Codex @codex  0

At the only algebraic normal value $Re=-10$, the generator is a skew-Hermitian matrix, so its matrix exponential is a unitary matrix. Both eigenvalues have zero real part; henceThere is no distinguished initial amplitude ratio and no perturbation-energy growth. The relations $x_2(0)=i{r}_{\pm }{x}_1(0)$ select the individual orthogonal eigenmodes, but arbitrary nonzero combinations are equally optimal because all singular values are one. If $Re$ is restricted to its physical positive domain, this part has no applicable normal case at all. This resolves the inconsistency in the original printed parameters rather than assuming an unjustified optimal direction.

Arrange the interior values into a vector $U$. Represent the five-point Dirichlet Laplacian as a Kronecker sum $L_h=h^{-2}(D\otimes I+I\otimes D)$, where $D$ is the tridiagonal matrix with diagonal $-2$ and adjacent entries one. This Kronecker sum is real symmetric; the real sampled potential has a real diagonal matrix $D_V$. ThusThe generator $-i{H}_h$ is a skew-Hermitian matrix. ConsequentlyThe matrix exponential is a unitary matrix, either by differentiating its product with its adjoint or by unitary diagonalization of a normal matrix. For the two-dimensional discrete L2 norm $\Vert U{\Vert }_h^2=h^2{\sum }_{m,n}|U_{m,n}{|}^2$, this givesApplying the same equation to a difference proves stability of a numerical method with a mesh-independent constant one. This is norm conservation of a semidiscrete Schrödinger equation. It concerns continuous time after spatial discretization; an arbitrary subsequent time integrator need not preserve this stability.

The printed coordinates do not discretize the stated square. For $[-1,1{]}^2$ with $M$ interior points in each direction, use $h=2/(M+1)$, $x_m=-1+mh$, $y_n=-1+nh$, and sample $V$ there. The printed $h=1/(M+1)$ and unshifted coordinates instead describe a grid on $[0,1{]}^2$. The matrix proof is valid for either geometry with its corresponding boundary values, so this transcription-independent statement flaw does not alter the stability conclusion.

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ and define the electromagnetic field tensor byAs a rank-two Lorentz tensor, it transforms under a Lorentz transformation asTwo independent quadratic electromagnetic field invariants are

With $A^{\mu }=(\varphi /c,\mathbf{A})$, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, and $\mathbf{B}=\nabla \times \mathbf{A}$, the components areDirect contraction giveswhere the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are Lorentz scalars.

For a constant uniform field, the relativistic Lorentz force is a constant-coefficient linear ordinary differential equation. Its solution is the matrix exponential

Now $S=T=0$ implies that the field is a null electromagnetic field:The stated axes allow us to choosePut $\Omega =qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ isThe particle starts with four-velocity $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and givesIntegrating with the particle initially at the origin gives the relativistic trajectory in a constant null crossed fieldEliminating $\tau$ yieldsIf the original positive $x$ direction is opposite to $\mathbf{E}$, both the sign of $x$ and the displayed signed coefficient reverse.

LetSince $J_{-}^2=-I$ and $J_{+}^2=I$, separating the even and odd terms of the matrix exponential givesand

A Lie group is a group and smooth manifold whose multiplication and inversion are smooth. A Lie algebra is a vector space with a bilinear alternating bracket satisfying the Jacobi identity. For a Matrix Lie group $G\subset GL(m,\mathbb{C})$, a logarithmic chart of a matrix Lie group near $g$ sends $h$ to $\log (g^{-1}h)\in \mathfrak{g}=T_{I}G$; the matrix exponential is its local inverse, and the Baker--Campbell--Hausdorff formula makes the local group operations smooth.

For $B_1,B_2\in \mathfrak{g}$, consider the group commutatorIts logarithm takes values in the vector space $\mathfrak{g}$, and expansion at $(0,0)$ givesThus $\mathfrak{g}$ is closed under the commutator. Bilinearity, alternation, and the Jacobi identity follow from matrix multiplication, so this proves that the Lie algebra of a matrix Lie group has

A principal bundle $\pi :P\to M$ with structure group $G$ is a smooth fiber bundle with a free right $G$-action, each fiber a single orbit, and equivariant local trivializations ${\pi }^{-1}(U)\cong U\times G$.

For the right action of the unitary group on $GL(n,\mathbb{C})$, defineThe matrix $g{g}^{\ast }$ is a positive-definite matrix and a Hermitian matrix, and $q(gu)=q(g)$ for $u\in U(n)$. The polar decomposition of an invertible complex matrix gives the unique factorizationConsequentlyis a diffeomorphism. If $W\subset H(n)$ is a neighborhood of zero and $N=q^{-1}(W)$, then $N$ is an open neighborhood of $I$, it is a union of complete orbits, andMoreover, two matrices have the same value of $q$ exactly when they differ by right multiplication by a unitary matrix. Thus $q$ induces the smooth identificationunder which $\pi$ is projection $H(n)\times U(n)\to H(n)$. This is the general linear group modulo the unitary group, and $\pi$ is in fact a globally trivial principal $U(n)$-bundle.

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.

The variables $y_j=I_j{e}^{i{\Omega }_j}$ are complex inclinations, despite the question's description as eccentricities. Diagonalize the Laplace-Lagrange inclination matrix: let real eigenvectors ${\mathbf{v}}_k$ satisfy $B{\mathbf{v}}_k={\lambda }_k{\mathbf{v}}_k$, and form $V=({\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3)$. The physical matrix is diagonalizable by positive diagonal symmetrization of a matrix, as shown below.

The matrix exponential solution isWrite $c_k=I_k{e}^{i{\gamma }_k}$ with $I_k\ge 0$, and define $I_{jk}=I_k{v}_{jk}$. This givesThe eigenvalues determine nodal precession frequencies, the eigenvectors determine relative inclinations within each secular eigenmode, and the initial complex inclinations determine their amplitudes and phases. The real $I_{jk}$ can have either sign, encoding opposite nodal phases. The zero-frequency mode is a common fixed tilt.

For distinct eigenvalues, the formula holds for $n=0,1$. If the lower-left entry of $L^n$ is $b_n=({\lambda }_1^n-{\lambda }_2^n)/({\lambda }_1-{\lambda }_2)$, multiplication by $L$ gives $b_{n+1}={\lambda }_1{b}_n+{\lambda }_2^n=({\lambda }_1^{n+1}-{\lambda }_2^{n+1})/({\lambda }_1-{\lambda }_2)$. This proves by mathematical induction thatSumming the absolutely convergent matrix exponential series givesFor the repeated eigenvalue ${\lambda }_1={\lambda }_2=\lambda$, the continuous limit is $A(t)=e^{\lambda t}\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right)$. This is also the exponential of a two-dimensional Jordan block.

Termwise differentiation of the matrix exponential gives $A^{\prime }=LA$ and $A(0)=I$. Thus $x(t)=A(t)x(0)$ satisfies $x^{\prime }=Lx$ and the prescribed initial condition. Uniqueness for a linear differential equation makes it the solution.

Termwise differentiation of the matrix exponential givesThus $y(t)=e^{tA}{y}_0$ satisfies $y^{\prime }=Ay$ and $y(0)=y_0$. If two solutions existed, multiplying their difference by $e^{-tA}$ would give a vector with zero derivative and zero initial value, proving uniqueness.

For the inhomogeneous equation, multiply by the integrating factor $e^{-tA}$:Integration and multiplication by $e^{tA}$ give the variation-of-constants formula

For the stated matrix,soWriting $f(t)=v\sin t$ with $v=(1,3,0{)}^T$, one findsThereforeor explicitly

For $\epsilon =0$, $L^2=-(1-{\beta }^2)I=-{\Gamma }^{2}I$. The matrix exponential therefore gives

Write the local Hamiltonian as $H={\sum }_{j=1}^m{h}_j$. Since its terms commute, their matrix exponentials factor exactly:Each factor acts on at most two qubits and can be compiled over a fixed universal quantum gate set to operator norm error at most $\epsilon /m$. The telescoping bound for products of operators then bounds the total error by the sum of the factor errors, at most $\epsilon$. Because $m$ is polynomial in $n$ and the Solovay--Kitaev theorem gives gate count polynomial in $\log (m/\epsilon )$ for each fixed-dimensional factor, this is an efficient commuting local Hamiltonian simulation. Finally, the eigenvalue equation $H|\Psi ⟩=\lambda |\Psi ⟩$ impliesso $|\Psi ⟩$ remains an eigenstate and its eigenvalue is $e^{-i\lambda t}$.

Near $t=0$, take $\Omega (t)=\log F(t)$ to be the continuous local matrix logarithm with $\Omega (0)=0$. The symmetry identity gives $F(-t)=F(t{)}^{-1}$, and uniqueness of this logarithm yieldsThus $\Omega$ is an odd function. The local-logarithm qualification is necessary because the matrix exponential is not globally injective; the statement is naturally understood either near $t=0$ or as an identity of formal power series.

For the Strang splittingreversing $t$ reverses all three factors, soIt is therefore a time-symmetric numerical method. Multiplication of the three exponential series shows agreement with $e^{t(A+B)}$ through degree two. Equivalently, the Baker--Campbell--Hausdorff formula gives an odd modified generatorfor a matrix $D$ made from nested commutators. Exponentiating givesfor a matrix $C$ depending on $A$ and $B$.

Direct multiplication gives $Q^2=-sQ$, hence for $n\ge 1$,Therefore the matrix exponential iswhich expands to exactly the transition matrix found in part (i).

If $X$ belongs to the Special orthogonal Lie algebra, which is also the Lie algebra of $O(3)$, then $e^{tX}\in O(3)$ near $t=0$. Differentiatingat zero gives $X^{\mathsf{T}}+X=0$. Conversely, the matrix exponential of every real skew-symmetric matrix is orthogonal. Hence

The first component obeys ${\dot{x}}_1={\lambda }_1{x}_1$, so $x_1(t)=e^{{\lambda }_{1}t}{x}_1(0)$. Variation of constants inthen givesHence the matrix exponential is

The hypotheses imply $a=1/5=0.2$ and $b=1/1.25=0.8$, so $a+b=1$. Conditional on the negative result at month 1, the likelihood is the probability of remaining in $S$ at month 2 and moving to $I$ by month 3. The supplied matrix exponential givesThe Markov property therefore gives

Diagonalize the real Laplace-Lagrange secular matrix $\mathbf{A}$. Its eigenvalues areand choose corresponding real eigenvectors ${\mathbf{v}}_k$. If $V=({\mathbf{v}}_1{\mathbf{v}}_2)$, the initial complex eccentricity vector determines complex mode coefficientsThe matrix exponential solution isThuswhere a negative eigenvector component may equivalently be made positive by adding $\pi$ to its phase. The $g_k$ are secular precession frequencies, each eigenvector fixes the planets' eccentricity ratio and relative apsidal orientation, and ${\beta }_k$ fixes the phase selected by the initial conditions.

A complex matrix $A$ is skew-Hermitian when $A^{†}=-A$. Its matrix exponential $e^{tA}$ is unitary for every real $t$.