Why? does the observing unit change the atmosphere in which the electron is passing through?
If the observing unit was a distance away but magnified to see the single slit as if it were at the same distance how far would the observing unit have to be to create the interferance pattern? Or would there be no change?
The most lucid explanation of this experiment is in the first chapter of the third volume of
Feynman's Lectures on Physics. I recommend anyone to pick it up in their local library. It doesn't require any physics background to follow. The following is a summary of this chapter.
Let us start by this two observations:
- If the electrons are not seen when passing through the slits, we get an interference pattern, such as would be made by light.
- If the electrons are seen, we don't get interference. This is qualitatively the same result as the one we would get doing the experiment with bullets, balls or some other macroscopic object
Why is this so? The observing unit is some kind of light source. When the electron passes through the slit, it scatters light and we see a flash. Ceritus asks how far the light source would have to be to create interference. And it is a good question. Let us phrase it in a slightly different way: we will use an increasingly dimmer light source and see what happens. Assume also that each time an electron passes through the slits and hits the target, we hear a 'click'.
Our intuition tells us that a very dim light source would eventually have a negligible effect. But we are talking QM here. The flashes are always the same size, no matter what the intensity of the light source! But, if the latter is very weak, sometimes we will hear a click but see no flash. The explanation is that light comes in photons and the detection consists on an electron colliding with one photon, so it is always the same. If the light has a low intensity, it will have fewer photons and sometimes we won't see a collision. The result is that those electrons that have been seen do not interfere with one another, while those that have passed undetected form an interference pattern.[1]
The reason why in one situation we see interference and in the other we don't is the
uncertainty principle. It is impossible to use a system that will observe the electrons and at the same time preserve the inference. I will now give a qualitative explanation.
The precise form of the uncertainty relations for Q and P (position and momentum) is
[latex]\footnotesize \Delta P \cdot \Delta Q \geq \frac{\hslash}{2}[/latex]
This is a
postulate of quantum mechanics.[2]
We will use a modified version of our experiment. Now the plate with the slits is mounted on rollers. If the detector is situated equidistant to the two slits, an electron passing through the upper (lower) slit will have to be deflected downwards (upwards) to reach the detector. The plate with the slits will get an equal momentum in the opposite direction. For any position of the detector, the recoil of the plate from an electron passing through each slit will be different. We can now determine which slit the electron passed without disturbing them at all! But wait: we haven't really cheated the uncertainty principle. To know the momentum of the plate we must measure it. Measuring it means that we no longer now with complete precision the position of the plate. So we don't know the exact location of the holes. So the centre of the pattern will be different for each electron. The fringes are blurred. What's more, if we measure the momentum of the plate with enough precision to determine the slit, the pattern will be shifted just enough for a maximum of the original pattern to coincide with a minimum of the new one. This effectively destroys any hope to see interference.
For a semiquantitaive example of the uncertainty relations, we will consider a horizontal plane wave of particles. Classically, this just means a lot of particles with horizontal mometum p
x = p
0 and vertical momentum p
y = 0. We don't know the vertical position of the particles. If we set a hole of diameter B we can determine the position of the particle with uncertainty ±B. So Delta y ~ B and Delta p
y = 0 (classically), because the momentum is horizontal. But in QM, by determining the vertical position we have destroyed our knowledge of the vertical momentum. Classically, when light goes throughn a small aperture (of the order of the wavelength), it experiments
diffraction, the wave is spread and no longer plane. In QM, this happens with particles too, because each particle has an associated wavelength via λ = h/p. If the particle experiments diffraction, there is a chance it will we deflected in a different direction, with a nonzero vertical component.
Let Δθ be the angular spread. Then Δ p
y = p
0 Δθ. To estimate Δθ we notice that the first mininum occurs at an angle Δθ such that the waves from one edge have travelled one more wavelength than the waves from the opposing edge.[3] So Δθ = λ/B and Δ p
y = p
0 λ / B. A small B means good measurement of y and bad knowledge of its momentum. In fact, we have
Δp
y = p
0 λ / B
Δy = B
So Δp
y Δy = p
0 λ. But p λ = h:
Δp
yΔy = h.
_____________
[1] You can propose using longer wavelengths, with 'weaker' photons. But if the wavelength is bigger than the distance between the slits, it won't let us decide where has the electron been.
[2] Not exactly. The actual postulate is [Q,P] = i hbar. We also now (from a different postulate) that (Delta A) (Delta B) >= -i/2 [A,B], for any observables A,B.
[3] This is classical optics. Destructive interference (minimum) occurs when the difference of travelled paths is equal to an odd number of wavelengths.
