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.