For equal masses with positions $\mathbf r_1,\mathbf r_2$, the two-body <Schrodinger equation> is
$$
\boxed{i\hbar\frac{\partial\Psi}{\partial t}
=\left[-\frac{\hbar^2}{2m}(\nabla_1^2+\nabla_2^2)
-\frac{Gm^2}{|\mathbf r_1-\mathbf r_2|}\right]\Psi}.
$$
Neglecting wave-packet spreading and overlap, write $|a\rangle_i$ for the packet $\psi_{ai}$. The initial state and its branchwise gravitational evolution are
$$
|\Psi(0)\rangle=\frac1n\sum_{a,b=1}^n|a\rangle_1|b\rangle_2,
\qquad
|\Psi(t)\rangle\simeq\frac1n\sum_{a,b=1}^n
e^{iGm^2t/(\hbar d_{ab})}|a\rangle_1|b\rangle_2,
$$
up to local kinetic phases. The branch dependence of the <Newtonian gravitational potential energy> is the source of <gravitationally induced entanglement>.

Under the approximation stated in the question, only the matching pairs $a=b$ acquire an appreciable common phase
$$
\phi=\frac{Gm^2t}{\hbar d}.
$$
The coefficient matrix of the resulting bipartite state is
$$
C=\frac1n[J+(e^{i\phi}-1)I],
$$
where $J$ is the all-ones matrix. Therefore
$$
\rho_1=CC^\dagger
=\frac1{n^2}\left(
[n+2(\cos\phi-1)]J+2(1-\cos\phi)I
\right).
$$
A <maximally entangled state> would require $\rho_1=I/n$. This demands
$$
n+2(\cos\phi-1)=0,
\qquad\text{or}\qquad
\cos\phi=1-\frac n2.
$$
That equation has no solution for $n>4$. Thus the requested large-$n$ conclusion does not follow from the assumptions printed in the paper: within the stated approximation, the system never becomes maximally entangled in the large-$n$ limit. For completeness, maximal entanglement is possible for $n=3$ at $\phi=2\pi/3$ and for $n=4$ at $\phi=\pi$; for $n=2$ it occurs at $\phi=\pi/2$ or $3\pi/2$. The $n=4$ time would be
$$
t=\frac{\pi\hbar d}{Gm^2}=\frac{hd}{2Gm^2}
\simeq9.9\ \mathrm{s},
$$
but it is not a large-$n$ answer.

The system also does not remain entangled for every $t>0$. Whenever $\phi=2\pi k$, the phase matrix again factorizes and the state returns to its initial <product state>. The revival period is
$$
\boxed{T=\frac{2\pi\hbar d}{Gm^2}=\frac{hd}{Gm^2}\simeq19.7\ \mathrm{s}}.
$$
For $n>2$ it is entangled at all intervening times; for $n=2$ there is the additional product-state revival at $\phi=\pi$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 325 / 3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-325/3/solution