Heisenberg uncertainty principle In the double slit experiment, an electron interferes with itself and creates the pattern because it is in a superposition, traveling through both slits. If we place a detector at one slit, the wave function collapses and we lose the interference pattern. 
If we place the detector over one slit then we know exactly where the electron was as it went through the slit. However, if we know exactly what the position is, shouldn't the uncertainty of momentum be infinite? As uncertainty of position approaches zero, uncertainty of momentum must approach infinity to be equal to the reduced plank constant. Couldn't it even shoot backwards? But instead, we don't see this. It places a clump behind the screen, acting just as we would expect marbles to be.
 A: The slit is not of zero width, so there is still considerable uncertainty in the position of the particle.
Moreover, the narrowing of the slit does lead to diffraction (i.e. a spread in the momentum).
A: I think part of the solution to the conceptual problem is also to recognize that as we make the slits narrower, ever fewer particles actually make it trough. The ideal Gedanken experiment doesn't contain the necessary derivation, it always assumes that all the quanta appear behind the screen. 
That's not the case in a real experiment, though. The material of the slits represent a physical barrier for the quantum fields. They can be made reflective, in which case most of the quanta end up in the opposite direction of the screen (we just don't place detectors there) or they can be absorptive, and the screens heat up (a little). Very little of the source radiation actually ends up behind the screen in a real experiment.  How much can be theoretically calculated based on the width of the slits and the material they are made from, but it is not a totally trivial calculation, which is why it is usually omitted. 
The detector, for the "surviving" quanta that make it behind the slits is essentially modifying the possible paths that can be taken. If the detector is absorptive, if will take them out of the field when the path in the path integral intersects with the detector surface. This removes half the intensity behind the slit, in addition to removing the interference. This is more obvious in the path integral formulation than it is in the Schroedinger equation, which we usually do not teach and use with absorptive boundary conditions. 
