Qm Math Controlled Dynamics — Examples

The two scalar eigenvalues in [F1] have distance $2\mathrm{\mid }s\mathrm{\mid }$, which tends to zero at the crossing. For $s<0$ the larger one is $-s$ with eigenspace $\mathbb{C}{e}_{\downarrow }$; for $s>0$ it is $s$ with eigenspace $\mathbb{C}{e}_{\uparrow }$. Their orthogonal projections differ by $\mathrm{diag}(1,-1)$ in norm $1$, so no choice at $s=0$ gives a continuous ordered-upper projection. The zero matrix at $0$ has only the full eigenspace ${\mathbb{C}}^2$ as its nonzero spectral projection.

Set $V_T(s)=\mathrm{diag}(e^{-iT(s^2-1)\mathrm{/}2},e^{iT(s^2-1)\mathrm{/}2})$. Entrywise differentiation gives $i{V}_T^{\mathrm{\prime }}=Ts{\sigma }_z{V}_T$ and $V_T(-1)=I$. Hence [F2] identifies $U_T=V_T$. At $s=1$ both phases are $1$, and $(I-P_{+}(1))U_T(1)P_{+}(-1)=\mathrm{diag}(0,1)$, an orthogonal projection of norm $1$. This proves the full leakage assertion for every $T>0$.

The constant projection $P_{\downarrow }$ has $H(s)P_{\downarrow }=-s{P}_{\downarrow }$ by [F1], commutes with $U_T(s)$ by step 2.1, and gives zero leakage out of its fixed range. It therefore exhibits the continuous eigenvector counterbranch through the crossing and disproves the old unqualified no-continuous-rank-one-branch assertion, while preserving the ordered-upper failure just proved.

[Explicit operator and spectral measure] The $L^2((1,2))$ completeness and pairing in [F1] make it a complex Hilbert space. If $c_n\oplus f_n$ is Cauchy in the stated direct-sum norm, each component is Cauchy; completeness of $\mathbb{C}$ and this $L^2$ space gives limits $c,f$, and the norm identity gives convergence to $c\oplus f$. Thus $H$ is complete before any Hilbert-space theorem or PVM definition is used. Since $1<x<2$, $M_x$ is bounded on all $L^2(1,2)$ with norm $2$, and real multiplication gives $⟨M_{x}f,h⟩=⟨f,M_{x}h⟩$. A bounded symmetric operator on the whole Hilbert space equals its adjoint directly by the defining pairing, so $H_0$ is self-adjoint with domain $H$. For every real Borel $B$ define $\mathcal{E}(B)(c\oplus f)={\mathbf{1}}_B(0)c\oplus {\mathbf{1}}_B(x)f(x)$. These are orthogonal projections, $\mathcal{E}(\mathbb{R})=I$, and their products are $\mathcal{E}(B\cap C)$. On a fixed vector, disjoint countable additivity follows from pointwise indicator sums and the integrable squared-norm majorant $\mathrm{\mid }f{\mathrm{\mid }}^2$ (with the scalar component evaluated at $0$), so this is a PVM. Its scalar measures are $\mathrm{\mid }c{\mathrm{\mid }}^2{\delta }_0(B)+{\int }_{B\cap (1,2)}\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx$. Integrating bounded simple approximations to the coordinate function on $\{0\}\cup [1,2]$ yields $H_0=\int td\mathcal{E}(t)$ on all $H$. Thus the representing PVM and its bounded multiplier realization are explicit.

[Spectrum, gap and absence of complementary eigenvectors] For $z\mathrm{\notin }\{0\}\cup [1,2]$, $(z-H_0{)}^{-1}(c\oplus f)=c\mathrm{/}z\oplus f\mathrm{/}(z-x)$ is a bounded two-sided inverse on $H$. For any $t\in [1,2]$ and $\epsilon >0$, the interval $J=(1,2)\cap (t-\epsilon ,t+\epsilon )$ has positive length. The vector $0\oplus {\mathbf{1}}_J\mathrm{/}\mathrm{\mid }J{\mathrm{\mid }}^{1\mathrm{/}2}$ is unit and has $\mathrm{\parallel }(H_0-t)u\mathrm{\parallel }\le \epsilon$, excluding a bounded inverse at $t$. Finally $\ker H_0=\mathbb{C}\oplus \{0\}$, so $0$ is an isolated simple eigenvalue and its distance to the remaining spectrum is $g=1$. The equation $(x-t)f=0$ forces $f=0$ in $L^2(1,2)$ because a singleton has Lebesgue measure zero. Consequently the complementary block has no eigenvectors and all its spectral measures are absolutely continuous with densities $\mathrm{\mid }f{\mathrm{\mid }}^2$ on $(1,2)$, including zero atom at the endpoints. In particular $P_0=\mathcal{E}(\{0\})$ and ${\psi }_0=1\oplus 0$ has norm one.

[Perturbation hypotheses and the restricted inverse] Let $I(f)={\int }_1^{2}f(x)dx=⟨f,\mathbf{1}⟩$. Cauchy-Schwarz and the interval's length one give $\mathrm{\mid }I(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$. Therefore $\mathrm{\parallel }V(c\oplus f){\mathrm{\parallel }}^2=\mathrm{\mid }I(f){\mathrm{\mid }}^2+\mathrm{\mid }c{\mathrm{\mid }}^2\le \mathrm{\parallel }c\oplus f{\mathrm{\parallel }}^2$; equality at ${\psi }_0$ gives $v=\mathrm{\parallel }V\mathrm{\parallel }=1>0$. For $c\oplus f,d\oplus h$, the first-variable-linear pairing gives $⟨V(c\oplus f),d\oplus h⟩=I(f)\bar{d}+c\overline{I(h)}=⟨c\oplus f,V(d\oplus h)⟩$, so $V=V^{\ast }$. Moreover $V{\psi }_0=0\oplus \mathbf{1}$ and the first-order pairing is zero. On $QH=0\oplus L^2(1,2)$, $H_Q-E=M_x$ has inverse $M_{1\mathrm{/}x}$, bounded on that exact whole complementary domain with norm one. Hence $S(c\oplus f)=0\oplus (f\mathrm{/}x)$, $S{P}_0=P_{0}S=0$ and $(H_0-E)S=S(H_0-E)=Q$ on $H$. No global inverse of $H_0-E$ is used. The preceding operator, spectral and perturbation computations verify every concrete hypothesis of [F2], with inherited CC and $q=2\mathrm{\mid }\lambda \mathrm{\mid }$.

[Exact coefficient and its continuous origin] The explicit reduced inverse gives $SV{\psi }_0=0\oplus (1\mathrm{/}x)$ and $VSV{\psi }_0=({\int }_1^{2}dx\mathrm{/}x)\oplus 0$. Pairing with the unit ${\psi }_0$ and applying [F2], [F3] yields $E^{\mathrm{\prime }\mathrm{\prime }}(0)\mathrm{/}2=-{\int }_1^{2}dx\mathrm{/}x=-\log 2<0$. The scalar spectral measure of $V{\psi }_0=0\oplus \mathbf{1}$ is exactly ${\mathbf{1}}_{(1,2)}(x)dx$ by step 1.1, so the entire coefficient integrates $1\mathrm{/}x$ over the continuous complementary spectrum. Step 2.1 shows that the only bound eigenvector belongs to $P_{0}H$, with zero coupling matrix element; there are no complementary bound eigenvectors. A sum restricted to bound states therefore gives zero and omits the exact negative coefficient. This proves all original example claims, its normalization and class, with the correct restricted inverse and explicit inherited CC.

The original displayed diagonal exponentials differentiate to $i{U}^{\mathrm{\prime }}=(1+2\cos t){\sigma }_{z}U$, $i{U}_{\mathrm{a}\mathrm{v}\mathrm{g}}^{\mathrm{\prime }}={\sigma }_z{U}_{\mathrm{a}\mathrm{v}\mathrm{g}}$, with initial value $I$. They are therefore the required propagators by [F2]. Left multiplication of their difference by $e^{it{\sigma }_z}$ leaves its norm unchanged and gives diagonal entries $e^{\mp 2i\sin t}-1$, whose common modulus is $2\mathrm{\mid }\sin (\sin t)\mathrm{\mid }$. Substitution at $\pi \mathrm{/}2+2\pi n$ proves the original positive nondecaying subsequence.

Define $W(t)=\exp (i\omega t{\sigma }_z\mathrm{/}2)$ and $G_n={\epsilon }_n{\sigma }_x+(\omega \mathrm{/}2){\sigma }_z$. The diagonal formula for $W$ gives $W{\sigma }_{+}{W}^{\ast }=e^{i\omega t}{\sigma }_{+}$ and $W{\sigma }_{-}{W}^{\ast }=e^{-i\omega t}{\sigma }_{-}$. Hence $H_n(t)={\epsilon }_{n}W(t){\sigma }_{x}W(t{)}^{\ast }$. Put $V_n(t)=W(t)e^{-it{G}_n}$. Both factors are unitary, $V_n(0)=I$, and differentiating gives $i{V}_n^{\mathrm{\prime }}=-(\omega \mathrm{/}2){\sigma }_z{V}_n+W{G}_n{e}^{-it{G}_n}={\epsilon }_{n}W{\sigma }_x{W}^{\ast }{V}_n=H_n{V}_n$, because $W$ commutes with ${\sigma }_z$. By [F2], $U_n=V_n$. The sign of $W$ and the positive $\omega {\sigma }_z\mathrm{/}2$ term in $G_n$ are essential in this cancellation.

For the commuting extension with real $\beta$, integrate its diagonal scalar entries to obtain $U(t)=\exp [-i(t+2\beta \sin (\omega t)\mathrm{/}\omega ){\sigma }_z]$. The same diagonal norm calculation gives $2\mathrm{\mid }\sin ((\beta \mathrm{/}\omega )\sin \omega t)\mathrm{\mid }\le 2\mathrm{\mid }\beta \mathrm{\mid }\mathrm{/}\omega$. Thus the original commuting nondecay example admits a time-uniform small-ratio estimate; the stronger failure requires the separate noncommuting construction.

By [F1], $G_n^2=({\epsilon }_n^2+{\omega }^2\mathrm{/}4)I={\nu }_n^{2}I$, where ${\nu }_n=(\omega \mathrm{/}2)(1+1\mathrm{/}n)>0$ from the exact definition of ${\epsilon }_n$. Splitting the absolutely convergent matrix exponential into even and odd powers gives $e^{-it{G}_n}=\cos ({\nu }_{n}t)I-i\sin ({\nu }_{n}t)G_n\mathrm{/}{\nu }_n$. At ${\tau }_n=2\pi n\mathrm{/}\omega$, the angles are $\omega {\tau }_n\mathrm{/}2=\pi n$ and ${\nu }_n{\tau }_n=\pi (n+1)$. Therefore $W({\tau }_n)=(-1{)}^{n}I$ and $e^{-i{\tau }_n{G}_n}=(-1{)}^{n+1}I$, so $U_n({\tau }_n)=-I$. Since $A_n=0$, $U_{\mathrm{a}\mathrm{v}\mathrm{g},n}(t)=I$ for every $t$, and the norm difference at ${\tau }_n$ is $\mathrm{\parallel }-2I\mathrm{\parallel }=2$.

The matrix ${\sigma }_{+}$ sends $(x,y)$ to $(y,0)$, so its operator norm is exactly $1$; hence $\mathrm{\parallel }{B}_n\mathrm{\parallel }\mathrm{/}\omega =\sqrt{1\mathrm{/}(2n)+1\mathrm{/}(4n^2)}\to 0$. Also ${\tau }_n\to \mathrm{\infty }$ for fixed $\omega$. For noncommutativity, the products ${\sigma }_{+}{\sigma }_{-}=\mathrm{diag}(1,0)$ and ${\sigma }_{-}{\sigma }_{+}=\mathrm{diag}(0,1)$ give $[H_n(t),H_n(u)]={\epsilon }_n^2(e^{i\omega (t-u)}-e^{-i\omega (t-u)}){\sigma }_z=2i{\epsilon }_n^2\sin (\omega (t-u)){\sigma }_z$. The proposed general function $f(r)\to 0$ would give $2\le f(\mathrm{\parallel }{B}_n\mathrm{\parallel }\mathrm{/}\omega )\to 0$, a contradiction. This proves the full intended no-general-uniform-smallness conclusion while retaining the distinct commuting nondecay example.