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Hello,

On the image above you'll find the display for the double slit "delayed choice quantum eraser" by Scully. I use this experiment for my question as it cames through trying to understand it.

The last part of the experiment on the right of the picture is a quantum eraser setup. When a Photon reach it the "which path information" is supposed to be lost.

But let be honest, it is impossible to make something absolutely perfectly symetrical. One of the red or blue path should be longer then the other (even by a tiny tiny tiny bit) For the purpose of this question let imagine the red is longer then the blue path by a time of T+x (where T is the time along the blue path).

If the photon take T time to reach D2 we know it was following the blue path, however if it takes T+x time we know it followed the red path... therefore the "which path information" is not really lost....

So how come that such quantum eraser still works?

agone07
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  • Closely related question here. The basic point is that all times in this problem are fuzzy, so even if one path is a bit longer, it doesn't destroy the coherence. – knzhou Nov 04 '18 at 21:43
  • thank you for the link, however I don't understand anything about this other post. Can you explain a bit more what you mean by "fuzzy" because _I don't really see why it doesn't give up the iformation (if you had a measurement device precise enough to measure such difference of time, or if the red path was much longer)

    If the time difference was below the planck time I could easily understand ... but we are far very far from it.

     – agone07 Nov 04 '18 at 21:50
  • The time for going through one path is $10 \pm 6$. The time for going through the other is $12 \pm 5$. You measure $11$. So tell me, which path was taken? – knzhou Nov 04 '18 at 22:31

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A photon is an excitation of a field mode. In experiments like this the field modes are extended over some length $L$, just like a plane wave in classical electromagnetism. It is called the coherence length (and there is an associated coherence time $L/c$). To eliminate the which-path information perfectly, the lengths of the different paths traveled would have to agree exactly. But if they disagree a little, it doesn't matter. Roughly speaking, a path length difference by $\delta x$ would give which-path information at the level $\delta x / L$. To be precise, the interference fringe contrast would be reduced from $1$ to $1 - \delta x / L$. Therefore it is sufficient if $\delta x \ll L$.

Andrew Steane
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