For a neutron in the Earth's <Newtonian gravitational potential> $\Phi_E$, the <Schrodinger equation> is
$$
\boxed{
i\hbar\frac{\partial\psi}{\partial t}
=\left[-\frac{\hbar^2}{2m_n}\nabla^2+m_n\Phi_E\right]\psi}.
$$
Near the surface, $\Phi_E(z)=gz$ up to an additive constant.

The <Colella–Overhauser–Werner experiment> uses crystal slabs as coherent beam splitters and mirrors in a <neutron interferometer>. The two paths contain horizontal segments of length $s$ separated vertically by $r$, then recombine. The <Newtonian gravitational potential energy>[gravitational potential energy] difference is $m_ngr$. A neutron of speed $v$ spends time $s/v$ on a horizontal segment, so the gravitationally induced relative <quantum phase> is
$$
|\Delta\phi|
=\frac{m_ngr}{\hbar}\frac{s}{v}.
$$
Using the <de Broglie wavelength> $\lambda=h/(m_nv)$ and $\hbar=h/(2\pi)$ gives
$$
\boxed{|\Delta\phi|
=\frac{2\pi m_n^2grs\lambda}{h^2}}.
$$
Rotating the apparatus changes the vertical projection of its enclosed area from zero to $rs$. The number of full interference oscillations is therefore
$$
N=\frac{|\Delta\phi_{\max}|}{2\pi}
=\frac{m_n^2grs\lambda}{h^2}.
$$
Here $rs=(\sqrt{10}\,\mathrm{cm})^2=10^{-3}\,\mathrm{m}^2$, so
$$
N=\frac{(1.67\times10^{-27})^2(9.8)(10^{-3})(1.42\times10^{-10})}
{(6.6\times10^{-34})^2}
\simeq8.9.
$$
The expected change is therefore $\boxed{9}$ oscillations to the nearest integer.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 325 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-325/3/b/solution