3
$\begingroup$

Talking about this experiment https://en.wikipedia.org/wiki/Delayed_choice_quantum_eraser#The_experiment_of_Kim_et_al._.282000.29

I am assuming that anyone able to answer my question is already somewhat familiar with the experiment and the wiki page explains it better than I can. But here is the gist of it according to my understanding: you send photons through a double slit then split them into two entangled photons. One goes to D0 and the other goes to a setup of half mirrors and detectors where there is a 50/50 change of erasing or keeping information relating to the path or "which slit" the photon passed through.

Setup of the delayed choice quantum eraser experiment of Kim et al. Detector D0 is movable

While the collective pattern of all photons hitting D0 is just a blurred line you can pick out subsets R1 R2 R3 and R4 using the coincidence counter. That is R1 fx shows only photons hitting D0 which also had a "sister" particle hitting D1. Now R1 and R2 shows an interference pattern, phase shifted in relation to each other such that they cancel out in the collective image of D0.

Distribution of signal photons at D0 can be compared with distribution of bulbs on digital billboard. When all the bulbs are lit, billboard won't reveal any pattern of image, which can be 'recovered' only by switching-off some bulbs. Likewise interference pattern or no-interference pattern among signal photons at D0 can be recovered only after 'switching-off' (or ignoring) some signal photons and which signal photons should be ignored to recover pattern, this information can be gained only by looking at corresponding entangled idler photons in detectors D1 to D4.

x axis: position of D0. y axis: joint detection rates between D0 and D1, D2, D3, D4 (R01, R02, R03, R04). R04 is not provided in the Kim article, and is supplied according to their verbal description

Now here is what I don't understand: why is there a difference between D1 and D2? On the graphs R1 has a peak in the middle while R2 has a valley, why not reversed? What makes D1 and D2 different? What decides which one gets peaks at certain points and the other gets the opposite?

Also does D1 and D2 always have a peak or valley in the middle? or are they simply always phase shifted pi in relation to each other but not in relation to the middle? Either way, what I really want to know is why/how does R1 and R2 turn out phase shifted?

Improve this question
$\endgroup$
1
  • $\begingroup$ Giving the details here rather than pointing people to wikipedia, and explaining which point exactly you don't understand about it (and which part you do unterstand) will likely increase your chance of getting an answer. $\endgroup$
    – Norbert Schuch
    Commented Feb 10, 2016 at 12:25

2 Answers 2

Reset to default
2
$\begingroup$

Delayed_choice_quantum_eraser#The_experiment_of_Kim_et_al. [...]
Now R1 and R2 shows an interference pattern

These two patterns of coincidence counts both arise as a function of "x" (see the labelled arrow next to detector D0 in the schematic); i.e. as detector D0 is put in different places of the plane of convergence where the "Lens" produces a suitably sharp image of the BBO crystal with the two slits.

Different "image places x" correspond to (ever so slightly) different ray trajectories from the BBO crystal to the lens; i.e. the upper red or blue traces differing (ever so slightly) depending on (the value of) "x".

What makes D1 and D2 different?

Importantly, due to the properties of the BBO crystal, photon by photon, there is a very precise relation between the exact ray trajectories from the BBO crystal to the lens and the exact ray trajectories from the BBO crystal to the prism "PS". As the former change depending on (the value of) "x", the latter do as well. Therefore the exact light path lengths from from either slit to deterctors D1 or D2 may change (ever so slightly) depending on (the value of) "x". Consequently, separately for detectors D1 or D2, the difference of pathlengths for the two slits may vary, leading to interference changing between constructive and destructive, as (the value of) "x" is being changed in the course of taking data.

Of course, the exact arrangement of experimental components had been optimized (and kept stable enough) for the interference effect to appear prominently.

Also: If it is required (for the setup to be said having "functioned properly and as specified") that the combined counting rate of D1 and D2 (in coincidence with D0) remained constant and independent of (the value of) "x" then the interference patters of D1 and of D2 necessarily appear complementary.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Does this imply that the detectors D1-4 must have a much bigger opening than D0, in order to catch all photons over a wide range of x? Or does the lens do something for that? I would actually expect that the lens would actually compress the range of x, requiring an even smaller opening on D0 $\endgroup$
    – fishinear
    Commented Jan 31, 2019 at 15:08
  • $\begingroup$ @fishinear: "Does this imply that the detectors D1-4 must have a much bigger opening than D0, in order to catch all photons over a wide range of x?" -- That's what I tend to expect, too. Indeed: the openings of detectors D1-4 might be more comparable to the lens diameter. "I would actually expect that the lens would actually compress the range of x, requiring an even smaller opening on D0" -- Well, it's a convex lens. Whose effect is nicely illustrated thus ... $\endgroup$
    – user12262
    Commented Jan 31, 2019 at 21:43
1
$\begingroup$

If we think the phenomenon in reverse an advance wave leaves either d1 or d2. The wave leaving from d1 strikes BSc and splits into blue and red, but the blue wave is phase shifted wrt to red due to reflection at BSc. similarly for wave leaving from d2,in this case red beam is phase shifted ..

Improve this answer
$\endgroup$
2
  • $\begingroup$ I hadn't thought about that, still don't quite get it though. So we have a phase shift, but how does that propagate to R01 and R02 subsets? Those are subsets of D0 right? $\endgroup$
    – RHawkeyed
    Commented Oct 17, 2016 at 9:38
  • $\begingroup$ another advance wave leaves D0 ( from R01 or R02 part) meets advance wave from D1 or D2, at BBO they merge and go to laser source. $\endgroup$
    – naive
    Commented Oct 18, 2016 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.