Is the entanglement of photons necessary for interference?

[alexepascual]

I have read an article were it is claimed that if when two photons are entangled by parametric down-conversion, you don't observe interference in the first photon (after going through a double slit) if you don't collapse the second in such a way that which-way information is destroyed.
The reasoning is that if the second photon is still alive, in principle you can recover this which-way information.
This line of reasoning does not seem very clear to me.
What do you think?

 

[jambaugh]

Could you cite the article?

You should note that you do not observe interference in a single photon. Only when you do many repeated experiments will an interference pattern emerge.

This question discussed in some detail in the thread:
https://www.physicsforums.com/showthread.php?t=108015"

 

[jambaugh]

Let me add that the article is correct in its claim but I'm not sure about the reasoning as you state it. The short answer is that half of an entangled pair is not by itself in a sharp mode and so will behave "noisily" i.e. with non-zero entropy. Nothing you do to the second photon of the pair will change affect the outcome of the experiment on the first photon.

However you can select a subset of all the pairs based on measurements made on the second photon and if it is the correct type of measurement then the subset you get will demonstrate an interference pattern. Note that this is not a "spooky action at a distance" but rather a screening out of a portion of the original set of photons.

You could also for example "cause" the set of first photons in the pairs to form an image of a smiley face after passing through a double split by only selecting those pairs where the second photons land in a parallel smiley face region after going through an isomorphic double slit.

It is just a matter of masking out cases not effecting the path of the photons directly.
EDIT: (Notice the imporance then of my earlier point that you don't see interference of single photons but only for a large aggregate even though this aggregate is formed one photon at a time over many experiments.)

 

[DrChinese]

Could you cite the article?

I think he is referring to something like this. Zeilinger, page 290, figure 2, there is no direct interference pattern for entangled photons:

Experiment and the foundations of quantum physics (1999)

If there were, you know either beat the HUP or send signals FTL. So there is in fact a method which can detect an entangled particle. If only there was some way to use that information...

 

[Cthugha]

The reasoning is that if the second photon is still alive, in principle you can recover this which-way information.
This line of reasoning does not seem very clear to me.

This is a rather...media-compatible line of reasoning, maybe pretty good to get funding. ;)

A more direct line of reasoning is found in the so called Dopfer thesis (one of Zeilingers PhD students). Unfortunately as far as I know it is only available in German.

However, the argument is, that the needed prerequisites to see single photon interference and entanglement effects are complementary. The prerequisite to see single photon interference in a double slit is simple: the spatial coherence of light must be large enough so that the coherence length is larger than the slit distance. The spatial coherence is mostly determined by the spread of wavevectors present in the emitted light. A small spread corresponding to a small angular size of the light source will result in high spatial coherence.

To demonstrate entanglement effects you can use two-photon interference (like in DCQE experiments) or similar stuff. In two-photon interference experiments the situation is different: Entanglement relies on the sum of the wavevectors of the two emitted photons being well defined, while the wavevectors in a single arms have a large spread. To achieve good interference visibility in two-photon interference experiments you will need this large spread. If you use a filter in k-space and filter out some of the possible wavevectors and therefore also some entangled photon pairs, the two-photon interference pattern will become narrower and fringe visibility will go down.

Now the requirement for two-photon-interference, which is one measure of entanglement, is to have a large spread in wavevectors, while the requirement for single-photon interference is to have a small spread in wavevectors. Apparently this is a contradiction. This can be tested pretty easily. All you need to do is to vary the distance between your pdc crystal and the double slit. If you place it far away, you increase spatial coherence because the range of k-vectors, which reaches the slits is rather narrow and you will see a usual double slit interference pattern. However the range of k-vectors is now so small that entanglement does not mean much anymore and you will see no two-photon interference. If you decrease the distance the reversed situation occurs.

 

[alexepascual]

Could you cite the article?

You should note that you do not observe interference in a single photon. Only when you do many repeated experiments will an interference pattern emerge.

This question discussed in some detail in the thread:
https://www.physicsforums.com/showthread.php?t=108015"

Thanks for pointing me to that thread.
I'll think about what you said about not being able to detect interference with a single photon. It appears though that we should be able to detect the lack of interference using a single photon.

 

[alexepascual]

I think he is referring to something like this. Zeilinger, page 290, figure 2, there is no direct interference pattern for entangled photons:

Experiment and the foundations of quantum physics (1999)

If there were, you know either beat the HUP or send signals FTL. So there is in fact a method which can detect an entangled particle. If only there was some way to use that information...

Yes. That's the article I read.
Zeilinger has his interpretation of Dopfer's experiment. But I don't know if his view is generally accepted or not.

 

[alexepascual]

This is a rather...media-compatible line of reasoning, maybe pretty good to get funding. ;)
A more direct line of reasoning is found in the so called Dopfer thesis (one of Zeilingers PhD students). Unfortunately as far as I know it is only available in German.
To demonstrate entanglement effects you can use two-photon interference (like in DCQE experiments) or similar stuff..

Yes, I had read Zeiliger's paper. It appears that Dopfer interpreted her own experiment in a slighty different way.
I'll give some thought to what you say in your post and look into the DCQE experiments. Thanks for your help.

 

[alexepascual]

This is a rather...media-compatible line of reasoning, maybe pretty good to get funding. ;)
A more direct line of reasoning is found in the so called Dopfer thesis (one of Zeilingers PhD students). Unfortunately as far as I know it is only available in German.

Reading your post again, it looks that you imply that Zeilinger's view is the one that is media-compatible and Dopfer's view is more direct. As you say, Dopfer's thesis is in German, so I have only read articles about her view but nothing actually written by her.

Now, here is what Zeiliger says (pg 290) and I am wondering there is something objectable in it:

"Will we now observe an interference pattern for particle 1 behind its double slit? The answer has again to be negative because by simply placing detectors in the beams b and b' of particle 2 we can determine which path particle 1 took.
Formally speaking, the states |a>1 and |a'>1 again cannot be coherently superposed because they are entangled with the two orthogonal states |b>2 and |b'>2 .
Obviously, the interference pattern can be obtained if one applies a so-called quantum eraser which completely erases the path information carried by particle 2.
That is, one has to measure particle 2 in such a way that it is not possible, even in principle, to know from the measurement which path it took, b or b'."

Note that in the obove I corrected two mistakes in the original article. My version here seems to make more sense and I think this is what Zeilinger intended to say.
I changed "|b'>2 and |b'>2 ." to: "|b>2 and |b'>2"
I also changed ( a' or b' ) to ( b or b' )
If I made a mistake in correcting Zeilinger please let me know. On the other hand I think that this idea may be objectable.

 

[alexepascual]

Let me add that the article is correct in its claim but I'm not sure about the reasoning as you state it. The short answer is that half of an entangled pair is not by itself in a sharp mode and so will behave "noisily" i.e. with non-zero entropy. Nothing you do to the second photon of the pair will change affect the outcome of the experiment on the first photon.

However you can select a subset of all the pairs based on measurements made on the second photon and if it is the correct type of measurement then the subset you get will demonstrate an interference pattern. Note that this is not a "spooky action at a distance" but rather a screening out of a portion of the original set of photons.

You could also for example "cause" the set of first photons in the pairs to form an image of a smiley face after passing through a double split by only selecting those pairs where the second photons land in a parallel smiley face region after going through an isomorphic double slit.

It is just a matter of masking out cases not effecting the path of the photons directly.
EDIT: (Notice the imporance then of my earlier point that you don't see interference of single photons but only for a large aggregate even though this aggregate is formed one photon at a time over many experiments.)

I'll have to think about this. Thanks jambaugh.

 

[alexepascual]

Could you cite the article?

You should note that you do not observe interference in a single photon. Only when you do many repeated experiments will an interference pattern emerge.

This question discussed in some detail in the thread:
https://www.physicsforums.com/showthread.php?t=108015"

As Dr. Chinesse listed in his post, this is the article:
http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf"

With respect to the link you included to another thread in this forum, I wish I had seen it before starting this thread. That thread was abandoned in 2006, but what we are talking about here is the same thing. It would be nice if both threads could be merged.
Or perhaps we could abandon this thread revive that one?
What do you think?

 

[DrChinese]

As Dr. Chinesse listed in his post, this is the article:
http://www.hep.yorku.ca/menary/courses/phys2040/misc/foundations.pdf"

With respect to the link you included to another thread in this forum, I wish I had seen it before starting this thread. That thread was abandoned in 2006, but what we are talking about here is the same thing. It would be nice if both threads could be merged.
Or perhaps we could abandon this thread revive that one?
What do you think?

I say proceed from here...

By the way, there are experiments that have been done which demonstrate the effect. They show that there is NO interference for entangled photons in the "normal" case.

http://grad.physics.sunysb.edu/~amarch/Walborn.pdf

I think it is figure 3 for above.

 

[alexepascual]

I say proceed from here...
I think it is figure 3 for above.

What do you mean by "proceed from here?"
I'll take a second look at the Walborn experiment but I don't think it adds anything. As far as I remember, it adds another level of complexity by giving you fringes and anti-fringes.

 

[DrChinese]

What do you mean by "proceed from here?"
I'll take a second look at the Walborn experiment but I don't think it adds anything. As far as I remember, it adds another level of complexity by giving you fringes and anti-fringes.

I meant: ignore the old thread and discuss in this one.

I thought one of the figures had the effect of showing fringes and anti-fringes added together so that you are essentially seeing without coincidence counting (you are effectively looking at 100% in other words). I.e. you are not looking at a subsample. But on further review, I am not sure. Makes my head hurt looking at them all...

 

[alexepascual]

I meant: ignore the old thread and discuss in this one.

I thought one of the figures had the effect of showing fringes and anti-fringes added together so that you are essentially seeing without coincidence counting (you are effectively looking at 100% in other words). I.e. you are not looking at a subsample. But on further review, I am not sure. Makes my head hurt looking at them all...

Maybe you are now under hypnotic trance and I can brain-wash you!
It takes time to think about these things. I'll see if I have a chance tonight. I think I had it printed out.
In Dopfer's experiment, the coincidence counter is needed because a lot of photons going towards the double slit in the lower path get absorved by the plate that contains the slits so that many of the photons found in the upper path don't correspond to any of the photons detected in the lower path. But I have seen another proposed design of this experiment, where the upper path has a double-slit that corresponds exactly to the path that only the entangled photons take. So it screens-out the unentangled photons. In that case you don't need the coincidence counter.
Some people claim that the coinidence counter is the reason you see interference, but I can't see a reason for this. At least in principle. It may be that something in the setup was not considered carefully and the coincidence counter has this side-effect though.
But before getting into an analyisis of the influence of the coincidence counter, which I think could be ommited anyway, I would like to analyze the case where a photon at a time is sent (no counter) and the upper photon is not measured. (Let's send it on it's way to Andromeda).

 

[Cthugha]

Reading your post again, it looks that you imply that Zeilinger's view is the one that is media-compatible and Dopfer's view is more direct.

Well I think both statements are correct (and I am in no position to tell Zeilinger how he should present his results ;) ).

In Dopfer's experiment, the coincidence counter is needed because a lot of photons going towards the double slit in the lower path get absorved by the plate that contains the slits so that many of the photons found in the upper path don't correspond to any of the photons detected in the lower path. But I have seen another proposed design of this experiment, where the upper path has a double-slit that corresponds exactly to the path that only the entangled photons take. So it screens-out the unentangled photons. In that case you don't need the coincidence counter.
Some people claim that the coinidence counter is the reason you see interference, but I can't see a reason for this.

The coincidence counter is indeed the reason, why you see an interference pattern. Without it, there will be no pattern. In the Dopfer experiment for example you will never see an interference pattern in D1 alone because there is no double slit in that path. Also you will never see an interference pattern in D2 because it is a bucket detector and not position sensitive, so there is no possibility to get an interference pattern here. Of course you could exchange this bucket detector for a small detector like D1. Now, however, you are in a regime similar to the Walborn experiment, where you can have fringes and anti-fringes showing up in coincidence counting. In fact, you will then notice that the situation is symmetric: You will se an interference pattern if you move D1 and keep D2 fixed and you will see another pattern if you move D2 and keep D1 fixed.

The reason for that is not too difficult. To see any kind of interference pattern you need a fixed phase relationship. In the original Young double slit experiment this is achieved by putting a single slit in front of the double slit. The single slit acts as a point source. There is a fixed phase relationship between the fields at the two slits and accordingly also a fixed phase relationship between the two fields originating from the slits at each point of the detection screen.

Entangled photons do not have this property. The phase in a single arm changes extremely fast due to the PDC process. If such light passes a double slit, there will be no fixed phase relationship and accordingly no interference. But the two-photon state has a very well defined phase. This means that although the phase of a single field changes pretty randomly, the phase difference between the two fields resulting from the PDC process is very well defined. So if you put D1 in a position, where the detection of a photon gives you information about the momentum of that photon two things happen:

1) By doing coincidence counting out of all the photons arriving at D2 only those having the same small momentum range as measured at D1 are filtered out. So the photons at D2 are also in a momentum eigenstate and which way information can not be present. This assures that that the photon at D2 could have gone through both slits of the double slit, which is the first prerequisite for interference to occur.

2) Due to the fixed two-photon state phase relationship this also means that there is now a fixed phase relationship between the fields at both slits of the double slit and accordingly also of the two fields originating from the two slits and arriving at D2. Interference can occur.

If you did not use coincidence counting, you will measure the superposition of all interference patterns of all possible momentum eigenstates instead of just the pattern of a single eigenstate. As a consequence the superposition of all of these patterns cancels out leaving you with no interference pattern at all. Coincidence counting is not just a means of reducing erroneous counts occurring due to stray light or other error sources, but indeed elementary.

 

[alexepascual]

A more direct line of reasoning is found in the so called Dopfer thesis (one of Zeilingers PhD students). Unfortunately as far as I know it is only available in German.

Do you have a copy (electronic) of the paper?. I read that Zeilinger had it in his university website but it looks like it has been removed now. If you have it I would appreciate it if you attached it to this thread or sent it to me by email.

However, the argument is, that the needed prerequisites to see single photon interference and entanglement effects are complementary.

I didn't know this.

The prerequisite to see single photon interference in a double slit is simple: the spatial coherence of light must be large enough so that the coherence length is larger than the slit distance.

I guess you mean "distance between slits" right?
I haven't gone back to take a look at my optics book, but it looks to me that this is an approximation. With the screen at a reasonable distance from the slits, the coherence length should be able to be smaller than the distance between slits and be limited by the difference in path. (in order to get interference, of course)

The spatial coherence is mostly determined by the spread of wavevectors present in the emitted light.

I would guess you are talking about longitudinal coherence right?

A small spread corresponding to a small angular size of the light source will result in high spatial coherence.

I assume here you are referring exclusively about the SPDC process. The angle of emission for each photon is related to the magnitude of it's momentum. Therefore, to small difference in angle corresponds small spread in momentum. Did I interpret you correctly?

To demonstrate entanglement effects you can use two-photon interference (like in DCQE experiments) or similar stuff.

Looking at Scully's paper, I can see (as you say) that this is different from Dopfer's experiment because you are mixing photons comming from different atoms.
I think this is a more complicated setup. I have printed out the paper and I will continue to think about it, but I would rather concentrate first on a Dopfer's kind of setup. I am not sure that an analyisis of Scully's setup will help in understanding Dopfer's experiment.

 

[alexepascual]

The coincidence counter is indeed the reason, why you see an interference pattern. Without it, there will be no pattern. In the Dopfer experiment for example you will never see an interference pattern in D1 alone because there is no double slit in that path.

According to Dopfer, you do see an interference pattern in D1 when you place the detector at the focal distance of the lens. You loose which-way information. On the other hand, if you place the detector at a distance of 2f, it is as if you were looking directly at the slits on the other arm. Then you get which-way information and interference is lost.
The reason you do see interference on D1 is that you are looking at the photons that are entangled with those that went through the slit on the other side. If you wanted, you could screen out all the photons in the upper arm that are entangled with photons in the lower arm that did not make it through the slits.

Also you will never see an interference pattern in D2 because it is a bucket detector and not position sensitive, so there is no possibility to get an interference pattern here. Of course you could exchange this bucket detector for a small detector like D1. Now, however, you are in a regime similar to the Walborn experiment, where you can have fringes and anti-fringes showing up in coincidence counting. In fact, you will then notice that the situation is symmetric: You will se an interference pattern if you move D1 and keep D2 fixed and you will see another pattern if you move D2 and keep D1 fixed

I'll try to skip Walborn's experiment for the moment. But I'll keep it in mind and come back to it if I feel it is necessary.

The reason for that is not too difficult. To see any kind of interference pattern you need a fixed phase relationship. In the original Young double slit experiment this is achieved by putting a single slit in front of the double slit. The single slit acts as a point source. There is a fixed phase relationship between the fields at the two slits and accordingly also a fixed phase relationship between the two fields originating from the slits at each point of the detection screen.

OK, I understand this.

Entangled photons do not have this property. The phase in a single arm changes extremely fast due to the PDC process. If such light passes a double slit, there will be no fixed phase relationship and accordingly no interference.

The frequency is proportional to the photon's energy. Now, if we consider both photons (entangled) as a single thing; should the frequency be that obtained by adding the energy of both photons?. I don't think so, unless the assumption that each photon's wavelength is twice that of the original photon is wrong. Now, if the frequency of each photon is what we would expect from its energy, why would the phase in a single arm change very fast?

I am not ignoring the rest of your post. But I would rather clarify the above issues before I go on to analyze the rest.

 

[Cthugha]

Do you have a copy (electronic) of the paper?. I read that Zeilinger had it in his university website but it looks like it has been removed now. If you have it I would appreciate it if you attached it to this thread or sent it to me by email.

Well, no. I always had a look at it at the Homepages of the universities of Vienna or Innsbruck. Maybe I have a copy somewhere on my office computer. I will check. Unfortunately I am attending a conference next week, so it might take some time.

I guess you mean "distance between slits" right?
I haven't gone back to take a look at my optics book, but it looks to me that this is an approximation. With the screen at a reasonable distance from the slits, the coherence length should be able to be smaller than the distance between slits and be limited by the difference in path. (in order to get interference, of course)

Yes, distance between slits. Also maybe my usage of coherence length was a bit unclear I suppose as this is sometimes also defined has coherence time times speed of light. So to be more exact, both slits must lie within the coherence volume of the light. Btw, this does not depend on the distance between slits and screen, but between the light source and the slits.

I would guess you are talking about longitudinal coherence right?

No, longitudinal/temporal coherence are defined by the spectral width of the light. Coherence time is the decay time of the autocorrelation. The autocorrelation is the Fourier transform of the spectral power density.

Spatial coherence (or transverse coherence) are defined by the spread of wavevectors. Collecting a small solid angle of a light source therefore leads to high spatial coherence. This is foe example, why starlight has rather high spatial coherence. This property was used to measure star diameters back in the 50s.

I assume here you are referring exclusively about the SPDC process. The angle of emission for each photon is related to the magnitude of it's momentum. Therefore, to small difference in angle corresponds small spread in momentum. Did I interpret you correctly?

Yes, but this is not limited to SPDC processes.

According to Dopfer, you do see an interference pattern in D1 when you place the detector at the focal distance of the lens. You loose which-way information. On the other hand, if you place the detector at a distance of 2f, it is as if you were looking directly at the slits on the other arm. Then you get which-way information and interference is lost.
The reason you do see interference on D1 is that you are looking at the photons that are entangled with those that went through the slit on the other side. If you wanted, you could screen out all the photons in the upper arm that are entangled with photons in the lower arm that did not make it through the slits.

Well, you do not see a pattern in D1, you see one in the coincidence counts.
Just screening out those photons, which do not make it through the slits, is not enough. The important, but often overlooked fact is, that D1 is a small detector. It is smaller than the beam is, so there are also photons, which do not hit D1. If you replace it by a larger detector, also the interference pattern in the coincidences will vanish. In the focal plane each possible position of the detector corresponds to a small spread of momentum eigenstates. This small spread shows interference in coincidence counts. The whole spread does not (due to low spatial coherence) and therefore there is also no interference in a single arm.

The frequency is proportional to the photon's energy. Now, if we consider both photons (entangled) as a single thing; should the frequency be that obtained by adding the energy of both photons?. I don't think so, unless the assumption that each photon's wavelength is twice that of the original photon is wrong. Now, if the frequency of each photon is what we would expect from its energy, why would the phase in a single arm change very fast?

I am not sure, why you bring up frequency here. However, indeed the wavelength is not necessarily exactly doubled. You get a rather broad peak of resulting wavelengths. The peak is of course indeed ad the doubled wavelength. The SPDC process can be considered as a process, which is stimulated by random vacuum fluctuations and is comparable to spontaneous emission in these terms. Therefore the phase changes just as fast as in any spontaneous emission process. In processes, where you have a well defined phase (lasing for example) most of the emission processes are stimulated emission processes. You have one well defined field and any stimulated emission process is caused by this field and "happens in phase with it". In spontaneous emission processes you have some field, but all subsequent emission processes are independent of each other because the "stimulation" happens by vacuum field fluctuations here. These happen at a random time with a random phase. As a result, the emitted light is pretty incoherent. Try for example to feed from a light bulb to a double slit and see interference. You will not see anything unless you use another single slit in front of the double slit to increase spatial coherence.

 

[alexepascual]

Well, no. I always had a look at it at the Homepages of the universities of Vienna or Innsbruck. Maybe I have a copy somewhere on my office computer. I will check. Unfortunately I am attending a conference next week, so it might take some time.

Thanks, I'll appreciate it

Yes, but this is not limited to SPDC processes.

Many of the things I said were a result of my misinterpreting the meaning of "spatial coherence" now I know it relates to the transverse component.

Well, you do not see a pattern in D1, you see one in the coincidence counts.
Just screening out those photons, which do not make it through the slits, is not enough. The important, but often overlooked fact is, that D1 is a small detector. It is smaller than the beam is, so there are also photons, which do not hit D1. If you replace it by a larger detector, also the interference pattern in the coincidences will vanish. In the focal plane each possible position of the detector corresponds to a small spread of momentum eigenstates. This small spread shows interference in coincidence counts. The whole spread does not (due to low spatial coherence) and therefore there is also no interference in a single arm.

Let's consider the following modification to the experiment:
(1) You screen out all the photons in the upper arm that don't correspond to those that made it through the slits in the lower arm. (this is just a geometrical thing).
(2) Instead of a small detector for D1, or a small slit that moves to different positions and scans the interference area, just use a very good resolution CCD of the right size.
(3) Consider the efficieny of the detectors 100% (at least for the thought experiment)

With these modifications, the number of photons at each detector is the same. To each photon that is detected in D1 there is a corresponding entangled photon detected in D2.
If you introduce the coincience counter now, it'll allow you to know exactly which photon is entangled with which on the other side. I don't see how you can use this information to filter out some data and obtain an interference pattern after the fact. Maybe I am missing something.

I am not sure, why you bring up frequency here. However, indeed the wavelength is not necessarily exactly doubled. You get a rather broad peak of resulting wavelengths. The peak is of course indeed ad the doubled wavelength. The SPDC process can be considered as a process, which is stimulated by random vacuum fluctuations and is comparable to spontaneous emission in these terms. Therefore the phase changes just as fast as in any spontaneous emission process. In processes, where you have a well defined phase (lasing for example) most of the emission processes are stimulated emission processes. You have one well defined field and any stimulated emission process is caused by this field and "happens in phase with it". In spontaneous emission processes you have some field, but all subsequent emission processes are independent of each other because the "stimulation" happens by vacuum field fluctuations here. These happen at a random time with a random phase. As a result, the emitted light is pretty incoherent. Try for example to feed from a light bulb to a double slit and see interference. You will not see anything unless you use another single slit in front of the double slit to increase spatial coherence.

My mention of energy was again due to my misinterpretation of what you had said. Still, if you consider a single photon emission, I don't see how you would get a phase difference between one slit and the other. Would the different waves that correspond to one photon be comming from different points on the crystal far appart enough to affect spatial coherence?

 

[Cthugha]

With these modifications, the number of photons at each detector is the same. To each photon that is detected in D1 there is a corresponding entangled photon detected in D2.
If you introduce the coincience counter now, it'll allow you to know exactly which photon is entangled with which on the other side. I don't see how you can use this information to filter out some data and obtain an interference pattern after the fact. Maybe I am missing something.

In fact, you will not be able to filter out an interference pattern. You will be able to filter out many. If you have a look at the coincidence counts between a small portion of the CCD and the other detector, you will see an interference pattern. If you look at the coincidence counts between another portion of the CCD and the other detector, you will see another interference pattern, but the position of the fringes and antifringes will be shifted with respect to the first interference pattern. All of these patterns sum up to no pattern at all, just a broad peak. Each position on the CCD will correspond to a different momentum eigenstate. This slightly different momentum also leads to this slight shift of the patterns on the other side.

My mention of energy was again due to my misinterpretation of what you had said. Still, if you consider a single photon emission, I don't see how you would get a phase difference between one slit and the other. Would the different waves that correspond to one photon be comming from different points on the crystal far appart enough to affect spatial coherence?

Well, if you are talking about single photons from SPDC processes, you do not know much about your single photon going to the double slit. It can have a very wide range of wavevectors (and can in principle come from a lot of different positions of the SPDC crystal). All of the probability amplitudes for these processes already interfere at the slits. As the emission processes are not coherent, this leads to a rapidly changing phase of the light field in this arm of the experiment.
However, having just a single well defined photon emission process would even make the situation worse. If you really had a deterministic single photon emission process at a given time (something like a ballistic photon), you would never see any interference in this arm because you would have which-way information (the photon has to hit the slit at a certain point) and the light field would be as incoherent as it gets. Coherence is a measure of how well you can predict the field at some time t, if you know its value now. In such a ballistic emission process you would just have a delta peak now and no information about the field at later times.

 

[alexepascual]

In fact, you will not be able to filter out an interference pattern. You will be able to filter out many. If you have a look at the coincidence counts between a small portion of the CCD and the other detector, you will see an interference pattern. If you look at the coincidence counts between another portion of the CCD and the other detector, you will see another interference pattern, but the position of the fringes and antifringes will be shifted with respect to the first interference pattern. All of these patterns sum up to no pattern at all, just a broad peak. Each position on the CCD will correspond to a different momentum eigenstate. This slightly different momentum also leads to this slight shift of the patterns on the other side.

I didn't know you could get anti-fringes in a Dopfer-type experiment. This is type-I SPDC, isn't it?. The Wallborn experiment on the other hand uses type-II. Am I right?
If I expect to see (or not see) an interference pattern in an extended CCD D1, I don't see a reason to look at a small portion of the CCD. I should take a note of all the hits on D1 and the (x,y) coordinates on the surface of the detector, and keep track of the coincidence information. Acording to your suggestion, all the hits on D1 would not from an interference pattern. Then somehow you use the coincidence count and suddenly the interference appears. In order to do this, you should select only a portion o the hits on D2, but what criterium would you use?.

Well, if you are talking about single photons from SPDC processes, you do not know much about your single photon going to the double slit. It can have a very wide range of wavevectors (and can in principle come from a lot of different positions of the SPDC crystal). All of the probability amplitudes for these processes already interfere at the slits. As the emission processes are not coherent, this leads to a rapidly changing phase of the light field in this arm of the experiment.

If you make the crystal small enough and set it at a reasonable distance from the slits, This shoud not be a problem. Otherwise the experiment is flawed from the start. Each photon will have a random phase with respect to other photons, but they are being produced one at a time, and each photon interferes with itself at the slits. The phase at each of the slits for a simgle photon should be considered roughly the same.

However, having just a single well defined photon emission process would even make the situation worse. If you really had a deterministic single photon emission process at a given time (something like a ballistic photon), you would never see any interference in this arm because you would have which-way information (the photon has to hit the slit at a certain point) and the light field would be as incoherent as it gets. Coherence is a measure of how well you can predict the field at some time t, if you know its value now. In such a ballistic emission process you would just have a delta peak now and no information about the field at later times.

When I talk about a single photon, I mean a single quantum. I am not saying that I see the photon as a little ball. The picture I have of the photon when in transit is still that of a superposition of waves of slightly different wavelengths around a central wavelength (a wave packet).

 

[alexepascual]

I think he is referring to something like this. Zeilinger, page 290, figure 2, there is no direct interference pattern for entangled photons:

Experiment and the foundations of quantum physics (1999)

If there were, you know either beat the HUP or send signals FTL. So there is in fact a method which can detect an entangled particle. If only there was some way to use that information...

Would you say you agree with Zeilinger's analysis fo Dopfer's experiment?

 

[DrChinese]

Would you say you agree with Zeilinger's analysis fo Dopfer's experiment?

Absolutely.

 

[alexepascual]

Absolutely.

DrChinese:
According to most descriptions of entanglement, once the particles are entangled, you can't talk about the state of a single particle but about the state of the composite system. That is, you consider a Hilbert space which is the vector product of the Hilbert spaces of the separate particles.
Now, let's assume I am trying to make a computer simulation of the photon in Dopfer's lower arm going throught the slits. I want to be able to assign a value to the wave function at each point in space (..and time, which makes it an animation). I am not really planning to do this because I don't know of a good (and cheap) tool to do it, but it would definitely be something interesting to do. What factors should I consider in deriving the expressions for this wave function?. Even if there are factors that may render this problem hard to solve, I would like to make reasonable simplifications for a first approximation. Maybe we could also ignore the fact that the waves comming from each slit do not have a sharp momentum due to the finite size of the slit, which gives you a small but finite range of angles for the line-of-sight between the source and a single slit. So, I would try to ignore subtleties as much as possible and concentrate on the most fundamental problems. The main problem here I guess would be that the wave function is not that of a single photon but that of both photons. I should nevertheless be able to assign a single complex number to the wavefunction at a single point for the lower photon because I should be able to use it to calculate the probability of finding the photon at each point in space. To analyze the behavior after the slits I am OK with leaving out the portion of the wave that strikes the surface of the plate that contains the slits. I am not asking you to give me detailed expressions, that would be too much. But maybe you can give me some ideas, such as how to derive the phase for the wave function. I guess in this case there is no interaction Hamiltonian, so there should be some Hamiltonian that depends on the energy of the particles right?. I could I guess consider each combination of positions for both photons and multiply their complex coefficients (functions of time in the Schrodinger picture) from the separate Hilbert spaces, but I am not sure this is valid. Again, I would appreciate your opinions about this, as it would help me understand better the experiment and the entangled state itself.
I would also be grateful for any ideas from other posters.
Oh! one more thing.. I am assuming the photons on the upper arm have not been detected and maybe will never be detected.

 

[jambaugh]

...The main problem here I guess would be that the wave function is not that of a single photon but that of both photons. I should nevertheless be able to assign a single complex number to the wavefunction at a single point for the lower photon because I should be able to use it to calculate the probability of finding the photon at each point in space...

No. You can at best assign only the probability value at that point.
What I think would be of interest is to plot the lateral component of each particle's wave-function as the product evolves.

The composite wave function for the entangled pair is a function over a six dimensional space (the three coordinates of each particle) evolving over time. What you're trying to do would be to average over half those dimensions (and ignore a couple say to keep things simple).

What I imagine being interesting is an animation showing the lateral component of the wave-function for both particles (one dimension of each x1 and x2 excluding y1,y2,z1, and z2). Since their position is anti-correlated you'd see the only non-zero component lies on the line L <--> x1+x2=0.

Initially you'd get a peak at the origin corresponding to the entangled pair's creation. Then it would evolve outward along the line L just like a single particle wave-function evolving from a localized point delta-function. Then as one of the particles approaches the double-slit you'd get a all but the components at the two slit positions projected out.

Remember that in most cases the particle approaching the slits never makes it through but rather strikes the barrier missing both slits. Since when we speak of the one particle passing through we are assuming it didn't miss we are excluding the cases for both particles where the one doesn't make it through. Since their positions are anti-correlated (entanglement) it is exactly the same as if both particles had been screened by corresponding double-slits.

This is why it seems like acting on one affects the other, this implicit rejection of cases counter to our assumption.

So back to our animation... as the one particle passes the double slits we impose the "passage assumption" and the composite wave-function is projected onto the two slits positions where it intersects the correlation line L. Now in spite of my earlier "smiley face" claim I'm not sure if the correlation in position is preserved exactly past this point. I'll have to think about it a bit (I just awoke from a nap and my head's a little foggy). I'll get back on that later. The behavior of this animation after the slit event is basically the main question being discussed here.

... I think not since that would imply you still get an interference pattern. Projecting an interference pattern wave-function along the line L onto the x1 axis would yield again an interference pattern corresponding to what particle 1 might do as it hits the final screen. The application of the composite Schrodinger equation should be straightforward. My first guess is you get what looks like a double pin-hole interference pattern in this x1,x2 plane.
(Actually this is my second guess and my first guess, the "smiley face" post would then be wrong!)

 

[alexepascual]

Welcome back to the thread.

No. You can at best assign only the probability value at that point.

Are you sure? Don't you need some wave function in order to get the probability?

What I think would be of interest is to plot the lateral component of each particle's wave-function as the product evolves.
The composite wave function for the entangled pair is a function over a six dimensional space (the three coordinates of each particle) evolving over time. What you're trying to do would be to average over half those dimensions (and ignore a couple say to keep things simple).
What I imagine being interesting is an animation showing the lateral component of the wave-function for both particles (one dimension of each x1 and x2 excluding y1,y2,z1, and z2). Since their position is anti-correlated you'd see the only non-zero component lies on the line L <--> x1+x2=0.

I think keeping 2 space dimensions for each particle would be intuitively better and perhaps calculationaly necessary (due to the necessity to consider different angles).

Initially you'd get a peak at the origin corresponding to the entangled pair's creation. Then it would evolve outward along the line L just like a single particle wave-function evolving from a localized point delta-function. Then as one of the particles approaches the double-slit you'd get a all but the components at the two slit positions projected out.
Remember that in most cases the particle approaching the slits never makes it through but rather strikes the barrier missing both slits. Since when we speak of the one particle passing through we are assuming it didn't miss we are excluding the cases for both particles where the one doesn't make it through. Since their positions are anti-correlated (entanglement) it is exactly the same as if both particles had been screened by corresponding double-slits.

This is clear.

This is why it seems like acting on one affects the other, this implicit rejection of cases counter to our assumption.
So back to our animation... as the one particle passes the double slits we impose the "passage assumption" and the composite wave-function is projected onto the two slits positions where it intersects the correlation line L. Now in spite of my earlier "smiley face" claim I'm not sure if the correlation in position is preserved exactly past this point. I'll have to think about it a bit (I just awoke from a nap and my head's a little foggy). I'll get back on that later. The behavior of this animation after the slit event is basically the main question being discussed here.

I think the smiley face copying does not present a paradox and could be done with momentum-correlated classical particles. On the upper detector you only get all the particles hitting it and there is no smiley face because these photons correspond to both the ones that went through the smiley on the lower path and the ones that didn't.
Once you do coincidence counting, you can figure out which of the upper detector hits correspond to the ones that did hit the lower detector and discard the others. This gives you a smiley face on the upper path. This is more like Bertlmann's socks than Aspect's experiment. Oh! there is a version of this in Penrose's "The Road to Reality" pg. 605. He does admit this is not essentially "quantum mechanical".

... I think not since that would imply you still get an interference pattern. Projecting an interference pattern wave-function along the line L onto the x1 axis would yield again an interference pattern corresponding to what particle 1 might do as it hits the final screen. The application of the composite Schrodinger equation should be straightforward. My first guess is you get what looks like a double pin-hole interference pattern in this x1,x2 plane.
(Actually this is my second guess and my first guess, the "smiley face" post would then be wrong!)

We could just try to figure out the correct way to do the simulation according to the Rules of Quantum Mechanics and see what we get.

 

[alexepascual]

... I think not since that would imply you still get an interference pattern. Projecting an interference pattern wave-function along the line L onto the x1 axis would yield again an interference pattern corresponding to what particle 1 might do as it hits the final screen. The application of the composite Schrodinger equation should be straightforward. My first guess is you get what looks like a double pin-hole interference pattern in this x1,x2 plane.
(Actually this is my second guess and my first guess, the "smiley face" post would then be wrong!)

Do you know if there is a way to paste images on a post?. I tried it but it appears that this requires having the image at some website. It would be good to be able to have some pictures here. (Not of smiley faces but of a double slit, the waves, etc)

 

[jambaugh]

Do you know if there is a way to paste images on a post?. I tried it but it appears that this requires having the image at some website. It would be good to be able to have some pictures here. (Not of smiley faces but of a double slit, the waves, etc)

When replying go to advanced mode. Below in Additional Options you can Manage Attachments. This allows you to upload files (e.g. images) and also paste in the link to the attachment if you want the image to appear in-line.

 

[alexepascual]

When replying go to advanced mode. Below in Additional Options you can Manage Attachments. This allows you to upload files (e.g. images) and also paste in the link to the attachment if you want the image to appear in-line.

Thanks. I'll try some time today to insert an image of the setup. I am thinking about something similar to the way the Dopfer experiment is usually represented but without lens or detector in the upper arm. The line followed by the beam that strikes the crystal would coincide with the x-axis. The y-axis would be a space-axis perpendicular to this.
I couldn't quite picture the representation that you had in mind. I tried to think along those lines (using only one space dimension) but I found some problems. Of course we can't make a picture of the Hilbert space. But at least having a 2-space-dimensions picture might help.
With respect to the phase, I have a problem. It appears that if the Hilbert space being considered is the compound Hilbert space, then the frequency should be determined by the energy of both photons, which would give a wavelength 1/2 the regular wavelength. I have seen a power point presentation where someone talks about this, and using it to make a better resolution microscope using entangled photons (using the shorter wavelength). But I don't know if this is just a person's idea or something proven. The strange thing would be that then the frequency of each photon would not agree with it's own energy. If we say that there are two types of frequency, that of the composite system and that of each individual phton, it is not very clear to me how you put both together. I guess for any simulation this should be understood before getting started.

 

[jambaugh]

With respect to the phase, I have a problem. It appears that if the Hilbert space being considered is the compound Hilbert space, then the frequency should be determined by the energy of both photons, which would give a wavelength 1/2 the regular wavelength.

[/quote]...The strange thing would be that then the frequency of each photon would not agree with it's own energy. If we say that there are two types of frequency, that of the composite system and that of each individual phton, it is not very clear to me how you put both together. I guess for any simulation this should be understood before getting started.[/QUOTE]

Here is an example where I feel being a "good Copenhagenist" helps one understand what is going on. Remember the wave functions are not the physical systems but represent knowledge about the systems.
(Think strongly with the analogy of a probability distribution.)

Don't be too stuck on the idea of a physical wave-function "out there" be it for a single particle or for the entangled pair. The wave function expresses probability amplitudes for potential measurement events. In the two particle case they would be measurements of each at two independently chosen locations. Remember you can also speak of the wave-function on momentum space... or any other cross-section of classical phase-space representing a manifold of possible mutually compatible observables.

If you want to be more substantive you can work instead in the field theory where you are talking about particle number densities with the global constraint of total particle number = 2 but you must then promote your hilbert space vectors to operators. Also you are likely to need intuition built upon understanding fully the QM version before you can derive much insight from the QFT version.

But mainly I suggest you recall that the Hilbert space of the composite is a tensor product of Hilbert spaces and remember the product to sum trigonometric identities which derive from the rules for (imaginary) exponentials:

$$e^{i\omega_1 t} e^{i\omega_2 t} = e^{i(\omega_1 + \omega_2)t}$$

So the energy of the composite is simply the sum of the energies of its components (ignoring interactions) which is nicely consistent with expressing composites as products.

 

[alexepascual]

Here is an example where I feel being a "good Copenhagenist" helps one understand what is going on. Remember the wave functions are not the physical systems but represent knowledge about the systems.
(Think strongly with the analogy of a probability distribution.)
Don't be too stuck on the idea of a physical wave-function "out there" be it for a single particle or for the entangled pair. The wave function expresses probability amplitudes for potential measurement events. In the two particle case they would be measurements of each at two independently chosen locations. Remember you can also speak of the wave-function on momentum space... or any other cross-section of classical phase-space representing a manifold of possible mutually compatible observables

OK. The squared modulus of the wave function gives you the probability distribution. And we need that probability distribution in order to find out what the pattern on the screen is going to be. I am not trying to give the wave function an interpretation here. I just want to use it to calculate results. I should be able to calculate the sum of the waves that went through each slit at each point on the screen.

If you want to be more substantive you can work instead in the field theory where you are talking about particle number densities with the global constraint of total particle number = 2 but you must then promote your hilbert space vectors to operators. Also you are likely to need intuition built upon understanding fully the QM version before you can derive much insight from the QFT version.

No, I don't want to get into QFT here. If it were absolutely necessary I would, but I don't think it is.

But mainly I suggest you recall that the Hilbert space of the composite is a tensor product of Hilbert spaces and remember the product to sum trigonometric identities which derive from the rules for (imaginary) exponentials:
$$e^{i\omega_1 t} e^{i\omega_2 t} = e^{i(\omega_1 + \omega_2)t}$$

Well, that's exactly what I was saying. You just expressed it in terms of angular velocity of the phase (You are showing the time component).

So the energy of the composite is simply the sum of the energies of its components (ignoring interactions) which is nicely consistent with expressing composites as products.

I agree. Now the question is: When I write the expression for the phase as function of time in the lower arm, I need to know what energy to use. On one hand I have read that once the particles are entangled you can't consider the separate Hilbert spaces anymore. In that case I would like to use the sum of the energies of both particles, even if I am looking at only one particle there. On the other hand I couls say that because I am considering only one particle in the lower arm, I should consider only the energy for that particle. In one case the frequency is twice what it would be in the other case. So, even if I believe that the wave funtion only exists in my brain, I need numbers to plug in for p and E.
Oh! by the way.. Do you use some editor for LateX?

 

[jambaugh]

I agree. Now the question is: When I write the expression for the phase as function of time in the lower arm, I need to know what energy to use. On one hand I have read that once the particles are entangled you can't consider the separate Hilbert spaces anymore. In that case I would like to use the sum of the energies of both particles, even if I am looking at only one particle there. On the other hand I couls say that because I am considering only one particle in the lower arm, I should consider only the energy for that particle. In one case the frequency is twice what it would be in the other case. So, even if I believe that the wave funtion only exists in my brain, I need numbers to plug in for p and E.

I'm still not quite clear on what you're asking. The E and |p| are as you would expect and the same as would be defined classically. You just won't be able to write down a wave-function where E reflects the frequency.

(Below I am using A and B to indicate the A particle and B particle of an entangled pair)

The proper way to talk about one half of an entangled pair is with the reduced density operator. In the continuous representation (i.e. with wave-functions) this will take the form of a function of two positions:
$$\rho_A(\vec{x},\vec{y})$$
(also a function of time) (Note this is another doubling of the variables)
Think of the vectors x and y as row and column indices for a matrix.
The bifunction forms act as operators on functions via:
$$T[f]=g \quad \equiv \quad g(x) = \int T(x,y)f(y)dy$$

The full density operator for the entangled pair will then have four vector variables:
$$\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)$$
and the reduced density operator is obtained by "tracing over" the variables corresponding to the other half of the pair:

$$\rho_A(\vec{x},\vec{y}) = \iint \delta^3(\vec{x}_B-\vec{y}_B)\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)d^3x_B d^3 y_B$$

In the case of a sharp mode (for the whole entangled pair) this density operator will take the form:
$$\rho_{AB}( = \psi(x_A,x_B)\psi^\dag(y_A,y_B)$$
(henceforth assume x's and y's are vectors.)

But once you take the partial trace to get the reduced density operator you won't have such a factorable form. (It is not in a sharp mode)

Now I get to the point... At best you can write the reduced density operator as a
(classical) probability weighted average of such sharp density operators.
$$\rho_A(x,y) = \sum_{k}p_k\rho_k(x,y) =\sum_k p_k \psi_k(x)\psi^\dag_k(y)$$
Each such will be a product of wave-functions as above where each of those wave-functions has the same energy and thus frequency.
$$\psi_k(x) = \phi_k(x)e^{i\omega t}$$

You can't speak meaningfully about the "phase of the lower (A) particle" at a given point or even at a given time. When you consider only one half of an entangled pair you get something similar to a particle from a noisy source and you must work within the density operator format.

And this is my point about being "a good Copenhagenist". Get out of the habit of thinking in terms of "the phase of a particle". Its the wave-function which has a phase and without a wave-function (since the mode is not a sharp one) we can't define phase.

As the ole codger on the porch says... "Ya caint get there from here!"

Oh! by the way.. Do you use some editor for LateX?

I use both Led and WinShell (both free). Sometimes I prefer one sometimes I prefer the other. Right now Led is the main one I use.

 

[alexepascual]

I'm still not quite clear on what you're asking. The E and |p| are as you would expect and the same as would be defined classically. You just won't be able to write down a wave-function where E reflects the frequency.

(Below I am using A and B to indicate the A particle and B particle of an entangled pair)

The proper way to talk about one half of an entangled pair is with the reduced density operator. In the continuous representation (i.e. with wave-functions) this will take the form of a function of two positions:
$$\rho_A(\vec{x},\vec{y})$$
(also a function of time) (Note this is another doubling of the variables)
Think of the vectors x and y as row and column indices for a matrix.
The bifunction forms act as operators on functions via:
$$T[f]=g \quad \equiv \quad g(x) = \int T(x,y)f(y)dy$$

The full density operator for the entangled pair will then have four vector variables:
$$\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)$$
and the reduced density operator is obtained by "tracing over" the variables corresponding to the other half of the pair:

$$\rho_A(\vec{x},\vec{y}) = \iint \delta^3(\vec{x}_B-\vec{y}_B)\rho_{AB}(\vec{x}_A,\vec{y}_A,\vec{x}_B,\vec{y}_B)d^3x_B d^3 y_B$$

In the case of a sharp mode (for the whole entangled pair) this density operator will take the form:
$$\rho_{AB}( = \psi(x_A,x_B)\psi^\dag(y_A,y_B)$$
(henceforth assume x's and y's are vectors.)

But once you take the partial trace to get the reduced density operator you won't have such a factorable form. (It is not in a sharp mode)

Now I get to the point... At best you can write the reduced density operator as a
(classical) probability weighted average of such sharp density operators.
$$\rho_A(x,y) = \sum_{k}p_k\rho_k(x,y) =\sum_k p_k \psi_k(x)\psi^\dag_k(y)$$
Each such will be a product of wave-functions as above where each of those wave-functions has the same energy and thus frequency.
$$\psi_k(x) = \phi_k(x)e^{i\omega t}$$

You can't speak meaningfully about the "phase of the lower (A) particle" at a given point or even at a given time. When you consider only one half of an entangled pair you get something similar to a particle from a noisy source and you must work within the density operator format.

And this is my point about being "a good Copenhagenist". Get out of the habit of thinking in terms of "the phase of a particle". Its the wave-function which has a phase and without a wave-function (since the mode is not a sharp one) we can't define phase.

As the ole codger on the porch says... "Ya caint get there from here!"


I use both Led and WinShell (both free). Sometimes I prefer one sometimes I prefer the other. Right now Led is the main one I use.

I realize that I don't have a full understanding of the situation. However, I am not convinced about your approach. The Copenhagen interpretation does no forbid using the wave function to describe the evolution of a system before it interacts with a macroscopic apparatus. Now, perhaps I don't understand very well the wave function of a pair of entangled particles, but I looks to me that using the density matrix would be a little like sweeping the problem under the rug. After the photons have been created, their evolution in time should be described by the wave function. After one of the photons has been detected, then maybe you should consider tracing over the density matrix. I will definitely have to think about this, but I think it may be a little premature right now. I could instead focus on what happens before detection.
I am trying to see how I can make sense of your argument without including the density matrix part. Let me see if the following agrees with the way you are thinking:
The wave function of the pair of photons is a function of the space coordinates for both photons and time coordinate (in a non-relativistic approach). So, for a particular combination of positions for both particles we have a single complex number which evolves in time with a frequency determined by the sum of the energies of both photons. If we look at the photon that goes towards the slits, and we want to assign to each point in space for that photon a wave function value which we can use to determine the probability of finding the particle at a particular position, we run into the problem that we don't know exactly where the other particle is.
Now, doing a trace on the density matrix would give you a mixture, but here the composite state is still a pure state evolving unitarily. If you had the degrees of freedom of the other photon getting entangled with the environment, according to the decoherence theory, you would have to do a trace. But in this case we don't have that situation yet.
Going back to the state vector, I would reconsider what I said about not knowing where the other particle is. There is some uncertainty along the path of the particle (perpendicular to the waves) and this needs to be considered. But if we ignore that for a first aproximation, then we have to consider any lateral uncertainty in the other particle. But it happens that due to the momentum entanglement, the lateral position of the second particle should be correlated to the position of the first. With respect to the position along the path, we could just use the speed of light x time (ignoring uncertainty).
In an x-y graph (configuration space) where x and y are the 1-D positions of both particles (I think this may be closer to your original idea for a simulation) If we ignore uncertainties and consider propagation along rays (This could be objected but I don't think it matters here), then we would have a parametric curve where each point corresponds to the configuration at a particular time. And there should be value for the wave function at each point. Does this make sense?
Thanks for the tip about the LateX editor.

 

[jambaugh]

I can only repeat what I've said before and I will try to with better examples but think about this some more before we go on.