Past exam of the mathematics course of the University of Cambridge 2013 ib Paper 1 1E Solution Created 2026-09-24 Updated 2026-10-07
The adjugate matrix is the transpose of the cofactor matrix: . Cofactor expansion gives the adjugate identityFor an invertible matrix this yields . Integer entries of imply integer cofactors, so makes its inverse integral. Conversely, if and both have integer entries, their determinants are integers and . The only integer units are . Hence an invertible integer matrix has an integer inverse exactly when it is a unimodular matrix, with determinant . Invertibility is implicit whenever is used.
Past exam of the mathematics course of the University of Cambridge 2013 ib Paper 2 1E Solution Created 2026-09-24 Updated 2026-10-07
Write for the conjugate transpose. Taking determinants in gives . Since the Hermitian matrix is invertible, every such is invertible and . The identity belongs to ; if , then ; and multiplying by and gives . Associativity comes from matrix multiplication, so this is a group.
The Hermitian form obeys precisely when . Thus is the group of linear isometries of this nondegenerate Hermitian form; positivity of is unnecessary.
For , is a unitary matrix. Here is a direct diagonalization of a matrix proof. Over , its characteristic polynomial has a root, giving a unit eigenvector with . Preservation of the inner product gives . If , then , so is invariant. The restriction is again a unitary matrix. Induction on the dimension supplies an orthonormal basis of eigenvectors. In that basis, is diagonal, with diagonal entries on the unit circle.
Past exam of the mathematics course of the University of Cambridge 2013 ib Paper 4 16A Solution Created 2026-09-24 Updated 2026-10-07
For a variation preserving both endpoint values and slopes, at the endpoints. The first variation isIntegrate the last term twice and the middle term once. All boundary terms vanish, givingThe fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equationFor the present functional this becomes . Its general solution is . The left endpoint gives and . The right endpoint then givesThe coefficient determinant is , so , . The stationary function is .
For any admissible perturbation , exact quadratic expansion, followed by the stationarity equation and integration by parts, givesApply the supplied Poincare inequality first to , then to , whose endpoint values also vanish:Thus for every admissible perturbation, proving a global maximum through the clamped second-variation inequality.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 15 2 Solution Created 2026-10-03 Updated 2026-10-07
A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms on the tangent spaces, equivalently a smooth section of with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix , and its inverse has matrix . Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on , giving a smooth vector bundle isomorphism.
For local flattening of a Riemannian metric, choose a manifold chart neighborhood around with closure contained in , and let on . Choose a smooth bump function , with compact support in and equal to one on a smaller neighborhood of . SetThe formulas glue smoothly because the modification is supported strictly inside . A convex combination of positive-definite forms is positive definite. On the metric is exactly , so gives an isometry to a Euclidean open subset. Outside the original metric remains unchanged.
Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.
The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, with , and satisfy those hypotheses. The geodesic reaches the missing unit sphere at and cannot continue within . Thus the printed assumptions alone do not prove completeness.
Here is the intended compact-core completeness for a Euclidean end. Write for the end coordinates and assume additionally thatis compact for every sufficiently large . Choose such an beyond the metric transition and large enough to include the initial point of a geodesic. Its speed is constant. On every segment outside , the Euclidean radius changes at a rate at most , so over a finite parameter interval it cannot exceed . The full segment therefore stays in the compact set . Its velocities also stay in a compact subset of , since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.
For upward stability of Riemannian completeness, lengths of all curves satisfy , hence . A -Cauchy sequence is therefore -Cauchy and has a limit because is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around , has a bounded largest matrix eigenvalue, so the coordinate straight segment gives . Thus convergence to is also in . That distance is complete, and Hopf-Rinow yieldsFor disconnected , apply this argument on each component.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 17 2 Solution Created 2026-10-03 Updated 2026-10-07
A connection on a vector bundle is a complex-linear map from smooth sections to vector-bundle-valued one-forms satisfyingChoose a locally finite trivializing cover, its componentwise flat connections on a vector bundle , and a subordinate smooth partition of unity . The formula is globally meaningful: each weighted term extends by zero outside its chart. Its Leibniz rule follows from , proving every smooth complex vector bundle admits a connection.
Extend the connection on a vector bundle to bundle-valued differential forms byThe curvature form of a connection is . Applying this twice to shows that the terms cancel, so is tensorial and defines an -valued two-form. With column coordinates for sections and connection one-form , the local expression is . Direct expansion on an arbitrary bundle-valued form givesThe terms containing a derivative of the argument cancel by the graded Leibniz rule. Thus Cartan curvature matrix equation isHere multiplication includes matrix multiplication and the exterior product of form entries. For a frame change , the connection one-form and curvature form of a connection transform asThe trace is consequently frame independent. Alsothe diagonal terms vanish and the off-diagonal terms cancel in pairs. Locally , and hence . The local primitives need not agree, but the two-form does. We obtain a global closed two-form .
The determinant connection on is defined intrinsically bywhere the differential-form coefficient of is placed first. In a local frame, only the diagonal components contribute to the derivative of , so its connection one-form is . A line-bundle connection has curvature , because a scalar one-form wedges with itself to zero. ThereforeTo see independence of the de Rham cohomology class, write , a globally defined endomorphism-valued one-form. The curvature difference formula and the same trace cancellations give the trace curvature transgressionTheir difference is globally an exact differential form, so their de Rham cohomology classes coincide.
For the Hermitian metric on a holomorphic vector bundle, use the displayed ordering of the local matrices and put . The identity givesThe plus sign comes from applying the graded Leibniz rule to the one-form . By the derivative formula for a determinant,The Hermitian positive-definite matrix has positive real determinant, so this logarithm is an ordinary smooth real function. The trace of Chern curvature in these conventions is thereforeIt is locally an exact differential form and in particular closed.
A holomorphic local frame change multiplies by . A nowhere-zero holomorphic function has a local holomorphic logarithm, so . Thus the preceding expression for is independent of the frame and glues globally. For two Hermitian metrics on a holomorphic vector bundle, the functionis globally defined because the frame-change factors cancel. Their two-forms differ by . Hence the class of is independent of the metric. The two-form is generally imaginary-valued; the metric independence statement is in complex de Rham cohomology, or equivalently for the real form .
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 44 1 Solution Created 2026-10-03 Updated 2026-10-07
Write , with . Independent variations of and give the Euler-Lagrange equation . Away from its unique endpoint solution isIntegration by parts cancels the linear fluctuation terms and gives . On the classical solution, the action is the boundary term . Substituting the endpoint derivatives therefore givesThe remaining Gaussian path integral contains two real fluctuation coordinates per mode, so it contributes an inverse functional determinant, rather than its inverse square root. Thus , with the measure normalization fixing the otherwise arbitrary constant.
For Dirichlet boundary conditions, the normalized sine modes have eigenvalues , . Their determinant ratio is the convergent Dirichlet oscillator determinant ratioThe last equality is the sine infinite product. With the prescribed free determinant this givesThe kernel uses the Feynman i-epsilon prescription; at a caustic it is a distributional limit, not an ordinary finite function. Its limit is .
There is a sign error in the printed complex-integral hint. For a positive damping parameter , polar integration gives the regulated complex Fresnel integralThis regulated value, together with the stated free-kernel normalization, fixes the phase consistently.
For and positive imaginary-time length , Wick rotation givesPut with and use the convergent real Gaussian integral. Writing givesThis is the parity-twisted oscillator thermal trace. The complex coordinate describes two independent real quantum harmonic oscillators, each with mass two in these units. Their total energy is ; level has degeneracy , and spatial inversion has parity operator eigenvalue . The plus sign is ; the minus sign is . The alternating trace inserts parity into a bosonic system; it does not change the oscillators into fermions.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 50 2 Solution Created 2026-10-03 Updated 2026-10-07
The geometric object underlying an gauge field is a principal connection on a principal bundle with structure group SU(2). The base is spacetime, or an oriented Riemannian manifold in the Euclidean formulation. A choice of local section identifies each fibre with SU(2) and produces a local gauge potential. Such a choice is a choice of internal frame; changing it produces a gauge equivalence of principal connections. The bundle specifies the global gluing of these frames, while the principal connection specifies how frames at neighbouring points are compared.
The SU(2) group consists of unitary matrices of determinant one. Its Lie algebra consists of traceless Skew-Hermitian matrices. For definiteness usewhere the are Pauli matrices. The commutator fixes all signs below. The positive invariant inner product on this Lie algebra is . Absorb the coupling into the gauge potential, so the gauge covariant derivative in the defining two-dimensional representation is with . A convention using Hermitian Pauli matrices instead moves factors of and the coupling into ; the underlying principal connection is the same geometric data.
A principal connection can be given by a Lie algebra-valued differential one-form on satisfyingHere is the vertical vector generated by the right group action. The first identity recovers vertical motion and the second makes the construction independent of a fibre frame. Its kernel is the horizontal distribution of a principal connection, complementary to the vertical tangent spaces. A curve is horizontal when annihilates its velocity, so a principal connection defines parallel transport along curves in .
For a local section , the local principal connection form is . Change the section to , with . Differentiating has a translated horizontal part and a vertical part from . The two defining identities for the principal connection therefore giveOn overlapping trivializations this is exactly the compatibility law for the local principal connection forms. In particular, the gauge potentials need not glue as ordinary globally defined differential one-forms on . On a nontrivial principal bundle there need not be any global section with respect to which one could write a single matrix-valued .
An active gauge equivalence of principal connections is generated by a bundle automorphism covering the identity on and commuting with the right SU(2) action. Its local description has the same transformation formula, but its local group-valued functions must satisfy transition compatibility on overlaps. This distinguishes the structure group SU(2) from the full gauge group of a fixed bundle. Two descriptions related by a passive change of section encode the same principal connection; two connections related by such an automorphism lie in the same physical gauge orbit. The space of principal connections on a fixed bundle is affine: the difference of two connection forms is a differential one-form with values in the adjoint bundle, since their inhomogeneous transformation terms cancel.
The curvature of a principal connection isIt is horizontal and equivariant, hence descends to a differential two-form on valued in the adjoint bundle. In a local section the gauge curvature becomesThe product combines the exterior product with matrix multiplication. The commutator term is the specifically non-Abelian interaction. Substitution of the transformation of , using , cancels all derivatives of and givesThus the gauge curvature transforms tensorially even though the gauge potential does not. Geometrically, the curvature of a principal connection measures the failure of the horizontal distribution of a principal connection to be integrable; physically it is the non-Abelian gauge field strength.
Matter fields are sections of associated vector bundles. A field in the defining representation of SU(2) is a section of . Its two local components satisfy under the section convention above. The induced gauge covariant derivative obeysThis covariance is why replacing ordinary derivatives by gauge covariant derivatives makes the kinetic terms compatible with changes of internal frame. Applying the gauge covariant derivative twice yields , so the gauge curvature also measures the noncommutativity of covariant differentiation. Other group representations give other associated vector bundles and replace by its image under the differential of the representation.
For an adjoint-valued differential form of degree , setExpansion of , using , proves the gauge-theory Bianchi identityThis is an identity for every principal connection, independent of any field equation. It should be distinguished from the dynamical Yang-Mills equations.
To obtain dynamics, equip with a Riemannian metric and an orientation, and let be its Hodge star operator. With the Skew-Hermitian normalization above, the positive Euclidean Yang-Mills action isThe matrix trace and the Hodge star operator make this independent of the choice of local section. The minus sign compensates for the negative trace form on Skew-Hermitian matrices. Varying the principal connection gives , and thereforeFor compactly supported variations, or on a closed base, integration by parts gives the Yang-Mills equationsIn components these are . The gauge-theory Bianchi identity provides the complementary geometric identity . Coupling matter adds the associated gauge current to the dynamical equation; the transformation law of the gauge covariant derivative ensures covariance of that current as well.
The principal connection also supplies nonlocal observables. Along a path , parallel transport is determined byThe ordering is necessary because the Lie algebra matrices at different points need not commute. The resulting Wilson line transforms as . For a closed path, its matrix trace is a Wilson loop, independent of the frame at the basepoint. These observables record the holonomy of a connection. A flat principal connection has trivial holonomy of a connection around sufficiently small contractible loops but may have nontrivial holonomy of a connection around noncontractible loops. Thus vanishing local gauge curvature need not eliminate global gauge information.
The global topology becomes particularly visible in four dimensions. For a closed oriented four-dimensional base, Chern-Weil theory gives the Second Chern number of the defining associated vector bundle:The plus sign here uses Skew-Hermitian curvature and the mathematical second-Chern convention: expanding with gives precisely the displayed Second Chern form. A convention defining a physics topological charge with a leading minus sign instead calls that charge . Neither choice changes the absolute-value action bound.
The Second Chern number is independent of the principal connection on a fixed bundle. Indeed, the gauge-theory Bianchi identity implieswhose integral vanishes on a closed base by Stokes theorem. A nonzero Second Chern number obstructs a global trivialization: for a globally defined , , so its integral on a closed base would be zero. On , gluing two trivial bundles over four-balls uses a transition map . Since SU(2) as the three-sphere identifies the target with , such gluing maps have integer degree of a map between oriented manifolds, illustrating how distinct topological sectors arise.
In four Euclidean dimensions, the Hodge star operator squares to one on differential two-forms. Decompose with . Define the nonnegative squared norms . Orthogonality givesEquality means that one component vanishes, giving the self-dual Yang-Mills equations or the Anti-self-dual Yang-Mills equations. Such finite-action solutions are Yang-Mills instantons. They solve the full Yang-Mills equations because . With the second-Chern convention just specified, anti-self-dual curvature has and self-dual curvature has . On noncompact spacetime one must impose suitable decay or boundary conditions before treating the topological integral as an integer and using the same bound without boundary terms.
Local gauge potentials, matter gauge covariant derivatives, gauge field strength, parallel transport and Second Chern number are different manifestations of one principal connection. The principal bundle preserves the global information that a single local gauge potential cannot capture; the Yang-Mills action then selects its physical dynamics.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 52 1 a Solution Created 2026-10-03 Updated 2026-10-07
Write . In the uniform equilibrium, the linearized ideal magnetohydrodynamic equations reduce toIntegrating from the undisplaced reference state, and using the uniform background magnetic field, givesThese are the perturbations induced by the fluid displacement; independent time-independent changes to the reference state are excluded. The linear Lorentz force density is . Its magnetic pressure and magnetic tension parts giveFor a Fourier mode, put , and . Then and . Substitution givesIf , the fluid displacement is perpendicular to both the wave vector and the magnetic field. Only magnetic tension restores it, andThis is the Alfvén wave. The Alfvén velocity is the field-directed propagation vector: . The signed phase velocity normal to the wavefront is ; the group velocity is . The distinction matters for oblique propagation. For generic directions the Alfvén wave has one transverse polarization, with .
To obtain the other magnetohydrodynamic waves, take the scalar products with and :The determinant condition for a nonzero pair isThus the fast magnetosonic wave and slow magnetosonic wave haveTheir fluid displacements lie in the plane of and and are generally compressive. In the fast magnetosonic wave, gas and magnetic pressure provide the stronger restoring combination; in the slow magnetosonic wave, their perturbations oppose one another. The phase speeds depend on direction. They satisfy and . For parallel propagation the two speeds are and , with a degeneracy between a transverse branch and the Alfvén wave. For perpendicular propagation and ; the latter is a nonpropagating limiting disturbance. These special directions require interpreting the polarizations by continuity rather than assuming three distinct nonzero frequencies.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 5 3 Solution Created 2026-10-03 Updated 2026-10-07
For a Young tableau , let permute the entries within its rows and within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power , the actions areThey commute because applying to every factor commutes with permuting the factors.
We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies , where is the nonzero hook product of a partition. Thus is a primitive idempotent. The standard-tableau decomposition of the right regular module is . Tensor this right-module direct sum with the left module . The map sending to is an isomorphism, with inverse . Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of -modules; individual summands need not be -invariant.
We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree . We also use the standard Schur algebra equivalence between its modules and homogeneous degree- polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent has one-dimensional image on and zero image on the other simple factors, so . This identifies the requested irreducibles. They classify the polynomial degree- representations in this tensor power, not all rational representations of every degree.
The length bound for a Schur module is if and only if . A column of length greater than antisymmetrizes more than vectors and gives zero. Conversely, for , fill every tensor position in row with the th basis vector of . Row symmetrization multiplies it by , and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.
A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For , its matrix entries belong to ; rational here permits determinant denominators but not arbitrary poles on . A one-dimensional rational character of is a Laurent polynomial with and . Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is , hence for .
Restrict a one-dimensional rational character of to its diagonal torus. The same argument in several variables gives . Conjugation by permutation matrices makes all equal. On every diagonalizable invertible matrix it consequently agrees with . The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.
For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over using normalized Haar measure. The orthogonal complement of an invariant subspace is then -invariant. Differentiating makes it invariant under and therefore under its complex span . The elementary unipotent matrices generate , so the complement is -invariant. This proves complete reducibility. Averaging over similarly gives complete reducibility for rational representations.
Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of into subspaces on which acts as , with and . These subspaces are invariant. For , choose with , set , and defineChanging to changes to , so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to on .
It is rational as well. Every matrix coefficient of has a polynomial representative on . Averaging that representative over the finite scalar center selects its homogeneous parts with . Substitution in the extension gives , a regular function on . Each -invariant subspace decomposes into its parts, and the extension acts on each part by a scalar times an action. Thus these subspaces are also invariant under the chosen extension. Consequently is irreducible if and only if this is irreducible. For , is trivial and the trivial extension supplies the same conclusion.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 5 4 Solution Created 2026-10-03 Updated 2026-10-07
For a decreasing integer tuple , let . Then is a partition and the highest-weight classification of rational GL representations definesThis is a determinant twist of a Schur module. Every irreducible rational representation becomes polynomial after multiplication by a sufficiently large positive determinant power, which clears all matrix-entry denominators. The polynomial degree decomposition and Schur–Weyl duality then identify it with a Schur module. Undoing the twist gives exactly one decreasing integer tuple . Distinct tuples have distinct highest torus weights, so these are the complete pairwise nonisomorphic irreducible rational representations.
The Weyl character formula specializes toIt is a symmetric Laurent polynomial in the eigenvalues. Equality extends from the dense set of diagonalizable matrices to all invertible matrices: both the character and the expression in the characteristic-polynomial coefficients are regular functions on . For a polynomial representation this also extends to every endomorphism of . For a general rational representation, the printed claim at singular endomorphisms needs this qualification: for example is undefined at a singular matrix. The displayed formula is valid on , and on all of when .
To compute the degree, set with distinct and take . For and , the leading coefficient of an exponential alternant isIndeed, expand every exponential in powers of . The first nonzero determinant uses the distinct powers ; its coefficient is the product of the two Vandermonde determinants divided by . Taking the same expansion in the denominator cancels the powers and the factors, giving the Weyl dimension formulaThis proof works for negative as well, since determinant twists have dimension one.
Every finite-dimensional rational module is completely reducible. The characters of its irreducible constituents are linearly independent: each Schur Laurent character has its highest dominant monomial with coefficient , and only lower weights besides it. In a finite relation, choose a lexicographically highest remaining weight; its coefficient must vanish, and iterate. Therefore equal characters give equal multiplicities of every irreducible constituent, proving rational modules with the same character are isomorphic.
Finally the symmetric algebra of has the formal torus characterEach factor sums the symmetric powers of a one-dimensional weight space; the exterior square has weights for . The permitted Schur identity makes this . In each fixed scalar degree there are only finitely many terms, so complete reducibility and character independence apply degree by degree without a convergence assumption. Thus the multiplicity-free symmetric-algebra model for polynomial GL representations contains each irreducible polynomial representation exactly once. The word irreducible is necessary: arbitrary reducible polynomial modules, such as two copies of the trivial module, do not each occur once in a multiplicity-free sum.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 6 1 d Solution Created 2026-10-03 Updated 2026-10-07
Differentiating the explicit characteristic flow map gives the block triangular Jacobian matrixIts Jacobian determinant is the product of its diagonal block determinants. Therefore
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 65 2 b i Solution Created 2026-10-03 Updated 2026-10-07
Take without loss of generality, so , and require decay as . The base velocity is everywhere; there is no jump in its derivative. Let and . A decaying representation that already solves the layer equations isIt is continuous at both interfaces. At the value is and the derivative jump is ; at they are and . The density anomaly drops by one across each interface. The jump conditions for stratified inviscid shear flow therefore reduce toFor nonreal the displacement denominators do not vanish. Neutral limiting values are obtained by continuation, rather than by dividing by zero at an interface.
Set . A nonzero mode requires the determinant to vanish:Expansion gives the dispersion relation for two density interfaces in uniform shear,Uniform shear throughout the exterior is important: replacing it with constant exterior velocities would introduce vorticity jumps and give a different dispersion relation.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 6 5 b Solution Created 2026-10-03 Updated 2026-10-07
The shear is a smooth diffeomorphism with determinant one and sends compact sets to compact sets. Thus are locally integrable functions. On a compact time interval and compact label set , the integral is finite. The Fubini theorem, followed by a countable exhaustion of time intervals and label sets, shows that for almost every .
To transform the weak equation, choose , where and . Including a velocity cutoff in ensures the required compact support. The transport derivative of this test function is . Change variables to the characteristic labels to obtainThe fundamental test-function identity makes the bracket zero almost everywhere. Choose a countable dense family of time test functions on each bounded interval and extend by continuity; there is then a single negligible label set outside which as one-dimensional distributions.
For each remaining label, subtract the primitive . The distributional derivative of is zero, so it is a constant almost everywhere. Choose the resulting absolutely continuous representative along free characteristics, . The fundamental theorem of calculus givesThe representative qualification is essential: an arbitrary locally integrable representative can be modified on one time slice without changing the distributional equation, and would not satisfy the identity for every pair of times.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 66 1 b Solution Created 2026-10-03 Updated 2026-10-07
For each of the four standard pairs, integration by parts givesConsequently the eigenvalues are nonnegative. For a positive eigenvalue , the eigenvalue equation is . Its four characteristic roots are , soHere are coefficients, avoiding a collision with the filament bending modulus . The regular finite-interval self-adjoint operator has compact resolvent; applying the spectral theorem for compact self-adjoint operators to a shifted inverse supplies a complete orthonormal basis of eigenfunctions.
For clamped boundary conditions at zero, a clamped--clamped bending mode can be writtenAt , writing , the two remaining boundary conditions areThe determinant is . Hence the positive wavenumbers obeyEquivalently, intersect with . Numerical root bracketing givesThe entire sequence has the useful large- descriptionIndeed, put in and use and .
The apparent root in the determinant equation is spurious for the clamped problem: at zero eigenvalue, is a cubic polynomial, and its four clamped conditions force . For other endpoint choices, zero-energy filament modes must be treated separately from the trigonometric formula. The free-free kernel consists of affine functions, the torqued-torqued kernel consists of constants, and the hinged-hinged kernel is trivial. Including those kernels is necessary for a complete eigenfunction expansion.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 75 3 b Solution Created 2026-10-03 Updated 2026-10-07
For a trajectory connecting the degenerate minima in infinite Euclidean time, the first integral gives . The quartic double-well instanton and its reverse areTheir action isThe instanton fluctuation prefactor has dimensions of inverse time. It incorporates nonzero Gaussian fluctuation eigenvalues and the translation zero-mode Jacobian; a conventional determinant expression iswith common endpoint regularization and the prime removing the translation mode. Only its positive quantum tunnelling rate is needed here.
In the dilute instanton gas, well-separated crossings alternate in direction. Integrating ordered centers gives . A path returning to the same well has an even number, and one connecting opposite wells has an odd number. The common leading harmonic endpoint factor follows from the supplied harmonic oscillator transition kernel:Expanding these functions gives the sums over even numbers and odd numbers, with the position-kernel normalization that its abbreviated formula suppresses. Validity requires , and , so typical crossing separation greatly exceeds an instanton width. Local anharmonic corrections replace the harmonic well energy by its perturbatively corrected value; the leading formula does not claim uniformly negligible relative error at arbitrarily large .
The two exponents identify the double-well tunneling splitting:The superposition with an even spatial wavefunction is the lower state. Coherent quantum tunnelling removes the classical degeneracy, and the exponentially small splitting sets the long quantum tunnelling timescale.