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Use rationalized electromagnetic units with $c=1$ and electrostatic constant $1/(4\pi )$. The four-potential $A^a=(\phi ,A_x,A_y,A_z)$ transforms by the same matrix as the coordinates:Write $r=\sqrt{y^2+z^2}$ and ${\mathbf{e}}_r=(0,y,z)/r$. Gauss's law for the line at rest gives ${\mathbf{E}}^{\prime }=\sigma {\mathbf{e}}_r/(2\pi r)$, ${\mathbf{B}}^{\prime }=0$. A suitable potential is ${\phi }^{\prime }=-\sigma \log (r/r_{\ast })/(2\pi )$, ${\mathbf{A}}^{\prime }=0$. The inverse Lorentz transformation gives $\phi =\gamma {\phi }^{\prime }$, $A_x=\gamma v{\phi }^{\prime }$. The fields from $\mathbf{E}=-\nabla \phi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$ are thereforeThe sign agrees with a line charge moving in the positive $x$ direction: its density is $\gamma \sigma$ and its current $\gamma v\sigma$. In SI units insert $1/{ϵ}_0$ in the electric expressions and replace $\mathbf{B}=\mathbf{v}\times \mathbf{E}$ by $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$.

The two invariants check explicitly:For $|v|\ll 1$, the electric field differs from the static field only at order $v^2$, while the magnetic field is of order $v$ and is the field of the moving line current. These fields are evaluated off the charged axis.

For unit incident plane-wave amplitude, define the scattering amplitude by $\psi \sim e^{ikz}+f(\theta )e^{ikr}/r$ at large $r$. Comparing outgoing radial flux with incident flux gives $d\sigma /d\Omega =|f(\theta ){|}^2$. Outside the potential, a real regular radial solution is proportional to $j_{\ell }(kr)\cos {\delta }_{\ell }-n_{\ell }(kr)\sin {\delta }_{\ell }$, with asymptotic phase $\sin (kr-\ell \pi /2+{\delta }_{\ell })/(kr)$. This defines the scattering phase shift modulo $\pi$.

The incident plane-wave expansion supplies coefficient $(2\ell +1)i^{\ell }$ for $j_{\ell }{P}_{\ell }$. To preserve its incoming part, replace $j_{\ell }$ by $e^{i{\delta }_{\ell }}(j_{\ell }\cos {\delta }_{\ell }-n_{\ell }\sin {\delta }_{\ell })$. Writing the asymptotic sine in exponentials shows that its incoming coefficient is unchanged and its outgoing coefficient is multiplied by $e^{2i{\delta }_{\ell }}$. The outgoing difference from the plane wave therefore gives

Put $\rho =ka$, $t=\tan {\delta }_{\ell }$, and abbreviate $j=j_{\ell }(\rho )$, $n=n_{\ell }(\rho )$. Logarithmic-derivative matching gives $a{R}_{\ell }^{\prime }(a)/R_{\ell }(a)=\rho (j^{\prime }-t{n}^{\prime })/(j-tn)$. The supplied asymptotics determine the constant Wronskian: ${\rho }^2(j{n}^{\prime }-j^{\prime }n)=1$. ConsequentlySolving for $t$ givesAt zeros of denominators it is interpreted by limits; the logarithmic-derivative definition itself presumes its stated denominators nonzero.

Near a dominant resonant partial wave, ${\sin }^2{\delta }_{\ell }\approx {\gamma }^2/[(k-k_0{)}^2+{\gamma }^2]$. Neglecting the small other partial waves yieldsThis is a resonant peak of width $2|\gamma |$ in $k$, with the partial-wave angular pattern and maximum allowed ${\sin }^2{\delta }_{\ell }=1$ at resonance. Integrating gives $\sigma \approx 4\pi (2\ell +1){\gamma }^2/[k^2((k-k_0{)}^2+{\gamma }^2)]$. A small background may matter at angular zeros of the dominant wave.

Generalized position eigenstates satisfy $⟨x|x^{\prime }⟩=\delta (x-x^{\prime })$ and $\int |x⟩⟨x|dx=I$. Define the position-space wavefunction $\psi (x)=⟨x|\psi ⟩$. Completeness then gives $⟨\psi |\psi ⟩=\int |\psi (x){|}^{2}dx$. Position acts by multiplication, $⟨x|\hat{x}|\psi ⟩=x\psi (x)$. Using $⟨x|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipx/\hslash }$ and momentum completeness, differentiating the Fourier integral gives $⟨x|\hat{p}|\psi ⟩=-i\hslash {\psi }^{\prime }(x)$ on the momentum operator's domain.

For the unit-mass, unit-frequency oscillator defineSince $[H,a]=-\hslash a$, a simultaneous energy eigenstate and $a$-eigenstate would satisfy $[H,a]|{\psi }_{\alpha }⟩=0=-\hslash \alpha |{\psi }_{\alpha }⟩$, forcing $\alpha =0$. For $\alpha =0$, the energy is $\hslash /2$, the smallest possible because $⟨a^{†}a⟩=\Vert a\psi {\Vert }^2\ge 0$.

The position-space equation is $(x+\hslash {\partial }_x){\psi }_{\alpha }=\sqrt{2\hslash }\alpha {\psi }_{\alpha }$. Solving and normalizing, with $\alpha =u+iv$, gives up to a constant phaseIts squared modulus is a normalized Gaussian, so every complex $\alpha$ gives a normalizable coherent state.

For a creation eigenstate, $(x-\hslash {\partial }_x)\psi =\sqrt{2\hslash }\beta \psi$ instead gives $\psi =C\exp [x^2/(2\hslash )-\sqrt{2/\hslash }\beta x]$. Every nonzero such function grows quadratically exponentially at infinity and fails square integrability. Hence there are no nonzero normalizable eigenstates of $a^{†}$. This last symbol is $a^{†}$ in the original PDF, although the converted TeX drops its dagger.

For the Laplace method, under the usual regularity ($p$ continuous at zero and $q$ sufficiently smooth near its isolated minimum), write $q(t)=q(0)+q^{\prime \prime }(0)t^2/2+o(t^2)$. On any interval bounded away from zero, continuity and uniqueness of the minimum give a positive gap in $q$, making that contribution exponentially smaller. Set $t=s/\sqrt{x}$ near zero. The leading integral isA local quadratic lower bound on $q-q(0)$ justifies dominated convergence after this rescaling. Thus, when $p(0)\ne 0$,If $p(0)=0$, the displayed coefficient vanishes and further local terms of $p$ determine the first nonzero asymptotic term; one must not interpret a zero coefficient as an asymptotic equivalence.

For the first Bessel integral, take $q(\theta )=-\cos \theta$, $p(\theta )=\cos (\nu \theta )$. The endpoint minimum is $q(0)=-1$, $q^{\prime \prime }(0)=1$, $p(0)=1$, so after dividing by $\pi$ its leading term is $e^x/\sqrt{2\pi x}$.

For the second integral, $\cosh t\ge 1+t^2/2$ gives, for fixed real $\nu$,Completing the square proves the last bound for large $x$. It is exponentially smaller than the first integral; the fixed prefactor $\sin (\nu \pi )$ does not change this. Therefore

The Brouwer fixed-point theorem says that every continuous map from a closed Euclidean ball to itself has a fixed point. For the closed disc $D\subset {\mathbb{R}}^2$ its equivalent no-retraction theorem says that there is no continuous $r:D\to \partial D$ satisfying $r(z)=z$ on $\partial D$.

Assume the fixed-point theorem and suppose such a retraction exists. Then $F(x)=-r(x)$ maps $D$ continuously into itself. A fixed point must lie on the boundary, where $r(x)=x$, giving $x=-x$, impossible on the unit circle.

Conversely suppose $F:D\to D$ is continuous and has no fixed point. From $F(x)$ draw the ray through $x$ and let $r(x)$ be its exit point on the circle. To check continuity rather than merely relying on the picture, put $v=x-F(x)\ne 0$. The exit point is $F(x)+t(x)v$, whereThis is continuous, lies on the circle and has $t(x)\ge 1$. On the boundary the exit point is $x$, so $r$ is a retraction, contradicting the assumed theorem. Hence the two disc formulations are equivalent; the same argument works for a ball in any dimension.