Experiments with Waves,

Balls and Electrons

A. Waves

Fig. 1: The diffraction of water waves
Fig. 2: The distribution φ of wave in the x-direction for two times t1 und t2.

Two water waves appear in windows 1 and 2 and next interfere on the detector D leading to the intensity I. If we close one of the windows, we will have only the wave: we readily recognize the sinus-like wave distribution in time t. The distance between two wave crests (or two minimums) is then the wavelength l. If we choose any point x on this wave, this point vibrates as φ from t again like sinus-wave having the vibration frequency v.
Detector D registers the Wave intensity I which is given by I = |φ|2.
Now we open both windows and watch time overlapping of the both waves φ1 and φ2 which must have the same frequency; it's clear they can be shifted to each other on phase j (see figure 2). So, in that way we will have:

f = f1+φ2 I = |φ1+φ2|2

More precise solution for the intensity will be as follows I = I1 + I2 + 2(I1I2)½.cos j. One can see that intensity also oscillates with cos j. These oscillations are especially strong for I1 = I2 = I0, since then all the intensity swings between 0 and 4I0: I = 2I0·(1+cos j). That is to say we will have the fourfold intensity in the extreme case. This is certainly the case for each point x, however having different phases in either point x.

B. Balls

Fig. 3: The experiment with balls.

When we held the "Interference" experiment with balls, we can find the particular distribution of probability P1 for the the balls to pass through slit 1 whether we keep slit 2 closed. And then we can find the probability P2 for the balls to pass through slit 2 if we keep slit 1 closed. If the two slits are opened one can find the total probability that is equal to:

P12 = P1 + P2

When the slits are situated far away from each other then there is no total probability like in the previous case and we will have the particular probabilities P1 and P2. Here one can see this situation in more rude form:

C. Electrons

Fig. 4: We will always have P12 ¹ P 1+ P2, even in the case when we can provide only one electron after another to be detected by our detection device.

In the similar experiment with electrons we can state that the total probability P12 is not equal to the sum of particular probabilities. And this is so even in the case when we can provide only one electron after another to be detected by our detection device:

P12 ¹ P1 + P2

The interference pattern which we observe on our screen corresponds to the behavior of wave interference which we can describe in the best way (let's hope so). Then we can describe our result applying two complex numbers φ1 and φ2. The square of absolute φ1 value gives the probability P1 whether slit 1 is opened: P1 = |φ1|2. Similar to previous statement one will have for slit 2 (when slit 1 is closed): P2 = |φ2|2. And if both slits are opened however we will have in this case:

P12 = |φ1+φ2|2¹ |φ1|2 + |φ2|2

In order to describe our observations in right way we must use complex numbers means in our calculations. In general, what will happen when we add a single electron that has passed through slit 1 or 2 ?

In order to get know it, we must carry out the experiment in which electrons are observed together with the light source:

Fig. 5: Experiment with electrons that are passing through one slit which is lighted by a lamp.

The result makes everything clear: the observed electron distribution on the screen matches exactly with the alternately closed slit. And in that way one will have

P12 = P1+P2 or P12 = |φ1|2+|φ2|2

Finally: If we carry out an experiment in which we can determine the way an electron moves, then we can say (but saying the truth with definite accuracy) has this electron taken this way or another one but however the interference pattern disappears. And if we don't know the path of electron has reached the screen then the statement that an electron moves by path 1 or 2 has no sense. So, we can't even approve that an electron has passed through this or other slit.

What will happen whether an experimenter who is observing a particle and waiting until this particle passes through the doubled slit has decided to continue with measurements (and moreover the two slits are situated pretty far away from each other)?

In this experiment the laser beam is directed into the beamsplitter and finally the splitted laser beams are collected by detector on which one can observe the interference pattern. Now we direct the part of initial beam into the Pockels Cell. And we can direct photons into the auxiliary detector when applying different voltages. And when we have zero voltage everything stays the same as it's been before. The random-signal generator switches on the cell only after the beam has passed the beamsplitter, but certainly before its reaching of the main detector. And it appears again that photon can take either one way or another. And if the Pockels cell is turned off then we will the interference pattern on the main detector.

This experiment with double slit shows that the two paths can be located as far away from each other as it is possible. Generally both splitted beams can travel for a one light year in opposite directions before they combine with each other to get interference pattern.

Efforts with atoms

Fig. 6: The double slit experiment with atoms passing through system of diaphragms and normal laser beam. Then atoms lose an energy falling into lower energy state and in that way producing photons which are kept by cavity. Since this process doesn't influence on the atomic movement, we won't take the uncertainty principle into account here. And in that way this experiment analysis shows that there is no possibility to determine paths of atoms from the interference pattern.