For positions $\mathbf r_1,\mathbf r_2$, the two-particle <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 packet spreading and writing $|a\rangle_i$ for $\psi_{ai}$, branchwise evolution gives
$$
|\Psi(t)\rangle\simeq\frac13\sum_{a,b=0}^2
\exp\left(\frac{iGm^2t}{\hbar d_{ab}}\right)
|a\rangle_1|b\rangle_2,
$$
up to local kinetic phases. This branch-dependent <Newtonian gravitational potential energy> produces <gravitationally induced entanglement>.

Under the stated distance approximation, only the three branches $a=b$ acquire an appreciable common phase
$$
\phi=\frac{Gm^2t}{\hbar d}.
$$
The coefficient matrix is
$$
C=\frac13[J+(e^{i\phi}-1)I],
$$
where $J$ is the $3\times3$ all-ones matrix. The <reduced density matrix> is
$$
\rho_1=CC^\dagger
=\frac19\left[(1+2\cos\phi)J
+2(1-\cos\phi)I\right].
$$
It equals $I/3$ when $\cos\phi=-1/2$, so the first maximally entangled state occurs at $\phi=2\pi/3$. Therefore
$$
\boxed{t_{\rm ent}
=\frac{2\pi\hbar d}{3Gm^2}
=\frac{hd}{3Gm^2}}
\simeq6.6\ {\rm s}.
$$

The state does not remain entangled for every $t>0$. Whenever $\phi=2\pi k$, all branch phases again agree and the state returns to its initial <product state>. The revival period is
$$
\boxed{T=\frac{2\pi\hbar d}{Gm^2}
=\frac{hd}{Gm^2}\simeq19.7\ {\rm s}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 325 / 3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-325/3/solution