 Solution

Using $m\simeq 4m_p$ and $T=15\mathrm{meV}$, the nonrelativistic de Broglie wavelength and speed areFor the square interferometer, the enclosed area projected into a vertical plane is $a^2\sin \theta$. The gravitational phase shift in matter-wave interferometry is $\Delta \phi =mg{a}^2\sin \theta /(\hslash v)$ to leading order in the small gravitational energy change. Thus rotation from horizontal to vertical changes the fringe count byOne way to see the phase is that the horizontal arms differ in gravitational potential energy by $mga\sin \theta$ for a flight time $a/v$. Which-path detection entangles the atom with distinguishable detector states. If those states are orthogonal and the path record is ignored, their coherence is lost and the interference fringes disappear. This is a loss of interference visibility, not a change in the atom's de Broglie wavelength.