Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 49 (Solitons and Instantons)

Substitution of the stationary ansatz into the focusing nonlinear Schrodinger equation gives the real ordinary differential equationMultiply by $2f^{\prime }$ and integrate to obtain the first integralThe finite L2 norm forces the relevant orbit to approach $(f,f^{\prime })=(0,0)$ and hence $C=0$. To justify this rather than assume pointwise decay, the conserved expression first bounds $f$ and $f^{\prime }$, since $f^4+2E{f}^2$ tends to infinity with $|f|$. Thus $f^2$ is uniformly continuous and integrable, and tends to zero at both ends. If $C>0$, then $|f^{\prime }|$ would eventually be bounded away from zero with fixed sign, contradicting $f\to 0$; $C<0$ is incompatible with $(f^{\prime }{)}^2=C-2E{f}^2-f^4$ near zero. Therefore $C=0$ and $f^{\prime }\to 0$.

For $E\ge 0$, the identity $(f^{\prime }{)}^2=-2E{f}^2-f^4$ permits only $f=0$. For a nonzero localized profile put $\kappa =\sqrt{-2E}>0$. ThenA nonzero solution cannot cross zero, because at a zero it also has $f^{\prime }=0$, and uniqueness of the ordinary differential equation would make it identically zero. Its sign is fixed, its maximum absolute value is $\kappa$, and separation on either side of that maximum givesDirect differentiation verifies the profile and its squared L2 norm is $\int f^{2}dx=2\kappa$. The two signs can be incorporated into an arbitrary constant phase of the complex bright soliton of the focusing nonlinear Schrödinger equation.

For a Galilean boost of a nonlinear Schrödinger soliton, define ${\psi }_u(x,t)=e^{i(ux-u^{2}t/2)}{\psi }_0(x-ut,t)$. Differentiating shows that the terms proportional to $u{\partial }_x{\psi }_0$ cancel between $i{\partial }_t$ and $-{\partial }_x^2/2$, as do the added $u^2/2$ terms. The cubic nonlinearity is unchanged because the multiplying phase has modulus one. Hence the boosted field solves the same equation. In particular,whose centre follows $X(t)=ut+X_0$.

For the slowly varying external potential, use the collective-coordinate effective Lagrangian method: retain the localized free profile but let its centre, boost and overall phase vary with time, with its norm fixed. Substitute this finite-parameter ansatz into the field Lagrangian, integrate over space and derive the Euler-Lagrange equations for the parameters. The potential contribution is the external potential averaged over the soliton's density; if it varies little across the profile, it is approximately its value at the centre times the conserved norm. This gives approximate particle-like centre motion while neglecting radiation and profile deformation. No solution of the full perturbed field equation is required for this approximation.

For the harmonic trap, keep the assumed trapped stationary profile $f(y)$, and put $y=x-X(t)$. Take $\Theta (x,t)=v(t)x+\gamma (t)-Et$. Substituting $e^{i\Theta }f(y)$, the imaginary terms are $-i\dot{X}{f}^{\prime }$ on the left and $-iv{f}^{\prime }$ on the right, so choose $v=\dot{X}$. Using the profile equation, the remaining real equation isIt is satisfied providedSince $d(X\dot{X})/dt={\dot{X}}^2-X^2$ for this oscillator, an explicit phase and trajectory areThus harmonic-trap translation of a nonlinear Schrödinger soliton is exact: the trapped profile translates without changing its shape, even though its centre accelerates.

For the usual scalar field theory with a nonnegative potential $U$, shifted to zero in the vacuum, write its static energy in $d$ dimensions asApply Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$. The gradient acquires a factor $\lambda$ and the volume a factor ${\lambda }^{-d}$, givingA smooth finite-energy static solution is a stationary point of this energy, so its Derrick virial identity isFor $d\ge 3$, both terms are nonnegative, forcing $T=P=0$. The field is a constant vacuum: there is no nontrivial finite-energy static soliton in this scalar-gradient-plus-nonnegative-potential model. The assumptions matter: extra derivative terms, gauge fields, time dependence or a potential not bounded below by its vacuum value can change the argument.

For $d=2$, the relation becomes $P=0$ and says nothing further about $T$. A gradient-only energy is scale invariant, so Derrick theorem alone does not rule out nonconstant solutions; sigma-model lumps are examples. For a scalar field with isolated potential minima, $P=0$ constrains the continuous field to one vacuum and hence makes it constant. If the vacuum values form a nontrivial manifold, the scaling relation alone leaves open a nonconstant map into it. In neither case does the two-dimensional scaling argument itself prove existence.

The Derrick scaling of the Abelian Higgs energy must rescale the gauge potential as well as the Higgs field. SetThen $B_{\lambda }(x)={\lambda }^{2}B(\lambda x)$ andLet $E_B=\int B^2$, $E_D=\int |D\Phi {|}^2$ and $E_P=\frac{1}{4}\int (1-|\Phi {|}^2{)}^2$, with all integrals over the plane. Changing variables givesAt a static solution, $V^{\prime }(1)=2E_B-2E_P=0$. ConsequentlyThe covariant-gradient contribution is scale invariant, while the magnetic and potential terms balance. This is why the scalar-only nonexistence conclusion does not extend to the Abelian Higgs model.

At this critical coupling, construct solutions of the Bogomolny vortex equations rather than superpose isolated vortices, which would not respect the nonlinear field equations. Define $D_j={\partial }_j-i{A}_j$ and $j_k=\mathrm{Im}(\overline{\Phi }{D}_k\Phi )$. Direct differentiation using $[D_1,D_2]\Phi =-iB\Phi$ givesIntegrating the last divergence away under the usual finite-energy vortex boundary conditions and completing the magnetic square yieldsThus positive vortices saturate this Bogomolny bound whenSuch fields solve the static second-order equations: under any compactly supported variation the flux is unchanged, and the first variation of both zero squares vanishes. The bound here is $2\pi N$, twice the bound in the frequently used half-normalized Abelian Higgs model energy.

For the prescribed-zero construction of planar Abelian Higgs vortices, choose arbitrary points $z_a\in \mathbb{C}$ and positive integer multiplicities $m_a$, permitting coincident vortices. Away from the zeros, write $\Phi =e^{h/2+i\chi }$, where $h=\log |\Phi {|}^2$. The first Bogomolny vortex equation givesChoose the phase winding $\chi ={\sum }_a{m}_a\arg (z-z_a)$ locally. Distributionally its curl contributes $2\pi {\sum }_a{m}_a{\delta }_{z_a}$, so the second Bogomolny vortex equation becomes the Taubes equationNear each zero, require $h=2m_a\log |z-z_a|+$ a smooth remainder. The planar existence theorem for this equation gives a solution for every finite multiset of prescribed points, unique in this decaying class; see the original theorem. This elliptic existence result is the nontrivial analytic ingredient of the construction, not a linear superposition assumption.

Reconstruct $\Phi =e^{h/2+i\chi }$ and $A$ from the formulas above. The phase factors have integer winding, so $\Phi$ is single-valued. Near a zero, the modulus and phase combine into $(z-z_a{)}^{m_a}$ times a smooth nonvanishing factor. The singular part of $\nabla \chi$ cancels the rotated logarithmic-gradient part of $A$, so the physical fields extend smoothly through the cores. This produces the desired Abelian Higgs vortices at the selected positions. Their vortex number is $N={\sum }_a{m}_a$. Replacing $(A,\Phi )$ by $(-A,\overline{\Phi })$ constructs the negative-sign antivortex family; the positive construction does not assert arbitrary mixtures of positive and negative cores solve the same first-order equations.

Since $|\Phi |\to 1$, the Higgs field is nonzero outside a sufficiently large disk. On a large positively oriented circle write $\Phi =\rho e^{i\chi }$ locally, where $\rho >0$. Its normalized phase is a map from the circle to the unit circle. The phase need not be a globally single-valued real function, but its total change is $2\pi N$ for an integer winding number $N$. This integer is the same on all sufficiently large circles because the intervening annulus contains no zeros.

The gauge covariant derivative givesIntegrate around the circle. The assumed rapid decay makes the integral of the second term tend to zero. For example, uniform $|D\Phi |=o(R^{-1})$ and $\rho \to 1$ on radius-$R$ circles suffice. HenceBy Stokes theorem, ${\oint }_{|x|=R}A={\int }_{|x|\le R}B{d}^{2}x$. Integrability of $B$ allows the limit, giving the magnetic flux quantization of an Abelian Higgs vortex:The topological charge $N$ is the degree of $\Phi /|\Phi |$ at infinity, counting net phase winding around the cores. For isolated zeros it is the sum of their signed local winding numbers. In the positive Abelian Higgs vortex construction, each core of multiplicity $m_a$ has local phase $m_a\arg (z-z_a)$, sothe number of vortices counted with multiplicity. Antivortices have the opposite sign. The integer is invariant under smooth gauge transformations defined throughout the disk, whose boundary winding is zero.

Write $w=u+iv$ and $R=1+u^2+v^2$. Inverting the specified stereographic projection givesThe coordinate takes the value infinity at the south pole, so a general map is described using the extended complex plane and a second chart there. A pole of $w$ is not necessarily a singularity of $\varphi$.

Differentiate the inverse formulas:On squaring and summing, the terms containing $(dR{)}^2$ cancel, leaving $|d\varphi {|}^2=4(d{u}^2+d{v}^2)/R^2$. Thus the stereographic energy of the O3 sigma model isThe same derivatives give $\varphi \cdot ({\partial }_u\varphi \times {\partial }_v\varphi )=4/R^2$, with positive sign: at $u=v=0$ the ordered tangent vectors are $(2,0,0)$ and $(0,2,0)$ at the north pole. The chain rule therefore givesThis is the topological degree for the orientation in the question.

Put $z=x_1+i{x}_2$, $w_z=(w_{x_1}-i{w}_{x_2})/2$ and $w_{\bar{z}}=(w_{x_1}+i{w}_{x_2})/2$. Expanding the squares givesFor $N\ge 0$, subtracting the second expression times $8\pi$ givesfor $N\le 0$, adding $8\pi N$ gives the analogous square with $w_z$. ConsequentlyThe factor is $8\pi$, not $4\pi$, because the printed energy has no prefactor $1/2$. Equality requires a holomorphic map for positive degree, an antiholomorphic map for negative degree, or a constant map for zero energy.

Explicit examples for every integer degree areTo verify both degree and finite energy for $w=z^m$, $m>0$, use polar coordinates: $|w_z{|}^2=m^2{r}^{2m-2}$ and $w_{\bar{z}}=0$. Thenwhere $s=r^{2m}$. Its energy is $8\pi m$. Complex conjugation changes the degree's sign while preserving the energy, and the constant example has degree and energy zero. These sigma-model lumps are smooth sphere-valued maps; the coordinate infinity at spatial infinity is regular in the other target chart.

Take a smooth gauge transformation $g:{\mathbb{R}}^3\to SU(2)$. In the anti-Hermitian convention of the adjoint covariant derivative, setExpanding ${\partial }_j(g\Phi g^{-1})$ shows that its two differentiated-$g$ terms cancel with the extra terms in $[A_j^g,{\Phi }^g]$. Thus $D_j^g{\Phi }^g=g(D_j\Phi )g^{-1}$. The gauge field strength is $F_{jk}={\partial }_j{A}_k-{\partial }_k{A}_j+[A_j,A_k]$, and the same cancellation, or the commutator of covariant derivatives, gives $F_{jk}^g=g{F}_{jk}{g}^{-1}$.

The matrix trace is unchanged by conjugation, soEach field-strength norm, covariant-gradient norm and Higgs norm is therefore invariant pointwise. It follows that $V_{\lambda }(A^g,{\Phi }^g)=V_{\lambda }(A,\Phi )$. The displayed tensor made from these fields is also gauge invariant wherever $\Phi \ne 0$.

There is a normalization problem in the requested final identity. With the ordinary defining $2\times 2$ trace and the printed inner product, an explicit orthonormal basis isThe last identity follows from the Pauli matrix multiplication law. In particular it is not permissible to combine this orthonormality with unit structure constants of a Lie algebra.

Write $A_j=a_j^a{e}_a$ and $\Phi =\phi e_3$ with $\phi \ne 0$, and put $s=\mathrm{sgn}\phi$. Define $C_{jk}=a_j^1{a}_k^2-a_j^2{a}_k^1$. ThenOnly the bracket of the transverse terms contributes in the $e_3$ direction, givingHence the literal tensor printed in the question isThe transverse term does not cancel with this trace normalization of su(2).

A concrete counterexample to the printed closure assertion is $\Phi =e_3$, $A_1=e_1$, $A_2=x_3{e}_2$, $A_3=0$ on a ball. Directly, $F_{12}=\sqrt{2}{x}_3{e}_3$, $F_{23}=-e_2$ and $F_{31}=0$, while $D_1\Phi =-\sqrt{2}{e}_2$, $D_2\Phi =\sqrt{2}{x}_3{e}_1$ and $D_3\Phi =0$. It follows thatThe Higgs field never vanishes. If global finite energy is desired, multiply both nonzero gauge-potential components by a smooth compactly supported cutoff equal to one on this ball and keep $\Phi =e_3$ everywhere. The energy is finite and the same local counterexample remains. Thus the claimed closure is false with the ordinary trace normalization actually printed; it is not a consequence of gauge invariance alone.

The consistent 't Hooft electromagnetic tensor for the printed inner product is insteadIn the directional gauge, its commutator coefficient is $1/{\kappa }^2$, so the terms cancel exactly andOn a connected region where $\phi \ne 0$, $s$ is constant. Commutation of ordinary partial derivatives provesAt any nonzero-Higgs point, use the permitted local gauge transformation to align the field with a fixed internal direction. Since $\mathcal{F}$ is gauge invariant, the identity derived in that gauge holds in the original gauge on the same ball. These balls cover every nonzero-Higgs region, proving that the corrected tensor is a closed differential form there.

Alternatively, keeping the tensor's printed coefficient one but replacing the inner product by $-2\mathrm{tr}(XY)$ gives the orthonormal basis $e_a=-i{\sigma }_a/2$ with $\kappa =1$. The same calculation then gives $f_{jk}=s({\partial }_j{a}_k^3-{\partial }_k{a}_j^3)$ and the intended identity. Either repair makes the bracket and inner-product normalizations agree; they must not be silently mixed.