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For a real scalar field with Lagrangian density $\mathcal{L}={\dot{\varphi }}^2/2-|\nabla \varphi {|}^2/2-V(\varphi )$, the canonical momentum is $\pi =\dot{\varphi }$. Its Hamiltonian density is the sum of momentum energy, gradient energy and potential energy. Hamilton's equations recover the Euler-Lagrange equation.

For Minkowski metric $(+---)$ and $\mathcal{L}=-G_{\mu \nu }^{\ast }{G}^{\mu \nu }/2+M^2{B}_{\mu }^{\ast }{B}^{\mu }$, use coordinates $B^{\mu },B^{\ast \mu }$. The spatial canonical momenta are ${\pi }_i=G^{\ast 0i}$, ${\pi }_i^{\ast }=G^{0i}$; the temporal momenta vanish. Constraint preservation gives ${\partial }_i{\pi }_i=M^2{B}_0^{\ast }$. Eliminating $B_0$ and its conjugate gives the positive reduced Hamiltonian density displayed above, up to a spatial surface term. In particular the massive Gauss law has a mass-dependent right-hand side; setting it to zero removes a physical longitudinal particle polarization.

With $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, a free real scalar field has mode expansion of a free field $\varphi =\int d{\Pi }_p(a{e}^{-ipx}+a^{†}{e}^{ipx})$ and commutator $[a(p),a^{†}(q)]=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. Substitution into the quadratic Hamiltonian density gives $H=\frac{1}{2}\int d{\Pi }_p{E}_p(a^{†}a+a{a}^{†})$. The terms with two annihilation operators or two creation operators cancel by $E_p^2={\mathbf{p}}^2+m^2$. Normal ordering removes the constant vacuum energy. The commutators $[H,a(p)]=-E_{p}a(p)$ and $[H,a^{†}(p)]=E_p{a}^{†}(p)$ give the boson energy without changing under this subtraction. Replacing $a(p)$ by $\sqrt{2E_p}c(p)$ recovers the noncovariant normalization $[c(p),c^{†}(q)]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$.

Insert the real scalar field mode expansion into its quadratic Hamiltonian density and use the Fourier representation of the Dirac delta function. The two-annihilator and two-creator terms cancel by $E_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The remaining symmetric expression is $H=\frac{1}{2}\int d^{3}p{E}_{\mathbf{p}}(a^{†}a+a{a}^{†})/(2\pi {)}^3$. Normal ordering removes its constant vacuum energy, yielding the displayed operator. The canonical commutation relations give $[:H:,a_{\mathbf{p}}]=-E_{\mathbf{p}}{a}_{\mathbf{p}}$ and $[:H:,a_{\mathbf{p}}^{†}]=E_{\mathbf{p}}{a}_{\mathbf{p}}^{†}$.

For a repulsive contact-interacting Bose-Einstein condensate, the classical Hamiltonian density generating evolution in a frame rotating at the chemical potential is $\mathcal{K}={\hslash }^2|\nabla \psi {|}^2/(2M)-\mu |\psi {|}^2+g|\psi {|}^4/2$. Its integral is the grand canonical ensemble generator $H-\mu N$, not the physical energy $H$ alone. For $g,\mu >0$ the uniform mean-field theory minima satisfy $|{\psi }_0{|}^2=\mu /g$, with a freely chosen constant phase. In the physical frame, the corresponding field has time factor $e^{-i\mu t/\hslash }$.

A Hamiltonian density is a local integrand for the field-theory Hamiltonian. For canonical fields with invertible velocity-momentum relation it is the Legendre transform in mechanics of the Lagrangian density, summing over all conjugate field components. Gauge theories require constraints and multiplier terms as well.

Spatial integration of the free Hamiltonian density cancels particle-antiparticle pair terms through $E_p^2-{\mathbf{p}}^2-m^2=0$. The raw Hamiltonian is $\int d{\Pi }_p{E}_p(a^{†}a+b{b}^{†})$. Normal ordering removes the field-independent vacuum energy and gives $H=\int d{\Pi }_p{E}_p(a^{†}a+b^{†}b)$. Both creation operators increase the energy by $E_p$, so a negative-frequency field mode represents a positive-energy antiparticle, not a negative-energy state.

Treat $\varphi ,{\varphi }^{†}$ as independent variables when taking the Legendre transform. Their canonical momenta areThe canonical quantization of a complex scalar field imposes, at equal times,with $[\varphi ,{\varphi }^{†}]=[\varphi ,\varphi ]=[\pi ,{\pi }^{†}]=[\pi ,\pi ]=0$, together with their adjoints, and $[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$. These are canonical commutation relations for two real scalar degrees of freedom, written in a complex basis. Local fields are operator-valued distributions, so these identities are understood after smearing or with a regulator.

The Hamiltonian density is $\pi \dot{\varphi }+{\pi }^{†}{\dot{\varphi }}^{†}-\mathcal{L}$, givingThe Heisenberg equation of motion is $\dot{O}=i[H,O]$. For example, $i[\int {\pi }^{†}\pi ,\varphi ]={\pi }^{†}$, while commuting $\pi$ with the gradient term and integrating its derivative of a Dirac delta distribution gives ${\nabla }^2{\varphi }^{†}$. ThusBoth fields satisfy the Klein-Gordon equation. Their spatial Fourier modes therefore have frequencies $\pm E_p$, $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. A complex scalar field has independent coefficients for these two frequency sectors, so, defining the invariant measure $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, its mode expansion isOne can recover the mode operators at any fixed time using complex scalar mode inversion:Substitution of the expansion verifies these inverses: the spatial integral selects the appropriate momentum, while the two frequency sectors acquire factors $E_p\pm E_q$. The unwanted sector vanishes when the delta function sets $E_p=E_q$. The canonical commutation relations then give, for example,Likewise $[b(p),b^{†}(q)]$ has the same value. The cross commutators vanish; their coefficients contain $E_p-E_q$ on the corresponding momentum delta function. Conversely, these oscillator brackets reproduce the equal-time field brackets, confirming the normalization of the measure.

Insert the mode expansion into $H$ and integrate over space. The terms $a^{†}{b}^{†}$ and $ba$ enforce opposite spatial momenta; their coefficient is $-E_p^2+{\mathbf{p}}^2+m^2=0$. For the diagonal terms the coefficient is $E_p^2+{\mathbf{p}}^2+m^2=2E_p^2$. Combining this with the two invariant measures leavesIn a box, $E_0={\sum }_{\mathbf{p}}{E}_p$, one half-quantum for each of the two real oscillator species. It diverges as the cutoff is removed. Normal ordering removes this vacuum energy, yielding the normal-ordered Hamiltonian of a free complex scalar field:In this nongravitational free theory an additive constant in $H$ changes only an overall state phase, not Heisenberg equations of motion, energy differences or scattering probabilities. It is therefore consistent to choose the Fock vacuum energy as zero. Both creation operators raise the energy: $[H,a^{†}(p)]=E_p{a}^{†}(p)$ and $[H,b^{†}(p)]=E_p{b}^{†}(p)$. Repeated creation builds a bosonic Fock space of spin-zero particles and antiparticles, with the same positive mass and energy. The negative-frequency part of the field does not describe negative-energy physical particles.

The global phase symmetry of a complex scalar field, with orientation $\varphi \mapsto e^{-i\alpha }\varphi$, has the indicated Noether current. Taking its divergence, the two first-derivative terms cancel, givingThe same calculation holds for its normal-ordered version. Integrating its temporal component gives a conserved charge when spatial boundary flux vanishes.

Here the ordering convention is important. Direct substitution into the literally ordered product $i\int ({\varphi }^{†}\dot{\varphi }-{\dot{\varphi }}^{†}\varphi )d^{3}x$ yieldsThe pair terms again vanish, now because their coefficient is a difference of equal energies. Reordering the $b$ term produces a divergent vacuum constant: with a finite-mode regulator $Q_{\mathrm{bare}}=N_a-N_b-N_{\mathrm{modes}}$. Thus the requested finite charge formula uses zero vacuum charge, or equivalently the vacuum subtraction of a complex scalar charge. With that conventional subtraction,The subtraction changes neither current conservation nor the action of $Q$ on fields; without it, the unrenormalized product does not literally equal the displayed normal-ordered operator.

Finally the oscillator commutators giveor $\boxed{Q{a}^{†}=a^{†}(Q+1),Q{b}^{†}=b^{†}(Q-1)}$. Acting on a charge eigenstate, the first creator raises its charge by one and the second lowers it by one. Starting from the neutral Fock vacuum, their one-particle states therefore have charges $+1$ and $-1$ respectively. This is the complex scalar charge operator $N_a-N_b$, whereas the energy counts the sum of the two occupations. A multiparticle state has additive charge equal to the number of particles minus the number of antiparticles.

The real scalar field has canonical momentum $\pi =\dot{\varphi }$. Its Legendre transform givesThe first three terms constitute $H_0$, the free massive Hamiltonian operator. The remaining term is $H_I=\lambda \int {\varphi }^4{d}^{3}x/4!=-\int {\mathcal{L}}_I{d}^{3}x$. This sign is a consequence of having no time derivatives in the interaction: the canonical momentum is unchanged, while the interaction enters the Hamiltonian density with the opposite sign to the Lagrangian density.

In the interaction picture, operators evolve with $H_0$ and states evolve with $H_I(t)$ constructed from the free interaction-picture field. Their evolution operator satisfiesIntegrating once gives $U_I=I-i{\int }_{t_0}^t{H}_I(t_1)U_I(t_1,t_0)d{t}_1$. Repeated substitution orders the integrations as $t\ge t_1\ge \cdots \ge t_n\ge t_0$. Replacing the ordered simplex by the full integration cube introduces $1/n!$ and a time-ordered product, which proves the Dyson seriesTaking the asymptotic limits with the usual adiabatic or regulated scattering prescription yieldsThe field in this expression is the free field evolving in the interaction picture, not an independently prescribed classical field. For quantum fields the expansion is a regulated perturbation series; the time-ordering symbol cannot be dropped because the operators at different times need not commute.

Use relativistically normalized two-particle states $|p_1,p_2⟩=a^{†}(p_1)a^{†}(p_2)|0⟩$, with $⟨p|q⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. The connected first-order scattering contribution isHere the connected S-matrix uses a normalized vacuum and physical external one-particle states. At this order, normal ordering the vertex separates the four-external-leg contribution from vacuum and tadpole pieces; those disconnected or external-mass contributions are not the two-to-two scattering amplitude.

Two fields annihilate the incoming particles and two create the outgoing ones. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ choices of the annihilating fields, two assignments of the incoming momenta and two assignments of the outgoing momenta. Hence the total factor is $\left(\genfrac{}{}{0ex}{}{4}{2}\right)2!2!=4!$, the external-leg factorial cancellation at a quartic scalar vertex. Each attachment contributes its plane-wave factor, soThe spacetime integral is the four-momentum conservation delta function, givingIn particular, the tree scattering amplitude is independent of angle.

For completeness, derive the relativistic two-body phase space in the centre-of-mass frame. Write $\sqrt{s}=2E_{\ast }$, $p_{\ast }^2=E_{\ast }^2-m^2$. The spatial delta function sets the two final momenta opposite, and the energy delta function is $\delta (\sqrt{s}-2\sqrt{p^2+m^2})$. Its radial Jacobian is $E_{\ast }/(2p_{\ast })$. ThusThe invariant flux factor is $\mathcal{F}=4p_{\ast }\sqrt{s}$ for equal initial masses. Using $d\sigma =|T{|}^{2}d{\Phi }_2/\mathcal{F}$, the labeled angular density is ${\lambda }^2/(64{\pi }^{2}s)$. The outgoing quanta belong to one real scalar field, so they are identical. A full-sphere integration counts each event twice and requires the identical final-state symmetry factor $1/2!$. ConsequentlyEquivalently one may integrate the labeled density over a hemisphere representing each event once. The final-state factor is separate from the vertex factorial: applying it at the vertex instead would give the wrong elastic scattering from a quartic scalar contact interaction cross-section.

The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and total energy areThe Euler-Lagrange field equation is ${\varphi }_{tt}-{\varphi }_{xx}+U^{\prime }(\varphi )=0$. Multiplication by ${\varphi }_t$ gives the local conservation of energy identityThus $E$ is conserved when the energy flux vanishes at spatial infinity. For finite energy with asymptotically constant fields, the endpoints ${\varphi }_{\pm }$ must be scalar-field vacua, where $U=0$.

Choose an antiderivative $G^{\prime }(\varphi )=\sqrt{2U(\varphi )}$. For either sign $s=\pm 1$, completing the square givesChoosing $s$ to make the boundary term nonnegative proves the Bogomolny boundThese Bogomolny equations obtained by square completion for a one-dimensional kink describe minima within a fixed endpoint sector when a saturating configuration exists. Differentiating the spatial equation gives ${\varphi }_{xx}=U^{\prime }(\varphi )$ wherever $U>0$, with smooth extension to its limiting scalar-field vacua. The absolute minimum across all sectors is a constant zero-potential scalar-field vacuum when one exists; the nontrivial minima are sector-specific kinks.

For $\beta \ne 0$, write $b=|\beta |$. The scalar-field vacua are $-b,0,b$. A static finite-energy solution has the first integralbecause differentiating its left side gives ${\varphi }_x({\varphi }_{xx}-U^{\prime })=0$, and its value is zero at infinity. In either adjacent interval, $\sqrt{2U}=|\varphi |(b^2-{\varphi }^2)$. Set $y={\varphi }^2/b^2$. The first-order equation becomes $y_x=\eta 2b^{2}y(1-y)$, with $\eta =\pm 1$, and separation gives $\log [y/(1-y)]=2\eta b^2(x-x_0)$. HenceFor $\eta =1$ the endpoints are $0\to \sigma b$; for $\eta =-1$ they are $\sigma b\to 0$. Thus there are four oriented static kink families, up to translation, namely two increasing kinks and two decreasing antikinks. Each has rest energyA static field joining $-b$ directly to $b$ would have to pass through $\varphi =0$ at finite $x$. The first integral then gives ${\varphi }_x=0$, and the Picard-Lindelöf theorem applied to the smooth second-order equation with those initial conditions forces $\varphi \equiv 0$. Therefore there is no such additional kink; this is the intermediate-vacuum obstruction to a kink. The Bogomolny classification of a rescaled phi-six kink also shows that if $\beta =0$, only the scalar-field vacuum $0$ remains and there is no nontrivial finite-energy static kink. Indeed the first integral gives a monotone field wherever it is nonzero, precluding a return to the same scalar-field vacuum.

A Lorentz boost of the $0\to b$ profile gives a moving solution with arbitrary $|v|<1$:If $K$ denotes the static profile and $\xi =\gamma (x-vt-x_0)$, then ${\varphi }_{tt}-{\varphi }_{xx}={\gamma }^2(v^2-1)K^{\prime \prime }=-K^{\prime \prime }$, proving the Euler-Lagrange field equation directly. Its energy is $\gamma b^4/4$: since $K^{\prime 2}=2U(K)$, changing variable to $\xi$ in the energy integral gives the factor $\gamma$. Reversing $\sigma$ or $\eta$ yields the other moving elementary profiles.

A partition function sums the Boltzmann weights of all states: $Z={\sum }_s{e}^{-{\beta }_{T}H(s)}$ for discrete states, or the corresponding integral for continuous fields. It normalizes equilibrium probabilities. Its physical Helmholtz free energy is $\mathcal{F}=-k_{B}T\log Z$, and its free-energy density is $\mathcal{F}/V$. With a momentum ultraviolet cutoff ${\Lambda }_0$, a scalar-field example isThe normalization of the functional measure and field-independent energy terms must be tracked when discussing absolute free energy.

Choose a smaller cutoff $\Lambda$ and split the field into retained modes ${\varphi }_{<}$ with $|q|<\Lambda$ and eliminated modes ${\varphi }_{>}$ with $\Lambda <|q|<{\Lambda }_0$. Define the Wilsonian coarse-grained statistical Hamiltonian bySubstitution into the partition function proves $Z=\int \mathcal{D}{\varphi }_{<}{e}^{-{\beta }_T{H}_{\Lambda }[{\varphi }_{<}]}$ for every intermediate cutoff. A source retained in this definition similarly preserves long-wavelength connected correlation functions and response. Thus the coefficients in a Hamiltonian density must run with $\Lambda$: the dependence induced by eliminated modes compensates the dependence of the remaining field integral. Holding the original bare coefficients fixed while changing the cutoff would generally change the system. Exact integration generates all interactions allowed by symmetry, including higher powers, derivative terms and an additive constant; a local derivative expansion or a truncation to a few couplings is an extra approximation.

Cutoff independence makes arbitrary intermediate coarse-graining scales unphysical. It allows calculations at the scale of the observable, avoids repeatedly resolving irrelevant short-distance details and exposes the renormalization-group fixed points responsible for universality. A microscopic cutoff, such as a Bravais lattice spacing, can still label a physical microscopic model; taking it to infinity as a continuum limit requires tuning appropriate bare couplings. This is distinct from lowering an intermediate Wilsonian cutoff in a fixed model.

To interpret the zero-cutoff limit, define $M=V^{-1}\int \varphi (x)d^{D}x$, the spatially uniform mode of the order parameter. Keep this mode fixed and integrate the others, rather than integrating over $M$ too. Write $\varphi =M+\eta$, with $\int \eta =0$, and define the constrained order-parameter free energy byIn a finite periodic box, once $\Lambda$ is below the smallest nonzero momentum, all nonconstant modes have been eliminated. The remaining effective energy is $H_{\Lambda }[M]=V{U}_{\Lambda }(M)$, so its defining integral is exactly $Z_M$. Therefore, with consistent normalization,When $\mathcal{H}(\Lambda ,M)$ denotes this uniform effective Hamiltonian density, this is the stated identification of $F(M)$ with its zero-cutoff limit. It requires a free-energy density constrained at fixed $M$, not the unconstrained extensive free energy; all integrated fluctuations and additive constants are included, the zero mode in field theory is retained, and a density for the uniform configuration is well defined. The finite-volume argument avoids an interchange of the thermodynamic and cutoff limits without justification. On a homogeneous pure-phase branch one may use a local potential description, while a thermodynamic limit allowing phase separation can convexify the constrained potential. A bare nonconvex Landau polynomial is accordingly a homogeneous-branch approximation, not a claim that the exact equilibrium Legendre potential stays nonconvex.

In a field $h$, the remaining integral is $Z(h)=\int dM{e}^{-{\beta }_{T}V[F_c(M)-hM]}$, up to the consistently chosen zero-mode measure. The thermodynamic saddle then gives $f(h)={\inf }_M\{F_c(M)-hM\}$ and selects the equilibrium magnetization. Landau theory assumes that the relevant uniform potential can be approximated by a regular analytic expansion in $M$, with symmetry-constrained coefficients, and minimizes it. At tree level the integrated fluctuations are neglected, recovering the familiar bare polynomial. With fluctuations retained, the coefficients run and can become singular near criticality; the zero-cutoff identification of constrained free energy alone does not justify mean-field critical exponents.

For an ordinary short-range scalar thermodynamic critical point, the leading interaction is quartic. Absorb inverse temperature and the gradient coefficient into field normalization and writeScale $x=b{x}^{\prime }$ and $\varphi (x)=b^{-(D-2)/2}{\varphi }^{\prime }(x^{\prime })$ to leave the gradient term invariant. The quartic term changes to $g{b}^{4-D}\int {\varphi }^{\prime 4}/4!$, so its engineering scaling exponent is $4-D$. It is irrelevant at the Gaussian fixed point for $D>4$, relevant for $D<4$, and marginal for $D=4$. HenceThe loop expansion makes the breakdown concrete. A quartic-vertex correction is proportional to $g^2{I}_2(r)$, whereFor $0<D<4$, rescaling $q=\sqrt{r}k$ gives $I_2(r)\sim C_D{r}^{(D-4)/2}$, diverging as $r\to 0$. At $D=4$, radial integration giveswhich is logarithmically divergent. Above four dimensions its infrared limit is finite; the leading cutoff-dependent constant can be absorbed into a coupling renormalization. The effective expansion parameter is thus of order $g{\xi }^{4-D}$ below four, or $g\log (\Lambda \xi )$ at four, since $\xi \sim r^{-1/2}$. It eventually becomes large for any fixed bare interaction in the lower-dimensional critical limit, defeating the uncorrected mean-field expansion. The Ginzburg criterion gives the same threshold by comparing correlation-volume fluctuations with the squared mean order parameter.

Below four dimensions an interacting renormalization-group fixed point can replace the Gaussian fixed point and change critical exponents, provided a continuous transition exists. At four dimensions the interaction is marginal and produces logarithmic corrections: the leading power exponents can remain their mean-field values, so “breakdown” there means that pure mean-field power laws omit those corrections, rather than necessarily that every power exponent changes. Above four the stabilizing quartic interaction is dangerously irrelevant for the ordered phase, explaining why mean-field thermodynamic exponents need not obey naive hyperscaling with the spatial dimension.

The Euler-Lagrange equation is obtained by varying the scalar field, integrating the derivative terms by parts, and requiring the coefficient of an arbitrary compactly supported variation to vanish:The canonical momentum is $\pi ={\varphi }_t$, so the Hamiltonian density and conserved energy areIndeed,Thus $dE/dt=[{\varphi }_t{\varphi }_x{]}_{-\infty }^{+\infty }=0$ for the finite-energy boundary conditions with no flux at infinity. The scalar-field vacua are the constant fields $2\pi m$, $m\in \mathbb{Z}$.

For a static field, multiply ${\varphi }^{\prime \prime }=\sin \varphi$ by ${\varphi }^{\prime }$ to obtain $\frac{1}{2}{\varphi }^{\prime 2}+\cos \varphi =C$. The vacuum limits fix $C=1$, so on the increasing interval $0<\varphi <2\pi$,ConsequentlyIts derivative is $2\mathrm{sech}(x-a)$, and its energy is $\int 4{\mathrm{sech}}^2(x-a)dx=8$. The increasing sign cannot change inside $(0,2\pi )$ because the derivative does not vanish there. This proves uniqueness up to translation for the prescribed boundary conditions.

There are continuously many translated static solutions with those same endpoints, but only one modulo translations. In the full theory, the adjacent-vacuum kinks and antikinks areTheir endpoints are $2\pi m$ and $2\pi (m+\sigma )$. Thus there are countably many adjacent-vacuum boundary sectors, each with a continuous center parameter; integer field shifts and reflection relate their shapes. A Lorentz boost gives moving kinks, replacing $x-a$ by $(x-vt-a)/\sqrt{1-v^2}$, with energy $8/\sqrt{1-v^2}$. There is no finite static kink joining nonadjacent scalar-field vacua: reaching an intermediate vacuum at finite $x$ would give both the equilibrium value and zero derivative, forcing a constant solution by uniqueness of the second-order initial value problem. Dynamical multisolitons are a different matter.

The printed Sine-Gordon Bäcklund transformation has a normalization error. With precisely the printed coordinates $\tau =x+t$, $\rho =x-t$,is the field equation. Put $s=({\varphi }_1+{\varphi }_0)/2$, $d=({\varphi }_1-{\varphi }_0)/2$. The printed relations give $d_{\rho }=b\sin s$ and $s_{\tau }=b^{-1}\sin d$, henceAdding and subtracting yieldsThese are the equations with mass squared four, not the unit-mass equation derived from the Lagrangian. A concrete counterexample is the seed ${\varphi }_0=0$, parameter $b=1$, and transformed field ${\varphi }_1=4\arctan e^{2x}$. It obeys both printed first-order relations but satisfies ${\varphi }_{1,xx}=4\sin {\varphi }_1$, so its unit-mass field-equation residual is $-3\sin {\varphi }_1$, which is not identically zero. The requested preservation claim is false with the printed normalization.

There are two equivalent repairs. Keep the printed first-order coefficients but take $\tau =(x+t)/2$, $\rho =(x-t)/2$; then ${\partial }_{\tau }{\partial }_{\rho }={\partial }_x^2-{\partial }_t^2$ and the preceding calculation proves preservation. Alternatively keep the printed coordinates and replace the coefficients $2b,2b^{-1}$ by $b,b^{-1}$. Then $d_{\rho }=(b/2)\sin s$, $s_{\tau }=(b^{-1}/2)\sin d$, and the differentiated equations carry a factor $1/4$, giving exactly ${\varphi }_{j,\tau \rho }=\frac{1}{4}\sin {\varphi }_j$. This is the light-cone normalization of the sine-Gordon Bäcklund transformation. For a zero seed it gives ${\varphi }_1=4\arctan \exp [(b\rho +b^{-1}\tau )/2]$; $b=1$ recovers the unit-width static kink.

Use static gauge for a D3-brane in flat space and let $X(\mathbf{x})$ be one physical transverse displacement. Put $\mathbf{E}=2\pi {\alpha }^{\prime }{\mathbf{F}}_0$ and ${\mathbf{B}}_i=\pi {\alpha }^{\prime }{\epsilon }_{ijk}{F}_{jk}$, so the fields entering the DBI action are dimensionless. With no magnetic field, writing $\mathbf{v}=\nabla X$ givesIntroduce the dimensionless electric displacement in DBI theory, $\mathbf{d}=T_3^{-1}\partial \mathcal{L}/\partial \mathbf{E}$. Its Hamiltonian density obeys the DBI Hamiltonian bound for an electric spike:The electric bound is saturated when $\mathbf{d}=\pm \nabla X$. On that locus the constitutive relation also gives $\mathbf{E}=\mathbf{d}$. With the Gauss law constraint in gauge theory away from charges, the electric BPS equations for a one-scalar D3-brane are thereforeThe magnetic version is $\mathbf{B}=\pm \nabla X$, $\mathbf{E}=0$. More generally, one constant charge angle gives the aligned dyonic branch $\mathbf{E}=\cos \vartheta \nabla X$, $\mathbf{B}=\sin \vartheta \nabla X$ with $X$ harmonic. A single aligned electric, magnetic or dyonic spike preserves eight of the D3 vacuum's sixteen supersymmetry generators. Thus “one-half BPS” here means one-half of the linearly realized worldvolume supersymmetry; the string–D3 configuration preserves one-quarter of the thirty-two spacetime supersymmetries.

The electric BIon is the single-centre harmonic solutionThe sign fixes its electric charge and spike orientation. The scalar is an actual transverse coordinate, so $X\to \infty$ as $r\to 0$ describes a long narrow spike emerging from an asymptotically planar D3-brane. It carries $|N|$ fundamental strings ending on the brane. A magnetic BIon similarly represents D1-branes, and an aligned dyonic BIon represents a (p,q) string.

The string-tension interpretation of BIon energy follows explicitly. With $T_3=[(2\pi {)}^3{g}_s{\alpha }^{\prime 2}{]}^{-1}$, the energy above the planar-brane vacuum outside a small radius $r_0$ isIts divergence is the energy of a semi-infinite string, with the correct finite tension. For a magnetic spike, $c=\pi |N|{\alpha }^{\prime }$ instead, and the same calculation gives the D1-brane tension $1/(2\pi g_s{\alpha }^{\prime })$. The solution solves the full nonlinear DBI action, not just its weak-field expansion.

The validity of the abelian BIon approximation concerns its derivative expansion, string loops and gravitational backreaction. Large field strengths are included by the DBI action, but rapidly varying field strengths and embedding curvature require corrections. Far outside the singular core these corrections can be small. The transition region has $r\sim \sqrt{c}$; an electric throat can be parametrically controlled with $|N|g_s\gg 1$ while keeping gravitational backreaction small with $g_s^2|N|\ll 1$. For a magnetic throat the corresponding conditions are $|N|\gg 1$ and $g_s|N|\ll 1$. The weak-coupling and large-charge windows allow both requirements. The ideal $r\to 0$ tip is not uniformly controlled by the single abelian effective action, so one should not infer exact microscopic core physics solely from its formal singularity.

Less supersymmetry survives when independent, compatible geometric or charge projectors must hold simultaneously. A useful purely geometric family is a D3-brane whose spatial worldvolume is a special Lagrangian submanifold of ${\mathbb{C}}^3$. In a graph $z^i=x^i+i{Y}^i(\mathbf{x})$, the conditions areThe first condition makes ${\partial }_i{Y}^j$ symmetric, so locally $Y=\nabla f$. Pulling back $\Omega$ gives $\det (I+i{D}^{2}f)d{x}^1\wedge d{x}^2\wedge d{x}^3$. Hence the phase-zero special Lagrangian graph equation iswhere ${\sigma }_2$ is the sum of the pairwise products of the three Hessian eigenvalues. The positive real branch fixes the orientation. These curved D3 geometries minimize volume by their calibration $\mathrm{Re}\Omega$ and generically preserve four spacetime supersymmetries, one-quarter of the D3 worldvolume supersymmetry. Special flat graphs can preserve more.

The generic count follows from the common tangent-plane projectors. With complex coordinate pairs $(1,4),(2,5),(3,6)$, the phase-preserving rotations require $({\Gamma }_{14}-{\Gamma }_{25})ϵ=0$ and $({\Gamma }_{14}-{\Gamma }_{36})ϵ=0$. Multiplication by ${\Gamma }_{14}$ turns these into two independent commuting involution conditions with eigenvalue $-1$. Each halves the sixteen-dimensional planar D3 eigenspace, leaving $16/2^2=4$. This is an intersection of compatible kappa symmetry projectors, rather than a count based on the rank of a single projector at one point.

Charge-carrying examples include compatible networks of differently oriented (p,q) strings ending on D3-branes. At each string junction, charge vectors add to zero and tension vectors balance; their directions and charge phases must also give common kappa symmetry projectors. They can impose further independent conditions beyond the planar D3 projector. Geometrically, several transverse scalar profiles describe the corresponding spikes and intersections. Merely replacing one electric spike by an aligned dyonic spike does not lower its preserved fraction; additional independent conditions are what reduce the common Killing spinor space.

Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.