Light as a Probability Wave — Question 7

Question 7

A continuous detection density predicts probabilities through an integral

The transverse detection probability density is $p(x)=A\cos^2(\pi x/2L)$ for $|x|\le L$, and zero elsewhere. Here $L>0$ and A is real.

1. Normalize p and state the units of A.

2. Find the probability of a detection in the central interval $-L/2\le x\le L/2$.

Question 7 - Solution

A continuous detection density predicts probabilities through an integral

(a) Method and calculation

$$1=A\int_{-L}^{L}\cos^2\!\left(\frac{\pi x}{2L}\right)dx=AL.$$$$\boxed{A=1/L},\qquad[A]=\mathrm{length}^{-1}.$$

(b) Interpretation and further calculation

$$P_c=\frac1L\left[\frac x2+\frac{L}{2\pi}\sin\frac{\pi x}{L}\right]_{-L/2}^{L/2}.$$$$\boxed{P_c=\frac12+\frac1\pi=.818310}.$$

Physical check

A probability density can exceed one in a chosen unit system; its integral over a region is dimensionless and lies between zero and one.