Starship Asterisk*

The lecture video is embedded below.

Additionally, slides used in the lecture are embedded below, or can also be downloaded directly from here.

Questions after the lecture? Please feel free to post them in the same thread.




*********************************************************************************************

Hello,

after having viewed the video on the Delayed Choice Quantum Eraser, in the Classroom: Physics X, I got a question.

As a single photon laser beam hits the double slit it can either go through both A and B or only A or B. This is written here: http://en.wikipedia.org/wiki/Delayed_ch ... tum_eraser under "The experiment"

But, as I can see from the drawing of the experiment, if a photon goes through A and B we get 4 (single and idle X 2) new photons after the "motherphoton" hits the BBO-crystal, not 2 (single and idle X 1).

So, instead of having 1 motherphoton, which get separated into 2 new and entangled with half the oiginal frequence, we get 4 with 1/4 og the original frequence?
Note: I am aware of the fact that this would not be the case if we track the idle photons route, because then the wavefunction fails. My question is only relevant when trying to understand what happens as we get a interference pattern fom the idle photon.

I can't seem to find any answar to this, please help.

Regards from DK.

The 2 possible paths for each photon (signal and idler) do not represent different photons. They are merely possible paths for an individual photon. Just as in an ordinary double-slit experiment, one can draw 2 distinct possible paths for a single photon. There are actually 4 possible paths illustrated for the idler photon.

Please be advised that I am not a physicist nor am I at Michigan Tech.

Thank you for your reply.

Unfortunately it does not solve my question.

I believe you are mistakeing and that the signal and idler are indeed two different, but entangled, photons. Please note that it is the BBO-crystal that creates the two entagled photons.

The way the illustration is drawn, the BBO crystal does indeed create, for each original photon, one signal photon which goes to D0 and one idler photon which goes to the other part of the apparatus. The 2 photons are entangled. This interpretation is confirmed by an examination of the Wikipedia entry describing the experiment.

http://en.wikipedia.org/wiki/Delayed_ch ... tum_eraser

In the case of the original photon, as with the signal and idler photons, the 2 paths through the 2 different slits do not represent different photons. They are merely possible paths for a single photon. Even if the photon goes through BOTH slits it is still a single photon.

The fact that 1 photon will "become" 2 entangled photons each with 1/2 of the oiginal frequence of the "motherphoton" when hitting the BBO is surely correct

But how it exactly works when the motherphoton interferes with itself i'm still in doubt. It would be nice if Robert could come with an explanation. But anyway it is not necessary to understand this to understand the concepts in the experiment, because when the motherphoton interferes with itself, we namely don't know anything about its rute and therefoe we don't violate heisenbergs uncertainty principle - we just assume that is interferes by wave potentiale and therefore goes through both slits.

... and therefore it is indifferent if the motherphoton goes through A or B or A+B, because there is no why to know.

"How exactly it works" is a question that the best and the brightest haven't been able to agree on for 80 years. Some would argue that it is pointless to speculate about what it happening when no one is looking. Others would suggest that a specific particle follows a specific trajectory but is guided by this abstract thing called the wave function. And there are many other viewpoints. But clearly our common-sense notions of time are at risk, when we can predict a particular result based on something that happens later in the experiment.

"How exactly it works" - yeah that is the 1 million doller question and whay this expeiment has been done.

But what I really refered to with the phrase was: when we don't have which-slit information, the photon is allowed to act like af wave and therefore we get interferance pattern. But as it goes through both slits, then, from the drawing of the expeiment, we would get 2 idler and 2 single.
I don't think this is correct.

You state that even if the photon goes through both slits, then the idler ans single photon will still follow either rute A or B and not both. This I'm still in doubt.

It's easy to infer from the drawing that the 2 paths through the double slits represent 2 different photons. But this is not the case. Think of it this way. A point-source water wave can go through 2 slits at the same time. In fact it is taking numerous paths simultaneously. But it is still, in a sense, 1 wave. One the other side it interferes with itself. Similarly, a photon can go through 2 slits at once and still be 1 photon. The difference with the photon, unlike the water wave, is that we can't see it until it produces an effect, and it never gets absorbed at multiple locations. It always gets absorbed at one and only one location. But this location seems to be dictated probabilistically by equations describing wave behavior.

Another description of QM phenomena with photons, in the context of QED, can be found here (although it doesn't go into delayed choice possibilities):

http://www.youtube.com/user/cassiopeiap ... 1GdgD77AQ4

I appreciate that you bother replying, but you don't seem to understand my question - maybe my english skills are poor for which I'am sorry?

The contents of your last reply is basic knowlegde and I don't understand how you can assume that it would be helpful for me when taking our last messages in consideration.

My question can be simplified: what is the frequence of the photons detected in the experiment at D0 and other detectors (it dosen't matter which) when we get interference in the southern part given that it even is possible to meassure (which I think it is)?

The language barrier is indeed presenting some problems. You will have to clarify for me what you mean by "frequence."

1. Pattern of occurrence of hits at the detectors in relation to each other?

2. Frequency (inversely related to wavelength) of a photon?

Your allusion to "interference on the southern part" is somewhat confusing. The experimenters do not look for interference on the southern part. The relevant detectors (D1-D4) merely record their respective hits (or not). Only at D0 is the detector moved during the course of the experiment so that an interference pattern can be built up from the individual hits.

In mean frequency of a photon = light colour. So, what is the frequency of the sigal and idler photons:
1. when we get interference?
and
2. when we do not?

There can easily be build up interference in the souhtern part (D1-D4) also - but for sure the most interesting stuff occurs at D0. But it is important to understand what ever happens at D1-D4 aswell.

As I understand it, the frequencies of the signal and idler photons are identical and will not vary with the paths taken. The actual wavelengths used were 702.2 nm. This would correspond to about 427 THz by my calculation. Which-path info should have no effect on this - the photons that build up the interference patterns should have the same frequencies as the photons that don't.

Yes, the idler and signal are entangled so they are identical.
Your calculations is correct - the light is close to the infrared wavelengths.

"the photons that build up the interference patterns should have the same frequencies as the photons that don't." - Yeah maybe - non of us can tell for sure, but it would be interesting to know.

However, wherefrom do you have the knowledge of the wavelength? are you a scientist, student?

I am obtaining the wavelength from the original paper.

http://arxiv.org/PS_cache/quant-ph/pdf/ ... 3047v1.pdf

I am actually a biologist, but my second choice would have been physics.

Interesting