Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 (Theoretical Physics 2; 23 April 2014)

An initially occupied, isolated instantaneous energy eigenstate follows its changing eigenspace, acquiring a dynamical phase and generally a Berry phase. For smooth nondegenerate evolution, a useful local smallness condition isA nonvanishing spectral gap and suitably smooth, slow driving throughout the evolution are needed; the local inequality alone is not a universal long-time error bound. Gap closure invalidates the simple nondegenerate approximation. A separated degenerate eigenspace may mix internally even when transitions out of it are suppressed.

Set ${\vartheta }_{\alpha }(t)={\hslash }^{-1}{\int }^t{E}_{\alpha }(s)ds$. Differentiating the expansion produces ${\dot{a}}_{\beta }|{\phi }_{\beta }⟩+a_{\beta }|{\dot{\phi }}_{\beta }⟩-i{E}_{\beta }{a}_{\beta }|{\phi }_{\beta }⟩/\hslash$ for each term. The energy term cancels $H|\Psi ⟩$ in the Schrödinger equation. Projecting the remainder with $⟨{\phi }_{\alpha }|$ and using orthonormality yieldsThe diagonal term is purely imaginary and supplies the geometric phase. Off-diagonal instantaneous eigenstate couplings transfer population; differentiating the eigenvalue equation gives $⟨m|\dot{n}⟩=⟨m|\dot{H}|n⟩/(E_n-E_m)$ when the levels are distinct.

Use normalized instantaneous modes ${\phi }_p(x,t)=\sqrt{2/L}\sin (p\pi x/L)$ and expand $\Psi ={\sum }_p{a}_p{e}^{-i{\vartheta }_p}{\phi }_p$. Differentiating at fixed $x$ givesFor $n\ne p$, integration usesso the moving-wall derivative coupling isFor $n=p$ the normalization derivative and the remaining integral cancel exactly, giving zero. The moving integration endpoint contributes no boundary term because every mode vanishes there. ThereforeThe real coupling matrix is antisymmetric, as required by conservation of total probability.

For first-order excitation in a linearly expanding infinite well, retain only the ground-state amplitude on the right. For the second level, the coupling coefficient is $4\dot{L}/(3L)$. With $L=vt$,The relative dynamical phase is $-\alpha /t$ plus a constant. Up to an irrelevant phase,The lower endpoint is an oscillatory improper limit: substituting $u=\alpha /s$ gives ${\int }_{\alpha /t}^{\infty }{e}^{-iu}du/u$. A physical initial state cannot occupy a zero-width well. Prepare the ground state at $t_0>0$, integrate from $t_0$, and then take the formal $t_0\downarrow 0$ limit of the probability. For $t/\alpha \ll 1$, endpoint asymptotics give $P\sim 16t^2/(9{\alpha }^2)$; the first-order expression is reliable only while total excitation remains small.

The quantum propagator is the position-space kernel of the time-evolution operator: $\Psi (x,t)=\int K(x,t|x^{\prime },t^{\prime })\Psi (x^{\prime },t^{\prime })d{x}^{\prime }$. For a time-independent Quantum Hamiltonian, $U=e^{-iH(t-t^{\prime })/\hslash }$. Inserting a complete orthonormal energy eigenbasis givesA continuous spectrum requires the corresponding integral. At equal times, completeness makes this kernel a Dirac delta function.

Let $I=m{R}^2$, $T=t-t^{\prime }>0$, and $\Delta =\theta -{\theta }^{\prime }$. Periodic angular wave functions have modes $e^{il\theta }/\sqrt{2\pi }$ with integer $l$ and energies ${\hslash }^2{l}^2/(2I)$. Normalizing the kernel with respect to $d{\theta }^{\prime }$ givesIf the integration measure is arc length $Rd{\theta }^{\prime }$, the corresponding kernel is $K_{\theta }/R$. This measure choice accounts for the otherwise ambiguous radius prefactor.

Lift the circle to its real-valued angular covering coordinate. The free classical equation is $\ddot{\theta }=0$. For each winding number $p\in \mathbb{Z}$, the endpoint lift is $\theta +2\pi p$, andPositive and negative windings are distinct classical paths with the same physical endpoints. Reducing the angle modulo $2\pi$ returns their trajectories on the ring.

The angular action is $S=\int I{\dot{\theta }}^{2}ds/2$. In each winding sector, write $\theta ={\theta }_p+\xi$ with $\xi$ zero at both endpoints. The cross term vanishes and the Gaussian fluctuation integral is the free-particle prefactor with mass replaced by $I$. ThusEquivalently, multiply the line kernel for displacement $R(\Delta +2\pi p)$ by $R$ to change its measure from arc length to angle. The quadratic action makes the classical-path sum exact, rather than merely a stationary phase approximation. Real-time sums are understood with the usual negative imaginary part of $T$ and its boundary limit.

Take $\alpha =I/(i\hslash T)$ in the supplied Poisson summation identity, initially with a convergent regulator. Its left side is the winding sum. Multiplying its right side by the prefactor in part d givesConsequentlyRelabelling $q=-l$ gives exactly the spectral answer. The square-root branch is fixed by continuation from positive imaginary-time damping; the limiting kernel has the required periodic delta-function normalization.

For an interaction of finite range $b$, higher partial waves are suppressed at $kb\ll 1$ by the centrifugal barrier, away from exceptional higher-wave resonances. Outside the potential, the zero-energy radial S-wave scattering equation is $u^{\prime \prime }=0$, where $\psi =u/r$. Its solution is proportional to $r-a$, so normalization of the constant term givesThe scattering length $a$ is the extrapolated zero of $u(r)$. At nonzero small $k$, this form applies in the overlap region $b\ll r\ll k^{-1}$; it is not the far-field limit at fixed $k$. Equivalently, $k\cot {\delta }_0\to -1/a$ and $f_0\to -a$.

Put $d=|{\mathbf{r}}_1-{\mathbf{r}}_2|$, $g=e^{ikd}/d$ and $s_j=e^{i{\mathbf{k}}_i\cdot {\mathbf{r}}_j}$. In the range-to-wavelength overlap region around centre $j$, the outgoing wave's singular and constant terms are $c_j/(k{\rho }_j)+i{c}_j$. Matching their ratio to the Bethe-Peierls boundary condition yieldsDefine $A_j=1+ik{a}_j$ and $D=A_1{A}_2-a_1{a}_2{g}^2$. Solving the coupled system givesThis is the two-centre scattering-length model result, retaining the outgoing-wave $ik$ term and all repeated exchange between the two centres. If terms of order $k{a}_j$ are neglected in the local matching, set $A_1=A_2=1$ in these formulae; if also $kd\ll 1$, set $g\simeq 1/d$ and $s_j\simeq 1$. Away from a vanishing denominator this gives $c_1\simeq -k{a}_1(1-a_2/d)/(1-a_1{a}_2/d^2)$ and its exchanged partner. The isolated-centre limit is $c_j=-k{a}_j{s}_j/(1+ik{a}_j)$. Finite-range corrections require additional effective-range data, not supplied here.

At large $r$, $|\mathbf{r}-{\mathbf{r}}_j|=r-\hat{\mathbf{r}}\cdot {\mathbf{r}}_j+O(r^{-1})$. Retain the displacement in the phase, but replace the denominator by $r$. ThereforeThe differential scattering cross-section is $|f{|}^2$ for elastic scattering with this unit incident plane-wave normalization. interference depends on outgoing direction and centre separation as well as the incoming phases already contained in $c_j$.

Choose the polar axis along ${\mathbf{r}}_1-{\mathbf{r}}_2$. ThenUsing the angular interference integral for two point scatterers, integrating the two self terms and the interference terms givesFor $kd\ll 1$, the amplitudes add coherently and $\sigma \to 4\pi |c_1+c_2{|}^2/k^2$; when $kd\gg 1$ the angle-integrated interference is suppressed. These limits hold for fixed incoming direction; no incoming-orientation average has been silently imposed.

With $\beta =(k_{B}T{)}^{-1}$, the normalized canonical density operator isIts eigenvalues are the Boltzmann probabilities and its trace is one.

The density operator is diagonal in the oscillator number basis, so $⟨a^2⟩=⟨a^{†2}⟩=0$. With $\bar{n}=⟨a^{†}a⟩$ and $a{a}^{†}=a^{†}a+1$,HenceEquality holds even at zero temperature, when each contribution is $\hslash \omega /4$. Classical equipartition alone would miss this zero-point energy.

Set $q=e^{-\beta \hslash \omega }$. The oscillator populations are $(1-q)q^n$, and differentiating the geometric series gives $\bar{n}=q/(1-q)$. The canonical oscillator coordinate variance follows from $2\bar{n}+1=(1+q)/(1-q)=\coth (\beta \hslash \omega /2)$, andIt approaches $\hslash /(2m\omega )$ at zero temperature and $k_{B}T/(m{\omega }^2)$ at high temperature, agreeing respectively with ground-state fluctuations and classical equipartition.

Write $T_p=p^2/(2m)$ and $V=m{\omega }^2{x}^2/2$. Dropping commutators of order ${\beta }^2$ permits $e^{-\beta H}\simeq e^{-\beta V}{e}^{-\beta T_p}$. With $⟨x|p⟩=e^{ipx/\hslash }/\sqrt{2\pi \hslash }$, the momentum Gaussian givesIntegrating the diagonal yields $Z\simeq 1/(\beta \hslash \omega )$, so the leading high-temperature density kernel isThis unsymmetrized product has an apparent off-diagonal asymmetry at the omitted order; reversing the product puts $x^{\prime }$ in the potential factor. A manifestly Hermitian approximation uses $e^{-\beta V/2}{e}^{-\beta T_p}{e}^{-\beta V/2}$, replacing $\beta m{\omega }^2{x}^2/2$ by $\beta m{\omega }^2(x^2+x^{\prime 2})/4$ with the same leading normalization. The condition is $\beta \hslash \omega \ll 1$; the Gaussian off-diagonal width is of order $\hslash \sqrt{\beta /m}$.

orthogonality of angular modes makes the one-body term $\mathcal{E}{a}_1^{†}{a}_1$, whereThe contact-interaction integral supplies $(U/4\pi ){\delta }_{l_1+l_2,l_3+l_4}$, expressing angular momentum conservation. Within modes zero and one, the four mixed permutations each reduce to $a_1^{†}{a}_0^{†}{a}_0{a}_1$. The two remaining terms have all four indices equal. ThusThis is the projected two-mode Quantum Hamiltonian; the full interaction could scatter into excluded modes, so the truncation is an assumed approximation rather than an exact invariant subspace of the complete ring problem.

The occupation state is an eigenstate of the projected Quantum Hamiltonian. Since $a_j^{†}{a}_j^{†}{a}_j{a}_j=N_j(N_j-1)$,Equivalently, the bracket is $N(N-1)+2N_0{N}_1$, where $N=N_0+N_1$; there is no self-interaction for a single particle.

Put $C=\cos (\chi /2)$ and $S=\sin (\chi /2)$. Commutativity of the two creation operators and the binomial theorem giveThusThe state is a fixed-number two-mode condensate with binomial mode occupations; it is not a product of two independent Poisson-distributed Glauber coherent states.

Binomial factorial moments give $⟨N_1(N_1-1)⟩=N(N-1)p^2$, its mode-zero counterpart $N(N-1)(1-p{)}^2$, and $⟨N_0{N}_1⟩=N(N-1)p(1-p)$. Consequently the exact projected expectation isSince $p(1-p)={\sin }^2\chi /4$ and $n=N/(2\pi )$,The finite-$N$ correction is simply the replacement $N\mapsto N-1$ in the interaction term. Number correlations must be retained to obtain that correction.

With $J=\mathrm{diag}(1,-1)$ these matrices obey $M^{†}JM=J$ and $\det M=1$, identifying the group as SU(1,1). Products preserve the displayed conjugate-entry pattern and determinant; the identity belongs, andalso belongs. Multiplication and inversion are smooth. The real constraint $|\alpha {|}^2-|\beta {|}^2=1$ has nonzero gradient on its level set, so the matrices form a smooth real three-dimensional manifold. Associativity is inherited from matrix multiplication. These facts establish a Lie group, which is noncompact because $|\beta |$ is unbounded.

At first order write $\alpha =1+iz$ and $\beta =x+iy$ with real $x,y,z$. The constraint forces the real part of the change in $\alpha$ to vanish. ThenThus the three supplied generators span the tangent space over the reals. Multiplying their matrices givesFor example, $m_1={\sigma }_x/2$, $m_2={\sigma }_y/2$, $m_3=i{\sigma }_z/2$ makes the noncompact signs explicit.

Set $P=a^{†}{b}^{†}$ and $Q=ab$. Independent bosonic canonical commutation relations give $[Q,P]=N_a+N_b+1=2K_3$, while $[K_3,P]=P$ and $[K_3,Q]=-Q$. Substituting $K_1=(P+Q)/2$ and $K_2=(P-Q)/(2i)$ yieldsThereforewhich matches the matrix Lie algebra. This boson-pair realization of SU(1,1) differs from the number-conserving Schwinger boson representation of $SU(2)$.

The pair products are $PQ=N_a{N}_b$ and $QP=(N_a+1)(N_b+1)$. Hencewhereas $K_3=(N_a+N_b+1)/2$. Subtraction givesThe number difference commutes with pair creation, pair annihilation and $K_3$, since each pair operation changes both occupations equally. Its boson-pair Casimir operator $C$ therefore commutes with every generator. In a fixed-difference sector $N_a-N_b=d$, the Casimir eigenvalue is $(d^2-1)/4$, including $-1/4$ for equal occupations.