Atom Interferometry

de Broglie wavelength suggests that a particle with momentum $p$ has a wavelength given by ??, where $\lambda$ is wavelength, $h$ is Planck constant, and $p$ is momentum.

(1) $\begin{equation*}\lambda = \frac{h}{p}\end{equation*}$

Because momentum $p=mv$ , de Broglie wavelength can be rewritten as:

(2) $\begin{equation*}\lambda=\frac{h}{mv}\end{equation*}$

An interference has a maximum intensity when path different is an integral number of wavelengths, given by

(3) $\begin{equation*}d\sin\theta=n\lambda, \quad, n=\pm 1, \pm 2, \dots\end{equation*}$

Using small angle approximation, $\sin\theta\approx\tan\theta=x/L$ , where $x$ is the position of the maximum in the detection plane, and $L$ is the distance between silts and detection plane. Then, the distance between adjacent maximum is:

(4) $\begin{equation*}d\sin\theta\approx d\tan\theta=d\frac{x}{L}=n\lambda, n=1 \rightarrow x=\frac{L\lambda}{d}\end{equation*}$

If Slowly moving (helium) atoms interfere, who not the fast ones as well?
For interference to occur, the diffraction envelopes from each slit much overlap in space. Since the first zero of the single-slit diffraction envelope occurs where $\sin\theta=\lambda/a$ , where $a$ is the slit width, the width of the central maximum in the detection plane is $\lambda L/a$ . When atoms move slowly, they have a smaller magnitude of momentum $P$ and thus a larger de Broglie wavelength. Therefore, the diffraction envelopes from the two slits overlap and interference occurs. For the faster-moving atoms, $\lambda$ is smaller. Thus, though there is measurable diffraction of the helium atoms from each slit, there is insufficient diffraction for the diffraction envelope from each slit to overlap and hence produce an interference pattern.