Light as a Probability Wave — Question 8

Question 8

Finite detector pixels average a probability fringe before counting

Across a small region of a broad beam, the density is $p(x)=p_0[1+\cos(kx)]$, with constant envelope $p_0$ and $k>0$. A pixel of width d is centered at $x_0$ and lies entirely in that region. The full beam is normalized elsewhere.

1. Integrate p over the pixel and derive the visibility as its center is scanned.

2. Evaluate the visibility for $d=\pi/k$ and $d=2\pi/k$.

Question 8 - Solution

Finite detector pixels average a probability fringe before counting

(a) Method and calculation

$$P_{\mathrm{pix}}=\int_{x_0-d/2}^{x_0+d/2}p(x)dx=p_0d\left[1+\frac{\sin(kd/2)}{kd/2}\cos(kx_0)\right].$$$$\boxed{\mathcal V_{\mathrm{pix}}=\left|\frac{\sin(kd/2)}{kd/2}\right|}.$$

(b) Interpretation and further calculation

$$d=\pi/k:\quad\boxed{\mathcal V_{\mathrm{pix}}=2/\pi=.636620}.$$$$d=2\pi/k:\quad\boxed{\mathcal V_{\mathrm{pix}}=0}.$$

Physical check

Loss of measured contrast here comes from spatial averaging, not from loss of coherence of the incident photon state.