H = [ε (u^{2} - v^{2}) + 2Δ uv] (N_{c} + N_{d}) + [Δ (u^{2} - v^{2}) - 2 ε uv] P + 2 ε v^{2} - 2Δ uv .

(2)

To eliminate the pairing term, choose a canonical rotation of two fermionic modes satisfying

2 ε uv = Δ (u^{2} - v^{2}) .

(3)

With

u = cos θ

v = sin θ

, take

0 < θ < π /4

, so

tan 2 θ = Δ/ ε

. Setting

E = ε^{2} + Δ^{2} > 0

gives

u^{2} - v^{2} = \frac{ε}{E}, 2 uv = \frac{Δ}{E}, u = \frac{E + ε}{2 E}, v = \frac{E - ε}{2 E} .

(4)

This positive choice fixes a convenient convention for the vacuum state in part (d). It yields

H = E (N_{c} + N_{d}) + ε - E, λ = ε^{2} + Δ^{2}, μ = ε - ε^{2} + Δ^{2} .

(5)

Both number operators independently have eigenvalues

0, 1

. Hence the energy eigenvalues and state degeneracies are

N_{c} + N_{d} 012 energy ε - E ε ε + E degeneracy 121

(6)

There is one fermionic Bogoliubov quasiparticle of energy

E

per occupied transformed orbital. The two Nambu spinor matrix eigenvalues

\pm E

should not be mistaken for two independently occupiable levels of opposite energy; the normal ordering constant and the two positive quasiparticle excitation costs give the four physical energy eigenvalues above.

An independent diagonalization of a matrix checks both the pairing sign and the additive constant. In the ordered orthonormal basis

∣0 ⟩, ∣11 ⟩

, the even-fermion parity block is

H_{even} = (0 - Δ - Δ 2 ε),

(7)

because

ab ∣11 ⟩ = - ∣0 ⟩

and

b^{†} a^{†} ∣0 ⟩ = - ∣11 ⟩

. Its characteristic polynomial is

x^{2} - 2 ε x - Δ^{2}

, giving

x = ε \pm E

. The odd-fermion parity Fock states

a^{†} ∣0 ⟩, b^{†} ∣0 ⟩

are unaffected by pairing and both have energy

ε

Equivalently, with the Nambu spinor

Ψ = (a, b^{†})^{T}

H = Ψ^{†} h Ψ + ε, h = (ε - Δ - Δ - ε), R^{T} h R = diag (E, - E) .

(8)

Replacing

Ψ

R (c, d^{†})^{T}

gives

E c^{†} c - E d d^{†} + ε = E (N_{c} + N_{d}) + ε - E

, reproducing the result by normal ordering.

Figure 1.
Pairing splits the even-parity levels and mixes the original vacuum with the occupied pair
. Original exact energy eigenvalues in units of $ε$ , and the probability $v^{2}$ of finding both original orbitals occupied in the two-mode fermionic paired vacuum. The middle energy level remains twofold degenerate; both outside energy levels are nondegenerate. **Details**.

d

 0  0

Solution

 0  0

The canonical anticommutation relations make

(a^{†} b^{†})^{2} = 0

, so the proposed exponential function terminates exactly:

∣ 0 ⟩ = A (∣0 ⟩ + B ∣11 ⟩) .

(1)

Use the inverse Bogoliubov transformation

c = u a - v b^{†}

d = v a^{†} + u b

, and the actions

a ∣11 ⟩ = b^{†} ∣0 ⟩

and

b ∣11 ⟩ = - a^{†} ∣0 ⟩

. The two vacuum state conditions become

c ∣ 0 ⟩ = A (u B - v) b^{†} ∣0 ⟩ = 0, d ∣ 0 ⟩ = A (v - u B) a^{†} ∣0 ⟩ = 0.

(2)

Both therefore require

B = v / u

. Normalizing this quantum state gives

1 = A^{2} (1 + B^{2})

. Choosing positive

A

, which fixes an irrelevant overall global phase, yields

A = u

because

u^{2} + v^{2} = 1

. Thus the two-mode fermionic paired vacuum is

∣ 0 ⟩ = u ∣0 ⟩ + v a^{†} b^{†} ∣0 ⟩, A = \frac{E + ε}{2 E}, B = \frac{Δ}{E + ε} = \frac{E - ε}{E + ε}, E = ε^{2} + Δ^{2} .

(3)

The other sign for

A

describes the same physical quantum state. The positive relative coefficient is fixed by the paper's pairing order and the convention

∣11 ⟩ = a^{†} b^{†} ∣0 ⟩

As a direct eigenvector check,

Δ v = (E - ε) u

and

Δ u = (E + ε) v

imply

(0 - Δ - Δ 2 ε) (v u) = (ε - E) (v u) .

(4)

It is therefore the unique ground state, as also follows from positive quasiparticle excitation energy

E

. It has definite even fermion parity but is not an eigenstate of the original total number operator: the empty-pair and occupied-pair probabilities are

u^{2}

and

v^{2}

. In particular

⟨ N_{a} ⟩ = ⟨ N_{b} ⟩ = v^{2}

. The pair amplitude is

⟨ ab ⟩ = - uv

, consistently with

ab ∣11 ⟩ = - ∣0 ⟩

For the limiting check

Δ \to 0^{+}

at fixed positive

ε

A \to 1

B \to 0

, so

∣ 0 ⟩ \to ∣0 ⟩

. For

ε \to 0^{+}

at fixed positive

Δ

A \to 1/ 2

B \to 1

, giving equal amplitudes for the empty and occupied pair. These checks also fix which branch of the Bogoliubov transformation produces the ground state.

2024 (A15758W1)

 0  0

Source PDF: mmathphys.physics.ox.ac.uk/sites/g/files/vzsiac2991/files/mmathphys/documents/media/aqt_path_integrals_and_many-particle_physics.pdf

1

 0  0

a

 0  0

Solution

 0  0

Put

T = β ℏ > 0

, the interval of imaginary time, and assume

m, ω > 0

. For a variation

x_{0} + ϵη

with

η (0) = η (T) = 0

, the first variation of the Euclidean action is

δ S = m \int_{0}^{T} (\overset{x}{˙}_{0} \overset{η}{˙} + ω^{2} x_{0} η) d τ = m [\overset{x}{˙}_{0} η]_{0}^{T} + m \int_{0}^{T} (- \overset{x}{¨}_{0} + ω^{2} x_{0}) η d τ .

(1)

The boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation

\overset{x}{¨}_{0} = ω^{2} x_{0} .

(2)

Its sign differs from the real-time harmonic oscillator equation because the Euclidean action contains positive kinetic energy and positive potential energy. The general solution is

A cosh (ω τ) + B sinh (ω τ)

. Applying the two endpoint conditions gives

A = x

and

B = - x tanh (ω T /2)

. Equivalently, the equal-endpoint Euclidean oscillator path is

x_{0} (τ) = x \frac{cosh [ ω ( τ - T /2 )]}{cosh ( ω T /2 )} .

(3)

This form displays reflection about the midpoint and satisfies both endpoints immediately. On this path,

\frac{d}{d τ} (x_{0} \overset{x}{˙}_{0}) = \overset{x}{˙}_{0}^{2} + x_{0} \overset{x}{¨}_{0} = \overset{x}{˙}_{0}^{2} + ω^{2} x_{0}^{2} .

(4)

Thus the Euclidean action reduces to a boundary term. Since

\overset{x}{˙}_{0} (T) = ω x tanh (ω T /2)

and

\overset{x}{˙}_{0} (0) = - ω x tanh (ω T /2)

S [x_{0}] = \frac{m}{2} [x_{0} \overset{x}{˙}_{0}]_{0}^{T} = mω x^{2} tanh (\frac{β ℏ ω}{2}) .

(5)

The positive quadratic form

m \int_{0}^{T} (\overset{η}{˙}^{2} + ω^{2} η^{2}) d τ

is the second variation, so this stationary path is the unique minimum, rather than merely an arbitrary stationary point.

b

 0  0

Solution

 0  0

After the shift

x (τ) = x_{0} (τ) + y (τ)

, the endpoint values of

y

are both zero. Expanding the Euclidean action gives

S [x_{0} + y] = S [x_{0}] + S [y] + m \int_{0}^{T} (\overset{x}{˙}_{0} \overset{y}{˙} + ω^{2} x_{0} y) d τ .

(1)

By integration by parts, the cross term is

m [\overset{x}{˙}_{0} y]_{0}^{T} + m \int_{0}^{T} (- \overset{x}{¨}_{0} + ω^{2} x_{0}) y d τ = 0.

(2)

The first term vanishes by the homogeneous Dirichlet boundary conditions, and the second by the Euler-Lagrange equation. The translation has unit Jacobian determinant in the time-sliced configuration-space path integral, so

D x = D y

. Consequently classical-path factorization of a quadratic path integral is exact here:

⟨ x ∣ e^{- β H} ∣ x ⟩ = e^{- S [x_{0}] /ℏ} Z, Z = \int_{y (0) = y (T) = 0} D y exp [- \frac{m}{2ℏ} \int_{0}^{T} (\overset{y}{˙}^{2} + ω^{2} y^{2}) d τ] .

(3)

All dependence on the common endpoint

x

lies in

e^{- S [x_{0}] /ℏ} = exp [- \frac{mω x ^{2}}{ℏ} tanh (\frac{β ℏ ω}{2})], Z is independent of x .

(4)

The endpoint domain and the integrand defining

Z

are the same for every

x

. This

Z

is a zero-endpoint Gaussian path integral; the canonical partition function instead integrates the entire diagonal kernel over

x

. The final sentence of the printed part omits

/ℏ

from its exponential: the physical Boltzmann factor has the dimensionless exponent

- S /ℏ

An optional normalization check uses the Dirichlet oscillator fluctuation determinant. Its eigenvalues are

m [(nπ / T)^{2} + ω^{2}]

n \geq 1

. Relative to the free-particle Gaussian path integral, the hyperbolic-sine infinite product gives

\frac{Z}{m / ( 2 π ℏ T )} = \prod_{n = 1}^{\infty} [1 + \frac{ω ^{2} T ^{2}}{n ^{2} π ^{2}}]^{- 1/2} = \frac{ω T}{s i n h ( ω T )} .

(5)

Thus

Z = mω / [2 π ℏ sinh (β ℏ ω)]

. Integrating the resulting diagonal thermal kernel of a quantum harmonic oscillator by a Gaussian integral yields

Tr e^{- β H} = [2 sinh (β ℏ ω /2)]^{- 1}

, including the zero-point energy.

c

 0  0

Solution

 0  0

Because

S [x_{0}] = 0

x = 0

, the factor

Z

cancels from the ratio. Write

z = β ℏ ω

and

q = mω /ℏ x

. The exact answer is

R = exp [- q^{2} tanh (z /2)] .

(1)

At high temperature,

z ≪ 1

. Expanding the hyperbolic tangent gives

lo g R = - q^{2} (\frac{z}{2} - \frac{z ^{3}}{24} + O (z^{5})), R ≃ exp (- \frac{β m ω ^{2} x ^{2}}{2}) .

(2)

The displayed leading expression is the ratio of classical Boltzmann factors for potential energy

V (x) = m ω^{2} x^{2} /2

and

V (0) = 0

. Indeed the classical canonical ensemble gives

p_{cl} (x) \propto \int_{- \infty}^{\infty} d p e^{- β [p^{2} / (2 m) + V (x)]} \propto e^{- β V (x)} .

(3)

The momentum Gaussian integral is independent of

x

and cancels in the ratio. This is the classical phase-space limit of the thermal kernel. Here

β = 1/ (k_{B} T)

, with

k_{B}

the Boltzmann constant; the associated classical position variance is

1/ (β m ω^{2})

. At fixed

x

R

ultimately tends to one as temperature tends to infinity, while its width grows.

At low temperature,

z ≫ 1

. Now

tanh (z /2) = 1 - 2 e^{- z} + O (e^{- 2 z})

, so at fixed

q

R = e^{- q^{2}} [1 + 2 q^{2} e^{- z} + O (e^{- 2 z})], R ≃ exp (- \frac{mω x ^{2}}{ℏ}) .

(4)

Unlike the classical high-temperature expression, this limiting width remains nonzero at zero temperature, reflecting the ground state of the quantum harmonic oscillator.

Figure 1.
Imaginary-time stationary paths and the crossover of the diagonal thermal kernel
. Original plots of the equal-endpoint Euclidean oscillator path and the diagonal thermal kernel of a quantum harmonic oscillator, divided by its value at zero. Increasing $z = β ℏ ω$ lowers the temperature: the path spends more of its interior near zero, and the ratio approaches the finite-width ground state profile. **Details**.

d

 0  0

Solution

 0  0

Insert the complete orthonormal basis of energy eigenstates on either side of the Boltzmann factor. The spectral theorem gives

e^{- β H} = \sum_{n = 0}^{\infty} e^{- β ℏ ω (n + 1/2)} ∣ n ⟩ ⟨ n ∣, ⟨ x ∣ e^{- β H} ∣ x ⟩ = n = 0 \sum \infty e^{- β ℏ ω (n + 1/2)} ∣ φ_{n} (x) ∣^{2} .

(1)

At low temperature, the gap

ℏ ω

suppresses excited energy eigenstates. The harmonic oscillator eigenfunction of the ground state is

φ_{0} (x) = (\frac{mω}{π ℏ})^{1/4} e^{- mω x^{2} / (2ℏ)} .

(2)

Therefore

⟨ x ∣ e^{- β H} ∣ x ⟩ \sim e^{- β ℏ ω /2} ∣ φ_{0} (x) ∣^{2}, R ⟶ \frac{∣ φ _{0} ( x ) ∣ ^{2}}{∣ φ _{0} ( 0 ) ∣ ^{2}} = e^{- mω x^{2} /ℏ} .

(3)

The common zero-point energy factor cancels. Dividing the kernel by its trace would give the position probability density of the canonical ensemble; at zero temperature this tends to

∣ φ_{0} (x) ∣^{2}

, whose variance is

ℏ/ (2 mω)

. Thus zero temperature projects onto the ground state and retains its quantum spread, rather than concentrating at the minimum of the classical potential energy.

There is also a check on the first correction. With

q = mω /ℏ x

, the first excited harmonic oscillator eigenfunction satisfies

∣ φ_{1} (x) ∣^{2} = 2 q^{2} ∣ φ_{0} (x) ∣^{2}

and

φ_{1} (0) = 0

. Hence the spectral theorem gives

R = e^{- q^{2}} [1 + 2 q^{2} e^{- z} + O (e^{- 2 z})]

, in agreement with part (c).

2

 0  0

a

 0  0

Solution

 0  0

Let

V = L^{3}

be the box volume. The number operator is the sum of the occupation numbers of all momentum eigenstates, including the zero mode:

\hat{N} = \int_{V} d^{3} r ψ^{†} (r) ψ (r) = k \sum ψ_{k}^{†} ψ_{k} .

(1)

In the Fourier series for the contact interaction, the two creation operators supply phase

e^{- i (k_{1} + k_{2}) \cdot r}

and the two annihilation operators supply

e^{i (k_{3} + k_{4}) \cdot r}

. Fourier orthogonality with periodic boundary conditions gives

\int_{V} d^{3} r e^{i (k_{3} + k_{4} - k_{1} - k_{2}) \cdot r} = V δ_{k_{1} + k_{2}, k_{3} + k_{4}} .

(2)

The four normalization factors together contribute

V^{- 2}

, leaving the momentum-space contact-interaction Hamiltonian

H_{int} = \frac{u}{2 V} k_{1}, k_{2}, k_{3}, k_{4} \sum δ_{k_{1} + k_{2}, k_{3} + k_{4}} ψ_{k_{1}}^{†} ψ_{k_{2}}^{†} ψ_{k_{3}} ψ_{k_{4}} .

(3)

The Kronecker delta enforces momentum conservation. An equivalent expression that resolves the momentum transfer is

H_{int} = \frac{u}{2 V} \sum_{k, p, q} ψ_{k + q}^{†} ψ_{p - q}^{†} ψ_{p} ψ_{k} .

(4)

Both are already in normal ordering; no extra commutator constant is introduced.

b

 0  0

Solution

 0  0

Taking the Hermitian adjoint gives

φ_{k} = c_{k} ψ_{k} + s_{k} ψ_{- k}^{†}

. For nonzero wavevectors, the canonical commutation relations imply

[φ_{k}, φ_{q}^{†}] = c_{k} c_{q} [ψ_{k}, ψ_{q}^{†}] + s_{k} s_{q} [ψ_{- k}^{†}, ψ_{- q}] = (c_{k} c_{q} - s_{k} s_{q}) δ_{k, q} .

(1)

When the Kronecker delta is nonzero,

k = q

, so the hyperbolic Pythagorean identity

c_{k}^{2} - s_{k}^{2} = 1

gives

[φ_{k}, φ_{q}^{†}] = δ_{k, q} .

(2)

For the two annihilation operators, the only potentially nonzero terms are

[φ_{k}, φ_{q}] = c_{k} s_{q} [ψ_{k}, ψ_{- q}^{†}] + s_{k} c_{q} [ψ_{- k}^{†}, ψ_{q}] = (c_{k} s_{q} - s_{k} c_{q}) δ_{k, - q} = 0.

(3)

Here

q = - k

implies equal magnitudes and hence equal

c, s

, because

θ_{k}

depends only on

∣ k ∣

. Thus

[φ_{k}, φ_{q}] = 0.

(4)

The corresponding commutators of two creation operators vanish by the Hermitian adjoint. This verifies that the Bogoliubov transformation preserves the full bosonic canonical commutation relations.

c

 0  0

Solution

 0  0

Normalize the specified state by

⟨ 0∣0 ⟩ = 1

and write

⟨ A ⟩ = ⟨ 0∣ A ∣0 ⟩

. Its zero mode is a coherent state, so any normally ordered product of zero-mode creation operators and annihilation operators can be evaluated using

ψ_{0} \mapsto N_{0}

and

ψ_{0}^{†} \mapsto N_{0}

. In particular

⟨(ψ_{0}^{†})^{2} ψ_{0}^{2} ⟩ = N_{0}^{2}

, rather than

N_{0} (N_{0} - 1)

, which would apply to a zero-mode number state.

Sort the momentum-space contact-interaction Hamiltonian by its number of zero-mode factors. Four such factors give

u N_{0}^{2} / (2 V)

. Three zero-mode factors and one nonzero momentum violate momentum conservation, so no such term survives. With two zero-mode factors there are three possibilities:

Two zero-mode creation operators leave the opposite-momentum pair $ψ_{k} ψ_{- k}$ .
Two zero-mode annihilation operators leave $ψ_{k}^{†} ψ_{- k}^{†}$ .
One zero-mode factor on each side leaves $ψ_{k}^{†} ψ_{k}$ . Choosing which of the two creation operators and which of the two annihilation operators are zero gives four placements.

The sums over

k \neq = 0

include both members of each opposite-momentum pair, consistently with the overall factor

1/2

. Terms with one zero-mode factor contain three nonzero-mode factors; their expected values vanish in the centered multimode squeezed vacuum. Indeed each

ψ

is linear in

φ, φ^{†}

, and an odd product has zero vacuum expectation value. Terms with no zero-mode factors are independent of

N_{0}

at fixed

θ_{k}

. Therefore

⟨ H_{int} ⟩ = \frac{u N _{0}^{2}}{2 V} + \frac{u N _{0}}{2 V} k \neq = 0 \sum [⟨ ψ_{k} ψ_{- k} ⟩ + ⟨ ψ_{k}^{†} ψ_{- k}^{†} ⟩ + 4 ⟨ ψ_{k}^{†} ψ_{k} ⟩] + O (N_{0}^{0}) .

(1)

To evaluate these vacuum expectation values, use the inverse Bogoliubov transformation and its Hermitian adjoint:

ψ_{k} = c_{k} φ_{k} - s_{k} φ_{- k}^{†}, ψ_{k}^{†} = c_{k} φ_{k}^{†} - s_{k} φ_{- k} .

(2)

Since

φ_{k} ∣0 ⟩ = 0

, the only nonzero two-factor vacuum expectation value is

⟨ φ_{k} φ_{q}^{†} ⟩ = δ_{k, q}

. The bosonic squeezed-vacuum contractions are consequently

⟨ ψ_{k}^{†} ψ_{q} ⟩ ⟨ ψ_{k} ψ_{q} ⟩ ⟨ ψ_{k}^{†} ψ_{q}^{†} ⟩ = s_{k}^{2} δ_{k, q}, = - c_{k} s_{k} δ_{k, - q}, = - c_{k} s_{k} δ_{k, - q} .

(3)

In particular the three requested quantities are

- sinh θ_{k} cosh θ_{k}

- sinh θ_{k} cosh θ_{k}

and

sinh^{2} θ_{k}

, respectively. The minus sign in the pair vacuum expectation values comes from the inverse Bogoliubov transformation and is essential to lowering the interaction energy. Substitution yields

⟨ H_{int} ⟩ = \frac{u N _{0}^{2}}{2 V} + \frac{u N _{0}}{V} k \neq = 0 \sum (2 s_{k}^{2} - c_{k} s_{k}) + O (N_{0}^{0}) .

(4)

d

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Solution

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The coherent state in the zero mode has mean occupation number

N_{0}

. The other occupation numbers follow from part (c), giving exactly

⟨ \hat{N} ⟩ = N_{0} + D, D \equiv k \neq = 0 \sum sinh^{2} θ_{k} .

(1)

Impose the fixed mean particle number

⟨ \hat{N} ⟩ = N

by choosing

N_{0} = N - D .

(2)

Here

D

is the quantum depletion of a Bose condensate. The approximation requires

0 \leq D ≪ N

, so

N_{0}

remains positive and macroscopically large. The specified coherent state and multimode squeezed vacuum are not exact eigenstates of the total number operator;

N

in this variational construction is their prescribed mean particle number.

Write

P = \sum_{k \neq = 0} s_{k} c_{k}

. Substituting into the retained part of the interaction energy before truncating gives

⟨ H_{int} ⟩ ≃ \frac{u}{2 V} (N - D)^{2} + \frac{u}{V} (N - D) (2 D - P) = \frac{u N ^{2}}{2 V} + \frac{u N}{V} (D - P) + \frac{u}{V} (- \frac{3}{2} D^{2} + D P) .

(3)

The last two terms quadratic in

D, P

have no explicit large-

N

factor when the Bogoliubov transformation parameters are held fixed, and are discarded along with the nonzero-mode quartic contribution already omitted in part (c). This is the same truncation that underlies the number-conserving Bogoliubov quadratic Hamiltonian. Thus, defining the physical number density correctly as

ρ = N / V = N / L^{3}

⟨ H_{int} ⟩ ≃ \frac{u ρ}{2} N + u ρ k \neq = 0 \sum (sinh^{2} θ_{k} - sinh θ_{k} cosh θ_{k}) .

(4)

Replacing

N_{0}

N

at the outset would give the wrong coefficient

2 sinh^{2} θ_{k}

: expanding the leading condensate interaction energy supplies the necessary subtraction.

The printed definition $ρ = N / L$ is an error. In three dimensions the contact interaction coupling

u

has units of energy times volume, so

u N / L

does not even have units of energy. There is a direct counterexample with no depletion: when

θ_{k} = 0

, the specified coherent state has exactly

⟨ H_{int} ⟩ = u N^{2} / (2 L^{3})

. In fixed length and energy units,

L = 2

u = 1

and

N = 100

give

625

, whereas the printed

u ρN /2

with

ρ = N / L

gives

2500

. The requested formula and the minimization in part (e) hold with

ρ = N / L^{3}

e

 0  0

Solution

 0  0

Put

ϵ_{k} = ℏ^{2} k^{2} / (2 m) > 0

for

k \neq = 0

and

g = u ρ

, using

ρ = N / L^{3}

. Since the zero mode has zero kinetic energy, the exact kinetic energy in this trial state is

⟨ H_{KE} ⟩ = k \neq = 0 \sum ϵ_{k} sinh^{2} θ_{k} .

(1)

Together with part (d), the fixed-mean-number Bogoliubov variational energy is

E [θ] ≃ \frac{g N}{2} + \sum_{k \neq = 0} f_{k} (θ_{k}), f_{k} (θ) = (ϵ_{k} + g) sinh^{2} θ - g sinh θ cosh θ .

(2)

For repulsive contact interaction,

u > 0

, each

ϵ_{k} + g > g > 0

. Using the double-angle hyperbolic-function identities,

f_{k} (θ) = \frac{ϵ _{k} + g}{2} (cosh 2 θ - 1) - \frac{g}{2} sinh 2 θ,

(3)

f_{k}^{'} (θ) = (ϵ_{k} + g) sinh 2 θ - g cosh 2 θ .

(4)

The stationary point therefore satisfies

tanh 2 θ_{k} = \frac{g}{ϵ _{k} + g} = \frac{u ρ}{u ρ + ℏ ^{2} k ^{2} / ( 2 m )} .

(5)

This is a minimum, not just a vanishing derivative. Define the positive Bogoliubov quasiparticle dispersion

E_{k} = (ϵ_{k} + g)^{2} - g^{2} = ϵ_{k} (ϵ_{k} + 2 g) .

(6)

At the displayed solution

θ_{k}^{*}

cosh 2 θ_{k}^{*} = (ϵ_{k} + g) / E_{k}

and

sinh 2 θ_{k}^{*} = g / E_{k}

. Hence

f_{k} (θ) = \frac{E _{k}}{2} cosh [2 (θ - θ_{k}^{*})] - \frac{ϵ _{k} + g}{2}, f_{k}^{''} (θ_{k}^{*}) = 2 E_{k} > 0.

(7)

The hyperbolic cosine has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude wavevectors together simply multiplies the same function by their multiplicity. At a fixed ultraviolet cutoff, the resulting energy change from

θ = 0

E_{m i n} - E [0] = \frac{1}{2} k \neq = 0 \sum (E_{k} - ϵ_{k} - g) < 0 (g > 0) .

(8)

Indeed

E_{k} < ϵ_{k} + g

. More directly, for small positive

θ

f_{k} (θ) = - g θ + (ϵ_{k} + g) θ^{2} + O (θ^{3}) .

(9)

The pair vacuum expectation value lowers the interaction energy at first order in

θ

, whereas the nonzero occupation number and its kinetic energy cost start at second order. The two-mode squeezed vacuum admixes opposite-momentum pairs into the condensate, introducing spatial correlations that reduce the chance of two bosons coinciding. These correlations lower repulsive contact interaction enough to outweigh the added kinetic energy. A small quantum depletion of a Bose condensate can therefore lower the energy below the uncorrelated condensate value.

Two qualifications delimit this result. First, the paper does not state the sign of

u

. More generally a finite stable minimum exists only when

ϵ_{k} + g > ∣ g ∣

for every retained nonzero mode; repulsion ensures this. For example, with

g = - ϵ_{k}

in some mode,

f_{k} (θ) = ϵ_{k} sinh (2 θ) /2

tends to

- \infty

θ \to - \infty

, so an unconditional minimum claim for arbitrary

u

is false. Also the formula minimizes the truncated Bogoliubov theory energy, not the exact quartic Hamiltonian including its discarded terms, and it must remain in the regime of small quantum depletion of a Bose condensate.

Second, a three-dimensional unregulated contact interaction cannot give a finite absolute beyond-leading-order ground-state energy by this bare sum. For large

k

E_{k} - ϵ_{k} - g = - g^{2} / (2 ϵ_{k}) + O (ϵ_{k}^{- 2})

, whose three-dimensional sum has an ultraviolet divergence. A finite ultraviolet cutoff, or renormalization of a three-dimensional dilute-gas contact coupling matched to a physical scattering length, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.

2023 (A15757W1)

 0  0

Source PDF: mmathphys.web.ox.ac.uk/sitefiles/advanced-quantum-theory.pdf

1

 0  0

a

 0  0

i

 0  0

Solution

 0  0

Assume

Q \geq 2

and

L \geq 2

; the trivial exceptions are described below. In this periodic Potts chain, every satisfied bond contributes

- J

to the energy, and no bond can contribute less. All

L

bonds can be satisfied simultaneously by choosing the same state at every site. Conversely, satisfying every bond forces this uniform configuration. Therefore the ground state energy and degeneracy factor are

E_{0} = - J L, d_{0} = Q .

(1)

The

Q

choices are distinct configurations on the labelled lattice: a global relabelling of the states is a symmetry, not an identification of microstates.

ii

 0  0

Solution

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Let

w

count the unequal neighbouring pairs, or domain walls. The energy is

E = - J L + J w

. A single domain wall is impossible with periodic boundary conditions: following all the other, equal-state bonds around the ring would force its endpoints to be equal too. Two domain walls are possible, splitting the ring into two nonempty uniform arcs carrying different states. Thus the first excited state has

E_{1} = - J L + 2 J, E_{1} - E_{0} = 2 J .

(1)

Choose the two wall bonds in

(2 L)

ways. For that choice, identify one of the resulting arcs by its smallest site label. Its state has

Q

choices and the other arc has

Q - 1

. This counts each configuration exactly once, giving the two-domain-wall degeneracy of a Potts ring

d_{1} = (2 L) Q (Q - 1) .

(2)

For example,

Q = 2, L = 4

gives

d_{0} = 2

and

d_{1} = 12

; the remaining two configurations are alternating and have four domain walls. The formula also covers

L = 2

, where the specified ring energy counts the two directed neighbouring bonds, and an unequal pair breaks both. If

Q = 1

L = 1

, every allowed configuration is a ground state and there is no first excited state. For

Q \geq 3

, the number of domain walls need not be even: three distinct states on a three-site ring give three, although one remains impossible.

b

 0  0

Solution

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The Boltzmann factor factorizes over bonds. Define the spin-chain transfer matrix by

T_{ab} = e^{β J δ_{ab}} = 1 + (e^{β J} - 1) δ_{ab}, a, b = 1, \dots, Q .

(1)

Here

δ_{ab}

is the Kronecker delta. Summing the product of these matrix elements over all configurations, and closing the final bond by periodic boundary conditions, gives the canonical partition function

Z = \sum_{σ_{1}, \dots, σ_{L}} T_{σ_{1} σ_{2}} \dots T_{σ_{L} σ_{1}} = Tr T^{L} .

(2)

Hence, writing

I

for the identity matrix and

P

for the all-ones matrix,

T = g (β J) I + P, g (β J) = e^{β J} - 1.

(3)

Every element of

P^{2}

is a sum of

Q

ones, while every diagonal element of

P

is one. Thus

P^{2} = QP, Tr P = Q .

(4)

More explicitly, for

u = (1, \dots, 1)^{T}

P u = Q u

. If

\sum_{a} v_{a} = 0

, then

P v = 0

. These complementary eigenspaces have dimensions

1

and

Q - 1

. Equivalently,

P / Q

is the orthogonal projector onto the linear span of

u

. The eigenvalues and their eigenvalue multiplicities are therefore

P : T : Q (1), 0 (Q - 1), λ_{+} = e^{β J} + Q - 1 (1), λ_{-} = e^{β J} - 1 (Q - 1) .

(5)

Since the two matrices share these eigenvectors, the exact finite-ring canonical partition function is

Z = (e^{β J} + Q - 1)^{L} + (Q - 1) (e^{β J} - 1)^{L} .

(6)

Both terms matter in the finite-ring low-temperature limit: discarding the second would lose the correct ground state degeneracy factor and introduce a spurious single-wall contribution.

c

 0  0

Solution

 0  0

Put

x = e^{- β J}

. Factoring out the ground state Boltzmann factor from the exact canonical partition function gives

Z = e^{β J L} {[1 + (Q - 1) x]^{L} + (Q - 1) (1 - x)^{L}} .

(1)

At fixed finite

L

, the binomial theorem gives

[1 + (Q - 1) x]^{L} + (Q - 1) (1 - x)^{L} = 1 + (Q - 1) + [L (Q - 1) - (Q - 1) L] x + (2 L) [(Q - 1)^{2} + Q - 1] x^{2} + O_{L, Q} (x^{3}) = Q [1 + (2 L) (Q - 1) x^{2} + O_{L, Q} (x^{3})] .

(2)

Taking the logarithm yields the finite-ring low-temperature expansion of the Helmholtz free energy,

F = - J L - \frac{lo g Q}{β} - \frac{L ( L - 1 ) ( Q - 1 )}{2 β} e^{- 2 β J} + O_{L, Q} (\frac{e ^{- 3 β J}}{β}) .

(3)

The constant term is the ground state energy. The term

- β^{- 1} lo g Q

is the contribution of the ground state degeneracy factor. The first thermal correction follows directly by retaining the first excited state manifold:

Z = e^{- β E_{0}} [d_{0} + d_{1} e^{- 2 β J} + \dots], \frac{d _{1}}{d _{0}} = (2 L) (Q - 1) .

(4)

Each pair of domain walls costs

2 J

, explaining the absence of an

e^{- β J}

term. For

Q = 2

, every configuration has an even number of domain walls, so the displayed remainder improves to

O_{L} (e^{- 4 β J} / β)

This is an expansion for fixed

L

, not a uniform thermodynamic limit. In particular, the small correction requires

(2 L) (Q - 1) x^{2} ≪ 1

, with higher terms also controlled as

x \to 0

at fixed

L, Q

. Taking

L \to \infty

first instead selects

λ_{+}

and gives

lim_{L \to \infty} \frac{F}{L} = - J - \frac{1}{β} lo g [1 + (Q - 1) x] .

(5)

Its leading correction is of order

x

. The finite-ring pair constraint and the extensive population of domain walls in an arbitrarily long chain explain the different ordering of these two limits.

2

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a

 0  0

Solution

 0  0

Use

ℏ = 1

throughout. Let

N_{a} = a^{†} a

N_{b} = b^{†} b

and

N = N_{a} + N_{b}

. Operators belonging to different boson species commute. The proposed Schwinger boson representation gives

S^{+} S^{-} S^{-} S^{+} = a^{†} b b^{†} a = N_{a} (N_{b} + 1), = b^{†} a a^{†} b = N_{b} (N_{a} + 1) .

(1)

Consequently the first of the spin commutation relations is

[S^{+}, S^{-}] = N_{a} - N_{b} = 2 S^{z} .

(2)

The number operator commutators

[N_{a}, a^{†}] = a^{†}

[N_{a}, a] = - a

and their

b

counterparts give

[S^{z}, S^{+}] [S^{z}, S^{-}] = \frac{1}{2} ([N_{a}, a^{†}] b - a^{†} [N_{b}, b]) = a^{†} b = S^{+}, = - b^{†} a = - S^{-} .

(3)

Thus all the spin commutation relations hold. Also

[N, S^{\pm}] = [N, S^{z}] = 0

: each fixed-

N

sector is invariant. To identify its spin, calculate the Casimir operator:

S^{2} = (S^{z})^{2} + \frac{1}{2} (S^{+} S^{-} + S^{-} S^{+}) = \frac{1}{4} (N_{a} - N_{b})^{2} + N_{a} N_{b} + \frac{1}{2} (N_{a} + N_{b}) = \frac{N}{2} (\frac{N}{2} + 1) .

(4)

The irreducible spin representation with prescribed

s

therefore requires

N = 2 s

, and its normalized Fock basis correspondence is

∣ s, m ⟩ ⟷ ∣ s + m, s - m ⟩ = \frac{( a ^{†} ) ^{s + m} ( b ^{†} ) ^{s - m}}{( s + m )! ( s - m )!} ∣0, 0 ⟩ .

(5)

Conversely,

s = (n_{a} + n_{b}) /2

and

m = (n_{a} - n_{b}) /2

. The nonnegative integer occupation numbers give precisely

m = - s, - s + 1, \dots, s

, and there are

2 s + 1

such states. As a check, the spin raising operator acts as

a^{†} b ∣ s + m, s - m ⟩ = (s + m + 1) (s - m) ∣ s + m + 1, s - m - 1 ⟩,

(6)

which is the usual spin ladder operator coefficient

s (s + 1) - m (m + 1)

For a lattice of fixed spins, the local spin constraint in the Schwinger boson representation is essential:

(a_{n}^{†} a_{n} + b_{n}^{†} b_{n}) ∣ ψ ⟩ = 2 s ∣ ψ ⟩ at every site n .

(7)

The full two-species bosonic Fock space also contains other local irreducible spin representations; fixing only the total boson number does not remove them.

b

 0  0

i

 0  0

Solution

 0  0

For the specified positive-exponent discrete Fourier transform, the inverse discrete Fourier transforms have negative exponents:

a_{n} = \frac{1}{L} q \sum e^{- i q n} A_{q}, b_{n} = \frac{1}{L} q \sum e^{- i q n} B_{q} .

(1)

Every sum runs over

q = 2 πj / L

j = 0, \dots, L - 1

. Indeed, orthogonality of roots of unity gives

\frac{1}{L} \sum_{q} e^{i q (r - n)} = δ_{r n},

(2)

so substituting the forward discrete Fourier transform into either proposed inverse recovers the original operator. Taking Hermitian conjugates supplies the inverse formulas for the creation operators.

Multiplying the inverse discrete Fourier transforms gives the local number operator combination

\frac{1}{2} (a_{n}^{†} a_{n} + b_{n}^{†} b_{n}) = \frac{1}{2 L} q, k \sum e^{i (q - k) n} (A_{q}^{†} A_{k} + B_{q}^{†} B_{k}) .

(3)

This local expression includes off-diagonal momentum terms. Only after summing over sites does orthogonality of roots of unity give

\frac{1}{2} \sum_{n} (a_{n}^{†} a_{n} + b_{n}^{†} b_{n}) = \frac{1}{2} \sum_{q} (A_{q}^{†} A_{q} + B_{q}^{†} B_{q}) .

(4)

In a physical fixed-

s

state the first boxed operator acts as

s

at every site, by the local spin constraint in the Schwinger boson representation.

ii

 0  0

Solution

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Take

L \geq 2

, so neighbouring operators act at distinct sites. The identity

S_{n}^{x} S_{n + 1}^{x} + S_{n}^{y} S_{n + 1}^{y} = \frac{1}{2} (S_{n}^{+} S_{n + 1}^{-} + S_{n}^{-} S_{n + 1}^{+})

(1)

expresses the Hamiltonian operator of this quantum XY chain in spin ladder operators. Substituting the Schwinger boson representation gives the requested quartic expression:

H = - \frac{J}{2} n \sum (a_{n}^{†} b_{n} b_{n + 1}^{†} a_{n + 1} + b_{n}^{†} a_{n} a_{n + 1}^{†} b_{n + 1}) .

(2)

Operators at different sites commute, allowing normal ordering without a correction here. For the first quartic term, the inverse discrete Fourier transforms give

n \sum a_{n}^{†} b_{n + 1}^{†} b_{n} a_{n + 1} = \frac{1}{L ^{2}} n, p, q, r, t \sum e^{i (p + q - r - t) n} e^{i (q - t)} A_{p}^{†} B_{q}^{†} B_{r} A_{t} = \frac{1}{L} p, q, r, t \sum δ_{p + q, r + t}^{(L)} e^{i (q - t)} A_{p}^{†} B_{q}^{†} B_{r} A_{t} .

(3)

Here

δ^{(L)}

is one when its two momenta agree modulo

2 π

and zero otherwise; it expresses lattice momentum conservation. The other quartic term has the same ordered operators with phase

e^{i (p - r)}

. On the support of this Kronecker delta,

p - r = t - q

modulo

2 π

, so the two phases sum to

2 cos (q - t)

. Thus the Fourier vertex of a Schwinger boson chain is

H = - \frac{J}{L} p, q, r, t \sum δ_{p + q, r + t}^{(L)} cos (q - t) A_{p}^{†} B_{q}^{†} B_{r} A_{t} .

(4)

An equivalent three-sum expression sets

t = p + q - r

modulo

2 π

, so its coefficient is

cos (r - p)

. The factor

1/ L

comes from four factors

L^{- 1/2}

and one site sum, while the factor of two from the two exchange terms cancels the original

1/2

. The formula is Hermitian: conjugation exchanges

p

with

t

and

q

with

r

, and the coefficient is unchanged after imposing lattice momentum conservation. It also preserves each local total boson number in its real-space form, hence preserves the physical constrained subspace.

c

 0  0

Solution

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Use

⟨ O ⟩ = ⟨ G ∣ O ∣ G ⟩

. The given state is a two-species uniform boson number state: only the

q = 0

mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while

A_{0}^{†} A_{0}

has eigenvalue

n_{A}

. Thus

⟨ A_{q}^{†} A_{k} ⟩ = n_{A} δ_{q 0} δ_{k 0}, ⟨ B_{q}^{†} B_{k} ⟩ = n_{B} δ_{q 0} δ_{k 0} .

(1)

The two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yield

⟨ a_{i}^{†} a_{j} ⟩ = \frac{n _{A}}{L}, ⟨ b_{i}^{†} b_{j} ⟩ = \frac{n _{B}}{L}, ⟨ a_{n}^{†} a_{n} + b_{n}^{†} b_{n} ⟩ = \frac{n _{A} + n _{B}}{L} .

(2)

In the momentum-space Hamiltonian operator, only

p = q = r = t = 0

contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. Hence

⟨ A_{p}^{†} B_{q}^{†} B_{r} A_{t} ⟩ = n_{A} n_{B} δ_{p 0} δ_{q 0} δ_{r 0} δ_{t 0}, ⟨ H ⟩ = - \frac{J}{L} n_{A} n_{B} .

(3)

There is no replacement of

n_{B}

n_{B} + 1

in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose

n_{A} + n_{B} = 2 s L =: N

. At fixed integer

N

, the energy is minimized by maximizing

n_{A} n_{B} = [N^{2} - (n_{A} - n_{B})^{2}] /4

. Consequently the best choices within this relaxed family are

{n_{A}, n_{B}} = {⌊ s L ⌋, ⌈ s L ⌉} .

(4)

When

N

is even,

n_{A} = n_{B} = s L

and

⟨ H ⟩ = - J s^{2} L

. When

N

is odd, either assignment of

(N - 1) /2, (N + 1) /2

gives

⟨ H ⟩ = - J (N^{2} - 1) / (4 L)

. Without fixing

N

, this auxiliary boson Hamiltonian operator has no lower bound: taking

n_{A} = n_{B} \to \infty

sends its expectation value to

- \infty

Next,

S_{i}^{x} = (a_{i}^{†} b_{i} + b_{i}^{†} a_{i}) /2

. Each term changes the two species' occupation numbers, producing a state orthogonal to

∣ G ⟩

. Therefore

⟨ S_{i}^{x} ⟩ = 0, ⟨ S_{i}^{y} ⟩ = 0, ⟨ S_{i}^{z} ⟩ = \frac{n _{A} - n _{B}}{2 L} .

(5)

For distinct sites

i, j

modulo

L

, expand the product of the two spin operators. The terms

S_{i}^{+} S_{j}^{+}

and

S_{i}^{-} S_{j}^{-}

again have zero expectation value. The remaining terms give

⟨ S_{i}^{+} S_{j}^{-} ⟩ ⟨ S_{i}^{-} S_{j}^{+} ⟩ = ⟨ a_{i}^{†} a_{j} ⟩ ⟨ b_{i} b_{j}^{†} ⟩ = \frac{n _{A}}{L} \frac{n _{B}}{L}, = \frac{n _{A} n _{B}}{L ^{2}} .

(6)

Here

⟨ b_{i} b_{j}^{†} ⟩ = n_{B} / L + δ_{ij}

and the Kronecker delta vanishes at distinct sites. Consequently

⟨ S_{m}^{x} S_{m + n}^{x} ⟩ = \frac{n _{A} n _{B}}{2 L ^{2}} (n \neq \equiv 0 (mod L)) .

(7)

The same holds for the

y

components. At the balanced choice and as

L \to \infty

, each transverse two-point correlation function tends to

s^{2} /2

independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total

S^{z} = (n_{A} - n_{B}) /2

and retains the global U(1) symmetry of rotations about

z

, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the

x

and

y

correlations tends to

s^{2}

, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and

L

growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-

s

Hilbert space for

s > 0, L > 1

. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with

L = 2, s = 1/2, n_{A} = n_{B} = 1

∣ G ⟩ = \frac{1}{2} (a_{0}^{†} + a_{1}^{†}) (b_{0}^{†} + b_{1}^{†}) ∣ vac ⟩ .

(8)

The nonzero component

a_{0}^{†} b_{0}^{†} ∣ vac ⟩ /2

has local total occupation numbers

(2, 0)

and hence local spins

(1, 0)

, instead of

(1/2, 1/2)

. More generally the local total number operator

N_{n}

has variance

Var (N_{n}) = \frac{n _{A} + n _{B}}{L} (1 - \frac{1}{L}) = 2 s (1 - \frac{1}{L}) > 0.

(9)

Each species contributes a binomial distribution at a given site, with success probability

1/ L

, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even has

⟨ S_{n}^{2} ⟩ = s (s + 1) + \frac{1}{4} Var (N_{n}),

(10)

by the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-

s

quantum XY chain. Projecting onto

N_{n} = 2 s

at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy

- J

, whereas the unprojected value is

- J /2

. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.

2019 (A15757W1)

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Source paper: www.whatdotheyknow.com/request/physics_past_papers_21/response/3037379/attach/2/FOI%20202505%20441%20Response.pdf?cookie_passthrough=1

TODO: Check complete coverage of this exact paper by the publicly advertised 2019 solution notes before writing new solutions. The available material has not established complete coverage.

Official course page advertising the 2019 solution notes.

2018 (A15757W1)

 0  0

Source paper: www.whatdotheyknow.com/request/physics_past_papers_21/response/3037379/attach/2/FOI%20202505%20441%20Response.pdf?cookie_passthrough=1. The held cover dates this paper to 12 June 2018. We use

ℏ = 1

, consistent with the source's momentum quantization, and take the pairing parameter

Δ

real so that the Hamiltonian is Hermitian.

1

 0  0

a

 0  0

Solution

 0  0

With no pairing, the Hamiltonian is diagonal in the fermionic occupation numbers. The Pauli exclusion principle allows each momentum mode to be either empty or occupied once. Without a fixed fermion number constraint, occupying a mode lowers the energy precisely when its single-particle energy is negative. Therefore, with

∣0 ⟩

the Fock vacuum, a ground state is

∣ GS ⟩ = p : ϵ (p) < 0 \prod c_{p}^{†} ∣0 ⟩, E_{GS} = p : ϵ (p) < 0 \sum ϵ (p) .

(1)

Choose a fixed ordering of the fermionic creation operators; changing it alters only the overall phase. Any zero-energy modes may be occupied or empty without changing the ground-state energy.

To identify the occupied modes, complete the square:

\frac{ϵ ( p )}{α} = (p^{2} - \frac{β}{2})^{2} + μ - \frac{β ^{2}}{4} .

(2)

Here

β

is the dispersion parameter, not inverse temperature. If

β^{2} < 4 μ

, every mode has positive energy and the unique ground state is the Fock vacuum, with

E_{GS} = 0

. If

β^{2} = 4 μ

, there are still no negative-energy modes, but

p = \pm β /2

have zero energy if they lie on the allowed momentum grid. Each such mode doubles the ground-state degeneracy.

For

β^{2} > 4 μ

, define

p_{-}^{2} = \frac{β - β ^{2} - 4 μ}{2}, p_{+}^{2} = \frac{β + β ^{2} - 4 μ}{2} .

(3)

Both are positive, and the occupied Fermi sea consists of the two intervals

p_{-} < ∣ p ∣ < p_{+}

. With

p = 2 πn / L

, the requested finite-size ground-state energy is explicitly

E_{GS} = 2 α n \in Z_{> 0} L p_{-} / (2 π) < n < L p_{+} / (2 π) \sum [(\frac{2 πn}{L})^{4} - β (\frac{2 πn}{L})^{2} + μ] .

(4)

The factor two counts opposite momenta. Endpoints are excluded from this sum because they have zero energy; when an endpoint is allowed, its optional occupation adds ground-state degeneracy. A narrow negative-energy interval can contain no allowed finite-size momenta, in which case the sum is empty. The zero-momentum mode is always empty since

ϵ (0) = αμ > 0

b

 0  0

Solution

 0  0

For real angles write

u_{p} = cos θ_{p}

and

v_{p} = sin θ_{p}

. For one independent pair

p > 0

, the transformation and its conjugate read

c_{p} = u_{p} α_{p} + i v_{p} α_{- p}^{†}, c_{- p} = u_{p} α_{- p} - i v_{p} α_{p}^{†} .

(1)

Using the canonical anticommutation relations for the

α

modes gives

{c_{p}, c_{p}^{†}} {c_{- p}, c_{- p}^{†}} {c_{p}, c_{- p}} {c_{p}, c_{- p}^{†}} = u_{p}^{2} + v_{p}^{2} = 1, = u_{p}^{2} + v_{p}^{2} = 1, = - i u_{p} v_{p} + i v_{p} u_{p} = 0, = 0.

(2)

The remaining anticommutators vanish for independent momentum pairs. The displayed two-by-two matrix is unitary and its inverse is obtained by changing

i

- i

When the same formula is indexed over both signs of momentum, its first row at

- p

must agree with the conjugate of its second row at

p

. Thus a consistent global convention requires

u_{- p} = u_{p}, v_{- p} = - v_{p}, θ_{- p} = - θ_{p} (mod 2 π) .

(3)

This is the odd-momentum fermionic Bogoliubov angle convention. If one checks only the first-row formulas, their cross anticommutator is

i sin (θ_{p} + θ_{- p})

; its vanishing gives the condition modulo

π

, with possible signs absorbable in the

α

modes. Requiring both rows to describe the same operators fixes the convention above. Merely checking a separate unitary matrix at each momentum misses this redundancy: choosing

θ_{p} = θ_{- p} = π /4

in both first rows would give

{c_{p}, c_{- p}} = i

p = 0

, the annihilation operator and its conjugate belong to one mode. The full displayed formula is consistent only with

v_{0} = 0

; choose

θ_{0} = 0

and

α_{0} = c_{0}

. This mode is not a two-mode pair.

c

 0  0

Solution

 0  0

Group the nonzero modes into independent pairs

p > 0

. For one pair set

ε = ϵ (p) = p^{2} / (2 m) + μ

, and introduce the Nambu spinor

Ψ_{p} = (c_{p}, c_{- p}^{†})^{T}

. Normal ordering using

c_{- p} c_{- p}^{†} = 1 - c_{- p}^{†} c_{- p}

gives

H_{p} = Ψ_{p}^{†} (ε - i Δ i Δ - ε) Ψ_{p} + ε .

(1)

This exactly contains both kinetic terms and the pairing terms once; there is no additional factor of one half because only

p > 0

is summed.

Insert

Ψ_{p} = U_{p} A_{p}

, where

A_{p} = (α_{p}, α_{- p}^{†})^{T}

and

U_{p}

is the Bogoliubov transformation matrix. Direct matrix multiplication yields

U_{p}^{†} h_{p} U_{p} = (ε cos 2 θ_{p} - Δ sin 2 θ_{p} - i (ε sin 2 θ_{p} + Δ cos 2 θ_{p}) i (ε sin 2 θ_{p} + Δ cos 2 θ_{p}) - ε cos 2 θ_{p} + Δ sin 2 θ_{p}) .

(2)

Choose the diagonalizing branch

cos 2 θ_{p} = \frac{ε}{E _{p}}, sin 2 θ_{p} = - \frac{Δ}{E _{p}}, E_{p} = ε^{2} + Δ^{2}, p > 0.

(3)

Then the off-diagonal terms vanish and the upper diagonal entry is

E_{p} > 0

. Extend the angle to negative momenta as in part b; the sine relation just written applies to the positive representative of each pair. The BCS coherence factors can be chosen as

u_{p} = \frac{E _{p} + ε}{2 E _{p}}, v_{p} = - \frac{Δ}{2 E _{p} ( E _{p} + ε )} .

(4)

Here

u_{p}

denotes

cos θ_{p}

, and

v_{p}

denotes

sin θ_{p}

; these expressions are nonsingular because

μ > 0

and

m > 0

. The minus sign of

v_{p}

for positive

Δ

is fixed by the pairing phase in this Hamiltonian.

A second use of the canonical anticommutation relations gives

H_{p} = E_{p} (α_{p}^{†} α_{p} - α_{- p} α_{- p}^{†}) + ε = E_{p} (α_{p}^{†} α_{p} + α_{- p}^{†} α_{- p}) + ε - E_{p} .

(5)

The zero-momentum term was never paired and is simply

μ α_{0}^{†} α_{0}

. Hence

H = p \sum E (p) α_{p}^{†} α_{p} + C, C = p > 0 \sum [ϵ (p) - ϵ (p)^{2} + Δ^{2}],

(6)

with

E (0) = μ, E (p) = ϵ (p)^{2} + Δ^{2} (p \neq = 0) .

(7)

The constant is the ground-state energy. At fixed

L

, its ultraviolet tail converges:

ϵ (p) - E (p) = - Δ^{2} / [ϵ (p) + E (p)] \sim - m Δ^{2} / p^{2}

in one dimension. Replacing

E (0)

μ^{2} + Δ^{2}

would incorrectly pair a spinless mode with itself.

d

 0  0

Solution

 0  0

Since every quasiparticle energy is positive, the ground state is the vacuum of all the

α_{p}

annihilation operators, including

α_{0} = c_{0}

. Its zero-momentum mode is therefore empty. For each positive momentum, the inverse Bogoliubov transformation gives

α_{p} = u_{p} c_{p} - i v_{p} c_{- p}^{†}, α_{- p} = u_{p} c_{- p} + i v_{p} c_{p}^{†} .

(1)

Both annihilate the normalized pair state

(u_{p} + i v_{p} c_{p}^{†} c_{- p}^{†}) ∣0 ⟩

. For example, applying

α_{p}

produces the two cancelling terms

i u_{p} v_{p} c_{- p}^{†} ∣0 ⟩

. Thus

∣Ω ⟩ = p > 0 \prod (u_{p} + i v_{p} c_{p}^{†} c_{- p}^{†}) ∣0 ⟩, c_{0} ∣Ω ⟩ = 0.

(2)

The factors have even fermion parity and commute for disjoint momentum pairs. Each normalized factor is a superposition of zero and two fermions, whose pair-occupation probability is

v_{p}^{2}

. Therefore

⟨ Ω∣ N ∣Ω ⟩ = 2 p > 0 \sum v_{p}^{2} = p > 0 \sum (1 - \frac{ϵ ( p )}{E _{p}}) .

(3)

There is no zero-mode contribution. For fixed

L

, the sum converges since

v_{p}^{2} = O (p^{- 4})

A normalized simple excited state is

∣ k ⟩ = α_{k}^{†} ∣Ω ⟩

for

k > 0

. Its excitation energy is

E_{k}

, and its total momentum is

k

: both terms in

α_{k}^{†} = u_{k} c_{k}^{†} + i v_{k} c_{- k}

add momentum

k

. Acting on the

k, - k

pair gives

α_{k}^{†} (u_{k} + i v_{k} c_{k}^{†} c_{- k}^{†}) ∣0 ⟩ = (u_{k}^{2} + v_{k}^{2}) c_{k}^{†} ∣0 ⟩ = c_{k}^{†} ∣0 ⟩ .

(4)

The affected pair now contains exactly one fermion instead of mean number

2 v_{k}^{2}

; all other pairs are unchanged. Consequently

⟨ k ∣ N ∣ k ⟩ = ⟨ N ⟩_{Ω} + 1 - 2 v_{k}^{2} = ⟨ N ⟩_{Ω} + \frac{ϵ ( k )}{E _{k}} .

(5)

This is the fermion number of a Bogoliubov excitation. Its change in mean fermion number lies between zero and one for

μ > 0

, with the value one at

Δ = 0

. The Bogoliubov quasiparticle coherently mixes adding a fermion and removing the opposite-momentum fermion; it is not an eigen-excitation of the original fermion number operator. Indeed,

[N, H] = 2 i Δ \sum_{p > 0} (c_{p}^{†} c_{- p}^{†} + c_{- p} c_{p}),

(6)

so fermion number is not conserved when

Δ \neq = 0

, while fermion parity is conserved. The generally noninteger change is an expectation value in a coherent paired background, rather than a fractional eigenvalue of the fermion number operator. In the weak-pairing limit at fixed

k

, it tends to one, as expected for an ordinary added fermion.

2

 0  0

a

 0  0

Solution

 0  0

Let

β = 1/ (k_{B} T)

and use the two spin values

+ 1, - 1

as row and column labels. Define the spin-chain transfer matrix

W_{s t} = e^{- β E_{0} (s, t)}, s, t \in {+ 1, - 1} .

(1)

It is a real symmetric matrix because

E_{0} (s, t) = E_{0} (t, s)

; every entry is strictly positive for finite bond energies at

T > 0

. The Boltzmann factor of a spin configuration factorizes into the product

W_{σ_{1} σ_{2}} \dots W_{σ_{L} σ_{1}}

. Summing the intermediate spin labels is exactly matrix multiplication, and the final closure imposes periodic boundary conditions. Thus

Z_{L} = σ_{1}, \dots, σ_{L} \sum j = 1 \prod L W_{σ_{j} σ_{j + 1}} = Tr W^{L} .

(2)

Writing

W = (a b b d)

, its eigenvalues are

λ_{\pm} = \frac{a + d \pm ( a - d ) ^{2} + 4 b ^{2}}{2}, Z_{L} = λ_{+}^{L} + λ_{-}^{L} .

(3)

A positive-entry matrix need not have two positive eigenvalues;

λ_{-}

can be negative. The Perron–Frobenius theorem ensures

λ_{+} > ∣ λ_{-} ∣

here. Therefore the Helmholtz free energy is

F_{L} = - k_{B} T ln Z_{L}

, and its value per spin in the thermodynamic limit is

- k_{B} T ln λ_{+}

b

 0  0

Solution

 0  0

For an open chain there are

L - 1

bonds, with both endpoint Ising spins independently summed. Retain the same spin-chain transfer matrix, but replace the trace by endpoint vectors. With

1 = (1, 1)^{T}

Z_{L}^{open} = 1^{T} W^{L - 1} 1 .

(1)

This follows directly by expanding the product of

L - 1

Boltzmann factors. If

v_{\pm}

are orthonormal eigenvectors of the symmetric matrix

W

, the equivalent spectral decomposition is

Z_{L}^{open} = \sum_{ν = \pm} (1^{T} v_{ν})^{2} λ_{ν}^{L - 1} .

(2)

Fixed endpoint Ising spins replace the corresponding

1

by a basis vector; extra endpoint energies replace its entries by the corresponding Boltzmann factors. The leading bulk Helmholtz free energy per spin agrees with the periodic result, because the boundary changes only subextensive terms. In particular,

L = 1

gives

Z_{1}^{open} = 1^{T} 1 = 2

, as required for one free Ising spin.

c

 0  0

Solution

 0  0

Let

M = L /2

, so that the total number of Ising spins is

3 M

. Write

K = β J

and group each triangle's two base spins and its apex into one cell. The Ising chain with alternating triangular decorations has a triangle bond followed by an undecorated connecting bond. Summing an apex Ising spin first gives a decorated-bond Boltzmann weight

A_{s t} = \sum_{τ = \pm 1} e^{- K [s t + (s + t) τ]} = 2 e^{- Ks t} cosh [K (s + t)] .

(1)

The undecorated bridge has Boltzmann weight

B_{t s^{'}} = e^{- K t s^{'}}

. In the basis

+ 1, - 1

A = (e^{K} + e^{- 3 K} 2 e^{K} 2 e^{K} e^{K} + e^{- 3 K}), B = (e^{- K} e^{K} e^{K} e^{- K}) .

(2)

Each cell contributes

T = A B

, so periodic boundary conditions give

Z_{L} = Tr (A B)^{M} .

(3)

The two matrices have common normalized eigenvectors

(1, 1)^{T} / 2

and

(1, - 1)^{T} / 2

, expressing spin inversion symmetry. Their respective eigenvalues are

a_{+} = 3 e^{K} + e^{- 3 K}, a_{-} = e^{- 3 K} - e^{K}, b_{+} = e^{- K} + e^{K}, b_{-} = e^{- K} - e^{K} .

(4)

Consequently the cell transfer matrix has eigenvalues

λ_{+} λ_{-} = (3 e^{K} + e^{- 3 K}) (e^{- K} + e^{K}) = 3 + e^{- 4 K} + e^{- 2 K} + 3 e^{2 K}, = (e^{- 3 K} - e^{K}) (e^{- K} - e^{K}) = e^{- 4 K} - e^{- 2 K} - 1 + e^{2 K},

(5)

and the finite-size partition function is

Z_{L} = λ_{+}^{L /2} + λ_{-}^{L /2} .

(6)

For

K > 0

λ_{-} \geq 0

and

λ_{+} - λ_{-} = 4 + 2 e^{- 2 K} + 2 e^{2 K} > 0

, so only

λ_{+}

survives in the thermodynamic limit. Divide the Helmholtz free energy by

3 M

, the number of sites rather than the number of cells or base spins:

f (T) = - \frac{k _{B} T}{3} ln λ_{+} = - \frac{k _{B} T}{3} ln (3 + e^{- 4 β J} + e^{- 2 β J} + 3 e^{2 β J}) .

(7)

As a counting check,

K = 0

gives

λ_{+} = 8

λ_{-} = 0

, hence

Z_{L} = 8^{M} = 2^{3 M}

, exactly the number of independent Ising spin configurations. This model has separated decorated bonds and should not be confused with the corner-sharing sawtooth Ising chain, which has a different site count.

d

 0  0

Solution

 0  0

For low temperature, factor out the largest Boltzmann factor from the dominant eigenvalue:

λ_{+} = 3 e^{2 K} (1 + e^{- 2 K} + \frac{1}{3} e^{- 4 K} + \frac{1}{3} e^{- 6 K}) .

(1)

Thus the Helmholtz free energy per site has the expansion

f = - \frac{2 J}{3} - \frac{k _{B} T}{3} ln 3 - \frac{k _{B} T}{3} e^{- 2 K} + O (k_{B} T e^{- 4 K}) .

(2)

Differentiate at fixed

J

, remembering that

K = J / (k_{B} T)

. The entropy per site is

s = \frac{k _{B}}{3} ln 3 + \frac{k _{B}}{3} (1 + 2 K) e^{- 2 K} + O (k_{B} (1 + K) e^{- 4 K}), T \to 0^{+} lim s (T) = \frac{k _{B}}{3} ln 3.

(3)

The nonzero residual entropy comes from an exponentially large ground-state degeneracy, generated by geometrical frustration. A triangle with antiferromagnetic energy

J [s t + (s + t) τ]

has minimum energy

- J

: if

s = t

, the apex must have the opposite sign; if

s = - t

, either apex sign is permitted. The two all-equal configurations instead have energy

3 J

. Thus exactly one of a minimum-energy triangle's bonds is frustrated. The bridge can simultaneously minimize its energy at

- J

by requiring

t = - s^{'}

, where

s^{'}

is the next cell's first base spin.

For fixed

s

and

s^{'}

, the number of allowed apex choices is therefore two when

s = s^{'}

and one when

s = - s^{'}

. The zero-temperature counting transfer matrix on successive first base spins is

G = (2112), g_{M} = Tr G^{M} = 3^{M} + 1, E_{GS} = - 2 J M = - J L .

(4)

The eigenvalues of

G

are

3

and

1

. Dividing

k_{B} ln g_{M}

by the

3 M

sites and taking the thermodynamic limit gives

k_{B} ln 3/3

, independently confirming the derivative calculation. This does not assert a unique ordered ground state: the extensive choice of frustrated bonds is precisely what leaves entropy at zero temperature. At high temperature, all Ising spins become independent and

s \to k_{B} ln 2

per site.

Figure 1.
Entropy of the Ising chain with alternating triangular decorations and its ground-state degeneracy
. **Details**.

2017 (A15757W1)

 0  0

Source paper: www.whatdotheyknow.com/request/physics_past_papers_21/response/3037379/attach/2/FOI%20202505%20441%20Response.pdf?cookie_passthrough=1. The original cover dates this paper to 13 June 2017. We use

ℏ = 1

and lattice spacing one. Both printed interaction sums omit grouping parentheses; the solutions explicitly use the symmetric nearest-neighbour interpretation. The unused

h

in question 1 is not treated as an additional interaction.

1

 0  0

a

 0  0

Solution

 0  0

Put

β = (k_{B} T)^{- 1}

and label the two spin values in the order

+ 1, - 1

. Assign each bond its Boltzmann weight,

V_{s t} = e^{- β E_{0} (s, t)}, V = (a b b d), s, t \in {+ 1, - 1} .

(1)

The symmetry of the bond energy makes

V

a real symmetric matrix. For a fixed configuration the product of successive matrix entries is its Boltzmann weight. Summing over all spins and identifying the last index with the first therefore gives

Z_{L} = σ_{1}, \dots, σ_{L} \sum V_{σ_{1} σ_{2}} \dots V_{σ_{L} σ_{1}} = tr V^{L} = λ_{+}^{L} + λ_{-}^{L},

(2)

where

λ_{\pm} = \frac{a + d \pm ( a - d ) ^{2} + 4 b ^{2}}{2} .

(3)

This replaces a sum over

2^{L}

configurations by the powers of a two-by-two matrix. For finite bond energies at

T > 0

, every matrix entry is positive, so the Perron–Frobenius theorem gives

λ_{+} > 0

and

∣ λ_{-} ∣ < λ_{+}

. Hence the Helmholtz free energy per spin tends to

- k_{B} T lo g λ_{+}

. A diagonal local observable

O

can be inserted at one site:

⟨ O_{1} ⟩ = tr (O V^{L}) / Z_{L}

. Its limiting average is the expectation in the normalized eigenvector of

λ_{+}

b

 0  0

Solution

 0  0

The sum needs a stated interpretation. The original equation leaves the second bond outside explicit parentheses, with its index then formally free. The natural translation-invariant, equal-bond interpretation used here is the energy

- J \sum_{j} τ_{j} (σ_{j} + σ_{j + 1})

. The subsequently mentioned

h

has no term in that energy, so no magnetic field or single-ion anisotropy is added. Such an extra term would define another model and change the answer.

Let

K = β J

. Sum over each decorating spin before summing over the remaining spins. The decimation of an alternating mixed-spin chain gives the effective transfer matrix:

V_{s t} = \sum_{t_{0} = - 1}^{1} e^{K t_{0} (s + t)} = 1 + 2 cosh [K (s + t)] = (1 + 2 cosh 2 K 3 3 1 + 2 cosh 2 K)_{s t} .

(1)

No on-site entropy has been omitted: the term

1

is the weight of the zero decorating spin. The normalized eigenvectors are

v_{\pm} = (1, \pm 1)^{T} / 2

, with eigenvalues

λ_{+} = 4 + 2 cosh 2 K = 2 + 4 cosh^{2} K, λ_{-} = 2 cosh 2 K - 2 = 4 sinh^{2} K .

(2)

There are

L

transfer steps but

2 L

physical sites, so

Z_{L} = λ_{+}^{L} + λ_{-}^{L}, f = L \to \infty lim \frac{- k _{B} T lo g Z _{L}}{2 L} = - \frac{k _{B} T}{2} lo g (4 + 2 cosh 2 β J) .

(3)

The free energy per two-site unit cell is twice this value. At

J = 0

, the formula gives

Z_{L} = 6^{L}

, as required by two states on one sublattice and three on the other. As

T \to 0

, the free energy per physical site tends to

- J

, the energy of an aligned chain with two satisfied bonds per unit cell. These checks also fix the factor of two in the site count.

c

 0  0

Solution

 0  0

Insert

D = diag (1, - 1)

at the first undecorated site. In the eigenbasis of the transfer matrix,

D

exchanges

v_{+}

and

v_{-}

, and both diagonal matrix elements vanish. Explicitly,

⟨ σ_{1} ⟩_{L} = \frac{tr ( D V ^{L} )}{tr V ^{L}} = \frac{λ _{+}^{L} v _{+}^{T} D v _{+} + λ _{-}^{L} v _{-}^{T} D v _{-}}{λ _{+}^{L} + λ _{-}^{L}} = 0.

(1)

Thus its large-

L

limit is zero at every fixed

T > 0

. The result is exact even before taking that limit. Equivalently, simultaneous reversal of all spins pairs configurations of equal energy and opposite

σ_{1}

. Positivity of the transfer matrix at fixed positive temperature prevents a degenerate dominant eigenvalue and spontaneous magnetization in this one-dimensional zero-field chain. Selecting one aligned state at strictly zero temperature is a different order of limits.

d

 0  0

Solution

 0  0

Insert the decorating-spin observable into the one bond that contains it. The decorated-spin occupation insertion uses the matrix:

Q_{s t} = \sum_{t_{0} = - 1}^{1} t_{0}^{2} e^{K t_{0} (s + t)} = 2 cosh [K (s + t)] = (2 cosh 2 K 2 2 2 cosh 2 K)_{s t} .

(1)

The partition function still contains

L

decorating-spin sums; one of them has been replaced by this weighted sum. Hence

⟨ τ_{1}^{2} ⟩_{L} = \frac{tr ( Q V ^{L - 1} )}{Z _{L}} = \frac{q _{+} λ _{+}^{L - 1} + q _{-} λ _{-}^{L - 1}}{λ _{+}^{L} + λ _{-}^{L}}, q_{+} = 2 cosh 2 K + 2, q_{-} = 2 cosh 2 K - 2.

(2)

Taking the dominant eigenvector gives

L \to \infty lim ⟨ τ_{1}^{2} ⟩_{L} = \frac{q _{+}}{λ _{+}} = \frac{cosh 2 K + 1}{cosh 2 K + 2} = \frac{2 cosh ^{2} K}{1 + 2 cosh ^{2} K} .

(3)

This is the probability that the decorating site is in either nonzero state. It tends to

2/3

at high temperature, when its three states are equally likely, and to one at low temperature, when the nonzero aligned states gain bond energy. Unlike the magnetization, this even observable is not forced to vanish by spin reversal.

- J

. Write

H_{f} = ∣ h ∣

and, for

h \neq = 0

s = sgn h

. The source explicitly fixes

J > 0

but does not separately constrain

h

in this question; keeping

H_{f}

covers either field direction.

For classical vectors of length

S

, define

q_{j} = (S_{j}^{x}, S_{j}^{y})

. The Cauchy-Schwarz inequality gives

q_{j} \cdot q_{j + 1} \leq ∣ q_{j} ∣∣ q_{j + 1} ∣ \leq S^{2}

, and

h S_{j}^{x} \leq H_{f} S

. Each term is therefore bounded below, so

E_{cl} \geq - L J S^{2} - L H_{f} S^{2} .

(1)

Uniform alignment along the field saturates every bound:

S_{j} = (s S, 0, 0), E_{cl, min} = - L (J + H_{f}) S^{2} (h \neq = 0) .

(2)

Equality requires all planar vectors to be parallel with length

S

, and the nonzero field then fixes their direction. At

h = 0

, every uniformly oriented vector in the

x y

plane is a classical ground state:

S_{j} = S (cos ϕ, sin ϕ, 0), E_{cl, min} = - L J S^{2} .

(3)

A nonuniform configuration cannot improve these simultaneously saturated bond bounds, including on the periodic ring.

b

 0  0

Solution

 0  0

The Holstein–Primakoff transformation realizes a spin algebra using bosonic creation operators and bosonic annihilation operators. A boson counts a unit of spin deviation from the local positive quantization axis:

S^{z} = S - n

with

n = a^{†} a

. The square root in

S^{+}

supplies the correct spin-ladder matrix elements. In particular,

S^{+} ∣ n ⟩ = n (2 S - n + 1) ∣ n - 1 ⟩, S^{-} ∣ n ⟩ = (n + 1) (2 S - n) ∣ n + 1 ⟩ .

(1)

Their commutator is

2 (S - n)

on every physical number state, reproducing the Spin commutation relations. Although the oscillator Fock space is infinite, the physical subspace is only

n = 0, 1, \dots, 2 S .

(2)

The order of the square root and oscillator operator matters in obtaining these coefficients.

When the spin deviations are dilute, expand the square root in

n / (2 S)

. The leading transverse operators are linear in the bosonic creation operators and bosonic annihilation operators and turn exchange interactions into a quadratic Hamiltonian, which can be diagonalized. Higher powers describe interactions between spin waves. The method therefore organizes corrections by inverse powers of

S

about an ordered classical configuration. Large

S

alone is insufficient: the average depletion must remain small compared with

S

. Possible problems include the occupation constraint after truncation, noncollinear order requiring site-dependent axis rotations, zero modes, and infrared fluctuations that destroy the assumed dilute expansion.

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