Question #43581

In a Young's Double Slit Experiment, a beam of photons of wavelength $\lambda$ is incident on two slits separated by a distance $d$. The interference pattern is observed on a screen at a distance $D$ ($D \gg d$). To obtain 'which-path' information, a quantum non-destructive detector is placed immediately after the slits to measure the transverse position $y$ of each photon. The measurement has a Gaussian uncertainty profile with standard deviation $\Delta y = a$. According to the Heisenberg uncertainty principle, this measurement imparts a random transverse momentum kick $\Delta p_y$ to the photon, which follows a Gaussian distribution with standard deviation $\sigma_p = \frac{h}{4\pi a}$ (where $h$ is Planck's constant). If the original interference pattern without the detector has a visibility of $1$, what is the visibility $V$ of the fringes on the screen with the detector in place?

(Hint: The convolution of $f(y) = \cos(ky)$ with a Gaussian probability density function $P(y) = \frac{1}{\sqrt{2\pi}\sigma}e^{-y^2/(2\sigma^2)}$ is $\cos(ky)e^{-k^2\sigma^2/2}$)