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Before its second decay A moves uniformly with $x=-Vt$. Its Lorentz factor isThe proper time along its world line satisfies $d\tau =dt/{\gamma }_A$. Since the first decay occurs at the spacetime origin, integration givesEquivalently these are the components of A's constant four-velocity times $\tau$. The assumption $m>0$ is needed for a daughter rest frame and its proper-time clock; a massless daughter has neither.

Conservation of momentum in the parent's rest frame gives $p_A+p_B=0$. Equal daughter masses and the relativistic energy-momentum relation then give equal daughter energies. Energy conservation fixesUsing $E^2=p^2{c}^2+m^2{c}^4$ and the specified direction of A,The velocities follow from $v=p{c}^2/E$:These are spatial momenta in the original inertial frame. For positive daughter rest mass, $m<M/2$ gives $0<V<c$.

Regularity at $r=0$ excludes every negative radial power and the logarithmic zero mode. Thus the general regular single-valued harmonic function on the disk has the formThe coefficients must give a series converging locally in the disk; if finiteness and continuity on its boundary are required, the boundary Fourier series must represent the prescribed finite continuous boundary values. Conversely every such locally convergent expansion is harmonic, and the Fourier coefficients on circles show that it includes every regular solution. There is no angular term linear in $\theta$, because it would be multivalued.

In the two-dimensional span of the first two energy eigenstates, the measured observable has matrix $S=\left(\begin{array}{cc}7& 24\\ 24& -7\end{array}\right)$. Its square is $625I$, and its trace is zero, giving eigenvalues $25,-25$. The orthogonal complement has eigenvalue zero. Thus the full spectrum is $\boxed{\{-25,0,25\}}$.

The normalized eigenstates for $-25$ and $25$ are respectivelyThe lowest measurement outcome prepares ${\chi }_{-}$ up to an irrelevant overall phase. The subsequent unitary time evolution givesBy the Born rule, the probability of measuring the lowest eigenvalue again is the squared projection onto its one-dimensional eigenspace:This is the two-level quantum return probability with $q=9/25$. At $t=0$ the probability is one; if the two energies happen to be equal it remains one for every time.

Expand the wavefunction in the complete orthonormal energy eigenstates. The Time-dependent Schrödinger equation makes each coefficient satisfy $i\hslash {\dot{a}}_n(t)=E_n{a}_n(t)$, soThe series is interpreted in the Hilbert space norm. Its norm is conserved because the phase factors have modulus one and the energy eigenstates are orthonormal. For initial data in the domain of the Hamiltonian operator, this is a strong solution of the equation; for any square-summable coefficients it gives the corresponding unitary time evolution.

Expand using the bilinearity and symmetry of the real inner product:Addition proves the parallelogram law,For the maximum norm on ${\mathbb{R}}^2$, take $x=(1,0)$ and $y=(0,1)$. All four vectors $x,y,x+y,x-y$ have maximum norm one. The left side of the parallelogram law is then $2$, whereas the right side is $4$. The maximum norm cannot come from an inner product.

Here equivalent metrics means that the two metrics induce the same topology, and Lipschitz equivalent norms means that there are constants $c,C>0$ such that $cp(x)\le q(x)\le Cp(x)$ for every $x$.

Suppose first that the inequalities hold. Since they also hold for $x-y$, the identity maps between the two metric spaces are Lipschitz continuous. In particular, each is continuous, so the topologies agree.

Conversely, equality of the topologies makes the identity from the $p$-norm to the $q$-norm continuous at zero. There is therefore $\delta >0$ such that $p(x)<\delta$ implies $q(x)<1$. For nonzero $x$, apply this to $\delta x/(2p(x))$ and use the absolute homogeneity of both norms:The reverse identity is also continuous, so there is $\epsilon >0$ with $q(x)<\epsilon \Rightarrow p(x)<1$. Rescaling again gives $p(x)<2q(x)/\epsilon$. ThusThe inequalities also hold at zero. This is topology determines norm equivalence; it works in arbitrary dimension because the rescaling argument needs neither a basis nor compactness of a unit sphere.

Let $d_n(x)=\det (xI-A_n)$, and set $d_0=1$, $d_1=x-a_1$. Expansion of the tridiagonal determinant along its last row givesFor the second term, the last-row entry is $-b_n$, its cofactor has negative sign, and the remaining minor has last-column entry $-b_n$ multiplying $d_{n-2}$; their product is $-b_n^2{d}_{n-2}$. Thus $d_n$ and $p_n$ have identical initial values and recurrence, and induction provesThe eigenvalues of the Jacobi matrix $A_n$ are precisely the roots of this characteristic polynomial. Part(a) therefore gives all $n$ eigenvalues simple and lying strictly in $(a,b)$. This conclusion follows from orthogonality as well as symmetry; symmetry alone would establish real eigenvalues but not their interval location.

The monic orthogonal polynomial $p_n$ is orthogonal to every polynomial of degree less than $n$, since $p_0,\dots ,p_{n-1}$ form a basis of that polynomial space. Let $r_1,\dots ,r_k$ be its distinct interior roots at which its sign changes, and put $q(x)={\prod }_{j=1}^k(x-r_j)$, with $q=1$ if $k=0$. The real polynomial $p_{n}q$ has constant sign throughout $(a,b)$: multiplication by each linear factor removes its corresponding sign change, while even-multiplicity roots do not change sign. It is not identically zero. Since $w>0$, its weighted integral is nonzero.

If $k<n$, orthogonality instead gives ${\int }_a^b{p}_{n}qwdx=0$, a contradiction. Thus $k\ge n$. Degree $n$ also gives $k\le n$, so $k=n$ and all roots have multiplicity one, with no roots left outside the interval. Therefore there are exactly $n$ distinct simple zeros in $(a,b)$. The weighted inner products are understood to exist, as required by the stipulated orthogonality.

If two hyperbolic lines intersect, move their intersection to the centre of the Poincaré disk model. Lines through the centre are diameters. Reflection in a diameter making Euclidean angle $\alpha$ with the real axis is $z\mapsto e^{2i\alpha }\overline{z}$. The composition of two such reflections is the rotation $z\mapsto e^{2i(\alpha -\beta )}z$. The Poincaré disk model is conformal, so the Euclidean angle difference is the hyperbolic intersection angle $\theta$, up to sign and replacing it by $\pi -\theta$, neither of which changes rationality of $\theta /\pi$.

The rotation has finite order exactly when $m\theta \in \pi \mathbb{Z}$ for some positive integer $m$, equivalently $\theta /\pi \in \mathbb{Q}$. If $\theta /\pi =r/s$ in lowest terms, its order is $s$. Conversely, the nonintersecting cases in part(a) have infinite order. Therefore

Parallel hyperbolic lines have one common ideal endpoint and no interior intersection. Move the common endpoint to infinity by a Möbius transformation. The two lines are then distinct vertical lines $x=a,x=b$. Their hyperbolic reflections have productwhose $m$th power is $z+2m(a-b)$. Since $a\ne b$, no positive power is the identity.

Ultraparallel hyperbolic lines have no common endpoint or intersection. To see their common perpendicular explicitly, first map one line to the imaginary axis. The other has two boundary endpoints on the same side, say $0<r<s$ after possibly reflecting. It is a semicircle of centre $(r+s)/2$ and radius $(s-r)/2$. The circle centred at zero with radius $\sqrt{rs}$ meets it orthogonally, since $((r+s)/2{)}^2=rs+((s-r)/2{)}^2$, and also meets the imaginary axis orthogonally. It supplies the common perpendicular. Map that perpendicular to the imaginary axis. Orthogonality then makes both lines semicircles centred at zero, of distinct radii $r,s$. This normal form can also be obtained by a real Möbius transformation carrying the endpoints of the common perpendicular to $0,\infty$. Their reflections are $z\mapsto r^2/\overline{z}$ and $z\mapsto s^2/\overline{z}$, so their product isNo positive power of this dilation is the identity because $r^2/s^2>0$ and $r\ne s$. Thus the product has infinite order in both nonintersecting cases. Geometrically the ultraparallel product is a translation along the common perpendicular through twice the separation of the lines.

Every orbital angular momentum operator annihilates a radial function: substituting ${\partial }_{i}f(r)=f^{\prime }(r)x_i/r$ makes the two terms cancel. It also annihilates $r^2$. Thus the radial factor can be carried through the angular operators. Since $x_3^2$ is independent of $x_1,x_2$, $L_3\psi =0$ for every value of $a$.

Direct differentiation givesConsequently $L^2{x}_3^2={\hslash }^2(6x_3^2-2r^2)$ and $L^2{r}^2=0$. For a nonzero admissible radial factor, comparison with $\lambda (x_3^2+a{r}^2)f(r)$ forces $\lambda =6{\hslash }^2$ and $6a=-2$. Therefore the simultaneous eigenfunction occurs atIt is the axially symmetric quadratic orbital eigenfunction, with quantum numbers $l=2$, $m=0$ and angular dependence proportional to the spherical harmonic $Y_{20}$. The subtraction of $r^2/3$ removes its rotationally invariant scalar component.

The orbital angular momentum operators areOn smooth wavefunctions, $[x_i,p_j]=i\hslash {\delta }_{ij}$, while position operators commute with each other and momentum operators commute with each other. Applying the product rule for commutators givesUse it on the four terms in $[L_1,L_2]$: only $[x_2{p}_3,x_3{p}_1]=-i\hslash x_2{p}_1$ and $[x_3{p}_2,x_1{p}_3]=i\hslash x_1{p}_2$ survive, with the latter entering positively from the two minus signs. ThereforeThe cyclic versions similarly give $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. Thus, for $L_{\pm }=L_1\pm i{L}_2$,Now define $L^2=L_1^2+L_2^2+L_3^2$. For each $i$,because the Levi-Civita symbol is antisymmetric in $j,k$ and the operator pair is symmetric. Linearity then gives $\boxed{[L^2,L_{\pm }]=0}$. The PDF superscript here is the squared total angular momentum, not the component $L_2$.

Full column rank makes $B=A^{T}A$ positive definite: for nonzero $v$, $v^{T}Bv=\Vert Av{\Vert }^2>0$. Let $B=R^{T}R$ be its unique upper-triangular Cholesky decomposition with positive diagonal. Defining $Q=A{R}^{-1}$ givesThus the skinny QR decomposition exists. If $A=\tilde{Q}\tilde{R}$ is another such decomposition, orthonormality gives $A^{T}A={\tilde{R}}^T\tilde{R}$. Cholesky uniqueness forces $\tilde{R}=R$, and then $\tilde{Q}=A{R}^{-1}=Q$. Therefore the skinny factorization with positive diagonal is unique. This is the thin QR factorization from Cholesky decomposition; no inverse of the rectangular $Q$ is assumed.

For a real symmetric positive-definite matrix, the Cholesky decomposition is $B=R^{T}R$, where $R$ is upper triangular with positive diagonal; equivalently $B=L{L}^T$ with $L=R^T$ lower triangular.

Existence can be proved by induction. Write $B=\left(\begin{array}{cc}b& v^T\\ v& C\end{array}\right)$, where $b>0$. For a nonzero vector $z$, evaluate the quadratic form at $(-v^{T}z/b,z{)}^T$ to obtain $z^T(C-v{v}^T/b)z>0$. The Schur complement is therefore positive definite. By induction it has factor $R_0^T{R}_0$, andhas $R^{T}R=B$. The one-dimensional starting case is the positive square root.

For uniqueness, if $R_1^T{R}_1=R_2^T{R}_2=B$, then $U=R_2{R}_1^{-1}$ is upper triangular and satisfies $U^{T}U=I$. Since $U^{-1}$ is upper triangular and equals $U^T$, which is lower triangular, $U$ is diagonal. Orthogonality makes its diagonal entries $\pm 1$, and positivity of the diagonals of $R_1,R_2$ forces $+1$. Hence $U=I$ and $\boxed{R_1=R_2}$.

Set $F(z)=\exp [-(1-i)g(z)]$. It is entire, and $\mathrm{Re}[(1-i)g]=u+v$ makes $|F|=e^{-(u+v)}\le 1$. By the just-proved Liouville theorem, $F$ is constant. Differentiation gives $0=F^{\prime }=-(1-i)g^{\prime }F$. The exponential never vanishes, so $g^{\prime }=0$ everywhere and $g$ is constant. This is the entire function confined to a half-plane is constant principle.

Uniformly continuous. For $x\ge 1$, the derivative satisfiesThe mean value theorem therefore gives a Lipschitz condition with constant two, hence uniform continuity.

Not uniformly continuous. ChooseTheir separation is $\pi /(x_n+y_n)\to 0$, while $\sin (x_n^2)=1$ and $\sin (y_n^2)=-1$. Thus the function values stay distance two apart, contradicting uniform continuity. Boundedness of the function does not prevent increasingly rapid oscillations.

Not uniformly continuous. Take $x_n=n$, $y_n=n+1/n$. Then $|x_n-y_n|\to 0$, butThis contradicts the defining implication for uniform continuity, for example with $\epsilon =1$. The sequential criterion for uniform continuity makes explicit why arbitrarily close arguments at large distances from the origin matter.

The initial wavefunction has unit norm, since ${\int }_0^a(2/a){\sin }^2(\pi x/a)dx=1$. On its nonzero half, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E_{\ast }\psi$ with $E_{\ast }={\hslash }^2{\pi }^2/(2m{a}^2)$. A small domain qualification is important: the derivative jumps at zero, so the state is not in the operator domain of the Dirichlet Hamiltonian operator. Its finite energy expectation value is nevertheless defined by the quadratic form of a positive quantum Hamiltonian. The energy form of a half-well sine state givesThis is also the rigorous weak interpretation of the printed $\int \psi H\psi$. Indeed the distributional second derivative is ${\psi }^{\prime \prime }=-(\pi /a{)}^2\psi +\sqrt{2/a}(\pi /a){\delta }_0$ inside the well. The Dirac delta term pairs to zero because $\psi (0)=0$. One should not instead conclude that this state is an energy eigenstate from its equation on just the positive half.

Its coefficients in the full-well energy eigenstates areFor even $n$, sine orthogonality gives $a_2=-1/\sqrt{2}$ and all other even coefficients zero. For $n=2p+1$, set $u=x/a$ and use $\sin [n\pi (u+1)/2]=(-1{)}^p\cos (n\pi u/2)$. Product-to-sum integration then givesSince these coefficients are $O(n^{-2})$, ${\sum }_n{E}_n|a_n{|}^2$ converges, confirming that the state has finite energy. With $E_n=n^2{E}_1$, the spectral expression isThe kinetic-energy quadratic form equals this spectral sum: expanding the weak derivative in the corresponding orthonormal cosine modes and using Parseval's identity gives $\int |{\psi }^{\prime }{|}^2={\sum }_n{k}_n^2|a_n{|}^2$. Equating it to the directly integrated value $4E_1$ provesIn contrast, ${\sum }_n{E}_n^2|a_n{|}^2$ diverges, consistently showing why $H\psi$ is not an ordinary square-integrable vector even though the energy expectation is finite.