tim.tdj Posted July 29, 2018 Posted July 29, 2018 (edited) Hi. Everyone I don't know for sure but I think I know why the interference pattern in the double slit experiment disappears when one of the slits is "observed" by an extra piece of apparatus. I would be very grateful for your comments. As I see it, the "observed" particles which create the double band pattern on the screen are just a subset of the other particles which do form the interference pattern and the "observed" particles happen to have a specific property which causes them to be "observed" by the extra piece of apparatus which does not observe all of the particles. When the extra piece of apparatus is removed and the interference pattern returns, the double band pattern of particles with the specific property is still there but it is totally camouflaged by the interference pattern made up by the other particles. Do you think I am right or wrong? EDIT: I think I should clarify something here. As I understand it, this experiment is usually conducted using a device called a "coincidence meter" which has two detectors attached to it. One of the detectors is positioned stationarily at the "observed" slit. The other detector is a moving detector positioned at the "screen". When a particle is detected in both channels of the coincidence meter, it scores a "hit". When a particle is detected in only one channel, it scores a "miss". The double band pattern is formed by the "hits". Thank you very much. Kind regards Tim Edited July 29, 2018 by tim.tdj
swansont Posted July 29, 2018 Posted July 29, 2018 Coincidence detection is used for some experiments, but the basic double slit does not require it. If you have two slits and do not know which slit the light goes through, you get interference. If you have "which path" information, you do not. (the simplest way of achieving this is to block one slit) QM explains this quite well. It is not a property of the particles; they are (or can be) identical.
tim.tdj Posted July 29, 2018 Author Posted July 29, 2018 45 minutes ago, swansont said: Coincidence detection is used for some experiments, but the basic double slit does not require it. If you have two slits and do not know which slit the light goes through, you get interference. If you have "which path" information, you do not. (the simplest way of achieving this is to block one slit) QM explains this quite well. It is not a property of the particles; they are (or can be) identical. Hi Swansont Thank you very much for your reply. One of the ways I have seen the "which path" experiment presented is that a photon-splitting crystal is positioned at one of the slits so that the photon can be detected going through that slit and the explanation given as to why the interference pattern disappears is that the photon "knows that it is being observed". I find this a difficult explanation to accept. If you just block one of the slits then the interference pattern will disappear for a very obvious reason so this does not get to the bottom of why it disappears when one of the slits is being monitored as opposed to just being blocked. I think that my original post might give a possible explanation for this but I may be wrong and would be very grateful to anyone who can verify this or otherwise comment on it. Thank you very much. Kind regards Tim
swansont Posted July 29, 2018 Posted July 29, 2018 28 minutes ago, tim.tdj said: Hi Swansont Thank you very much for your reply. One of the ways I have seen the "which path" experiment presented is that a photon-splitting crystal is positioned at one of the slits so that the photon can be detected going through that slit and the explanation given as to why the interference pattern disappears is that the photon "knows that it is being observed". I find this a difficult explanation to accept. If you just block one of the slits then the interference pattern will disappear for a very obvious reason so this does not get to the bottom of why it disappears when one of the slits is being monitored as opposed to just being blocked. I think that my original post might give a possible explanation for this but I may be wrong and would be very grateful to anyone who can verify this or otherwise comment on it. Thank you very much. Kind regards Tim As I said, QM already explains this. You need to create a model from your idea, and see what else it predicts.
Sensei Posted July 29, 2018 Posted July 29, 2018 Do you have double slit experiment apparatus.. ? To repeat it by yourself at home.. ? You should buy it. It's pretty inexpensive (as long as it's done with photons, not electrons, or other particles). It cost $20 here, for pair of laser + double slit filter with fixed holes. (it would be better to have red, green, and blue lasers, with couple different slits setups).
tim.tdj Posted July 29, 2018 Author Posted July 29, 2018 (edited) Hi 30 minutes ago, swansont said: As I said, QM already explains this. You need to create a model from your idea, and see what else it predicts. Hi Swansont Thank you very much for your reply. Unfortunately, I do not have access to any lab equipment to test this out and I don't have the expertise to develop much of a model around this. I would be very grateful if you could provide a link to a web page which gives the explanation that you mention. Thank you very much. Kind regards Tim 21 minutes ago, Sensei said: Do you have double slit experiment apparatus.. ? To repeat it by yourself at home.. ? You should buy it. It's pretty inexpensive (as long as it's done with photons, not electrons, or other particles). It cost $20 here, for pair of laser + double slit filter with fixed holes. (it would be better to have red, green, and blue lasers, with couple different slits setups). Hi Sensei Thank you very much for your reply. Do you know of a good home kit I can buy in order to do this which includes a photon-splitting crystal, two detectors and a coincidence meter so that I can test the specific scenarios I am talking about. To be honest, I am hoping that this thread will become a discussion not just of what happens but why it happens. Thank you very much. Kind regards Tim Edited July 30, 2018 by tim.tdj
Sensei Posted July 30, 2018 Posted July 30, 2018 (edited) I bought setup in local educational store. I don't want to encourage you to buy something I didn't test by myself. But here you have some double slit (x4 versions) on Amazon: https://www.amazon.com/Diaphragm-Double-Slits-Different-Spacings/dp/B00KWZ5WQ0 (very wide...) You will also need laser pointers red, green and blue. It would be also good to have diffraction grating filters, and polarization filters (at least couple). Search for it by yourself. Edited July 30, 2018 by Sensei
tim.tdj Posted July 30, 2018 Author Posted July 30, 2018 3 minutes ago, Sensei said: I bought setup in local educational store. I don't want to encourage you to buy something I didn't test by myself. But here you have some double slit (x4 versions) on Amazon: https://www.amazon.com/Diaphragm-Double-Slits-Different-Spacings/dp/B00KWZ5WQ0 You will also need laser pointers red, green and blue. It would be also good to have diffraction filters, and polarization filters (at least couple). Search for it by yourself. Hi Sensei Thank you very much for this. Kind regards Tim
Sensei Posted July 30, 2018 Posted July 30, 2018 Search Amazon, eBay, Google and Ali etc. for "optical bench".. It allows precise placement of optical elements, in exact positions, from light source, one by another, and observation and measurement what happens with light beam during passing through various filters which you put on laser/light path.
Strange Posted July 30, 2018 Posted July 30, 2018 (edited) 14 hours ago, tim.tdj said: As I see it, the "observed" particles which create the double band pattern on the screen are just a subset of the other particles which do form the interference pattern and the "observed" particles happen to have a specific property which causes them to be "observed" by the extra piece of apparatus which does not observe all of the particles. What is different between one subset of particles and the other? Can you suggest some (other) way of measuring this difference? 7 hours ago, tim.tdj said: the explanation given as to why the interference pattern disappears is that the photon "knows that it is being observed". I find this a difficult explanation to accept. That is a "popular science" (journalistic) explanation and only has a very indirect resemblance to reality. You should watch Feynman'sQED videos (or read the book) which will give you a better idea of how a photon (or electron or whatever) is affected by everything around it (including the presence of slits, detectors, etc) 7 hours ago, tim.tdj said: I think that my original post might give a possible explanation for this but I may be wrong and would be very grateful to anyone who can verify this or otherwise comment on it. All particles of a particular type (photon, electrons, etc) are identical so there is no known mechanism for some of them to interfere and others not (and for those that interfere to be not detected, etc). On the other hand, there is a known mechanism for the current explanation - the fact that the behaviour(and evolution over time) of "particles" is described by a wave function. The probability of a particle landing in a particular place (and hence contributing to the interference pattern or not) is determined by every possible path that the particle can take - through one or other slit, into the detector or not, and so on - once you calculate the sum of all those probabilities, you get the same result as the classical experiment. Edited July 30, 2018 by Strange 1
swansont Posted July 30, 2018 Posted July 30, 2018 10 hours ago, tim.tdj said: Hi Hi Swansont Thank you very much for your reply. Unfortunately, I do not have access to any lab equipment to test this out and I don't have the expertise to develop much of a model around this. I would be very grateful if you could provide a link to a web page which gives the explanation that you mention. http://physics.mq.edu.au/~jcresser/Phys201/LectureNotes/TwoSlitExpt.pdf
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now