Scalar-field kink

A scalar-field kink is a one-dimensional finite-energy static solution approaching different vacua as $x\to -\infty$ and $x\to +\infty$. Its boundary values define a topological sector.

An antikink is a scalar-field kink with the opposite orientation of its vacuum boundary values. If ${\varphi }_K(x)$ joins vacua ${\varphi }_{-}$ to ${\varphi }_{+}$, then ${\varphi }_K(-x)$ joins ${\varphi }_{+}$ to ${\varphi }_{-}$. For an even double-well potential and an odd centered phi-four kink, the centered antikink is $-{\varphi }_K(x)$. With $Q=[\varphi (+\infty )-\varphi (-\infty )]/({\varphi }_{+}-{\varphi }_{-})$, a kink has $Q=1$ and an antikink has $Q=-1$. Their localized energies coincide by spatial reflection.

For the static energythe kink joining the vacua $-c$ and $c$ isIt saturates the Bogomolny bound $E\ge 4c^3/3$ by satisfying the first-order equation ${\varphi }^{\prime }=c^2-{\varphi }^2$.

For the real scalar field with metric $(+,-)$ and potential $U={\kappa }^2({\varphi }^2-1{)}^2/2$, the static phi-four kink is ${\varphi }_0(x)=\tanh (\kappa x)$ and satisfies ${\varphi }_0^{\prime 2}=2U({\varphi }_0)$. Its mass is $M=\int {\varphi }_0^{\prime 2}dx=4\kappa /3$. A Lorentz boost gives $\varphi (t,x)={\varphi }_0(\gamma (x-vt))$. The stress-energy tensor gives $P=-\int {\varphi }_t{\varphi }_{x}dx=\gamma v\int {\varphi }_0^{\prime 2}dx=\gamma Mv$. Similarly, $E=\frac{1}{2}[\gamma (1+v^2)+{\gamma }^{-1}]M=\gamma M$, proving the energy–momentum relation $E^2-P^2=M^2$.

For $V(\varphi )=\lambda (m^2-{\varphi }^2{)}^2$ with $\lambda ,m>0$, set $\kappa =\sqrt{2\lambda }m$. A well-separated kink–antikink pair at $-d/2,d/2$ is approximated by $\varphi =m\tanh \kappa (x+d/2)-m\tanh \kappa (x-d/2)-m$. At its midpoint, ${\varphi }_x=0$ and $\varphi =m-4m{e}^{-\kappa d}+O(e^{-2\kappa d})$. The static stress-energy tensor component $T^{11}={\varphi }_x^2/2-V$ is therefore negative there. Momentum conservation gives the force on the left half-line as $T^{11}(-\infty )-T^{11}(0)$, which is positive. Its leading magnitude is $F\sim 32m^2{\kappa }^2{e}^{-2\kappa d}$. Thus the pair attracts. This is a controlled leading-tail estimate at large separation, not an exact superposed solution or a theorem about all subsequent collision outcomes.

For the initial field $\varphi (x)=\tanh (x+c)-\tanh (x-c)-1$ and zero initial time derivative, evaluate the scalar-field momentum flux at the midpoint: ${\varphi }_x(0)=0$ and $\varphi (0)=2\tanh c-1$. The force on the left kink is the displayed positive value, attractive toward the right antikink. This exact initial expression has the large-separation asymptotic $32e^{-2d}$ for $d=2c$. If the initial velocity at the cut is $\psi (0)$, subtract $\psi (0{)}^2/2$.

Linearizing about the phi-four kink gives the displayed one-dimensional Schrödinger operator in unit kinetic normalization. Its translational zero mode in field theory is proportional to ${\mathrm{sech}}^2(cx)$. The localized shape eigenfunction $\mathrm{sech}(cx)\tanh (cx)$ has squared frequency $3c^2$, while the continuum starts at $4c^2$. Substitution verifies both bound-state eigenfunctions. The Pöschl-Teller potential permits a short completeness argument for the bound modes. In $y=cx$ let $A_{\ell }={\partial }_y+\ell \tanh y$. Then $H_K/c^2=A_2^{†}{A}_2$, its partner is $A_2{A}_2^{†}=A_1^{†}{A}_1+3$, and $A_1{A}_1^{†}=-{\partial }_y^2+1$. The last operator has no bound states. The kernels of $A_1$ and $A_2$ give the two stated modes, while the partner spectra exclude any further normalizable bound modes.

For ${\varphi }_K(x-X)=c\tanh (c(x-X))$ in the unit kinetic normalization, $\int ({\varphi }_K^{\prime }{)}^{2}dx=M=4c^3/3$. Substitution of $X(t)$ therefore gives the free-particle collective-coordinate effective Lagrangian $-M+M{\dot{X}}^2/2$. Its quantization has momentum $p$ and energy $M+p^2/(2M)$ to this order. The exact uniformly moving classical kink is a Lorentz boost of the static one, with energy $M/\sqrt{1-v^2}$.

Kinks in a phi-six model commonly arise from a nonnegative degree-six potential with three degenerate vacua. For the unit-vacuum normalizationthe kink joining $0$ to $1$ isand has mass $1/4$. A rescaled model with $U(\varphi )={\varphi }^2({\varphi }^2-4{)}^2$ has the elementary kinkand its mass is $4\sqrt{2}$ in the normalization $E=\int [({\varphi }^{\prime }{)}^2/2+U]dx$.

For $U={\varphi }^2({\varphi }^2-b^2{)}^2/2$ and $b>0$, the three scalar-field vacua are $-b,0,b$. The static first integral is ${\varphi }_x^2=2U$. In either adjacent interval, $y={\varphi }^2/b^2$ obeys $y_x=\pm 2b^{2}y(1-y)$, giving the displayed four oriented kink profiles with $\sigma ,\eta \in \{1,-1\}$. Each has rest energy $b^4/4$. The intermediate-vacuum obstruction to a kink excludes a direct static connection from $-b$ to $b$: reaching zero would give zero derivative, and uniqueness of the smooth second-order field equation forces the zero solution. A Lorentz boost replaces $x-x_0$ by $(x-vt-x_0)/\sqrt{1-v^2}$.

// Destination: relativistic-quantum-field.bigb

If a proposed kink must cross a third degenerate vacuum, its first integral makes both the field derivative and potential vanish there. ODE uniqueness prevents it from crossing at finite distance, so the two elementary kink sectors can be joined only at infinite separation.