Module 5 : MODERN PHYSICS
Lecture 26 : Wave Nature of Particle - the de Broglie Hypothesis
Clearly, the result is absurd in the sense that it says that when both slits are open, there are particles which neither goes through the slit $ S_1$ nor goes through the slit $ S_2$. Let us try to look at the actual situation. To keep track of which electron came through which slit is easy enough. We put a source of light near each of the slits, so that when an electron passes through one of the slits, it scatters light and we can see a flash. If we do the experiment this way, keeping track of the particle, we find that $ p=p_1+p_2$ and there is no contradiction. However, if we do not keep track of which slit each electron goes through, we get the distribution pattern shown in curve C. What is even more funny is that in curve C there are some points (minima) where the number of particles is even less than that which reach these points when only one slit is kept open.
  We define a probability amplitude $ \psi$ such that $ p=\mid\psi\mid^2$. In terms of probability amplitude, it turns out that if we keep track of electrons by watching them,
 
$\displaystyle p=p_1+p_2=\mid\psi_1\mid^2+\mid\psi_2\mid^2\eqno(1)$
  However, if we do not observe the electrons the probability distribution is given by
 
$\displaystyle p=\mid\psi_1+\psi_2\mid^2\ne p_1+p_2\eqno(2)$
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