Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 (Theoretical Physics 2)

Let $\Delta ={\Omega }_0-\omega$ and $R(t)=e^{-i\omega t{S}_z/\hslash }$. Differentiating $|\Psi ⟩=R|{\Psi }_R⟩$ gives the rotating-frame Hamiltonian operator $R^{†}HR-i\hslash R^{†}\dot{R}$. The angular momentum commutator yields $R^{†}{S}_{\pm }R=e^{\pm i\omega t}{S}_{\pm }$, while $-i\hslash R^{†}\dot{R}=-\omega S_z$. Thus the rotating-frame Rabi Hamiltonian isThis transformation is exact for the circular drive; no rotating-wave approximation has been used.

Define $\Omega =\sqrt{{\Delta }^2+{\Omega }_R^2}$. Squaring the two-by-two matrix gives $H_R^2={\hslash }^2{\Omega }^{2}I/4$, soHere $|\pm {⟩}_R$ is the corresponding normalized eigenstate at $t=0$. In the rotating frame only its quantum phase changes. In the laboratory frame its spin direction also rotates about $z$ at angular velocity $\omega$: $⟨\mathbf{S}{⟩}_{\pm }=\pm \hslash ({\Omega }_R\cos \omega t,{\Omega }_R\sin \omega t,\Delta )/(2\Omega )$. An arbitrary initial state is the same fixed linear combination of these two solutions. If $\Omega =0$, the rotating-frame Hamiltonian operator vanishes and the entire doublet is degenerate; evolution is just $R(t)$.

Take $\omega >0$ and $T=2\pi /\omega$. A full spin one-half rotation gives $R(T)=-I$, an essential sign. The cyclic phase of a spin in a rotating field has factors and total quantum phasesTo separate the adiabatic dynamical phase, set $B=\sqrt{{\Omega }_0^2+{\Omega }_R^2}>0$ and $\cos \theta ={\Omega }_0/B$. Then $\Omega =B-\omega \cos \theta +O({\omega }^2/B)$. The instantaneous laboratory energies are $\pm \hslash B/2$, so the adiabatic dynamical phases are $\mp BT/2$. Subtracting them from ${\Phi }_{\pm }$ givesThe oriented solid angle $\mathcal{A}$ is swept out by the field direction. These are the Berry phases for the eigenstates parallel and antiparallel to that direction. Retaining the spinor sign makes the quantum phase vanish modulo $2\pi$ when the field does not trace a nontrivial loop. The assumption $\omega \ll B$ suffices for adiabatic following; near degeneracy it must not be silently dropped.^[ref]

The angular momentum commutators used in the frame transformation are representation-independent. Therefore $H_R=\Delta S_z+{\Omega }_R{S}_x$ still holds. Rotating its axis onto $z$ gives $2s+1$ eigenvalues and exact laboratory solutionsNow $R(T)=(-1{)}^{2s}I$, so $e^{i{\Phi }_{\mu }}=(-1{)}^{2s}{e}^{-i\mu \Omega T}$. In the adiabatic limit the dynamical phase is $-\mu BT$ andIndeed $2\pi s+2\pi \mu \cos \theta$ and $-2\pi \mu (1-\cos \theta )$ differ by $2\pi (s+\mu )$, an integer multiple of $2\pi$. Integer spins return with a positive rotation sign and half-integer spins with a negative one; the geometric phase need not vanish for either.

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

For the momentum-space propagator in a uniform force, the stated Fourier convention makes $x$ act on momentum wavefunctions as $i\hslash {\partial }_p$. HenceFor $T>0$, the characteristic has $p(s)=p^{\prime }-\alpha (s-t^{\prime })$. Along it the amplitude acquires quantum phase $-{\int }_{t^{\prime }}^{t}p(s{)}^{2}ds/(2m\hslash )$. momentum is translated without a Jacobian, givingThe cubic expression is evaluated on the delta-function support. There it equals $-i[p^{\prime 2}T-\alpha p^{\prime }{T}^2+{\alpha }^2{T}^3/3]/(2m\hslash )$, making both its sign and its continuous $\alpha \to 0$ limit clear. The jump at $T=0$ is the identity momentum kernel.

Use the Dirac delta function to set $p^{\prime }=p+\alpha T$, and write $X=x-x^{\prime }$. The remaining momentum-dependent quantum phase isShifting the integration variable by $\alpha T/2$ reduces the integral to the free Gaussian integral. ThereforeThe force is $-\alpha$, not $+\alpha$. Both the linear endpoint term and the cubic-time term disappear in the zero-force limit. The prefactor remains the free one.

Measure intermediate time $u$ from $t^{\prime }$. The classical path with the fixed endpoints isIt satisfies $m{\ddot{x}}_c=-\alpha$. Direct integration of $m{\dot{x}}_c^2/2-\alpha x_c$ givesFor an arbitrary path write $x=x_c+\eta$ with $\eta (0)=\eta (T)=0$. Integrating the cross term by parts gives $-\int (m{\ddot{x}}_c+\alpha )\eta du=0$. Thus $S[x]=S_c+\int m{\dot{\eta }}^{2}du/2$ exactly. The fluctuation path integral is the free endpoint-fixed one, so its determinant supplies $\sqrt{m/(2\pi i\hslash T)}$. Multiplication by $\theta (T)e^{i{S}_c/\hslash }$ reproduces the previous kernel. This exact classical-path factorization of a quadratic path integral is possible because a linear potential introduces no additional quadratic fluctuation term.

For elastic scattering by a real short-range scalar potential, write $C=m/(2\pi {\hslash }^2)$, ${\mathbf{k}}_i=\mathbf{k}$ and $|{\mathbf{k}}_f|=k$. In the far field, $|\mathbf{r}-{\mathbf{r}}^{\prime }|=r-\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }+O(r^{-1})$ and ${\mathbf{k}}_f=k\hat{\mathbf{r}}$. The outgoing spherical-wave coefficient isSubstitution of the incident plane wave inside this integral givesThe incoming and outgoing flux factors cancel because their speeds agree. Here $\mathbf{q}={\mathbf{k}}_f-{\mathbf{k}}_i$ is a wavevector transfer; the actual momentum transfer is $\hslash \mathbf{q}$. This distinction preserves dimensions. The approximation requires the scattered correction to the interior wave to be small, and does not in general apply to an unscreened long-range potential.

Expanding the modulus squared yields an outgoing-angle factor $e^{-i{\mathbf{k}}_f\cdot \mathbf{d}}$ and an incoming factor $e^{i{\mathbf{k}}_i\cdot \mathbf{d}}$, where $\mathbf{d}=\mathbf{r}-{\mathbf{r}}^{\prime }$. Choosing a polar axis along $\mathbf{d}$ givesConsequentlyThe apparent $d=0$ singularity is removable: $\sin (kd)/(kd)\to 1$. Exchanging $\mathbf{r}$ and ${\mathbf{r}}^{\prime }$ cancels the imaginary part, while nonnegativity follows directly from its origin as an integral of $|f^{(1)}{|}^2$. A complex absorbing potential would require $V({\mathbf{r}}^{\prime }{)}^{\ast }$ instead; the lossless optical theorem comparison below uses real $V$.

The first scattered correction to the incident wave isInserting it into the amplitude gives a second, positive contribution because the two minus signs multiply:Thus the amplitude through second Born order is $f^{(1)}+f^{(2)}+O(V^3)$. The intervening outgoing Green function for the three-dimensional Helmholtz equation describes propagation between the two interactions; replacing it by a real standing-wave kernel would discard the imaginary part needed for unitarity.

In the forward direction ${\mathbf{k}}_f={\mathbf{k}}_i$, the first Born amplitude $-C\int V$ is real. Symmetrize the second-order double integral under $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$ before taking its imaginary part. This replaces $e^{-i{\mathbf{k}}_i\cdot \mathbf{d}}$ by $\cos ({\mathbf{k}}_i\cdot \mathbf{d})$, and givesThe preceding total-cross-section integral equals the same real integral multiplied by $4\pi /k$. ThereforeThis is the optical theorem at second Born order. Comparing only the real first-order forward amplitude with a nonzero second-order scattering cross-section would mix perturbative orders. Terms of order $V^3$ require both the next amplitude order and interference in the scattering cross-section.

There is a state-assumption issue. For a number-conserving Gaussian quantum state with no anomalous contractions, put ${\psi }_i=\psi ({\mathbf{r}}_i)$, $n_i=⟨{\psi }_i^{†}{\psi }_i⟩$ and the one-body density matrix $g_{ij}=⟨{\psi }_i^{†}{\psi }_j⟩$. Wick's theorem pairs each creation operator with an annihilation operator. The direct pairing gives $n_1{n}_2$; the exchanged pairing gives $ϵg_{12}{g}_{21}$, with $ϵ=+1$ for bosons and $-1$ for fermions. Hencenormal ordering removes the equal-position delta-function self-counting term. Fermionic occupation-number product states are Slater determinants and obey this Wick theorem. Independent thermal bosonic modes also obey it, with $N_{\alpha }$ understood as mean occupations.

A bosonic Fock product state does not generally obey the asserted identity. The bosonic number-state corrections to Gaussian density correlations follow from $⟨a_{\alpha }^{†2}{a}_{\alpha }^2⟩=N_{\alpha }(N_{\alpha }-1)$ and giveFor one occupied mode this is $N(N-1)|{\phi }_1{|}^2|{\phi }_2{|}^2$, whereas the uncorrected Gaussian expression gives $2N^2|{\phi }_1{|}^2|{\phi }_2{|}^2$. This counterexample disproves the asserted identity for arbitrary bosonic number-product states. The formulas below give both the Gaussian result, with its additional state assumption, and the exact bosonic number-state alternative. The ensemble distinction is also explicit in a later primary treatment of ideal-gas correlations.^[ref]

Under the Gaussian assumption, $g(\mathbf{r},\mathbf{r})=n(\mathbf{r})$, so the coincident normally ordered density correlation is $(1+ϵ)n(\mathbf{r}{)}^2$. For a homogeneous nonzero density $n$, the large-separation limit is $n^2$, givingThese coincident-point bunching factors describe enhanced bosonic coincidence and excluded same-component fermionic coincidence. More generally the density-normalized ratio $C_{12}/(n_1{n}_2)$ tends to one at large separation. The unnormalized ratio printed in the paper additionally needs matching nonzero local and remote densities; $g_{12}\to 0$ alone does not ensure this. For a bosonic Fock product state, the local normalized value instead is $2-{\sum }_{\alpha }{N}_{\alpha }(N_{\alpha }+1)|{\phi }_{\alpha }(\mathbf{r}){|}^4/n(\mathbf{r}{)}^2$, and even a large-separation factorization must be checked separately.

The three-point density correlation of an ideal quantum gas follows first for a number-conserving Gaussian quantum state. There are six pairings, one for each permutation of three annihilation operators. The identity gives the product of densities; each transposition has sign $ϵ$; both three-cycles have positive sign. ThusThe last two terms are complex conjugates. They equal $2\mathrm{Re}(g_{12}{g}_{23}{g}_{31})$, not an additional signed pair term. This is the permanent of a matrix $(g_{ij})$ for bosons and its determinant for fermions. Wick's theorem supplies the six terms only with the state assumption stated in part (a).

For completeness an exact bosonic Fock-state alternative can be given without that assumption. Let $\mathbf{\ell }$ be a mode-occupation decrement with ${\sum }_{\alpha }{\ell }_{\alpha }=3$. Define $A_{\mathbf{\ell }}(1,2,3)$ as the sum of ${\prod }_{j=1}^3{\phi }_{{\alpha }_j}({\mathbf{r}}_j)$ over all distinct ordered triples having these mode multiplicities. Acting three annihilation fields on the number state leaves mutually orthogonal states for different $\mathbf{\ell }$. ThereforeHere $N^{\underline{\ell }}=N!/(N-\ell )!$, with zero when $\ell >N$. This expression includes repeated-mode occupation corrections absent from the Gaussian permanent; for a single mode it reduces to $N(N-1)(N-2){\prod }_j|\phi ({\mathbf{r}}_j){|}^2$.

At coincident points the Gaussian expression has $g_{ij}=n$. Its six contributions giveWhen all mutual separations become large and the density stays the same nonzero $n$, only $n_1{n}_2{n}_3=n^3$ remains. ConsequentlyThe bosonic value is $3!$ from Gaussian exchange pairings; the fermionic determinant vanishes when its rows coincide. Density-normalized correlations avoid the additional homogeneity assumption. For a specified bosonic number product, part (c)'s occupation sum must replace the six-pairing formula, so neither the universal factor six nor a universal printed ratio follows from that specification alone.

Use $[a_i,a_j^{†}]={\delta }_{ij}$ and define $E_{ij}=a_i^{†}{a}_j$. A direct multiplication givesWith $S_{+}=\hslash a_{\uparrow }^{†}{a}_{\downarrow }$, $S_{-}=\hslash a_{\downarrow }^{†}{a}_{\uparrow }$, this implies $[S_{+},S_{-}]=2\hslash S_z$ and $[S_z,S_{\pm }]=\pm \hslash S_{\pm }$. Resolving the raising and lowering combinations givesThese operators realize the Schwinger boson representation of angular momentum. Each commutes with $\hat{N}=n_{\uparrow }+n_{\downarrow }$, so fixed total occupation is an invariant subspace.

The bosonic commutator is important in $S_{+}{S}_{-}={\hslash }^2{n}_{\uparrow }(n_{\downarrow }+1)$ and $S_{-}{S}_{+}={\hslash }^2{n}_{\downarrow }(n_{\uparrow }+1)$. ThereforeOn the $N$-particle sector,For example $N=0,1,2$ gives spin $0,\frac{1}{2},1$. The total number operator can be retained in the same identity before restricting to a sector.

Total symmetry makes a component depend only on its number $n_{\uparrow }$ of up indices. The possible values are $0,1,\dots ,N$, givingA normalized basis is $|n_{\uparrow },N-n_{\uparrow }⟩$, with $S_z$ eigenvalue $\hslash (n_{\uparrow }-N/2)$. It is precisely one spin-$N/2$ multiplet. These are independent complex amplitudes before normalization and removal of a common quantum phase, rather than $N+1$ independent real physical parameters.

The defining matrix conditions areThe group operation is matrix multiplication. Equivalently $U=a_{0}I-i\mathbf{a}\cdot \mathbf{\sigma }$ with real $a_0,\mathbf{a}$ and $a_0^2+|\mathbf{a}{|}^2=1$, so it has three real parameters and is a compact connected simply connected Lie group. Its Hermitian spin generators obey the commutator from part (a); the corresponding Lie-algebra matrices themselves are traceless and anti-Hermitian.

The literal printed difference is zero, because the same spin-index product appears twice, and zero is invariant under every transformation. The intended SU(2)-invariant antisymmetric spinor contraction uses opposite indices in the second term. With $\epsilon =\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, define $I={\varphi }^T\epsilon \chi$. For any two-by-two matrix,Thus for $U\in SU(2)$,Expanding the determinant identity proves the result without complex conjugating either spinor. This is the invariant alternating tensor behind the spin singlet state. If the two spinors are identical, the contraction vanishes. An antisymmetric internal two-boson state also needs an antisymmetric additional degree of freedom for overall bosonic symmetry; it does not lie in the fully symmetric fixed-two-mode sector discussed above.

A constant electrostatic energy offset is removed locally by the quantum phase substitution $\psi =e^{-i{V}_{0}t/\hslash }{\psi }_0$, since $(i\hslash {\partial }_t-V_0)\psi =e^{-i{V}_{0}t/\hslash }i\hslash {\partial }_t{\psi }_0$. Therefore the two frequency branches becomeThe normalization uses the kinetic frequency ${\omega }_k$, while the canonical energy eigenvalues are $E=V_0\pm \hslash {\omega }_k$. The opposite-frequency branch is not obtained by replacing ${\omega }_k$ by the same shifted positive frequency in both terms. A globally constant offset produces no electric field, whereas different offsets on opposite sides of a step cannot be removed by a single smooth global phase.

Assume a positive-energy incoming particle, $E>m{c}^2$, and a nonzero upward step $V_0>0$. Set $M=m{c}^2$ and $k=\sqrt{E^2-M^2}/(\hslash c)>0$. With the common time factor $e^{-iEt/\hslash }$ removed, take ${\varphi }_{<}(x)=e^{ikx}+r{e}^{-ikx}$ and ${\varphi }_{>}(x)=t{e}^{iqx}$. The right-hand wavenumber satisfies $q^2=[(E-V_0{)}^2-M^2]/(\hslash c{)}^2$; its physical branch is specified in part (c). Integrating the stationary equation across the finite step requires continuity of $\varphi$ and ${\varphi }^{\prime }$, giving $1+r=t$ and $k(1-r)=qt$. ThusThese are wave amplitudes, not flux probabilities. The Noether current of scalar electrodynamics, in a convenient common normalization for a stationary mode, is ${\rho }_Q=(E-V)|\varphi {|}^2/(m{c}^2)$ and $j_Q=(\hslash /m)\mathrm{Im}({\varphi }^{\ast }{\partial }_x\varphi )$. For a real $q$, its charge-current ratios are $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(q/k)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The same statement uses $\mathrm{Re}q$ for a decaying solution. The electrostatic potential couples through the time derivative; it is not a Lorentz-scalar mass shift. A zero step is recovered separately with $q=k$, $r=0$ and $t=1$.

Let $z=E-V_0+i0$. The outgoing analytic branch of Klein-Gordon step momentum iswhere each square root is continued from the upper half-plane. It behaves as $z/(\hslash c)$ at large positive or negative real $z$. The branch points are $E=V_0\pm M$; replacing this product by a principal square root of $z^2-M^2$ can choose the wrong sign below the lower threshold.

I. For $E-V_0>M$, $q>0$ and both transmitted kinetic energy and group velocity are positive. Thus $0\le \mathcal{R}<1$ and $\mathcal{T}=4kq/(k+q{)}^2>0$; the transmission decreases to zero on approaching the upper threshold from above.

II. For $|E-V_0|<M$, write $q=i\kappa$ with $\kappa =\sqrt{M^2-(E-V_0{)}^2}/(\hslash c)>0$. The right wave decays, $r=(k-i\kappa )/(k+i\kappa )$ has modulus one, and $\mathcal{T}=0$. There is an evanescent tail but no flux transmitted to infinity. Its reflection quantum phase changes continuously between the two thresholds.

III. For $E-V_0<-M$, the analytic continuation gives $q=-Q$, where $Q=\sqrt{(E-V_0{)}^2-M^2}/(\hslash c)>0$. The group velocity $c^2\hslash q/(E-V_0)$ is again rightward, while the charge density and transmitted charge current are negative. ThereforeChoosing $q=+Q$ would describe a wavepacket arriving from the right, inconsistent with the chosen incoming channel. As the lower threshold is approached, $q\to 0^{-}$ and $\mathcal{R}\to 1$ from above. This Klein-Gordon step amplification has a real denominator zero at $E=V_0/2$ when $V_0>2M$: there $Q=k$, and no finite stationary solution with unit incoming amplitude satisfies both matching equations. The resulting formal pole must not be described as an infinite single-particle probability. Finite switching, wavepackets and the treatment of the external source require a multiparticle analysis.^[ref]

The negative-kinetic-energy branch is interpreted as an antiparticle channel. In the overlap $m{c}^2<E<V_0-m{c}^2$, particles can escape to the left and oppositely charged antiparticles to the right. This interval exists only when $V_0>2m{c}^2$. The external electrostatic source supplies the energy for pair creation; the reflected particle flux can exceed the incident one because the outgoing state need not contain only the incident particle.

The conserved quantity is signed charge current, not a positive one-particle probability in both regions. Negative $\mathcal{T}$ denotes rightward transport of opposite charge, so it is consistent with $\mathcal{R}>1$ and $\mathcal{R}+\mathcal{T}=1$. A quantized charged field with particle and antiparticle creation operators gives the appropriate interpretation of this Klein paradox. This bosonic result must not be replaced by a fermionic reflection law or interpreted as a failure of charge conservation.^[ref]