Question about Double Slit Experiment with Electrons

[AlexChandler]

In my modern physics course we learned that electrons, when shot through two close together slits, will produce an interference pattern just light light. However, when you set up an experiment to observe which slit the electron goes through, there is no longer an interference pattern. To my understanding, this is done by essentially putting a light source near the slits, and two microscope like devices to determine which slit the electron went through.

My question is: If we simply put a light source near the slits, but do not try to determine which slit each electron went through, will there be an interference pattern on the screen?

[silmaril89]

Hey Alex,

You asked this a while ago, so maybe you've gained some insight since then. But, from what I understand it doesn't matter if we also have the microscopes there to detect the electron. If a photon from the light bulb by the right slit interacts with the electron (i.e. is absorbed by the electron) it has been determined universally that it has traveled through the right slit and the interference pattern will not be created.

When I first learned about this, I just assumed it was subjective, that a scientist had to be there to "observe" it go through the right or the left slit. But, I'm quite certain it's not that way, if the interaction happens between the photon and the electron the path the electron took can be determined.

Another interesting thing I've read about this is observing the effect of lowering or increasing the intensity of the light source. If you lower it, the chances of an interaction between the electron and photon decreases and you get a mixture between an interference pattern and just two slits after many electrons are shot. Eventually, you can increase the intensity to a point where every electron will interact with a photon (or at least a huge majority) and you will always see the interference pattern.

Let me know if you think that's right or if you have anything to add to it.