Theoretical physics major option course of the University of Oxford (Physics Part C major option, C6)

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Source bundle pages 48–54; the cover identifies Trinity Term 2022 despite the attachment filename. Solutions use $\hslash =c=1$ except where $\hslash$ is displayed. In Questions 4–6 the Minkowski metric is $\eta =\mathrm{diag}(1,-1,-1,-1)$, and spatial surface terms vanish or cancel under periodic boundary conditions.

For the sawtooth Ising chain, there is a source inconsistency: the printed Ising spin values are $\{-1,r+1\}$, with no definition of $r$. The requested transfer matrix follows for $r=0$, namely the usual Ising model with values $\{-1,+1\}$. For arbitrary $r$, summing an apex spin actually givesFor example, $r=1$, $s=s^{\prime }=-1$ and $\beta \bar{J}=1$ give the bracket $e^{-2}+e^4$, whereas the claimed hyperbolic cosine gives $e^{-2}+e^2$. Thus the printed identity is false literally. The rest of Question 1 uses the intended symmetric Ising spin values, $r=0$.

For a fixed configuration of base spins, each apex spin appears in only one local Boltzmann factor. The apex spins are therefore conditionally independent, and their finite sums factorize:Only the base spins carry information from one triangle to the next. The apex spins have already been summed out, so no apex index remains in the transfer matrix; the final trace implements the periodic boundary condition.

In the $s=+1,-1$ basis the transfer matrix and its eigenvalues areFor finite positive temperature, $a,b>0$ and ${\lambda }_{+}>|{\lambda }_{-}|$. Consequently $Z={\lambda }_{+}^L+{\lambda }_{-}^L$ and $L^{-1}\log Z\to \log {\lambda }_{+}$ in the thermodynamic limit. There are $2L$ spins, so the Helmholtz free energy per spin isThe factor $1/2$ is essential: one transfer matrix step counts two spins. For completeness, if the undefined $r$ is retained, set $a_r=A_{-1,-1}^{(r)}$, $d_r=A_{r+1,r+1}^{(r)}$ and $b_r=A_{-1,r+1}^{(r)}$. The same thermodynamic limit gives $f_r=-(2\beta {)}^{-1}\log [(a_r+d_r+\sqrt{(a_r-d_r{)}^2+4b_r^2})/2]$ whenever these are two distinct spin values.

Differentiate the Helmholtz free energy at fixed couplings. With $\lambda ={\lambda }_{+}$ from solution, the entropy per spin isIt is particularly useful to writeAt $\bar{J}=J$, this is $3e^{\beta J}+e^{-3\beta J}$, so $f=-J/2-(\log 3)/(2\beta )+o({\beta }^{-1})$. For fixed $\bar{J}>J$, the first exponential dominates alone, giving $f=(J-2\bar{J})/2+o({\beta }^{-1})$. Hence the residual entropy isThe finite positive difference $\bar{J}-J$ must be kept fixed in the second limit; taking it to zero first gives the other answer.

At high temperature, $\lambda \to 4$ and $\beta {\partial }_{\beta }\log \lambda \to 0$, whenceAll $2^{2L}$ configurations become equally probable because their Boltzmann factors tend to one. The two available states per spin, rather than the coupling values, determine this entropy.

For one triangle, equal base spins $s=s^{\prime }=\sigma$ give energy $J+2\bar{J}\sigma \bar{s}$. Their minimum occurs at $\bar{s}=-\sigma$ and is $J-2\bar{J}$. Opposite base spins give energy $-J$, independently of the apex spin. The all-equal configurations have energy $J+2\bar{J}$.

When $\bar{J}=J$, the six configurations with precisely one spin different from the other two all have minimum energy $-J$. For a fixed first base spin, there is one minimum choice if the next base spin is the same, and two choices of apex if it is opposite. The ground-configuration counting transfer matrix is thereforeThis ground-state counting of the frustrated sawtooth Ising chain proves the extensive geometric frustration and the residual entropy $k_B{\lim }_{L\to \infty }\log {\Omega }_L/(2L)=(k_B/2)\log 3$.

When $\bar{J}>J$, $J-2\bar{J}<-J$. Every triangle must have equal base spins and the opposite apex spin. Shared base spins propagate that choice around the chain, leaving just two global ground states, of energy $L(J-2\bar{J})$. Their entropy per spin is $k_B\log 2/(2L)\to 0$.

For couplings favoring ferromagnetism, write $J=-K$, $\bar{J}=-\bar{K}$ with $K,\bar{K}>0$. Both orientations of an all-equal triangle minimize every bond simultaneously. The common base spins then make the whole chain uniform. There are two global ground states for either $K=\bar{K}$ or $K\ne \bar{K}$, so no corresponding distinction in residual entropy occurs. This statement assumes all bonds remain nonzero.

The continuous internal symmetry is a common rotation around $z$:Both the transverse exchange and the Dzyaloshinskii-Moriya interaction contain one raising and one lowering spin operator, so the phases cancel. Equivalently, $[H,{\sum }_j{S}_j^z]=0$ and the continuous group is U(1).

The discrete internal symmetry is time reversal, $\mathcal{T}{\mathbf{S}}_j{\mathcal{T}}^{-1}=-{\mathbf{S}}_j$, together with $\mathcal{T}i{\mathcal{T}}^{-1}=-i$. Each term in $H$ contains two spin operators, so it is invariant. The operation is antiunitary; treating the reversal of all three spin components as a unitary rotation would violate the spin commutation relations. It reverses the magnetization and interchanges the two easy-axis ordered orientations. A convenient equivalent representative, composed with a $z$ rotation, is $R_z(\pi )\mathcal{T}$, which reverses $S^z$ while preserving $S^x,S^y$ as operators.

The statement about one continuous and one discrete internal symmetry describes generic parameters. At $D=0$, additional unitary $\pi$ rotations about transverse axes are allowed; at $D=u=0$, the continuous internal symmetry enlarges to full spin rotation symmetry. Spatial translations of the uniform chain are separate symmetries of the sites.

The Holstein–Primakoff transformation maps the finite spin basis $|S,m⟩$ to bosonic Fock states $|n⟩$ with $n=S-m$. Thus $n=0$ is the fully polarized reference state, and creating a boson lowers $S^z$ by one. The physical subspace has the Holstein–Primakoff occupation constraint $0\le n\le 2S$, not the whole infinite oscillator Hilbert space.

The specified ordering of the square root givesThese are the exact spin ladder operator matrix elements; in particular, the lowering amplitude vanishes at $n=2S$. The underlying oscillator obeys its usual canonical commutation relation on the enlarged space, while the spin operators preserve the constrained physical subspace. Projecting the oscillator operators themselves to a finite space would change their commutator at its upper endpoint.

Expanding $\sqrt{2S-n}=\sqrt{2S}(1-n/(4S)+\cdots )$ converts an ordered spin problem to approximately free magnons, with interactions in higher orders of $1/S$. Two complications are the nonpolynomial, order-sensitive square roots, which generate interacting bosons, and loss of the occupation bound when a truncated expansion uses the full bosonic Fock space. Large depletion, high temperature, or soft modes near a phase transition can invalidate the assumption $⟨n⟩\ll S$ even when $S$ is large.

Expand around the $+z$ polarized ground state. With $n_j=a_j^{†}{a}_j$ fixed as $S\to \infty$, the Holstein–Primakoff transformation gives $S_j^{+}=\sqrt{2S}{a}_j+O(S^{-1/2})$, $S_j^{-}=\sqrt{2S}{a}_j^{†}+O(S^{-1/2})$ and $S_j^z=S-n_j$. The longitudinal exchange contributesUseThe transverse exchange then gives $-JS(a_{j+1}^{†}{a}_j+a_j^{†}{a}_{j+1})$, and the Dzyaloshinskii-Moriya interaction gives $-iDS(a_{j+1}^{†}{a}_j-a_j^{†}{a}_{j+1})$. Discarding the constant $-LJ(1+u)S^2$ and all $O(S^0)$ interactions yieldsHere periodic boundary conditions make each $n_j$ occur twice in the longitudinal sum. The approximation is the linear spin-wave approximation, controlled by small local magnon occupation compared with $S$.

Choose the unitary discrete Fourier transformThe finite-sum orthogonality identity gives $[b_k,b_q^{†}]={\delta }_{kq}$. It also gives ${\sum }_j{a}_{j+1}^{†}{a}_j={\sum }_k{e}^{-ik}{b}_k^{†}{b}_k$. Substituting $A,B$ from solution therefore gives the dispersion relationUsing the opposite sign in the Fourier transform reverses $k$ and the sine term together, without changing the spectrum.

For $u\gg 1$ every $ϵ(k)>0$, so the ground state of this quadratic Hamiltonian is the bosonic vacuum, $b_k|\mathrm{GS}⟩=0$ for every $k$. In the original spin variables it is the fully $+z$ polarized product state. Its energy is zero after the constant subtraction. Repeating the construction about the fully $-z$ polarized state describes the other easy-axis sector of the original Hamiltonian.

A one-magnon excitation $b_k^{†}|\mathrm{GS}⟩$ has energy $ϵ(k)$ above the reference ground state. A normalized many-magnon Fock state with occupations $n_k$ has excitation energy ${\sum }_k{n}_kϵ(k)$, within the dilute regime of the linear spin-wave approximation. The lowest excitation in this sector has the minimum allowed $ϵ(k)$.

Let $R=\sqrt{J^2+D^2}$ and $k_0=\arctan (D/J)$. Since $J\cos k+D\sin k=R\cos (k-k_0)$, the continuum minimum of the dispersion relation isFor $D>0$, the energy gap closes at a positive $u_c$, at wave vector $k_0$. Below it, the quadratic expansion about the polarized state has a negative mode and is no longer a stable ground state approximation; higher-order terms and a different ordered reference state are needed.

Two qualifications are required by the actual assumptions. First, $D=0$ is permitted, in which case $u_c=0$; it is a concrete counterexample to the printed strict inequality $u_c>0$. Second, a finite ring only has discrete momenta. Its finite-ring spin-wave threshold with antisymmetric exchange isThis approaches $u_c$ in the thermodynamic limit and equals it when $k_0$ is allowed. For example, $L=4$ and $0<D<J$ give $u_c(L)=0$, since the maximum of $1,D/J,-1,-D/J$ is one. Thus the continuum formula and its positive threshold require the large-ring interpretation. The two symmetry-related polarized ground states already have equal exact energy; the calculated energy gap refers to excitations about one chosen orientation.

Here $p_n(t)$ is the probability of occupying state $n$, and $W_{nm}$ for $n\ne m$ is the transition rate from $m$ to $n$. More precisely, over $dt>0$ that transition has probability $W_{nm}dt+o(dt)$. This is a time-homogeneous continuous-time Markov chain: the present state contains all information needed for the next transition, the rates do not depend on time, and finite rates make multiple jumps have probability $o(dt)$. In a physical rate description, untracked coherent dynamics or memory must be negligible on the timescale being modeled.

The master equation separates incoming and outgoing terms:Columns, rather than rows, sum to zero with this indexing. These conditions preserve normalization and nonnegative probabilities. For infinitely many states one also needs a well-defined nonexplosive process. Detailed balance is an additional equilibrium condition, not a general requirement on a master equation.

Write the probability column vector as $\mathbf{p}=(p_0,p_A,p_B{)}^T$, with $0$ denoting the uncharged state. The arrows in the original figure give transitions $0\to A$ at rate $f_1$, $A\to B$ at rate $f_2$, $B\to 0$ at rate $f_3$, and $B\to A$ at rate $f_4$. Exclusion of double occupation means injection occurs only from state $0$. Thus the master equation isFor example, ${\dot{p}}_B=f_2{p}_A-(f_3+f_4)p_B$. Every allowed jump adds to one probability and subtracts from another, which verifies the zero column sums.

Set the time derivatives in the master equation to zero. The first and third rows give $f_1{p}_0=f_3{p}_B$ and $f_2{p}_A=(f_3+f_4)p_B$. Normalization then gives, with $\mathcal{N}=f_2{f}_3+f_1(f_3+f_4)+f_1{f}_2$,These expressions assume positive rates, or a limiting choice for which $\mathcal{N}\ne 0$ and the stationary distribution is unique. If rates vanish so that multiple closed classes appear, the initial condition can matter.

The word “equilibrium” in this part must mean a stationary distribution. With positive $f_1,f_2,f_3$ the one-way contact transitions support a nonzero stationary electric current, so the process is not in thermodynamic equilibrium and does not satisfy detailed balance. In particular, $0\to A$ carries positive probability flow while the direct reverse rate is zero.

Count net transmitted charges per unit time, taking left-to-right as positive. The contact on the left, the internal bond, and the contact on the right give respectivelyThe backward rate $f_4$ must be subtracted in the internal count. Substitution of solution yieldsThe equalities also follow directly from ${\dot{p}}_0=I_R-I_L=0$ and ${\dot{p}}_B=I_{AB}-I_R=0$: there can be no stationary accumulation of electric charge. If each carrier has charge $q$, the electric current is $q{I}_R$. During a stationary interval of duration $t$, the expected net number crossing any one bond is $I_{R}t$. In particular, counting both directions of internal hopping as positive would count traffic, not transmitted charge.

For $f>0$ set $\tau =ft$ and $y(\tau )=p_0(t)$. The master equation becomesUsing $p_A=1-y-p_B$ and $p_B=(y^{\prime }+y)/2$ givesInitially the molecule occupies $B$, so $y(0)=0$ and $y^{\prime }(0)=2$. The characteristic roots are $-2,-3$; the constant particular solution is $1/3$. ConsequentlywhenceIt has the required initial slope $2f$, tends to the stationary uncharged probability $1/3$, and remains nonnegative. As a further check, $p_B=1/6-e^{-2ft}/2+4e^{-3ft}/3$ and $p_A=1/2-e^{-2ft}/2$ are nonnegative and sum with $p(t)$ to one. If $f=0$, no jump occurs and $p(t)=0$, also obtained directly from the boxed expression.

The free complex Proca field has Poincaré symmetry, with $B_{\mu }$ transforming as a spacetime four-vector, and the global internal U(1) symmetryThe free theory is also invariant under charge conjugation $B_{\mu }\leftrightarrow B_{\mu }^{\ast }$ and the usual parity and time reversal transformations of a vector field. The internal real-field description consists of two equal-mass real Proca fields, with an $O(2)$ symmetry containing the phase rotations and charge conjugation.

A gradient shift of $B_{\mu }$ leaves $G_{\mu \nu }$ unchanged, because partial derivatives commute. It changes the mass term byThis is generally nonzero: for example, $B_{\mu }=0$ and real $\lambda =at$ produce $\Delta \mathcal{L}=M^2{a}^2$. Thus a massive Proca field has no such gauge symmetry. The massless kinetic term alone would have it.

Treat $B_{\mu }$ and $B_{\mu }^{\ast }$ as independent variables in the Euler-Lagrange equation. Varying $B_{\nu }^{\ast }$ gives the Proca equationThe divergence of its first term vanishes by antisymmetry and commuting partial derivatives. Since $M\ne 0$, it follows that ${\partial }_{\nu }{B}^{\nu }=0$. Substitution into the original equation givesThe complex-conjugate variation gives ${\partial }_{\mu }{G}^{\ast \mu \nu }+M^2{B}^{\ast \nu }=0$, hence $(\square +M^2)B^{\ast \nu }=0$ and ${\partial }_{\nu }{B}^{\ast \nu }=0$.