Qm Math Magnetic And Detection — Examples

[Periodicity] If $q\mathrm{\ne }0$ and $\chi (x)-{\chi }^{\mathrm{\prime }}(x)=\frac{2\pi \mathrm{\hslash }}{q}k(x)$ with $k(x)\in \mathbb{Z}$, then $i\frac{q}{\mathrm{\hslash }}(\chi -{\chi }^{\mathrm{\prime }})(x)=2\pi ik(x)\in 2\pi i\mathbb{Z}$. By [F1], $e^{i\frac{q}{\mathrm{\hslash }}\chi (x)}=e^{i\frac{q}{\mathrm{\hslash }}{\chi }^{\mathrm{\prime }}(x)}$ pointwise, hence the multipliers agree; $\chi +2\pi \mathrm{\hslash }\mathrm{/}q$ is the case ${\chi }^{\mathrm{\prime }}=\chi$. If $q=0$ then $e^{i0\cdot \chi \mathrm{/}\mathrm{\hslash }}=1$. This is (a).

[Linear gauge conjugations] Take $\chi (x)=c\cdot x$, ${\mathrm{\partial }}_j\chi =c_j$ constant and ${\mathrm{\partial }}_t\chi =0$. By [F2] with $A=0$, $(P_j-q{c}_j)G=G{P}_j$ on $W^{1,2}$. For the inverse, compute on $f\in W^{1,2}$: $D_j(G^{-1}f)=G^{-1}(D_{j}f-i\frac{q{c}_j}{\mathrm{\hslash }}f)$, hence $P_j{G}^{-1}f=G^{-1}(P_j-q{c}_j)f$; multiplying by $G$ on the left gives $G{P}_j{G}^{-1}=P_j-q{c}_j$ on $W^{1,2}$.

[Operator identity and propagator] For $u\in W^{2,2}$, $(P_j-q{c}_j{)}^{2}Gu=(P_j-q{c}_j)G(P_{j}u)=G{P}_j(P_{j}u)$ by two uses of step 1.2, so $K_{c,0}G=G{K}_{0,0}$ on $W^{2,2}$; since $G$ is a bijection of $W^{2,2}$ onto itself ([F2]), this gives $K_{c,0}=G{K}_{0,0}{G}^{-1}$ with common domain $W^{2,2}$. Both operators are self-adjoint by [F3]; the conjugate of the spectral PVM of $K_{0,0}$ by $G$ is a regular PVM representing $K_{c,0}$, hence is its spectral PVM by the uniqueness clause of the spectral theorem, so every bounded Borel function satisfies $h(K_{c,0})=Gh(K_{0,0})G^{-1}$ and in particular $e^{-it{K}_{c,0}\mathrm{/}\mathrm{\hslash }}=G{e}^{-it{K}_{0,0}\mathrm{/}\mathrm{\hslash }}{G}^{-1}$ (Spectral theorem for unbounded self-adjoint operators (PVM form), A self-adjoint operator generates a strongly continuous unitary group, Integral of a measurable function against a projection-valued measure); the spectra coincide because $G$ carries the resolvent of $K_{0,0}$ to that of $K_{c,0}$, $(K_{c,0}-z{)}^{-1}=G(K_{0,0}-z{)}^{-1}{G}^{-1}$. This is (b).

[Dimension vector] With the stated vectors, $[q]+[\chi ]-[\mathrm{\hslash }]=(0+2-2,0+1-1,1-2+1,1-1-0)=(0,0,0,0)$, so the combination $q\chi \mathrm{/}\mathrm{\hslash }$ has zero dimension vector; this is a four-component integer computation. This is (c).

[The two laws] By [F2], $\mathrm{\parallel }a+b{\mathrm{\parallel }}^2={\int }_0^1(2+2\cos (2\pi x))dx=2$, so $p_{\mathrm{c}\mathrm{o}\mathrm{h}}(B)=\frac{1}{2}{\int }_B(2+2\cos (2\pi x))dx={\int }_B(1+\cos (2\pi x))dx$ by [F1]. Since $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }b\mathrm{\mid }=1$, $p_{\mathrm{m}\mathrm{i}\mathrm{x}}(B)=\frac{1}{2}\mathrm{\mid }B\mathrm{\mid }+\frac{1}{2}\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }B\mathrm{\mid }$. This is (a).

[The dark point and the dark bin] The density $1+\cos (2\pi x)$ equals $1+\cos \pi =0$ at $x_0=\frac{1}{2}$, a single point of Lebesgue measure zero. For $0<\epsilon <\frac{1}{2}$, [F2] gives ${\int }_{1\mathrm{/}2-\epsilon }^{1\mathrm{/}2+\epsilon }\cos (2\pi x)dx=\frac{\sin (2\pi x)}{2\pi }{\mid }_{1\mathrm{/}2-\epsilon }^{1\mathrm{/}2+\epsilon }=\frac{\sin (\pi +2\pi \epsilon )-\sin (\pi -2\pi \epsilon )}{2\pi }=\frac{-\sin (2\pi \epsilon )-\sin (2\pi \epsilon )}{2\pi }=-\frac{\sin (2\pi \epsilon )}{\pi }$, so $p_{\mathrm{c}\mathrm{o}\mathrm{h}}(B_{\epsilon })=2\epsilon -\frac{\sin (2\pi \epsilon )}{\pi }$. By [F2], $1+\cos (2\pi x)$ is nonnegative and vanishes on $[0,1)$ only at $x=\frac{1}{2}$. The point $x_{\ast }:=\frac{1}{2}+\frac{\epsilon }{2}$ lies inside $B_{\epsilon }$ and the density is positive there; continuity (A function differentiable at $c$ is continuous at $c$) gives a subinterval of positive length on which it is bounded below by a positive constant, so the bin probability is strictly positive; the Taylor expansion of $\sin$ gives $2\epsilon -\frac{\sin (2\pi \epsilon )}{\pi }=\frac{(2\pi \epsilon {)}^3}{6\pi }+O({\epsilon }^5)=\frac{4{\pi }^2}{3}{\epsilon }^3+O({\epsilon }^5)$. This is (b).

[Comparison and scope] By step 1.1, $p_{\mathrm{m}\mathrm{i}\mathrm{x}}(B_{\epsilon })=2\epsilon$, and $0<2\pi \epsilon <\pi$ implies $\sin (2\pi \epsilon )>0$ by Pi is the first positive zero of sine, so the exact formula of step 2.1 gives $p_{\mathrm{c}\mathrm{o}\mathrm{h}}(B_{\epsilon })<p_{\mathrm{m}\mathrm{i}\mathrm{x}}(B_{\epsilon })$; this is (c). The statements are properties of the two computed bin laws on the specified amplitudes, and no observed frequency is claimed.

[Normalisation and moments of the Gaussian] Substituting $x=\sigma t$ in ${\rho }_{\sigma }$ gives $\int {\rho }_{\sigma }=(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}\sigma \int e^{-t^2\mathrm{/}2}dt=1$ by [F1] with $a=1$. Hence $\mathrm{\parallel }{\psi }_{\sigma }\mathrm{\parallel }=1$. The mean $\int x{\rho }_{\sigma }(x)dx$ vanishes because the integrand is odd; the second moment is $\int x^2{\rho }_{\sigma }(x)dx=(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}{\sigma }^3\int t^2{e}^{-t^2\mathrm{/}2}dt=(2\pi {\sigma }^2{)}^{-1\mathrm{/}2}{\sigma }^3\sqrt{2\pi }={\sigma }^2$ by [F1].

[The additive-error law] Let $\rho$ be a density as in (b). Since $\mathrm{\mid }y\mathrm{\mid }\le 1+y^2$, its finite second moment implies $\int \mathrm{\mid }y\mathrm{\mid }\rho (y)dy<\mathrm{\infty }$, as required for the first-moment identities in [F3]. The same nonnegative change of variables gives $\int x^2\rho (x\pm e)dx=\int (y\mp e{)}^2\rho (y)dy\le 2\int y^2\rho (y)dy+2e^2<\mathrm{\infty }$; thus all means and variances below are finite. By [F3], $\int {\rho }_e=\frac{1}{2}\int \rho +\frac{1}{2}\int \rho =1$, so ${\rho }_e$ is a probability density; likewise $m({\rho }_e)=\frac{1}{2}\int x\rho (x+e)dx+\frac{1}{2}\int x\rho (x-e)dx=\frac{1}{2}\int (y-e)\rho (y)dy+\frac{1}{2}\int (y+e)\rho (y)dy=m(\rho )$, and the cross terms cancel: $V({\rho }_e)=\frac{1}{2}\int (y-e-m{)}^2\rho (y)dy+\frac{1}{2}\int (y+e-m{)}^2\rho (y)dy=\int (y-m{)}^2\rho (y)dy+e^2=V(\rho )+e^2$. This is (b).

[Momentum moments] By [F2], $\int {\psi }_{\sigma }^{\mathrm{\prime }}\overline{{\psi }_{\sigma }}dx=-\frac{1}{2{\sigma }^2}\int x{\rho }_{\sigma }(x)dx=0$, so $b=⟨P{\psi }_{\sigma },{\psi }_{\sigma }⟩=0$; and $\mathrm{\parallel }D{\psi }_{\sigma }{\mathrm{\parallel }}^2=\frac{1}{4{\sigma }^4}\int x^2{\rho }_{\sigma }(x)dx=\frac{1}{4{\sigma }^2}$, so $\mathrm{\parallel }P{\psi }_{\sigma }{\mathrm{\parallel }}^2={\mathrm{\hslash }}^2\mathrm{/}(4{\sigma }^2)$, i.e. $\mathrm{\Delta }P=\mathrm{\hslash }\mathrm{/}(2\sigma )$ and $\mathrm{\Delta }Q=\sigma$ by step 1.1. These are the maximal-domain quantities of Position-momentum uncertainty on the maximal domains of the line and give $\mathrm{\Delta }Q\mathrm{\Delta }P=\mathrm{\hslash }\mathrm{/}2$: the equality case of Covariance-refined preparation uncertainty without operator products clause (c), since $x_Q$ and $y_P$ are then linearly dependent (both proportional to $x{\psi }_{\sigma }(x)$). This is (a).

[Gaussian case] By step 1.1, $V({\rho }_{\sigma })={\sigma }^2$, and step 1.2 gives $V(({\rho }_{\sigma }{)}_e)={\sigma }^2+e^2$. The mixture is centered, and translation in each component gives its fourth central moment ${\int }_{\mathbb{R}}{x}^4({\rho }_{\sigma }{)}_e(x)dx=\frac{1}{2}{\int }_{\mathbb{R}}(y-e{)}^4{\rho }_{\sigma }(y)dy+\frac{1}{2}{\int }_{\mathbb{R}}(y+e{)}^4{\rho }_{\sigma }(y)dy=3{\sigma }^4+6{\sigma }^2{e}^2+e^4,$ using evenness, $\int y^2{\rho }_{\sigma }={\sigma }^2$, and $\int y^4{\rho }_{\sigma }=3{\sigma }^4$ from [F1]. Any $g_{\mu ,\tau }$ has mean $\mu$, variance ${\tau }^2$, and fourth central moment $3{\tau }^4$ by the same Gaussian moments and translation. Thus if $e>0$ and $({\rho }_{\sigma }{)}_e$ were a Gaussian density, its mean and variance would force $\mu =0$ and ${\tau }^2={\sigma }^2+e^2$, whose Gaussian fourth moment $3({\sigma }^2+e^2{)}^2$ differs from the displayed mixture fourth moment by $2e^4>0$, a contradiction. At $e=0$ the mixture is ${\rho }_{\sigma }$. Hence the model variance does not determine $\sigma$ when $e$ is unknown, although it implies the stated upper bound; this proves (c).