Does the observation instrument need to be conscious in order to change the interference pattern in the "Double slit experiment"? [closed]

Question

I am an absolute amateur in physics but some questions haunt us all, don't they? In some places I hear/read that the conscious observer changes the pattern (https://youtu.be/XDpurdHKpb8?t=1559). In other places, it is explicitly stated that "it is necessary to have a conscious observer - just a detector/measuring device/some-unconscious-gadget and the interference pattern is broken.

And finally this experiment shows that it is not the measuring gadget changing the pattern, but rather whether one could know the path of the particle.

What do you think?

Answer 1

The beautiful thing about QM is that if the setup is right, the interference pattern is there, whether you observe it or not.

If by changing the interference pattern you mean making it disappear, then again if it is done right (for example by putting a detector on one of the slits), the interference pattern will disappear, whether you observe it or not.

It is very important to understand that the appearance of the interference pattern (caused by the superposing of the waves from the slits) is observer independent, and only depends on the setup, boundary conditions.

The light waves from each of the slits superpose (interfere) and produce an interference pattern.

Fringe width and spacing and number of slits in diffraction experiments

Now you are asking whether you need a conscious observer to make the interference pattern disappear. It is very important to understand that if you put a detector at one of the slits, the interaction between the photons and the detector (the boundary conditions will change, the photons will be out of phase and the scattering angle will be random) will cause the interference pattern to disappear. This interaction is observer independent too.

So the answer to your question is no, the setup itself (if it is right) makes the interference pattern appear or disappear, no conscious observer is needed, it is all observer independent.

Answer 2

Please look on the screenshot I made from the video you mentioned:

Take any point on the screen and imagine what happen with water waves the next view seconds. To me, neighboring dark and white point on the screen will be „exposured“ nearly by the same amount of energy from the two waves. In the white positions periodically with twice the amplitude from the waves. And in between the drawn peaks periodically with following one by one amplitudes from the two waves.

Energy distribution from water waves and from light

The peaks on the screen are a time dependent moment only of what happens over time. Summing up all points, we wall experience a continually falling amount of energy to the left and the right of the central point (due to the raising distance from the source to the screen). Youngs sketch from water waves wasn’t animated, but today we have the animation and should care about what is obvious.

So we clearly we have a difference of the energy distribution that struck the wall for water waves and for light. Because, without any doubt, EM radiation has a swelling energy distribution behind edges.

Interference or no interference?

EM radiation consists of photons. Photons are indivisible quanta and (nearly) not interact with each other. They hit the screen or they do not. Cancelling interference simply does not exist for two photons in the case of a photometer or a silver emulsion screen or a CCD chip. The only conclusion could be that photons get deflected by the interaction with the slits and get directed with a periodicity to the screen.

Slit or edge deflection of photons?

Surprisingly even behind single edges the intensity distribution of light takes place. For this reason too, the assumed interference of light from the two edges of a slit or from double slits is a good way to calculate the phenomenon, but for the understanding what happens in reality, it is missleading.

Long story short:

1. The intensity distribution behind slits is the sum of the interaction between EM radiation and every involved edge.

2. The fringes are not the result of interference but of different deflection as a function of the distance of the photon to the edge.

The role of the measuring instrument

Closing slits is part of a measuring instrument (together with the screen).

If you close one slit, it is not true, that the fringes disappear. They still there for the remaining slit. Moving apart the edges of the single slit, it is not true, that the fringes disappear. They still there for both the single edges.

Another popular idea ist the measuring of the flying photon (or electron) from the side. It is obvious that this try changes the trajectory of the particle and the intensity distribution on the screen gets lost.

Science has made itself comfortable with Young's explanation and the mystification of the interference assertion makes every novice shiver with fear. And once learnt, any argumentation about it will be fought.