Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 52 2 Solution Created 2026-10-03 Updated 2026-10-14
A useful model of spontaneous breaking of a Z2 scalar symmetry is the four-dimensional real scalar theoryThe action is invariant under the Z2 symmetry . Its classical vacuum manifold consists of the two values . In the broken quantum phase, choose a pure vacuum with ; the symmetry maps it to a different vacuum with opposite expectation value, although it leaves the action unchanged. To distinguish quantum symmetry breaking from simply minimizing a classical potential, introduce a small source and take the infinite-volume limit before . A finite-volume symmetry-preserving ground state can remain an even superposition because of tunneling, whereas the selected infinite-volume pure vacua have nonzero vacuum expectation values. The parameter below is the tree-level expectation, with its quantum value determined by renormalized parameters.
Writing givesThe original sign symmetry relates expansions about the two vacua; in one expansion it acts as . It is not a symmetry that fixes the chosen vacuum. Since the broken group is discrete, there is no continuous broken generator and no required Goldstone boson.
With relativistically normalized external states, the general two-body partial decay width isHere the bar sums final spins and averages initial spins only if the initial ensemble is unpolarized. The identical final-state symmetry factor is for two identical daughters and one for distinguishable daughters. Integrating the three-momentum delta function in the parent rest frame leaves . The energy delta function fixes and has radial Jacobian . Hence the two-body decay phase space isConsequentlywhen the spin-summed amplitude is angle independent.
For Higgs decay to two W bosons, the scalar parent has no spin average and the amplitude, up to an irrelevant phase, is . Use the massive vector polarization sum with :Momentum conservation gives . With the PDF's ratio , this becomesThe charged daughters are distinguishable. Combining the last two equations with givesFor Higgs decay to two Z bosons, the Higgs boson coupling to Z bosons is , using . This compensates the changed powers of in the polarization contraction, giving the same normalization of the squared amplitude with replaced by . The two Z bosons are identical, so the phase-space symmetry factor supplies an additional half:Both expressions have mass dimension one, vanish at their two-body thresholds and approach the ratio for . The cubic large-mass behavior comes from longitudinal-vector polarizations. These are on-shell tree-level widths under the stated heavy-Higgs hypothesis; below threshold the physical off-shell multi-particle decays require a different calculation.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 50 2 Solution Created 2026-10-03 Updated 2026-10-14
Use the metric signature , so , and the weight . The momentum-space Feynman rules for the massless Phi-fourth theory are: an internal scalar line contributes ; a quartic interaction vertex contributes with its momentum-conservation delta function; every independent loop is integrated with ; and each graph is multiplied by its Feynman-diagram symmetry factor. All external momenta can be taken incoming. These rules follow by expanding the interaction exponential and contracting free fields with the Wick theorem.
For a connected graph with internal lines, interaction vertices and loops,Here is the superficial degree of divergence. In four dimensions,Odd vanish by Z2 symmetry. Thus only the nonvacuum two-point and four-point one-particle-irreducible Feynman diagrams can have overall ultraviolet divergences: degree two for , degree zero for . Vacuum graphs with may also diverge but cancel from normalized correlators. This is an overall power-counting statement; subgraphs must be subtracted before using it to conclude finiteness for .
For the full one-particle-irreducible vertices, the tree-level quadratic kernel and quartic interaction giveIf is instead reserved for interaction-generated self-energy insertions, its tree value is zero and is displayed separately as the free inverse kernel. The distinction is only that convention; the loop insertions below are unchanged.
The requested one-loop diagrams are the two-point tadpole diagram and the three four-point bubble diagrams. Each has symmetry factor . The three bubble channels correspond to , , .
The tadpole insertion obeysIn massless dimensional regularization this scaleless integral is zero. A mass or cutoff regulator exposes the ultraviolet divergence allowed by power counting; its vanishing in the scaleless convention does not remove the need to include this diagram.
For a bubble, the vertex and quantum field theory propagator factors give . The PDF defines with the factor outside the Minkowski integral, so the integral itself is . ThereforeThe converted TeX misplaces this as though it were part of the exponent of ; keeping the printed is essential to this real pole coefficient.
For nonexceptional Euclidean external momentum, Wick rotation cancels that factor . Use a Feynman parameter and translate the integration variable:For example, represent the inverse square as , do the Gaussian momentum integral, and then use the Gamma function. This also derives the formula rather than assuming a loop-integration table. At , and the parameter integral tends to one, givingTake before continuation back to Minkowski momenta. At , a massless scaleless integral mixes ultraviolet and infrared issues, so it cannot be used to read off this pole.
Introduce a renormalization scale and a quartic counterterm. With denoting a dimensionless renormalized coupling constant,After factoring out from the vertex, each loop contains . The counterterm contributes to and cancels the pole from all three channels. More explicitly,In modified minimal subtraction, replace the pole-only counterterm bywhich subtracts the first three terms inside the brackets. The finite renormalized quartic scalar vertex is thenAnother renormalization condition can change the finite constant, but not the pole cancellation or the momentum-dependent logarithms. The corresponding timelike expressions follow by the same Feynman continuation.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 46 1 b Solution Created 2026-10-03 Updated 2026-10-14
Let the shell-restricted scalar propagator beWrite and average with the normalized Gaussian shell measure. The cumulant expansion givesThe subtraction removes disconnected Feynman diagrams; the logarithm retains connected Feynman diagrams made from shell contractions. This is the connected shell-contraction expansion. An external line denotes the background low field, not an additional low-momentum propagator.
Connected quartic shell diagrams through second order, including vacuum terms and the two momentum-support zeros
. Details. At order , panel A is the existing four-field interaction vertex. The new shell contractions are B, a tadpole diagram contributing a two-field vertex, and C, a Vacuum Feynman diagram contributing a constant. The Gaussian functional determinant in panel L is another field-independent term, of order ; the original quadratic kinetic term and mass term are retained as well.
For completeness, all two-vertex topologies from Wick contractions at order are shown. Let be the number of shell lines joining the vertices and their numbers of self-contractions. The external-field counts areInterchanging the two vertices identifies the same topology. Enumerating these conditions gives exactly the following eight possibilities:
- D: , with ; a six-field vertex kernel.
- E: , with ; a formal four-field kernel, which vanishes for the sharp shell split.
- F: , with ; a formal two-field kernel, which also vanishes for the sharp shell split.
- G: , with ; a four-field vertex kernel.
- H: , with ; a two-field vertex kernel, containing a tadpole diagram.
- I: , with ; a connected Vacuum Feynman diagram.
- J: , with ; the two-field sunset diagram kernel.
- K: , with ; a connected Vacuum Feynman diagram with four joining lines.
The support qualification is important. In E or F, a vertex with one external low field and one bridge also has a tadpole whose two momenta cancel. Conservation forces the bridge momentum to equal that single low momentum, outside the shell. Equivalently, convolution by annihilates . This is momentum-support exclusion for Wilsonian bridge diagrams. D is different: its bridge carries the sum of three low momenta, which can lie in the shell. Its kernel is proportional to and need not vanish. Thus one-particle-reducible Feynman diagrams must not be excluded merely because the object being computed is an effective action.
These kernels need not be local before a derivative expansion. Their momentum dependence generates derivative interactions where such an expansion is valid. If all external momenta are sufficiently far below , D vanishes too; in particular it does not produce a zero-momentum local coupling at order . A local six-field term is allowed and is generated at higher orders, for example by a three-vertex shell triangle.
To all orders, expect every scalar interaction allowed by the original symmetries: arbitrary even powers of the field and their allowed derivative couplings, together with vacuum terms. The Z2 symmetry excludes odd-field vertices. The original quartic form is therefore not closed under exact Wilsonian integration.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 303 1 Solution Created 2026-10-03 Updated 2026-10-05
Use the usual homogeneous, local, analytic Landau-Ginzburg theory; rotational symmetry and Z2 symmetry alone do not specify locality, analyticity, or the signs of the coefficients. The local derivative expansion contains every rotational scalar with an even total number of fields:The omitted terms include all allowed higher field powers and contracted spatial derivatives. Terms differing by a total derivative are equivalent for bulk behavior with suitable boundary conditions. The displayed gradient energy assumes spatial homogeneity. If only rotations about an origin are imposed, rotationally invariant position-dependent coefficients are also possible.
In the mean-field approximation, take a spatially uniform order parameter and minimize its Landau free energy. For , and with , keeping the quadratic and quartic terms givesThe single minimum splits into two minima as the temperature passes below the critical temperature. Choosing one minimum gives spontaneous symmetry breaking of the Z2 symmetry and a continuous phase transition. The order parameter is the expectation in a selected ordered state; averaging equally over both states gives zero. A Z2 symmetry does not guarantee a continuous phase transition: a negative quartic coefficient stabilized by a positive sextic coefficient can instead give a first-order phase transition.
Relaxing the Z2 symmetry permits odd terms. A nonzero linear source generally rounds this ordinary continuous phase transition. A cubic term can instead give a first-order phase transition: for , and , equal minima occur atThe order parameter jumps from zero to . A continuous phase transition without an exact Z2 symmetry is still possible with additional tuning; odd terms do not logically exclude it.
For the multicritical even Landau potential, set and assume an integer , and . For , all lower even interactions must be absent or tuned to zero to obtain these mean-field critical exponents. The stationary equation isIts nonzero solutions are minima because . Thus the order-parameter critical exponent isSubstituting into the minimum free energy density givesThe singular heat capacity per unit volume is . Smooth variation of the other coefficients does not change the leading power. ConsequentlyFor , this means a finite heat capacity jump, with ; for , the tricritical point has and . The case would contain only a quadratic potential and would not stabilize an ordered phase.
The Gaussian fluctuation correction near a critical point is more singular than the mean-field approximation contribution whenThis is the upper critical dimension of an even scalar interaction. Equivalently, its coupling has engineering dimension , positive below . Dominant fluctuations signal breakdown of the mean-field approximation; the Gaussian expression is not then a calculation of the exact interacting heat-capacity critical exponent. At , equality of the powers allows logarithmic corrections.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 303 3 a Solution Created 2026-10-03 Updated 2026-10-14
Write and assume , as required for quartic stability. In the mean-field approximation, the two-component radial quartic Landau potential isSet , . At fixed , the quartic term is fixed and the quadratic term is minimized by assigning all of to the component with smaller . Minimizing over then givesThe resulting phase diagram has these regions:
- If and , : the disordered phase preserves both Z2 symmetries.
- If , and : the first Z2 symmetry undergoes spontaneous symmetry breaking.
- If , and : the second Z2 symmetry undergoes spontaneous symmetry breaking.
On , the symmetry is enhanced to the orthogonal group , and the minima form the circleMixed-component minima occur on this line, rather than in an open coexistence region. Crossing the negative diagonal swaps the selected component discontinuously: this is a first-order phase transition, also called a spin-flop transition. For example, holding fixed and moving across , the derivative jumps from to zero. The angular degeneracy exactly on the line does not remove this discontinuity.
The positive half-axis , is a continuous phase transition, as is , . Their endpoint at is a bicritical point, where the two continuous boundaries meet the first-order negative diagonal. Approaching the origin radially makes the ordered amplitude vanish continuously. The negative half-axes lie inside ordered phases and are not additional transition lines. There are two continuous half-axis boundaries and one first-order negative diagonal, meeting at the origin.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 303 2 e ii Solution Created 2026-09-24 Updated 2026-10-03
No. With only , the free energy has an exact Z2 symmetry . Integrating out fast modes and rescaling preserve that symmetry, whereas is odd. Therefore no five-point vertex and no correction to can be generated at any order in .
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 303 3 b iii Solution Created 2026-09-24 Updated 2026-10-03
No. Even when the two masses differ, the free energy retains an independent Z2 symmetry for each component: and . The bilinear is odd under either transformation, so integrating out modes cannot generate it.
Spontaneous breaking of a Z2 scalar symmetry 2026-10-14
A real scalar field with the displayed potential is invariant under the Z2 symmetry , but a chosen broken-phase pure vacuum has nonzero vacuum expectation value of one sign. Expanding about gives . The excitation is massive, and no continuous broken generator requires a Goldstone boson. In a finite-volume theory with symmetry-preserving conditions, tunneling can produce a symmetric ground state. A broken pure vacuum is selected by taking the infinite-volume limit before removing an infinitesimal source, or by specifying appropriate boundary conditions. The two signs are related by the symmetry, although the individual pure vacua are not invariant under it.


