Year 4 of the mathematics course of the University of Oxford (Part C, MMath)

Eight to ten units, including a compulsory two-unit dissertation; shared teaching with the MSc in mathematical sciences course of the University of Oxford.

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

For identical bosons the full wave function is symmetric under the particle exchange operator:In the given position-only model this is the spatial parity under the particle exchange operator. If additional spin variables are present, the symmetry applies to the complete state including them.

For identical fermions the full wave function is antisymmetric:In particular it vanishes when the two complete one-particle arguments coincide. Applying this to the position-only model assumes that no independent spin symmetry changes the required spatial parity.

The orthogonal coordinate change preserves the kinetic energy and takes the particle exchange operator to $(x_{+},x_{-})\mapsto (x_{+},-x_{-})$. Thus the spatial wave function is even in $x_{-}$ for bosons and odd for fermions.

The total potential energy and Hamiltonian separate asAssume $1+2\kappa >0$ so the relative oscillator is confining. Its spectrum is $E_{n_{+},n_{-}}=\hslash \omega (n_{+}+1/2)+\hslash \Omega (n_{-}+1/2)$, and its parity in the two-particle relative coordinate is $(-1{)}^{n_{-}}$.

The lowest boson state uses $n_{+}=n_{-}=0$, whereas the lowest fermion state uses $n_{+}=0$, $n_{-}=1$. ThereforeFor an oscillator of frequency $\Omega$, $⟨x_{-}^2{⟩}_n=(n+1/2)\hslash /(m\Omega )$. Since $(x_1-x_2{)}^2=2x_{-}^2$,The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied oscillator wave functions correspond to kinetic energy $P^2/(2m)$; the extra factor $1/2$ printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional kinetic energy dictated by particle mass $m$.

Use the normalized Rayleigh quotient $R(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$. Its real directional derivative in an arbitrary complex direction $\eta$ isIt vanishes for every direction precisely when $(H-R)\psi =0$, since testing $\eta =(H-R)\psi$ gives its squared norm. Conversely an eigenvector makes every derivative zero. Stationarity over all states is stronger than stationarity in a restricted trial family.

Normalize the Gaussian as ${\psi }_{\alpha }=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, with $\alpha >0$. Its moments are $⟨X^2⟩=1/(2\alpha )$, $⟨X^4⟩=3/(4{\alpha }^2)$ and $⟨P^2⟩={\hslash }^2\alpha /2$. The printed potential energy consequently givesPut $a=\alpha \hslash /(m\omega )$. Then $\mathcal{E}/(\hslash \omega )=a/4+1/(4a)+15/(256a^2)$, whose second derivative is $1/(2a^3)+45/(128a^4)>0$. It diverges at both positive endpoints, so its sole stationary point is the unique global minimum. Its equation isThe variational principle therefore yields the optimized upper boundThe printed minimizer is inconsistent with the printed quartic coefficient. At $a=3/2$ the derivative is $5/48>0$, so this is not the stationary Gaussian for that potential energy. It is nevertheless a legitimate trial vector and gives the simple, weaker bound $\boxed{E_0\le 109\hslash \omega /192}$. If the quartic coefficient were instead $5m^2{\omega }^3/(16\hslash )$, four times the printed value, then $a=3/2$ would indeed minimize the family and its energy would be $31\hslash \omega /48$. This possible coefficient correction is identified explicitly rather than silently used. The reusable calculation is Gaussian trial for a quadratic-plus-quartic oscillator.

Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.

The no-cloning theorem states that no fixed deterministic quantum operation can perfectly duplicate every unknown pure state. For a unitary copier with a fixed blank, suppose it cloned two normalized vectors $|a⟩,|b⟩$. Preservation of inner products would give $⟨a|b⟩=⟨a|b{⟩}^2$, impossible for a nonorthogonal distinct pair with overlap strictly between zero and one in modulus. Including an environment does not help: its final overlap has modulus at most one, so one would require $|⟨a|b⟩|\le |⟨a|b⟩{|}^2$. Thus a universal perfect copier is impossible. Orthogonal known alternatives do not contradict this restriction.

Label the three qubits as input $S$, Alice's entangled qubit $A$ and Bob's qubit $B$. Initially their joint state isThe controlled-NOT gate from $S$ to $A$ changes this toApply $H|0⟩=(|0⟩+|1⟩)/\sqrt{2}$ and $H|1⟩=(|0⟩-|1⟩)/\sqrt{2}$ to the first qubit and group by Alice's two computational-basis outcomes. The result isIn expanded form the last two Bob factors are ${\alpha }_0|0⟩-{\alpha }_1|1⟩$ and ${\alpha }_0|1⟩-{\alpha }_1|0⟩$. Their Alice labels must be $10$ and $11$; the repeated $00$ labels in the source are printing errors.

Alice measures her two qubits and sends the two classical bits $(a,b)$ to Bob. Each outcome has probability $1/4$ and Bob's conditional state is $X^b{Z}^a|\psi ⟩$. He applies $\boxed{Z^a{X}^b}$ to recover the input, up to physically irrelevant global phase. Explicitly the corrections for $00,01,10,11$ are $I,X,Z,ZX$. This is quantum teleportation: Alice's input is consumed by measurement, so no second surviving copy is created. Before her classical message, Bob's reduced density operator is $I/2$, independent of the input, so the protocol also supplies no faster-than-light communication.

Use metric signature $(+---)$ and the curvature sign fixed by the stated Newtonian limit. In flat spacetime the source-free Maxwell equations are ${\partial }_a{F}^{ab}=0$ and ${\partial }_{[a}{F}_{bc]}=0$. The stress-energy tensor in mixed form is $T^a{}_b=F^{ac}{F}_{cb}+{\delta }_b^a{F}_{cd}{F}^{cd}/4$. Its divergence isContracting the cyclic Maxwell equation with $F^{ac}$ gives $F^{ac}{\partial }_a{F}_{cb}=-{\partial }_b(F_{cd}{F}^{cd})/4$. The first term also vanishes, proving $\boxed{{\partial }_a{T}^{ab}=0}$. The component $b=0$ expresses local electromagnetic energy conservation, and the spatial components express momentum conservation. Integrating converts them into the usual energy and momentum flux balances.

In curved spacetime the tensor equations areThey agree with the flat equations in coordinates of a local inertial frame at a point, where the connection vanishes, and covariance then extends them to arbitrary coordinates. Antisymmetry makes the homogeneous equation equivalent to $dF=0$. Metric compatibility permits the same contraction proof, now giving $\boxed{{\nabla }_a{T}^{ab}=0}$. In general,which need not vanish. Covariant local balance remains valid; ordinary coordinate components do not provide a universal conserved global energy-momentum vector in a curved geometry. Additional spacetime symmetries can furnish conserved currents. A locally inertial frame removes the connection at a point, not the curvature or the global obstruction.

The alternating array with fixed component ${ϵ}_{0123}=1$ is appropriate in oriented inertial special-relativistic coordinates, whose proper Lorentz transformations have determinant one. It is not itself a covariant tensor under general coordinate changes. For an oriented Lorentzian metric, replace it by the metric metric volume tensor ${\epsilon }_{abcd}=\sqrt{-\det g}[abcd]$. The genuine dual isThis is the Hodge star; the determinant factor, orientation and raising conventions are essential. Even in flat spacetime arbitrary curvilinear coordinates require the metric volume tensor, so the distinction is about coordinate covariance rather than curvature alone.

The isolated equation for $R_{00}$ singles out a coordinate component and cannot be a generally covariant law. Replacing it by a proportionality $R_{ab}\propto T_{ab}$ still fails: the contracted Bianchi identity gives ${\nabla }^a{R}_{ab}={\nabla }_{b}R/2$, whereas stress-energy conservation requires ${\nabla }^a{T}_{ab}=0$. Such a law would force $R$ constant and hence impose an unphysical restriction on a varying relativistic dust density. Missing units or a coupling constant in the proposed proportionality would not repair this structural problem.

The Einstein tensor $G_{ab}=R_{ab}-R{g}_{ab}/2$ is divergence-free. Thereforeis consistent with covariant stress-energy conservation in the source's curvature convention. Its trace gives $R=8\pi GT$ and hence $R_{ab}=-8\pi G(T_{ab}-T{g}_{ab}/2)$. For slowly moving relativistic dust with $V^a{V}_a=1$, $T_{00}\simeq \rho$, $T\simeq \rho$, $g_{00}\simeq 1$, so $R_{00}\simeq -4\pi G\rho$, exactly matching $-\Delta \varphi$ and Poisson's equation.

Finally, electromagnetic stress-energy is trace-free: $T^a{}_a=F^{ac}{F}_{ca}+F_{cd}{F}^{cd}=0$ by antisymmetry. The traced field equation then gives $\boxed{R=0}$ for a purely electromagnetic source. This does not say $R_{ab}=0$ or that spacetime is flat; its trace-free Ricci curvature can still be nonzero.

Under a smooth coordinate change $x\mapsto x^{\prime }$, a contravariant vector obeysThe scalar-contraction hypothesis gives $U_a^{\prime }(\partial x^{\prime a}/\partial x^b)T^b=U_b{T}^b$ for every choice of $T^b$. Equality of all coefficients and multiplication by the inverse Jacobian matrix give $\boxed{U_a^{\prime }=(\partial x^b/\partial x^{\prime a})U_b}$, the covector law.

Let $V^a=d{x}^a/ds$ be a timelike geodesic tangent. Metric compatibility and the geodesic equation giveA Killing vector therefore gives a conserved contraction. Conversely, if that contraction is conserved along every timelike geodesic, then at any point its symmetric tensor $S_{ab}={\nabla }_{(a}{T}_{b)}$ satisfies $S_{ab}{V}^a{V}^b=0$ for every unit timelike initial tangent. Homogeneity extends this to the entire timelike cone, an open set of tangent vectors. A quadratic polynomial vanishing on an open set has every coefficient zero, so $\boxed{{\nabla }_{(a}{T}_{b)}=0}$. Considering a single selected geodesic would not suffice. This is the Converse geodesic characterization of a Killing vector.

For the given exterior transformation, differentiating at fixed $r$ givesThe transformation also gives $UV=(2m-r)e^{r/(2m)}$. This determines $r$ smoothly for $r>0$ throughout $UV<2m$. The flow of the displayed vector field rescales $U\mapsto e^{-\lambda /(4m)}U$, $V\mapsto e^{\lambda /(4m)}V$. It leaves $UV$, $r$, $dUdV$ and the angular metric unchanged. Thus it preserves the full metric everywhere in that region and extends as a global Killing field, not merely by a formal exterior coordinate substitution.

Its squared norm isIt is timelike for $UV<0$ ($r>2m$), spacelike for $0<UV<2m$ ($0<r<2m$), and a nonzero null vector on either horizon $UV=0$ away from $U=V=0$. At $U=V=0$ it is the zero vector, on the whole bifurcation surface of angular points. Zero must be distinguished from a nonzero null vector. This is the Schwarzschild time-translation Killing field in Kruskal coordinates.

The coordinate derivatives of the smooth two-parameter map commute. With a torsion-free Levi-Civita connection,because ${\Gamma }^a{}_{bc}$ is symmetric in its lower indices. Each member of the family is a geodesic, so $DV/Ds=0$. ConsequentlyThe commutator of covariant derivatives is the action of the Riemann curvature tensor. With the source's index and sign convention this givesThis is geodesic deviation: free fall relative acceleration measures curvature, even where an individual observer can eliminate the connection at one point.

For Earth's exterior Newtonian field, $\varphi =-GM/r$ and acceleration is $a_i=-{\partial }_i\varphi$. Nearby freely falling particles have relative acceleration $\delta a_i=-{\varphi }_{,ij}{Y}^j$. At the observation point align $z$ radially upward and $x,y$ horizontally. Since $GM/R_E^2=g$,The weak-field geodesic-deviation equation therefore identifiesIn these units time is measured in seconds as in the stated Newtonian comparison; if the zeroth coordinate is $ct$ instead, the curvature components acquire the corresponding $c^{-2}$ factor. A small freely falling cloud is squeezed in the two horizontal directions and stretched radially. These are tidal effects, not the uniform acceleration removed by a freely falling frame.

Their characteristic inverse-square time is $g/R_E$, so $\boxed{t_{\mathrm{tide}}=\sqrt{R_E/g}\simeq 8.1\times 10^2\mathrm{s}\simeq 13.4\min }$. One concrete thought experiment is an ideal satellite moving in a circular orbit just above a spherical Earth, ignoring atmosphere and terrain. Its orbital angular frequency is $\sqrt{GM/R_E^3}=\sqrt{g/R_E}$ and its period is $2\pi t_{\mathrm{tide}}\simeq 84$ minutes. Equivalently, initially nearby free fall trajectories develop order-one tidal changes on this characteristic scale. This is the surface tidal time scale of a spherical gravitating body.

Write $f(r)=1-2m/r$ and restore $c$ for the numerical estimate. A stationary clock at radius $r$ has $d\tau =\sqrt{f(r)}dt$. Let radii $r_1$ and $r_2=r_1-h$ denote the upper and lower computers. The metric is stationary and the same two fixed spatial points exchange both signals, so each signal has the same coordinate propagation time. Therefore the interval between reception events equals the interval between emission events in coordinate time: the common delay cancels.

For upper proper running time ${\tau }_1$, the lower computer has run forIn the weak gravitational field near Earth's surface, $m=GM/c^2$ andThe lower clock runs more slowly, so its remaining work isThus about 660 micro-operations remain; the proper-time shortfall is only about $6.6\times 10^{-10}$ seconds.

Successful comparison tests the gravitational dependence of stationary proper-clock rates, specifically the radial variation of $g_{tt}$ and its weak-field gravitational redshift prediction, assuming the computers execute equally per unit proper time. It also checks that this clock effect is independent of the chosen physical clock mechanism. It does not separately verify the Schwarzschild radial metric coefficient: although light propagation uses the null metric, the identical start and finish propagation delays cancel from this particular observable. This is Gravitational redshift measured by a computation counter.

Put $f=1-2m/r$, use proper time $s$, and denote its derivatives by dots. The timelike geodesic equations, away from the Schwarzschild coordinate singularity, arewith normalization $f{\dot{t}}^2-f^{-1}{\dot{r}}^2-r^2{\dot{\theta }}^2-r^2{\sin }^2\theta {\dot{\varphi }}^2=1$. Spherical symmetry permits an equatorial plane, $\theta =\pi /2$, and the cyclic coordinates give $\boxed{E=f\dot{t},J=r^2\dot{\varphi }}$. Normalization then gives ${\dot{r}}^2=E^2-f(1+J^2/r^2)$.

For $J\ne 0$ and $u=m/r$, $du/d\varphi =-m\dot{r}/J$. ThusAt $E=1$, $J=4m$, the right side factors as $2u(u-1/4{)}^2$. Set $q=2\sqrt{u}$, so $u=q^2/4$ and $(q^{\prime }{)}^2=(q^2-1{)}^2/8$. Integrating the regions $0<q<1$ and $q>1$ gives, up to angular shifts and time reversal,The constant $u=1/4$, or $r=4m$, is the corresponding circular timelike orbit. The hyperbolic-tangent branch lies at $r>4m$ and approaches that orbit from outside as $|\varphi |\to \infty$; the hyperbolic-cotangent branch lies at $0<r<4m$ and approaches it from inside. Near the limit $u-1/4$ decays exponentially as $e^{-|\varphi |/\sqrt{2}}$. Since $\dot{\varphi }$ tends to $J/(4m{)}^2=1/(4m)$, the limiting approach also takes infinite proper time. Neither nonconstant branch reaches and joins the circular orbit at a finite angular parameter.

Differentiating the orbit equation gives $u^{\prime \prime }=m^2/J^2-u+3u^2$. A perturbation about $u=1/4$ obeys $\delta u^{\prime \prime }=\delta u/2$, showing that the circular orbit is unstable. The two explicit curves are its separatrix branches, not ordinary small stable oscillations. For $E=1$ the energy is marginally bound, so the outer critical branch is the limiting orbit between scattering and Schwarzschild marginally bound capture from infinity. Time-reversed inner and outer branches describe the corresponding approach or departure from that unstable orbit. These are the Explicit separatrices of the marginally bound Schwarzschild orbit.

Lie Groups and Differentiable Manifolds, Mathematics Parts C and B; Wednesday 28 May 2008, 9.30am–12.30pm. The complete four-page source and both sections are privately transcribed from original Part C bundle pages 117–120, supplied through the past-papers disclosure. No exact worked-answer advertisement was found in the bounded availability check. Representations are finite-dimensional and continuous; manifolds have no boundary unless stated otherwise.

For a Matrix Lie group, $T_{g}G$ consists of derivatives ${\gamma }^{\prime }(0)$ of differentiable curves in $G$ with $\gamma (0)=g$, viewed inside $M_n(\mathbb{C})$ as a real vector space. At the identity, multiplying two such curves adds their derivatives, and reparametrizing a curve multiplies its derivative by any real scalar. Thus $L(G)=T_{I}G$ is a real vector subspace.

If $Y\in L(G)$ and $g\in G$, conjugating a curve with derivative $Y$ proves $gY{g}^{-1}\in L(G)$. Let $g(s)$ have $g(0)=I$ and $g^{\prime }(0)=X\in L(G)$. Differentiating the curve $g(s)Yg(s{)}^{-1}$ inside the finite-dimensional, hence closed, vector subspace $L(G)$ givesThe matrix commutator is real bilinear, antisymmetric, and satisfies the Jacobi identity by associativity of matrix multiplication. This proves that $L(G)$ is a real Lie algebra.

The exponential-map theorem for a linear Lie group says $\exp X\in G$ whenever $X\in L(G)$. Applying it to every real multiple of $X$ gives $\exp (tX)\in G$ for every real $t$. Conversely that curve has derivative $X$ at zero, so $X\in L(G)$. Hence

Here the orthogonal group is the real group $O_n(\mathbb{R})$. Differentiating $g(t{)}^{T}g(t)=I$ at zero gives $X^T+X=0$. Conversely a real skew-symmetric matrix satisfies $(e^{tX}{)}^T{e}^{tX}=e^{-tX}{e}^{tX}=I$. ThereforeFor $n\ge 2$ this is not a complex Lie subalgebra of $M_n(\mathbb{C})$: multiplying a nonzero real member by $i$ leaves this real subspace. For $n=1$ the zero algebra is the trivial complex Lie algebra as well. The complex orthogonal group would be a different group from the real orthogonal group used here.

Differentiating the affine block form givesConversely the exponential of such a block matrix has bottom row $(0,1)$ and invertible upper-left block $e^{tB}$, so it remains in the affine group. Its commutator corresponds toThe space is closed under multiplication by $i$ and its bracket is complex bilinear. Thus it is a complex Lie algebra of dimension $n^2+n$, and its real dimension is $2n^2+2n$.

The matrix exponential series differentiates term by term on bounded intervals and gives $(e^{tX}{)}^{\prime }=X{e}^{tX}$ and $e^{0X}=I$, proving existence. If $a$ is another differentiable solution, thenIts initial value is $I$, so $e^{-tX}a(t)=I$, giving $a(t)=e^{tX}$ uniquely.

The exponential satisfies the initial conditions and $e^{(s+t)X}=e^{sX}{e}^{tX}$ because the two matrix exponents commute. Conversely, differentiating $a(s+t)=a(s)a(t)$ with respect to $s$ at zero gives $a^{\prime }(t)=Xa(t)$. Even if only differentiability at zero was initially specified, the identity $a(t+h)=a(h)a(t)$ supplies differentiability at every $t$. The uniqueness result just proved yields $\boxed{a(t)=e^{tX}}$.

For fixed $X\in L(G)$, set $b(t)=\varphi (e^{tX})$. The homomorphism property gives $b(s+t)=b(s)b(t)$ and $b(0)=I$. Its initial derivative is $L(\varphi )(X)=D{\varphi }_I(X)$. The one-parameter subgroup uniqueness theorem givesFor the Adjoint representation, ${\mathrm{Ad}}_{g}Y=gY{g}^{-1}$. ThusThe differential $L(\varphi )$ is a real linear map of Lie algebras; the bracket identity follows by differentiating the compatibility of conjugation with $\varphi$.

Write ${\mathfrak{g}}_{\mathbb{C}}=L(G_{\mathbb{C}})$ and $\mathfrak{h}=L(H)$. A root is a nonzero weight $\alpha \in {\mathfrak{h}}^{\ast }$ of the adjoint action. Its root space isAt the group level these weights come from nontrivial characters of the Cartan subgroup acting by conjugation. The Cartan root-space decomposition isThe zero-weight space is $\mathfrak{h}$; for the complexification of a compact group the algebra is reductive, with any centre contained in this zero-weight space.

For the product in question,Take the diagonal Cartan subalgebra generated by $H_1=(\mathrm{diag}(1,-1),0)$ and $H_2=(0,\mathrm{diag}(1,-1))$. On $Z=x{H}_1+y{H}_2$, define ${\alpha }_1(Z)=2x$ and ${\alpha }_2(Z)=2y$. The roots are exactlyIf $E=E_{12}$ and $F=E_{21}$, their four root spaces are respectively $\mathbb{C}(E,0)$, $\mathbb{C}(F,0)$, $\mathbb{C}(0,E)$, and $\mathbb{C}(0,F)$. These four lines and the two-dimensional diagonal subalgebra form the full six-dimensional decomposition. There are no mixed roots: the two direct summands commute. This is the root system $A_1\times A_1$.

With the convention applicable also to disconnected compact groups, the Weyl group is $W=N_G(T)/T$, where the normalizer consists of elements taking $T$ to itself by conjugation. Put $\mathfrak{t}=L(T)$ and $\Lambda =\ker (\exp :\mathfrak{t}\to T)$. This is a full lattice with $T\cong \mathfrak{t}/\Lambda$.