What destroys interference: Possibility or actual measurment?

[greypilgrim]

The measurement of the photons occur when they hit the CCD and are irreversibly registered (and threreby absorbed and thus destroyed).

But in the usual delayed choice quantum eraser setups, isn't the claim that the choice can be made after the signal photon hits the detector?

 

[vanhees71]

It's not a claim, it's done, and the delayed choice is done using the fixed measurement protocol at any time after the measurement in the above sense is done! You just choose an appropriate subensemble of the full ensemble of registered photons. Please look at the very nicely written orginal paper:

https://arxiv.org/abs/quant-ph/0106078

 

[DrChinese]

It's not a claim, it's done, and the delayed choice is done using the fixed measurement protocol at any time after the measurement in the above sense is done! You just choose an appropriate subensemble of the full ensemble of registered photons. Please look at the very nicely written orginal paper:

https://arxiv.org/abs/quant-ph/0106078

That is not what is meant by "delayed choice" in this situation. From the paper (and relating to greypilgrim's question):

Delayed erasure

The possibility of obtaining which-path information after the detection of photon s leads to delayed choice [34]. Delayed choice creates situations in which it is important to have a clear notion of the physical significance of quantum mechanics. A good discussion can be found in references [11–14]. In as much as our quantum eraser does not allow the experimenter to choose to observe which-path information or an interference pattern after the detection of photon s , it does allow for the detection of photon s before photon p, a situation to which we refer to as delayed erasure. The question is: “Does the order of detection of the two photons affect the experimental results?” ... We use the term “delayed choice” loosely, in that in our experiment there is no “choice” available to the observer in the time period after the detection of photons s and before the detection of photon p.

As before, you can simply define "measurement" to mean "when both s and p have been detected" and then your view is fully descriptive (and nothing is considered as "delayed"). Or you can split the measurement into 2 parts, one for s and a separate one for p. Then the term "delayed" that the authors use is relevant, and they test that. Obviously, the order of detection of s and p does not change the results.

 

[vanhees71]

Ok, I thought the paper is clear enough, but obviously we have to discuss the details. One starts with a maximally polarization-entangled two-photon state from a appropriate parametric-down-conversion apparatus:
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|\hat{x} \rangle_s \otimes |\hat{y} \rangle_p-|\hat{y} \rangle_s \otimes |\hat{x} \rangle_p),$$
where $|\hat{x} \rangle_s$ means a single photon that is polarized in $\hat{x}$ direction. The index $s$ means the "signal photon", i.e., that it is the photon that runs through the double slit (with quarter-wave plates) and is finally detected on the screen, and $p$ means the "polarizer photon", which is send to a far-distant polarization analyzer.

The state of the photons immediately after the double slit is given by applying the operator describing the quarter-wave plates which are in $\pm 45^{\circ}$ direction in the $xy$ plane, which are
$$\hat{Q}_{+}=|L \rangle \langle \hat{x}|+\mathrm{i} |R \rangle \langle \hat{y},\\
\hat{Q}_{-}=|R \rangle \langle \hat{x}|-\mathrm{i} |L \rangle \langle \hat{y},$$
where $|R/L \rangle$ are the right-/left-circular polarization states (helicity eigenstates).

So when the $s$ photon would be detected right at the double slits 1 or 2, you'd have the two-photon state (we assume that there's no strange interaction of the quarter-wave plates with the $p$ photon)
$$|\Psi_1 \rangle=\hat{Q}_+ \otimes \hat{1} |\Psi \rangle, \quad |\Psi_2 \rangle=\hat{Q}_- \otimes \hat{1} |\Psi \rangle.$$
Now when the $s$ photon is observed at the far-distant screen, the state is
$$|\Psi' \rangle = \frac{1}{\sqrt{2}} (|\Psi_1 \rangle + \exp(\mathrm{i} \varphi(\vec{x}))|\Psi_2 \rangle),$$
where
$$\varphi(\vec{x})=\frac{2 \pi x d}{L \lambda},$$
with the geometry given at the figure on

http://theory.gsi.de/~vanhees/faq/qradierer/node3.html

Now if you measure all photons without taking care about anything concerning the $p$ photons, you get
$$\langle \Psi' | \Psi' \rangle=1,$$
because $\langle \Psi_1|\Psi_2 \rangle=0$. Indeed there's no interference pattern, because now you can determine uniquely through which slit each $s$ photon has gone by checking, whether they are in polarization state 1 or 2, which are true alternatives since the corresponding vectors are orthogonal (i.e., you loose the interference pattern completely).

Now to erase the which-way information to some degree or completely, you measure the polarization state of the $p$ photon with the polarizer set with an angle $\alpha$ wrt. the $x$ direction. Then you filter out only such photons for that the $p$ photon has gone through the polarizer. This is described by the projection operator $\hat{1} \otimes \hat{P}(\alpha)$, where
$$\hat{P}(\alpha)=|\hat \alpha \rangle \langle \hat{\alpha}|, \quad |\hat{\alpha} \rangle = \cos \alpha |\hat{x} \rangle + \sin \alpha |\hat{y} \rangle.$$
Applying this projector with $\alpha = \pm 45^{\circ}$ to $|\Psi' \rangle$ you get
$$|\Psi_{+45}'' \rangle=\frac{1-\mathrm{i} \exp(\mathrm{i} \varphi(x))}{2 \sqrt{2}} (|L \rangle_s \otimes |\hat{45} \rangle_p+\mathrm{i} |R \rangle_s \otimes |\hat{45} \rangle_p),$$
and thus for the corresponding partial ensemble the interference pattern
$$\langle \Psi_{+45}'' |\Psi_{45}'' \rangle=\frac{1}{2} [1+\sin(\varphi(x))].$$
The corresponding complementary part of the ensemble is given by considering only those $s$ photons whose entangle $p$ photon is not absorbed, i.e., which was polarized in direction $-45^{\circ}$. The corresponding application of the projector finally gives indeed the complementary interference pattern
$$\langle \Psi_{-45}'' |\Psi_{-45}'' \rangle=\frac{1}{2} [1-\sin(\varphi(x))].$$
As it must be both partial ensembles together give $1$ as it must be, since nothing has been done to the $s$ photons by measuring the polarization state of the $p$ photons.

If you choose any other angle $\alpha$ for the $s$ photons you get interference patterns with more or less contrast, and the complementary pattern for $\alpha'=\alpha \pm 90^{\circ}$. That's the demonstration of a delayed erasure of which-way information without ever interfering with the $s$ photon. That's made possible by the entanglement of the $s$ with the $p$ photon from the very beginning of their preparation and a specific quantum feature, i.e., here you need to really deal with the single biphotons and you have to keep the time of the measurement protocols at the screen and the polarizer (corresponding to the usual observers Alice and Bob :-)) accurately enough to know which $s$ photon was entangled with which $p$ photon.