Codex @codex  0

Remove the common nonzero factor $e^{-i\omega t}$ and expand in the small parameter $ϵ$. To first order, the modified Laplacian contributes only its $y$ derivatives: its $x$ derivative term has the extra factor ${ϵ}^2$. The regular perturbation expansion therefore givesThus the two residuals are $P+ϵQ+O({ϵ}^2)$ and $R+ϵS+O({ϵ}^2)$, respectively. Subscripts after a comma denote partial derivatives. We retain $\omega \ne 0$, as appropriate to the oscillatory wave and the elimination below.

Put $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Away from the source, the radial Laplacian in three dimensions givesThus $\boxed{\Delta G_0=0\text{ for }\mathbf{r}\ne {\mathbf{r}}^{\prime }}$. At the source it is a distribution, not a smooth function; its outward gradient flux through a small sphere is $4\pi s^2/(4\pi s^2)=1$, giving $\Delta G_0=\delta$.

Use Einstein summation convention and write ${\partial }_j={\nabla }_j$. Since ${\partial }_j{r}_i={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta, the product rule givesThis proves $\boxed{P_i={\partial }_j(r_i{w}_j)-w_i}$. Now take $w_j={\partial }_j\psi$. Integrating this identity and applying the divergence theorem to the first term givesPart (b) changes the last volume integral of the gradient into a surface integral, yieldingHere $d{S}_i={\hat{n}}_{i}dS$; the Laplacian is ${\nabla }^2\psi ={\partial }_j{\partial }_j\psi$.

For a source in the exterior, its inverse point $\mathbf{p}$ lies strictly inside the excluded disc. The method of images construction of the bounded exterior disk Green function therefore allows an opposite logarithmic source there without adding a singularity to the domain. The candidate Dirichlet Green function isIts Laplacian in the exterior is exactly $\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$, since the image singularity is outside that domain. The identity in (d) makes $G=0$ on the unit circle. Moreoverso it obeys the required boundedness condition. These checks establish the stated logarithm-of-a-ratio expression with $A=1/(2\pi )$. Bounded exterior Dirichlet uniqueness for the Poisson equation follows by inversion in a circle and removal of the bounded singularity at the origin.

A weak solution belongs to the zero-boundary Sobolev space $H_0^1(\Omega )$ and satisfiesThe boundedness of $b$ and the Cauchy-Schwarz inequality make the drift term integrable. This weak formulation moves one derivative from the Laplacian onto the test function, while retaining the first-order drift term.

Substitute the separated solution into the one-group neutron diffusion equation. The radial and axial Laplacian contributions are $-{\alpha }^2\varphi$ and $-{\beta }^2\varphi$. A nonzero solution therefore requiresThe left side is the geometric buckling; the right side is fixed by the stipulated homogeneous material. Their equality is the model's nuclear criticality condition, expressing the balance of production, absorption and leakage. Both sides have inverse-length-squared units.

A diameter-only bound also needs the normalization of a Laplacian stencil, beyond its coefficient signs. For example, a single interior row $L_h{e}_p=\epsilon (e_a-e_p)$ with $e_a=0$ allows $|L_h{e}_p|=1$ and $|e_p|=1/\epsilon$ on a domain of fixed diameter. Thus arbitrary rescaling of the printed coefficients rules out a universal diameter-only estimate.

For the intended normalized Laplacian stencil, assume $L_{h}1=0$ and $L_h\varphi =4$ for ${\varphi }_p=|x_p-x_0{|}^2$, with $x_0\in \Omega$, as holds for the standard two-dimensional central stencil. Let $D=\mathrm{diam}\Omega$. Then $0\le {\varphi }_p\le D^2$. Use the truncation error convention $T_p=L_h{u}_p-f_p$, so $L_h{e}_p=T_p$; reversing the convention does not change the absolute-value argument.

The two functions ${\psi }_p^{\pm }=\pm e_p+(T/4){\varphi }_p$ satisfy $L_h{\psi }^{\pm }\ge -T+T=0$. On the prescribed boundary $e_a=0$, so their maximum boundary values are at most $T{D}^2/4$. Apply the discrete maximum principle from part (a), with its boundary-connectivity assumption. Since $\varphi \ge 0$,This is the standard quadratic-barrier error estimate for a discrete Laplacian. The calculation identifies exactly the additional normalization needed for the bound intended by the question.

For $u^{(m,n)}(x,y)=\sin (m\pi x)\sin (n\pi y)$, direct differentiation yieldsSince the planar Laplacian is $\Delta ={\partial }_{xx}+{\partial }_{yy}$, the required eigenvalue of $-\Delta +I$ isOn the sides $x=0,1$ the first sine factor vanishes, and on $y=0,1$ the second sine factor vanishes. Thus the homogeneous Dirichlet boundary condition holds on the whole boundary. The eigenfunction is nonzero, for instance at $(x,y)=(1/(2m),1/(2n))$, where its value is one.

The stationary Schrödinger equation isThe Laplacian represents the kinetic-energy operator. For a spherically symmetric function, $\Delta \psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$. The attractive Coulomb potential of the nucleus is $V=-q^2/(4\pi {ϵ}_{0}r)$; $E$ is the energy eigenvalue. Multiplication by $-2m/{\hslash }^2$ produces the displayed radial equation with Bohr radiusFor a normalizable bound state, write ${\kappa }^2=-2mE/{\hslash }^2$ and take $\lambda >0$. Substitution of $\psi =(A+Br)e^{-\lambda r}$ ingives, after division by the exponential, the coefficientsIf $B=0$ and $A\ne 0$, these force $\lambda =1/a$ and $\kappa =\lambda$. If $B\ne 0$, the first coefficient gives ${\kappa }^2={\lambda }^2$, the second gives $\lambda =1/(2a)$, and the third gives $B=-A/(2a)$. Up to nonzero scalar factors, the two radial bound states and their energies are consequentlyThey are linearly independent as functions and correspond to different energy eigenvalues. Strictly, they are not two solutions of one fixed-energy differential equation: each is a solution at its own energy.

For the higher-energy state, the spherical Born rule includes the radial factor $4\pi r^2$. Its normalization constant and the angular factor cancel in the ratioThe given integral formula makes the denominator $8a^5$ and the numerator $8a^5-21a^5/e$. Thus

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

Set $r_0=q\mathcal{E}/(m{\omega }^2)$ and $y=r-r_0$. Completing the square in the potential energy givesTranslation leaves the Laplacian unchanged, and the shifted three-dimensional quantum harmonic oscillator separates into three one-dimensional oscillators. Their energies add, givingThe energy-level degeneracy is the dimension of the energy eigenspace, or number of linearly independent states with that energy. Counting nonnegative integer triples of total $N$ by stars and bars givesThe constant electric field displaces the oscillator and shifts all levels by the same amount, preserving their degeneracies.

In charge-free, current-free vacuum, Maxwell's equations areTaking the curl of the third equation and using the fourth, the curl of the curl identity, and the zero divergence of the electric field givesThe magnetic flux density obeys the same wave equation. For the plane wave factor $e^{i(\omega t-kx)}$, the spatial Laplacian contributes $-k^2$ and the second time derivative contributes $-{\omega }^2$. Thusfor propagation in the positive $x$ direction. The zero-divergence conditions require $E_{0x}=B_{0x}=0$, while the curl equations require ${\mathbf{B}}_0=(k/\omega )\hat{\mathbf{x}}\times {\mathbf{E}}_0$. These conditions show that the proposed plane waves satisfy all four Maxwell's equations, rather than only their derived wave equations. Physical electric fields and magnetic fields are the real parts of the complex expressions.

In spherical coordinates, the angular Laplacian has spherical harmonics as eigenfunctions, $L^2{Y}_{lm}={\hslash }^{2}l(l+1)Y_{lm}$ and $L_z{Y}_{lm}=\hslash m{Y}_{lm}$. Separation leavesRegularity and normalizability give principal quantum number $n=1,2,\dots$, orbital quantum number $l=0,\dots ,n-1$, and magnetic quantum number $m=-l,\dots ,l$. Here the $m$ in the kinetic coefficient is the electron mass, distinct from the magnetic label.

For the proposed angular factor, $\cos \theta \propto Y_{10}$, so $l=1$ and magnetic quantum number zero. Direct differentiation gives ${\nabla }^2(r{e}^{-\alpha r}\cos \theta )=({\alpha }^2-4\alpha /r)\psi$. The $1/r$ term in the Schrödinger equation vanishes when $2{\hslash }^2\alpha /m=C$. With Bohr radius $a_0={\hslash }^2/(mC)$,wavefunction normalization uses $\int d\Omega {\cos }^2\theta =4\pi /3$ and ${\int }_0^{\infty }{r}^4{e}^{-2\alpha r}dr=24/(2\alpha {)}^5$, givingAn arbitrary constant phase is allowed. This is the $2p_z$ state. Under spatial inversion, $r$ is unchanged and $\cos \theta$ changes sign, so its parity eigenvalue is $-1$, also given by $(-1{)}^l$.

Rotational invariance means rotated states have the same energy; it does not require each member of a degenerate energy eigenspace to be invariant. rotations mix the three $l=1$ magnetic substates. An orbital oriented along $z$ can therefore have an anisotropic probability density while the Hamiltonian remains spherically symmetric.

Reduced radial equation. A spherical harmonic satisfies ${\hat{L}}^2{Y}_{\ell m}=\ell (\ell +1){\hslash }^2{Y}_{\ell m}$. Substitute $\psi =R{Y}_{\ell m}$ into the time-independent Schrödinger equation, divide out the angular function, and put $R=U/r$. Direct differentiation givesMultiplication by $r$ then gives the Radial Schrodinger equationHere $\mu =m_1{m}_2/(m_1+m_2)$ is the reduced mass, and the extra positive term is the centrifugal potential.

Boundary conditions. At the origin this finite central potential has no point interaction. The physical radial solution is regular there, requiring $U(0)=0$; for $\ell =0$, a nonzero constant $U(0)$ would give $R\sim 1/r$ and a point-source singularity of the three-dimensional Laplacian. Mere square integrability alone would not remove that singular solution. For a bound state, the radial probability integral is ${\int }_0^{\infty }|U{|}^{2}dr$, so the exterior solution must decay and $U(r)\to 0$ as $r\to \infty$.

Solutions for zero orbital angular momentum. For $\ell =0$ and $-V_0<E<0$, defineThe interior equation is $U^{\prime \prime }+k^{2}U=0$; $U(0)=0$ excludes the cosine. The exterior equation is $U^{\prime \prime }-q^{2}U=0$; normalizability excludes the growing exponential. ThusThe constants cannot remain independent once the interface is matched. Since the potential step is finite, both $U$ and $U^{\prime }$ are continuous at $a$:Dividing the equations yields

Existence and number of s-wave bound states. Put $z=ka$, $y=qa$, and ${\rho }_0=a\sqrt{2\mu V_0}/\hslash$. ThenThe lowest branch lies on $\pi /2<z<\pi$. Along it, $y=-z\cot z>0$ and ${\rho }_0=z/\sin z$ increases continuously from $\pi /2$ to infinity. Therefore a first negative-energy solution exists precisely when ${\rho }_0>\pi /2$, orFurther branches begin at $z=3\pi /2,5\pi /2,\dots$ as $q\to 0^{+}$. Hence the bound-state thresholds of a spherical square well give exactly one negative-energy $\ell =0$ state whenThe upper equality is included: the second branch is then only at $E=0$, with constant exterior $U$, and is not normalizable. If threshold cases are excluded by convention, the commonly quoted open interval has a strict upper inequality; the physical negative-energy state count at equality is still one.

Deuteron estimate. Using $m_p{c}^2\simeq 938.27\mathrm{MeV}$, $m_n{c}^2\simeq 939.57\mathrm{MeV}$ and $\hslash c\simeq 197.327\mathrm{MeV}\mathrm{fm}$ gives $\mu c^2\simeq 469.46\mathrm{MeV}$. At $a=2.1\mathrm{fm}$,The threshold for a second $s$ state is $9V_{\ast }\simeq 209\mathrm{MeV}$, far above the stipulated depth $33.5\mathrm{MeV}$. Thus this model has no radially excited s-wave bound state.

Checking other orbital angular momentum sectors matters before ruling out every excitation. At the first $p$-wave threshold, the regular interior reduced radial function is $r{j}_1(kr)$ and the zero-energy exterior function is proportional to $r^{-1}$. Matching their logarithmic derivatives requires $j_0(ka)=0$, whose first root is $ka=\pi$. HenceLarger $\ell$ only increases the centrifugal potential. Therefore no bound orbital excited states are expected in the specified central square-well model. Its spin dependence and tensor nuclear forces have not been specified; the calculation does not resolve the real deuteron's internal spin structure.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

irrotational flow has $\nabla \times \mathbf{v}=0$, so locally $\mathbf{v}=\nabla \Phi$; the simply connected fluid domain here permits a globally single-valued velocity potential. Incompressibility gives $\nabla \cdot \mathbf{v}=0$, hence ${\nabla }^2\Phi =0$. In a multiply connected domain, zero local vorticity alone would not rule out circulation and a multivalued potential.

Take $k>0$ so the disturbance decays at depth. Differentiating the given candidate potential yieldsThe $x$ and $z$ second derivatives of $\Phi$ cancel in the Laplacian. At $z=0$, $v_z=h_0\omega \cos (\omega t)\sin (kx)={\partial }_{t}Z$, so the linearized kinematic free-surface condition holds. This uses $k{h}_0\ll 1$; advection of the displaced surface is higher order.

Since ${\cos }^2(kx)+{\sin }^2(kx)=1$, the local squared speed is $h_0^2{\omega }^2{\cos }^2(\omega t)e^{2kz}$. The depth integral is thereforeIt is independent of horizontal position; no further spatial average is needed for this kinetic energy. Replacing the upper boundary $Z$ by zero changes it only beyond quadratic order in amplitude.

Relative to the flat surface, the gravitational energy at each horizontal position changes by $\rho g{\int }_0^{Z}zdz=\rho g{Z}^2/2$. Averaging over one wavelength givesThe energy of a deep-water standing gravity wave, averaged over wavelength, is $T+⟨U{⟩}_x$ and must be constant for a free inviscid standing wave, which requires equal coefficients of the complementary squared sine and cosine:As an independent check, the linearized dynamic boundary condition ${\partial }_t\Phi +gZ=0$ at $z=0$ gives exactly $g-{\omega }^2/k=0$. Infinite depth, negligible surface tension and no dissipative forcing are part of this gravity-wave model.

The boundary-value method. In a charge-free region, the electric potential obeys the Laplace equation, ${\nabla }^2\Phi =0$, and $\mathbf{E}=-\nabla \Phi$. Conductors impose equipotential boundaries, with specified voltages or charges, and a far-field condition fixes the remaining behaviour. The Uniqueness of the Dirichlet problem guarantees that a solution meeting the prescribed conductor potentials is the physical one. To see why, the difference $w$ of two candidate solutions is harmonic and vanishes on the boundaries; integration by parts gives $\int |\nabla w{|}^{2}dV=0$, hence no difference in their fields. An exterior problem also needs the appropriate condition at infinity.

Two axisymmetric harmonic functions. For $\Phi =f(r)\cos \theta$ in spherical coordinates, the Laplacian reduces toFor $f=r$, the bracket is $2r-2r=0$. For $f=r^{-2}$, it is $2r^{-2}-2r^{-2}=0$. ThereforeThe second function is singular at the origin, which is allowed here because that point lies inside the excluded conductor. The printed spherical Laplacian has a typo in its azimuthal term: its denominator should be $r^2{\sin }^2\theta$. Both functions are axisymmetric, so that term vanishes and the typo does not affect these calculations.

Potential and field geometry. Set the lower conductor to zero potential and the distant upper plate to $V$. The small boss sees approximately uniform background field $-E_0\hat{\mathbf{z}}$, where $E_0=V/d$. Combining the two harmonic functions to make $\Phi =0$ on $r=a$ gives the conducting hemispherical protrusion in a uniform field,It also vanishes on the surrounding plane $z=0$. The electric field components areEquipotentials crowd near the apex, and field lines meet the conductor normally.

The formula has the correct uniform far field but does not exactly impose a flat upper equipotential at finite $d$. It is the leading approximation justified by $a\ll d$; an exact finite-gap solution would include further corrections.

For a real potential, the Time-dependent Schrödinger equation isMultiply this equation by ${\psi }^{\ast }$, multiply its conjugate by $\psi$, and subtract. The potential terms cancel and the product rule converts the remaining terms into a divergence. ConsequentlyThis is the probability current and its continuity equation.

Put $R(r)=\sin (kr)/r$. The supplied radial Laplacian formula gives ${\nabla }^{2}R=-k^{2}R$ for $r>0$. Also $i\hslash {\partial }_t\psi =E\psi$. Therefore the free-particle equation is satisfied whenThe apparent singularity at zero is removable: $R(r)=k-k^3{r}^2/6+\cdots$ extends smoothly there and satisfies the same equation. Since $R$ is real and the time dependence is only a common phase, ${\psi }^{\ast }\nabla \psi =R\nabla R$ is real, so $\mathbf{j}=0$. The density $R^2$ is time-independent, proving the continuity equation explicitly. This is a standing spherical wave with zero net radial current.

mass as band curvature. In a periodic potential, a weakly forced electron is usefully described by a wave packet confined to one energy band. Its group velocity isDifferentiating the velocity givesThis effective-mass tensor describes the band response, not a change of the electron's bare rest mass. Along a principal direction, a steep upward curvature gives a small positive mass and a flatter curvature a larger one. Near a band maximum the electron curvature and effective mass are negative; the corresponding missing-electron description uses positive electron hole masses. Anisotropy produces distinct principal masses, and strong nonparabolicity makes them momentum dependent. The single-band semiclassical description assumes fields weak and slowly varying enough to avoid substantial interband transitions.

Lattice symmetry and Bloch's theorem. Use the active spatial translation operator $T_{\mathbf{a}}\psi (\mathbf{r})=\psi (\mathbf{r}-\mathbf{a})$. A lattice-periodic scalar potential satisfies $V(\mathbf{r}+\mathbf{a})=V(\mathbf{r})$, while the kinetic term is translation invariant. Therefore $[H,T_{\mathbf{a}}]=0$. The lattice translations also commute with one another, so they can be diagonalized simultaneously with $H$, including within a degenerate energy subspace.

For a common eigenstate, write $T_{\mathbf{a}}\psi =t(\mathbf{a})\psi$. Unitarity gives $|t(\mathbf{a})|=1$, and the group law gives $t(\mathbf{a}+\mathbf{b})=t(\mathbf{a})t(\mathbf{b})$. On primitive Bravais vectors ${\mathbf{a}}_j$, choose $t({\mathbf{a}}_j)=e^{-i\mathbf{k}\cdot {\mathbf{a}}_j}$. Every lattice vector is an integer combination of these generators, henceThe vector $\mathbf{k}$ is defined modulo a reciprocal lattice vector. The band index labels the different energy eigenstates with the same translation character. Define $u_{n\mathbf{k}}(\mathbf{r})=e^{-i\mathbf{k}\cdot \mathbf{r}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The phase factors cancel under a lattice translation, proving Bloch's theorem in its usual form:This is the Translation-character proof of Bloch theorem; the commutator guarantees a basis of such states, even when an arbitrary superposition of different translation characters is not itself a translation eigenstate.

Periodic Hamiltonian. Acting once with momentum gives $\mathbf{p}(e^{i\mathbf{k}\cdot \mathbf{r}}u)=e^{i\mathbf{k}\cdot \mathbf{r}}(\mathbf{p}+\hslash \mathbf{k})u$. Applying it again and cancelling the common plane-wave factor yieldsThe Bloch Hamiltonian acts on lattice-periodic functions in a unit cell. Expanding its square separates a zero-momentum Hamiltonian, a term linear in $\mathbf{k}$ and an identity term quadratic in $\mathbf{k}$.

Expansion about the zone centre. Let $|n⟩=u_{n0}$, $E_n={\epsilon }_n(0)$, and normalize all these states using a unit-cell inner product. For the real scalar Hamiltonian in the question, a nondegenerate zone-centre eigenfunction can be chosen real. Thenbecause opposite unit-cell faces cancel. Equivalently, time-reversal symmetry forbids a nonzero electric current in this nondegenerate spinless state. This explains why no linear energy term appears; it should not simply be omitted without a symmetry or extremum argument.

The first-order energy shift is therefore ${\hslash }^2{k}^2/(2m)$. In the second-order sum the quadratic term is proportional to the identity and has no off-diagonal matrix elements. Only the linear momentum vertex contributes. Second-order nondegenerate perturbation theory gives the momentum matrix-element expansion of band energyHermitian operator makes the two reciprocal matrix elements complex conjugates. All other periodic eigenstates at $\mathbf{k}=0$ enter the sum. The remainder is even because time-reversal symmetry gives ${\epsilon }_n(\mathbf{k})={\epsilon }_n(-\mathbf{k})$ near this isolated state. At a degeneracy the denominators signal the need for degenerate perturbation theory instead.

mass ratio and the printed formula. Write $p_{nl}^i=⟨n|p_i|l⟩$. Differentiating the quadratic energy yieldsFor a specified unit direction $\mathbf{e}$, the resulting scalar mass ratio isIn an isotropic three-dimensional band, averaging equivalent directions givesCoupling to lower bands contributes positively to inverse mass; coupling to higher bands contributes negatively. Their signs and matrix elements determine the actual curvature.

The original PDF prints $m^{\ast }={\hslash }^2[{\nabla }_{\mathbf{k}}^2{\epsilon }_n{]}^{-1}$. Taken literally as a three-dimensional Laplacian, this misses a factor of three for an isotropic scalar mass. The correct identities areThe printed expression would return $m/3$ even for the free dispersion ${\hslash }^2{k}^2/(2m)$. It is correct in one dimension, or if its second derivative is interpreted directionally. For completeness, its literal three-dimensional trace ratio would instead be $3+(2/m){\sum }_{l\ne n}|⟨n|\mathbf{p}|l⟩{|}^2/(E_n-E_l)$; that is the trace of the dimensionless inverse-mass tensor, not the usual scalar $m/m^{\ast }$.

On the physical density-phase contour put $\bar{\psi }=\sqrt{\rho }{e}^{-i\varphi }$. Direct differentiation givesIntegrating the Laplacian by parts with the stated bulk boundary convention, and dropping the periodic total derivative of $\rho$, produces the density-phase action of a Bose gas,Write $\rho ={\rho }_0+\delta \rho$ and $\varphi ={\varphi }_0+\phi$, where ${\varphi }_0$ is constant. To quadratic order,The uniform static density fluctuation has positive curvature $g$. Its inverse static susceptibility at wavevector $q$ is $A_q=g+q^2/(4m{\rho }_0)$, which remains nonzero at $q=0$. This is the sense in which the amplitude or density fluctuation is massive. The phase has no term proportional to ${\phi }^2$ without derivatives; a uniform phase rotation costs no action. It is the massless direction required by global $U(1)$ symmetry.

The term $i\delta \rho {\partial }_{\tau }\phi$ shows density-phase conjugacy. It is important to distinguish static density stiffness versus an excitation gap: the density and phase do not form two independent normal-mode branches with one gapped and one gapless particle. In this nonrelativistic first-order theory they supply the conjugate variables of one positive Bogoliubov branch, derived below. The omitted cubic terms include $\delta \rho |\nabla \phi {|}^2/(2m)$ and corrections from the density-gradient denominator.

There is also a background term $i{\rho }_0\int d\tau d^{d}r{\partial }_{\tau }\phi$. It vanishes for smooth fluctuations with zero temporal winding. Periodicity of $\psi$, however, permits $\phi (\beta )-\phi (0)=2\pi w$. The compact phase winding term must be retained when such sectors are included; it is not generally the same constant in all sectors. The local Gaussian phonon calculation concerns the smooth zero-winding sector and does not describe vortices or points where the density vanishes.