Question 1.10.8: In a double-slit experiment with a source of monoenergetic e...

Using the integral \int_{-\infty }^{+\infty }{dy/(1+y^2)}=\pi ,   we can obtain the normalization constants at once: A_1=A_2=1/\sqrt{\pi };   hence \psi _1  and  \psi _2  become  \psi _1(y,t)=e^{-i(ky-\omega t)}/\sqrt{\pi (1+y^2)},\psi _2(y,t)=e^{-i(ky+\pi y-\omega t)}/\sqrt{\pi (1+y^2)}.

(a) When we use a light source to observe the electrons as they exit from the two slits on their way to the vertical screen, the total intensity recorded on the screen will be determined by a simple addition of the probability densities (or of the separate intensities):

I(y)=|\psi _1(y,t)|^2+|\psi _2(y,t)|^2=\frac{2}{\pi (1+y^2)}.            (1.190)

As depicted in Figure 1.17a, the shape of the total intensity displays no interference pattern.

Intruding on the electrons with the light source, we distort their motion.

(b) When no light source is used to observe the electrons, the motion will not be distorted and the total intensity will be determined by an addition of the amplitudes, not the intensities:

=\frac{4}{\pi (1+y^2)} \cos ^2\left(\frac{\pi }{2}y \right).              (1.191)

The shape of this intensity does display an interference pattern which, as shown in Figure 1.17b, results from an oscillating function, \cos ^2(\pi y/2) ,  modulated by 4/[\pi (1+y^2)].