Consider the standard double slit experiment with one photon at a time. Now, replace the photographic plate (or fine resolution single photon detector screen) with a converging lens, and move the single photon detector screen to the image location of the double slit screen.

Experimental setup

Suppose there are $M$ interference fringes. Cover the lens with $2M$ weak measurement photon detector strips — one for each crest and one for each trough — which always allow photons to pass through preserving their phases. The resolving power of each strip is the entire strip itself. Each strip is associated with a continuous coherent quantum pointer, and each time a photon passes through the strip, the pointer undergoes the map $|p\rangle \to |p+\Delta\rangle$. Each strip pointer is initialized to $\int \frac{dp}{(2\pi)^{1/4}} e^{-p^2/4} |p\rangle$. If $\Delta \ll 1$, we have a weak measurement. To avoid any confusion, by weak measurement, I just mean a Gaussian pointer state with tiny measurement shifts compared to the Gaussian standard deviation. No post-selection is involved here.

Let $N$ photons pass through, one at a time. The strip pointers are shielded from decoherence and not measured until all $N$ photons have passed through. For each and every photon however, measure and note down where it landed on the single photon detector screen right at the very back. Assume $\Delta \frac{N}{M} \gg 1$ but $\Delta \sqrt{N} \ll 1$. Then, the strip pointers will demonstrate an interference pattern.

The probability that none of the $N$ photons get disturbed — i.e. scattered/decohered — by the weak measurements goes as $1-\mathcal{O}\left(\Delta^2 N \right)$. In that case, the detector screen will only detect photons in two spots — the image of the double slits — and we can read off the which-way information of each and every photon! Even in case there is a measurement disturbance, the conditional probability that more than one out of the $N$ photons are scattered is only $\mathcal{O}\left(\Delta^2 N \right)$. We'd still have the correct which-way info for each of the unscattered photons. Also, the scattered photon would land uniformly across the detector screen, and most likely, it would land outside the two main spots and can be easily filtered out. Besides, by adjusting the values of $\Delta$ and $N$, the probability $\mathcal{O}\left( \Delta^2 N \right)$ can be made arbitrarily small.

Admittedly, the observed interference pattern is a collective property of many photons. That's the nature of cumulative weak measurements. Has wave-particle complementarity been circumvented? How does the Copenhagen interpretation handle this experiment? What would Niels Bohr say?

Note: This is nothing at all like Afshar's experiment. Afshar's experiment has wire strips performing strong measurements, and as such, the wires act as a diffraction grating with probability 1, washing out any which-way info.

  • $\begingroup$ "Cover the lens with 2M weak measurement photon detector strips which always allow photons to pass through preserving their phases." I'm not sure what you mean by that but I suspect what you're describing is impossible and the root of why you're coming up with weird hypothetical results. $\endgroup$ – DanielSank Jan 17 '15 at 9:04

No, complementarity hasn't been bypassed.

First, a superficial analysis which apparently suggests it's bypassed.

By adjusting the values of $\Delta$ and $N$, the probability $\mathcal{O}\left( \Delta^2 N \right)$ can be made arbitrarily small. In this limit, the consistency conditions of the consistent histories formalism will be satisfied, when the only projectors are the which-slit info for each of the $N$ photons and the landing spot of each of the $N$ photons on the final screen. Not only that, the consistent histories probability for even a single mismatch between which-slit and which-landing goes to zero in this limit. This defines a realm.

Another realm is gotten when the only projectors are the landing spots of each of the $N$ photons on the final screen, and the final (but not intermediate or initial) values of the $2M$ strip pointers. Once again, the consistency conditions are satisfied.

However, including both the which-slit projectors and the final pointer values for each of the strips violates the consistency conditions. So, the previous two realms are incompatible.


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