= Solution
Take $|L\rangle=|0\rangle$ and $|R\rangle=|1\rangle$. Initially the two masses are in the <product state>
$$
|\psi(0)\rangle=\frac12(|LL\rangle+|LR\rangle+|RL\rangle+|RR\rangle).
$$
The branch-dependent <Newtonian gravitational potential energy> is $-Gm^2/d_{ab}$. Under the stated approximation, only the $|RL\rangle$ branch acquires an appreciable relative phase, so after time $t$,
$$
|\psi(t)\rangle\simeq
\frac12(|LL\rangle+|LR\rangle+e^{i\phi}|RL\rangle+|RR\rangle),
\qquad
\phi=\frac{Gm^2t}{\hbar d}.
$$
The determinant of its two-by-two coefficient matrix is $(1-e^{i\phi})/4$, which is nonzero unless $\phi$ is a multiple of $2\pi$. Thus the state generally has <Schmidt rank> two: the branch-dependent gravitational phase creates <gravitationally induced entanglement>.
An <entanglement witness> has a bound obeyed by every <separable quantum state> and violated by at least one entangled state. For a product state with <Bloch vectors> $\mathbf r$ and $\mathbf s$,
$$
|\langle X\otimes Z+Y\otimes Y\rangle|
=|r_xs_z+r_ys_y|
\leq\sqrt{r_x^2+r_y^2}\sqrt{s_z^2+s_y^2}
\leq1
$$
by the <Cauchy-Schwarz inequality>. Convexity gives the same bound for every separable mixed state. Consequently $W>1$ certifies entanglement; in conventional operator form, one of $I\mp(X\otimes Z+Y\otimes Y)$ has negative expectation whenever the absolute-value criterion is violated.
For the state above, direct use of the <Pauli matrices> gives
$$
\langle X\otimes Z\rangle
=\langle Y\otimes Y\rangle
=\frac{\cos\phi-1}{2},
\qquad
W=1-\cos\phi.
$$
With the supplied values,
$$
\phi\simeq
\frac{(6.674\times10^{-11})(10^{-14})^2(10)}
{(1.054\times10^{-34})(2\times10^{-4})}
\simeq3.17,
$$
which is close to $\pi$. Hence $W\simeq2.00$, and to the nearest integer
$$
\boxed{W=2}.
$$
This is the operating principle of the <Bose--Marletto--Vedral experiment>.
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