Kim's delayed choice quantum eraser -- Is it really delayed?

[Derek Potter]

I am trying to work through the maths in the paper to understand how it predicts what was observed. I am finding it difficult because it uses different terminology from that with which I am familiar from papers like Bell's and those on some other double slit experiments (Aspect, Walborn).

In particular, on page 2, 2nd column, I am not familiar with the 'Glauber formula' on which it relies and I don't know what it means by "positive and negative-frequency components" (how can a frequency be negative?). Nor do I know the meaning of the 'creation operators' for photons, denoted by $a_i^\dagger$. The notation looks like that of raising and lowering operators used in solving harmonic oscillator problems, but I don't think that's what they mean.

Does anybody know of a less cryptic presentation of the mathematics, or can explain the above terms?

Thank you.

I can't follow the maths either, not being familiar with "standard quantum mechanical calculations" of any kind :) nor even creation operators, let alone the Glauber formula.

However I'll hazard a guess that the frequencies are spatial frequencies, ditto the phase shift of π, especially since the authors proceed to calculate the diffraction pattern using sinc functions: the far-field diffraction pattern is the Fourier transform of the diffractor aperture and the FT of a rectangular function is a sine(x)/x or "sinc" function). Negative frequencies are obvious for spatial sinusoids: you just reverse the sign of x e.g. sine(-x). This also makes sense since a standard FT produces components of both signs.

Yes, when the variable is time, a literal negative frequency is pretty meaningless, but you just move the sign around inside the term:
sin(ω.-t) = sin(-ωt) = -sin(ωt) i.e. sign reversal
cos(ω.-t) = cos(-ωt) = cos(ωt) i.e. no sign reversal

Notwithstanding my ignorance of second quantization, I'm pretty sure the dagger is a creation operator. I guess Kim et all want to introduce the random SPDC events properly and bundle them and the optics and everything else (except Schrodinger's cat for some reason) into one expression. It might be fun to see whether the creation simply propagates through as a "joint (potential) hit", meaning we could do as I do and ignore it, just assuming that photon pairs are created at random.

Unfortunately this kitchen sink approach obscures (to simple minds like mine) what is going on optically. Kim et al glibly take the FT of the two slits (which they do have the grace to put in quotation marks) but no-where do they show why this is appropriate. The photons are created one pair at a time in a probability distribution: either at A or at B as far as I can tell. They are not created behind a twin slit and there is no interference at D0 where the patterns are found.

I rather like the sound of that so I'll say it again. There is no interference at D0 where the patterns are found.

Instead, an interference pattern is created using a technique of Ghost Interference. I think the optical trick is to be seen by looking back at A and B from D1 and D2. As A and B only "flash" separately, there is no interference at D0 where the patterns are found. However, D1 and D2 see the two regions optically superimposed by the "half-silvered mirror" BSC. Hence the interference is A interfering with A, or B interfering with B, never A interfering with B. It occurs at D1 and D2 - not at D0 where the patterns are found*.

The question then is how the width of A, the pumped region, translates to a slit width. I have to assume that with SPDC, the photons are emitted across the entire excited region, all of it but no more - i.e. "as if" they were passing through a slit. That is a serious bit of physics to look into. Pity it gets glossed over.

Hence my concern as to whether the "slit" width is actually an artifact of the path aperture.

* Allegedly. In fact, spatial data is recorded there but the interference patten is encrypted by the D1-D4 data.

Does that help?

 

[Cthugha]

However I'll hazard a guess that the frequencies are spatial frequencies, ditto the phase shift of π, especially since the authors proceed to calculate the diffraction pattern using sinc functions: the far-field diffraction pattern is the Fourier transform of the diffractor aperture and the FT of a rectangular function is a sine(x)/x or "sinc" function). Negative frequencies are obvious for spatial sinusoids: you just reverse the sign of x e.g. sine(-x). This also makes sense since a standard FT produces components of both signs.

No,they are really talking about time. Take a simple cosine function and decompose it into complex exponentials:
\cos(\omega t)=\frac{1}{2}(e^{i \omega t}+ e^{-i \omega t})[\itex]<br /> There you already have a positive and a negative frequency component. One simply finds that negative components are associated with photon annihilation operators and positive components are associated with photon creation operators. However, I am not sure doing all this math really helps for the problem at hand if one is not familiar with the math. The experiment is really not as complicated as it seems.<br /> <br /> <blockquote data-attributes="member: 518005" data-quote="Derek Potter" data-source="post: 5178668" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Derek Potter said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The photons are created one pair at a time in a probability distribution: either at A or at B as far as I can tell. They are not created behind a twin slit and there is no interference at D0 where the patterns are found. </div> </div> </blockquote><br /> One very important point: This is exactly NOT what is happening. If they are created et either slit, the experiment will not work. Only if it is completely indistinguishable from which slit the photons originated, you will be able to see interference.<br /> <br /> <blockquote data-attributes="member: 518005" data-quote="Derek Potter" data-source="post: 5178668" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Derek Potter said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I rather like the sound of that so I&#039;ll say it again. There is no interference at D0 where the patterns are found. </div> </div> </blockquote><br /> Of course not. And you are right, this is a very impoirtant point. Light created by SPDC is about as incoherent as it gets. It is more or less like using unfiltered sunlight, which will also not show interference in a usual size double slit experiment. Many people are not aware that in a common double slit experiment you can easily switch between seeing interference and not seeing interference by just placing the light source closer to the double slit and that you can change the positions of the maxima and minima of the interference pattern by moving the light source parallel to the double slit. In my opinion understanding the &quot;simple&quot; double slit is the most important prerequisite for understanding the DCQE.<br /> <br /> <blockquote data-attributes="member: 518005" data-quote="Derek Potter" data-source="post: 5178668" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Derek Potter said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Instead, an interference pattern is created using a technique of Ghost Interference. I <i>think</i> the optical trick is to be seen by looking back at A and B from D1 and D2. As A and B only &quot;flash&quot; separately, there is no interference at D0 where the patterns are found. However, D1 and D2 see the two regions optically superimposed by the &quot;half-silvered mirror&quot; BSC. Hence the interference is A interfering with A, or B interfering with B, never A interfering with B. It occurs at D1 and D2 - not at D0 where the patterns are found*. </div> </div> </blockquote><br /> If that was the case,one would get the same result by aiming at one slit 50% of the time and aiming at the other also 50% of the time. That does not work.<br /> <br /> <blockquote data-attributes="member: 518005" data-quote="Derek Potter" data-source="post: 5178668" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Derek Potter said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> The question then is how the width of A, the pumped region, translates to a slit width. I have to assume that with SPDC, the photons are emitted across the entire excited region, all of it but no more - i.e. &quot;as if&quot; they were passing through a slit. That is a serious bit of physics to look into. Pity it gets glossed over. </div> </div> </blockquote><br /> Usually the crystal size more or less directly translates into a slit width. But yes, this is a nontrivial detail.

 

[Derek Potter]

No,they are really talking about time. Take a simple cosine function and decompose it into complex exponentials:
\cos(\omega t)=\frac{1}{2}(e^{i \omega t}+ e^{-i \omega t})[\itex]<br /> There you already have a positive and a negative frequency component. One simply finds that negative components are associated with photon annihilation operators and positive components are associated with photon creation operators. However, I am not sure doing all this math really helps for the problem at hand if one is not familiar with the math. The experiment is really not as complicated as it seems.

<br /> I don&#039;t see any negative frequencies there, I see imaginary ones :)<br /> But I&#039;m not familiar with the math (<i>of 2Q I mean</i>) so if you are absolutely sure that the analysis is about creation and annihilation operators then I will duck out and leave Andrew to unravel it.<br /> <blockquote data-attributes="member: 77153" data-quote="Cthugha" data-source="post: 5178715" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Cthugha said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> One very important point: This is exactly NOT what is happening. If they are created et either slit, the experiment will not work. </div> </div> </blockquote>According to the caption of fig 2 &quot;A pair of signal-idler photons is then generated from either the <i>A </i>or the <i>B </i>region.&quot; Why would the authors say this if it not true?<br /> If you insist that each photon is created across both psuedo-slits simultaneously then the big question is whether the part at A is created coherently with that at B or whether there is a random phase difference. If they were coherent then there would be interference at D0, regardless of coincidences. This does not happen. If there were a random phase difference then there could be no interference between the parts coming from A and from B, not even of the &quot;coincidence&quot; kind. So the premise must be false: the photon must be created in one or other of the psuedo-slits, just as Kim et al say.<br /> An important point is that D1 and D2 must be able to &quot;see&quot; the doubled (both paths) A and the doubled B. But if that were not the case then there would be no interference however you cut it.<br /> <blockquote data-attributes="member: 77153" data-quote="Cthugha" data-source="post: 5178715" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Cthugha said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Only if it is completely indistinguishable from which slit the photons originated, you will be able to see interference. </div> </div> </blockquote>I do not agree. If you have just one slit and divide the ray path into two before recombining them, so that it <i>looks like</i> a double slit from D1 or D2&#039;s viewpoint, then you will get interference. This is exactly what is done here.<br /> <blockquote data-attributes="member: 77153" data-quote="Cthugha" data-source="post: 5178715" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Cthugha said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> If that was the case, one would get the same result by aiming at one slit 50% of the time and aiming at the other also 50% of the time. That does not work. </div> </div> </blockquote>Why not? There would be no interference pattern at D0 - same result. There would be no interference at D0/D3 or D0/D4 - same result. You must mean there would be no interference at D0/D1 or D0/D2 which I am pretty sure is wrong for the reasons given.

 

[Cthugha]

I don't see any negative frequencies there, I see imaginary ones :)
But I'm not familiar with the math so if you are absolutely sure that the analysis is about creation and annihilation operators then I will duck out and leave Andrew to unravel it.

Yes. I am sure. I should retract some papers if the math was different. ;)
You can read up on that in pretty much every standard textbook on quantum optics in the chapter where quantization of the em field is introduced. Look it up in the Mandel/Wolf, Scully/Zubairy or in Schleich's book. It is really not as complicated as it sounds, though it is intimidating. :)

According to the caption of fig 2 "A pair of signal-idler photons is then generated from either the A or the B region." Why would the authors say this if it not true?

I do not see your point. Of course the caption is correct. The pair is created at the A or B region. No other regions are possible and the photons are emitted as a pair. Note that they do not say that it is necessarily defined from which region A or B the pair actually came.

If you insist that each photon is created across both psuedo-slits simultaneously then the big question is whether the part at A is created coherently with that at B or whether there is a random phase difference.

Well, if we have which-way information, we can of course say from which slit the pair was emitted. Otherwise, I have already explained the difference. Are you aware that two photon interference is a different concept than single photon interference? So, again: The two-photon state is coherent. The single photon states are of course not. They cannot be as explained in the paper I linked. Have you read it?

If they were coherent then there would be interference at D0, regardless of coincidences. This does not happen. If there were a random phase difference then there could be no interference between the parts coming from A and from B, not even of the "coincidence" kind. So the premise must be false: the photon must be created in one or other of the psuedo-slits, just as Kim et al say.

This is getting cumbersome. I explained already several times that coherence of the two-photon state does NOT imply coherence of the single photon states which make up the two-photon state. You are insisting that the mechanism of coherence for both is the same, which is wrong. I do not know, how to make this easier to understand besides linking to papers which describe the details as I do not know your level of expertise and how much of the concept is clear to you.
To boil it down: The two-photon probability amplitudes from A and B are of course coherent. The single photon probability amplitudes are not coherent. They vary from shot to shot. However, this description might be pointless, if you are not even aware of these concepts. It would be really helpful to know what level of expertise you have. That is why I asked, whether you are aware that even in a standard double slit not every light source will show an interference pattern.

I do not agree. If you have just one slit and divide the ray path into two before recombining them, so that it looks like a double slit from D1 or D2's viewpoint, then you will get interference. This is exactly what is done here.

You mean like in a Mach-Zehnder interferometer? That is somewhat similar, but usually rather used for measuring temporal coherence and gets more complicated as only spatial coherence matters in the DCQE. But it is remotely similar, yes.

Why not? There would be no interference pattern at D0 - same result. There would be no interference at D0/D3 or D0/D4 - same result. You must mean there would be no interference at D0/D1 or D0/D2 which I am pretty sure is wrong for the reasons given.

Indeed there would be no interference in the coincidence counts. This can be seen in the double slit version of the quantum eraser as discussed by Walborn (http://arxiv.org/abs/1010.1236). One can also see that in the quantum versions of DCQE (http://arxiv.org/abs/1206.4348), but I doubt that these are easy to understand without doing some math. The whole paper of Kim et al. is very clear about no interference being present as soon as there is which-way information, which you simply cannot erase if just one slit is pumped.

 

[Derek Potter]

Thanks Cthugha. We are somewhat at cross-purposes but the reason seems to be that I did not understand the actual optical set-up as I thought I did.
So, apologies to anyone whom I may have confused as a result.

 

[Derek Potter]

So, will you please tell me if I have at last understood the source of the entangled pairs correctly? Pardon the quantum baby-talk.

Photon pairs are created at random times throughout the pumped regions of the BBO but the phase of the pair's probability amplitude is determined by the phase of the pump excitation which is, of course, the same over all the BBO. Thus all the "possible" emissions are in superposition and add up to a proper, spatially-coherent two slit source. The rest follows :)

 

[andrewkirk]

Cthuga I have now carefully worked through your explanation in the other thread. I found it very helpful and feel that I now almost understand what's going on. The idea that there is coherence between the A and B idlers in relation to the corresponding arrivals at D0 (two-photon coherence, which I think of as 'four-photon coherence'), but no coherence between A and B signal photons (one-photon coherence, which I think of as 'two-photon coherence') seems to be the crux of it.

I don't quite understand the relationship between photons and light waves though, in particular when photons are created - or if that question even makes sense. The interpretation that I find easiest to grasp is that photons are created only when observed, so the slits, BBO, mirrors and prisms just shape a wavefunction that generates probabilities of photon detection at D0-D5. So although we say the BBO 'creates' a photon pair we actually mean it transforms the wavefunction into one that gives different probabilities of observing photons, in two different directions - one heading towards D0 and one heading towards D1-D4 in the diagram. Those probabilities only turn into actual photons at the detectors D0-D5.

I can write out a crude, fudgy bra-ket description of this that sort of makes sense to me, and includes the entanglement, and that seems to embody the principles you lay out in the other thread.

But I'm stuck on how one can represent the incoherent light. My representation describes a wavefunction, after exit from the BBO, with form $\frac{1}{\sqrt{2}}\big( |\psi(p_1,\phi_1)\rangle |\psi(p_2,\phi_2)\rangle+|\psi(p_2,\phi_1) |\psi(p_1,\phi_2)\rangle \big)$ where $|\psi(p,\phi\rangle$ denotes a wavefunction for a particle with phase $\phi$ and mean momentum $p$. But that description seems to be only for one 'thing' (photon? moment in time?). I'm not sure how to think about, let alone represent, there being millions of such things happening, all with different values of $\phi_1,\phi_2$. I feel that it's probably a sum or integral of ket expressions of the above type, maybe over time, but I'm not sure how to bring the time dimension into it.

I don't know if this question makes any sense. I'll think about it some more and see if I can express it better.

 

[Cthugha]

So, will you please tell me if I have at last understood the source of the entangled pairs correctly? Pardon the quantum baby-talk.

Photon pairs are created at random times throughout the pumped regions of the BBO but the phase of the pair's probability amplitude is determined by the phase of the pump excitation which is, of course, the same over all the BBO. Thus all the "possible" emissions are in superposition and add up to a proper, spatially-coherent two slit source. The rest follows :)

The phase is not necessarily the same, but it has some fixed and well defined relationship, which is just as good. I am not exactly sure, what you mean by a proper spatially coherent two slit source. I assume that you mean the two-photon state. If so, I guess you are getting the idea. I admit that it is hard to express what is really going on without "talking math".

I don't quite understand the relationship between photons and light waves though, in particular when photons are created - or if that question even makes sense. The interpretation that I find easiest to grasp is that photons are created only when observed, so the slits, BBO, mirrors and prisms just shape a wavefunction that generates probabilities of photon detection at D0-D5. So although we say the BBO 'creates' a photon pair we actually mean it transforms the wavefunction into one that gives different probabilities of observing photons, in two different directions - one heading towards D0 and one heading towards D1-D4 in the diagram. Those probabilities only turn into actual photons at the detectors D0-D5.

As a somewhat tongue-in-cheek response, let me present a quote I like:
"My complete answer to the late 19th century question "what is electrodynamics trying to tell us?" would simply be this: Fields in empty space have physical reality; the medium that supports them does not.
Having thus removed the mystery from electrodynamics, let me immediately do the same for quantum mechanics: Correlations have physical reality; that which they correlate, does not."
N. David Mermin, in "What is Quantum Mechanics Trying to Tell Us?"

As a somewhat more serious remark: One of the problems is that there are no wavefunctions in the common sense for photons. There are for example no position eigenstates for massless particles and when looking at it in more detail one finds that the common wavefunction description does not work. What we are left with are probability amplitudes for processes connecting some initial state and some final state. Indeed we do not know anything about what the photon actually does in between. There is some quantized excitation of the light field, which is the photon. However, it is a particle in the sense of qm, which should never be confused with the classical notion of a localized "ball-like" thing. The exact time of emission is as well defined as the coherence time of the light field. We do not know much more.

I can write out a crude, fudgy bra-ket description of this that sort of makes sense to me, and includes the entanglement, and that seems to embody the principles you lay out in the other thread.

But I'm stuck on how one can represent the incoherent light. My representation describes a wavefunction, after exit from the BBO, with form $\frac{1}{\sqrt{2}}\big( |\psi(p_1,\phi_1)\rangle |\psi(p_2,\phi_2)\rangle+|\psi(p_2,\phi_1) |\psi(p_1,\phi_2)\rangle \big)$ where $|\psi(p,\phi\rangle$ denotes a wavefunction for a particle with phase $\phi$ and mean momentum $p$. But that description seems to be only for one 'thing' (photon? moment in time?). I'm not sure how to think about, let alone represent, there being millions of such things happening, all with different values of $\phi_1,\phi_2$. I feel that it's probably a sum or integral of ket expressions of the above type, maybe over time, but I'm not sure how to bring the time dimension into it.

I don't know if this question makes any sense. I'll think about it some more and see if I can express it better.

Getting the math straight is not trivial. Maybe have a look at the Review paper by Walborn et al. I cited in post #34. Section 6.1. deals with conditional interference fringes and complementarity. This is not exactly the same setup as used by Kim, but there are similarities. Also chapter 4 might be helpful.

 

[Derek Potter]

I assume that you mean the two-photon state. If so, I guess you are getting the idea.

The two-photon state was never an issue for me. The issue was how the A and B regions could cooperate to create an output that would show (two photon) interference. The common-cause explanation (the laser) had eluded me.