Tuesday, March 13, 2018 ... /////

What erases and can restore the interference patterns

LHC anomaly: look at this fresh paper on Higgs decays, page 26 (28 of 35), top. The graph shows $\sigma_{VH}/\sigma_{SM}$ to be $12.88\pm 5$ instead of the expected $1$. A dramatic difference but it still translates to a 2.5-sigma deviation.
Under the latest discussion about the delayed choice quantum eraser experiment, the basic questions kept on coming. Can you replace the beam splitters by humans who press switches? Will the interference pattern reappear? Does the consciousness of these humans matter? And so on.

OK, let me discuss the experiment and its pieces again. On the picture above, you see the laser beam on the left upper side. The parent photon enters a double slit – there are two slits, the red slit (in San Francisco) and the light blue slit (in Los Angeles). If that parent photon just continued, it could contribute to an interference pattern on a photographic plate. Here I assume that you know the basic double slit experiment.

However, we want to make the experiment more complicated and combine it with an entanglement experiment. So the parent photon actually undergoes a process with a complicated name in the BBO crystal. That splits the parent photon to two photons of lower energy.

That splitting is applied to every parent photon. The upper daughter continues through some (yellow) lens towards the detector D0 where the coordinate $x$ is measured – just like in the simple double slit experiment. Some values of $x$ should be more likely (interference maxima), others should be very unlikely or prohibited (interference minima).

The lower daughter photon continues through a (white triangular) prism PS to a system of (light green) beam splitters (BS) and (light brown or grey) mirrors (M) to one of the four detectors, D1 or D2 or D3 or D4. In one of those detectors, the lower daughter photon is ultimately detected, and that's how we measure the discrete variable $y=1,2,3,4$ with the obvious meanings of the values.

Everything that can be predicted are the probabilities of various values of $x$, the position of the upper daughter photon on the photographic plate, and $y$, the integer-valued label of the detector where the lower daughter photon is detected. I think it's useful to start with the destination where we're going. So $x$ and $y$ aren't independent, they're correlated, and the probability distribution $\rho(x,y)$ for all possible values of $x$ and $y$ – which is a density relatively to $x$ because $x$ is continuous – is given by these four graphs which are equivalent to the graph of a function of two variables:

So when many parent photons are sent to this apparatus and split to daughter photons that are detected, we get random results but at the end, with many parent photons in the same initial state, we may measure the probabilistic distribution above. As I discussed before, there's no way in which the lower daughter photon "commands" the upper one or vice versa. They play an equally good role and the correlation between their properties is a qualitatively symmetric phenomenon. The reason of this correlation, i.e. of the fact that we can't write the equality$\rho(x,y) = \rho(x) \cdot P_y:\quad \text{No, it's }\neq$ is already the joint birth of the two daughter photons. The cause is not any communication at the later time of the measurements or something like that. The correlation – as described by the entanglement – was an inbred characteristic of the two photons.

OK, so why are the predictions what they are? At the beginning, right after it gets through the double slit, the wave function of the parent photon may be written as${\ket\psi}_1 = \frac{ \ket{SF}+\ket{LA} }{\sqrt{2}}$ So the parent photon has the same probability to be in either slit and the relative phase is known – in our conventions, it's naturally $+1$. This relative phase would determine the places of the interference maxima for $x$ if this parent photon were directly sent to measure its interference pattern. With our relative phase of $+1$, the interference maxima could be at $x\in\ZZ$, integer values of $x$. If we managed to change the relative phase, the set of interference maxima could be $x\in \ZZ+\phi/2\pi$.

Fine. But the BBO crystal splits the parent photon to two daughters. This process is done locally. So when the parent photon is in San Francisco (the red color – it's communist red, not Republican red), we get two daughter photons in San Francisco. When it's in Los Angeles (light blue), we get two daughter photons in Los Angeles. So after the parent photon gets through the BBO crystal, the state changes to${\ket\psi}_2 = \frac{ \ket{SF}\ket{SF}+\ket{LA}\ket{LA} }{\sqrt{2}}.$ By locality, there are no mixed terms such as $\ket{LA}\ket{SF}$ or vice versa. The crystal has either split the parent photon in San Francisco or in Los Angeles, it couldn't have created daughter photons at two faraway cities (slits) simultaneously, OK?

In the tensor products such as $\ket{SF}\ket{SF}=\ket{SF}\otimes \ket{SF}$, the first factor (written on the left side) corresponds to the photon that continues up towards D0, the second factor (written on the right side) corresponds to the photon that goes to the detectors D1,D2,D3,D4 through the triangular prism, beam splitters, and mirrors.

Now, we must always respect some phase convention for all the ket vectors but I have used the most natural convention – and kept the relative phase of the two terms as $+1$.

But a funny thing is that if you only study the "left factor", the photon that goes towards D0, the relative phase becomes sort of ambiguous. It's because the coefficients multiplying $\ket{SF}\otimes\dots$ and $\ket{LA}\otimes \dots$ include the factors of $\otimes \ket{SF}$ or $\otimes\ket{LA}$, respectively (those from the photon going to D1-D4), and these are "different kets" whose norm is equal to one, so one can't really say that they have the same phase.

For this reason, if you ignore the fate of the lower photon going towards the D1-D4 subapparatus, the interference pattern is guaranteed to disappear by now. You don't know whether the photon should selectively land near $x\in \ZZ$ or near $x\in \ZZ+1/2$ or any other collection of possible interference maxima because you don't really know the relative phase between the $\ket{SF}\otimes\dots$ term and the $\ket{LA}\otimes \dots$ term.

More systematically, this point may be made using the density matrices. If you lose your contact with the lower photon that goes towards D1-D4 (or if you decide not to care about the fate of that photon), and if you guarantee that the lower daughter photon won't ever influence the upper one (by getting in physical contact with it, directly or indirectly), all predictions for the upper photon going to D0 may be made using the density matrix for this upper photon. This density matrix may be obtained by tracing the full density matrix over the unobserved degrees of freedom – those of the lower photon traveling to D1-D4. This density matrix ends up being$\rho_{\rm upper} = \left( \begin{array}{cc} 1/2&\circ\\ \circ&1/2 \end{array} \right)$ where I wrote zero as $\circ$ for the nonzero entries to become more visible. The off-diagonal entries of this density matrix are zero exactly because the relative phase between the $\ket{SF}\otimes$ term and the $\ket{LA}\otimes \dots$ terms is unknown. Equivalently, the off-diagonal elements are zero because the states $\otimes \ket{SF}$ and $\otimes \ket{LA}$ of the lower photon (going to D1-D4) are orthogonal to each other.

(Note that while the diagonal elements of a density matrix are non-negative real numbers, the off-diagonal elements are general complex numbers and the phase matters. The off-diagonal elements remember whether or how much two amplitudes are coherent with one another – the absolute value – and what their relative phase is – the phase of the matrix element.)

You are supposed to learn how to compute the partial trace if you don't know it yet. It's a basic technical skill. OK, at any rate, the density matrix is proportional to the unit matrix which means that there is no knowledge about the relative phase. It follows that the upper photon – when it's always detected, without any "coincidence counter" that would care about the properties of the lower photon – will create a picture without any interference pattern.

That's true simply because the density matrix with $(1/2,1/2)$ on the diagonal is nothing else than the "classical statistical mixture" of the two slits, $\ket{SF}\otimes$ and $\ket{LA}\otimes$, which correspond to the columns (and rows) of this density matrix. So there's really a 50% probability that the upper photon is in the upper slit, and 50% that the upper photon is in the lower slit, and you just add the profiles for these two non-interference patterns that would be created by one open slit (either of them).

Again, I need to emphasize that by locality, whatever you do with the other photon – it can go through mirrors, prisms, beam splitters, magic sticks of Harry Potter, and it may be detected in any kind of a detector – cannot possibly affect the probabilities computed for all photons that continue towards D0 according to the old plan. The interference pattern for all D0 photons has been lost and by locality, no other photon in the Universe can possibly change the fact! I have mentioned Harry Potter but I am sure that 50 readers will suggest that if the lower photon is processed e.g. by Hillary Clinton instead of Harry Potter, it will affect the upper photon in D0. No, it won't. Not even Hillary Clinton can do anything. Neither can Barack Obama. Do I have to add 7.6 billion similar sentences? I am afraid that even that would fail to be enough for the stubborn, "creative" stupidity of some people.

But the upper daughter photon and the lower daughter photon are still entangled – i.e. their measurements will be correlated. So if you care about the full probabilities or probability densities like $\rho(x,y)$ of combined measurements of both daughter photons, interesting things may happen and the interference pattern may reappear in those.

Now, you should understand that because the relative phase for the upper photon in D0 has been made ambiguous, you absolutely need to consider properties of the lower photon – measure some of its properties – if you want to restore any interference pattern as a function of $x$, a coordinate of the upper photon measured at D0. It's pretty hard to restore the interference pattern at any rate. And what you basically need to do to restore the interference pattern is to "measure the relative phase" between the two slits, as remembered by the lower daughter photon. Let me remind you that${\ket\psi}_2 = \frac{ \ket{SF}\ket{SF}+\ket{LA}\ket{LA} }{\sqrt{2}}.$ and this expression is unchanged if the first $\ket{LA}\otimes$ factor is multiplied by $\exp(i\gamma)$ while the second $\otimes \ket{LA}$ factor is multiplied by $\exp(-i\gamma)$. So you may imagine that this ket vector has a fixed "total relative phase" between the two slits from both photons. You may imagine that the relative phase between the two slits for the upper photon is increased by $\gamma$ but the relative phase for the lower daughter photon is reduced by $\gamma$, and you still get exactly the same state. Yes, this also means that if you simply drew the interference pattern for the two daughter photons, the two-variable distribution $\rho(x_1,x_2)$ would have interference maxima for $x_1+x_2\in \ZZ$ or something like that.

So by learning something about the relative state between $\otimes \ket{SF}$ and $\otimes \ket{LA}$ for the second, lower photon, you also learn something about the relative phase for the upper photon, and that's how you can restore some interference pattern in $x$.

OK, so we must focus on the further adventures of the lower daughter photon. It goes through the triangular prism PS where the two possible beams get deflected to beam splitters BSa and BSb, respectively. A beam splitter (BS) is something that has the 50% chance to reflect the photon like a mirror, and the 50% chance to allow it to get through along a straight line, while keeping the relative phase between these two resulting beams known (say $+1$ with some phase conventions).

As you can see on the diagram, when either of these two beam splitters happens to act as a mirror on this lower daughter photon, the lower daughter photon immediately gets to the detectors D3 and D4, respectively. But when the lower daughter photon is found in D3 or D4, it means that it arose from the Los Angeles (light blue) and the San Francisco (red) slit, respectively! In effect, you have measured the "which slit" information of the lower daughter photon.

If an observer becomes aware of this result $y=3$ or $y=4$ – meaning that the lower daughter photon was caught in D3 or D4 – it means that he should collapse his wave function because the "already disproven" states can no longer affect further observations. So the previous state${\ket\psi}_2 = \frac{ \ket{SF}\ket{SF}+\ket{LA}\ket{LA} }{\sqrt{2}}$ collapses either to $\ket{SF}\ket{SF}$, or to $\ket{LA}\ket{LA}$, for $y=4$ and $y=3$, respectively. Sorry, the diagram chose D3 for the blue slit and I respect the convention. But when this "which slit" information for the lower daughter photon was measured, it means that we learned where the crystal created the upper daughter photon, too. So for $y=3$ and $y=4$, we know that the upper daughter photon also came from the blue slit and the red slit, respectively.

(I need to repeat that the collapse according to the measurement of $y$ is only relevant when the measurement of $x$ occurs first. When you first measure $x$, the wave function collapses according to the result of $x$, and gives predictions $P_1,P_2,P_3,P_4$ for the probabilities of the detectors D1,D2,D3,D4. The ordering plays no role for the overall predictions because physics predicts the whole $\rho(x,y)$.)

So what we predict for $x$ in the detector D0 are the two non-interference bumps that are located near the places you associate with the blue and red slits for $y=3$ and $y=4$, respectively. Perhaps, these two bumps should be slightly shifted to different places – the graphs of $\rho(x,y)$ that I have included above look the same for $y=3$ and $y=4$ but they shouldn't be quite the same. Both of them should be non-interference bumps, however.

Again, if you measure that $y=3$ or $y=4$, you have determined the which-slit information for the lower daughter photon, and therefore the upper daughter photon as well (because they were born as twins in one of the two cities). And when the which-slit information is actually known, it's obvious that there is no interference.

However, the probability is just 25% and 25% to get $y=3$ or $y=4$, respectively. With the probability 50%, the lower daughter photon gets through the first beam splitter (one of the two first beam splitters) instead of being reflected. And on this path, the other beam splitters and mirrors allow the sub-beams from the red and blue slits to re-interfere again.

In effect, if the lower daughter photon is detected at D1, it proves that its state was$\ket{D1} = \frac{\otimes \ket{SF} + \otimes \ket{LA}}{\sqrt{2}}$ where I include the sign $\otimes$ for tensor product to emphasize that we're talking about the lower daughter photon. Correspondingly, if the photon is detected at D2, it proves that the state of this lower daughter photon was$\ket{D2} = \frac{\otimes \ket{SF} - \otimes \ket{LA}}{\sqrt{2}},$ the analogous state with the relative minus sign. The last two displayed states are orthogonal to each other. But the logic is analogous to the logic above. Once we find out that the lower daughter photon is in one of these two states, we should collapse the two-photon entangled state to the corresponding states that have the form $\ket{\psi_1}\otimes \ket{D1}$ or $\ket{\psi_2}\otimes \ket{D2}$, respectively. With this collapse, you will find that the state $\ket{\psi_1}$ or $\ket{\psi_2}$ of the upper photon, after the detection of the lower photon at D1 or D2, will be$\ket{\psi_{1,2}} = \frac{\ket{SF}\otimes \pm \ket{LA}\otimes}{\sqrt{2}}$ where the plus and minus sign apply to the detection at D1 and D2, respectively. OK, that's why you get the usual interference pattern – or the complementary one – if you only collect the upper photons whose lower twin daughters were caught at D1 or D2, respectively. This basically concludes the derivation of the probability distributions for $\rho(x,y)$ that I started with.

So note that the interference pattern may always be there and is there as long as the information about the relative phase is preserved. The actual knowledge of the "which slit" information always kills the information about the relative phase. This relative phase is forgotten either because you are the actual observer who has already become aware of the which slit information and treats this measured value as a classical fact, and once there are classical facts, there is no interference between the amplitudes.

Or, more generally, the relative phase could have been made ambiguous even by the BBO crystal that splits the parent photon. If the which-slit information is "cloned" to additional degrees of freedom, in this case to the which-slit information about the lower daughter photon, then the relative phase for one of the subsystems (upper photon) also becomes undetermined. However, there's always a chance to "restore the knowledge" of that relative phase as long as the observer avoided perceiving the which-slit information himself.

In the standard delayed choice quantum eraser experiment, the relative phase was "restored" by measuring the lower daughter photon at D1 or D2, respectively. Just to be sure, no one could "order" the lower photon to be measured in a particular detector D1 or D2 or D3 or D4 – no one could even pick a subset of the detectors or add any bias. It was up to Mother Nature to choose which of the four results emerges, and the overall probability of each was 25%.

If we didn't construct the complicated "reinterference gadget" for the lower daughter photon that ends by the detection at D1 or D2, we couldn't learn anything about the relative phase for the upper photon, and the interference pattern would be lost. Also, if someone knew the which-slit information (which is the same for both photons), e.g. if you replaced the beam splitters by someone who splits the beams manually by switching a switch (which either does "reflect" or "transmit"), the which-slit information would be known, and the interference pattern for the upper daughter photon would be impossible to restore.

If you replaced those first beam splitter with another human (or a puppy) that presses the switch, his or its hypothetical awareness about the which-slit information wouldn't count as the observer's knowledge – because you haven't communicated with the human and puppy and only the true observer, you, is who plays a special role. (Whether other humans let alone puppies or molecules are conscious is a debatable question and you never need the answer to make quantum mechanical predictions! Quantum mechanical predictions are only supposed to be compared to your own perceptions and you know that you are conscious.) But it would still be true that the interference pattern would be lost in practice because that human or puppy would act just like another BBO crystal that has "cloned" some classical information, and therefore linked the relative phase to the relative phase in some additional degrees of freedom (in the human or puppy's brain).

But in principle, before some result is measured by the actual observer, namely you, there's always a potential for the options to "re-interfere" and for the interference pattern to be revived. All the evolution of the "observed systems" is unitary in quantum mechanics – and therefore reversible in principle. Only the actual observation is irreversible from your viewpoint!

So a sufficiently simple "puppy" that presses some switch could be dealt with much like the BBO crystal and the interference pattern, while lost, could be restored with a coincidence counter following a (much more complex) system of mirrors and beam splitters. A master observer would just carefully treat the photon and the puppy as two entangled subsystems. Again, all the coherence – the information about the relative phases – is in principle "restorable" before the actual observer makes an actual measurement.

Well, the actual observer could be the puppy as well. By considering himself to be the "actual observer", he or it uses the classical approximation for the measurements. If some re-interference takes place and leads to paradoxical conclusions (or the surprising interference pattern), those can be blamed on the inaccurate classical approximation – or, empirically speaking, on the imperfection of the puppy's consciousness. Needless to say, every observer's observations – which induce the collapse of the wave function – is just an approximation and the failure of this approximation could in principle be observed by another, more accurate observer who can carefully track the pure states of other human brains and the relative phases between the brains' states! In practice, there can exist no "truly perfect" observer in this sense. Observation is always a messy event in principle although, in practice, the inaccuracy of the classical approximation may be extremely, expo-exponentially tiny. If you saw an interference pattern proving the interference between two amplitudes in front of two states of your brain – and you believed that you have perceived one particular state among the two – you would consider this interference pattern to be a miracle.

Equivalently, the possibility that someone reconstructs the relative phases between some distinct states of a human brain by some "mirrors and beam splitters" is so tiny that we may ignore it – it belongs to the realm of science fiction. But again, in principle, every evolution of the observed part of the Universe may be reversed and every interference pattern may therefore be restored. More complex objects that are entangled with some quantum information make the reversal harder but for any finite system, the difficulty is finite.