Dear All,



Sorry for bringing back this old topic but I think I might have an explanation to satisfy the original query, which I believe was not conclusively put to rest in the end.



I think the problem that me and the original poster, Jacob, were having was that we were confusing energy with amplitude (at least I did). I.e., anomalous scattering affects/reduces the amplitude of the atomic form factor (or structure factor in case of a crystal), but not the energy (or wavelength) of the scattered photon, which is the same as that of the incident photon. Therefore, the anomalous scattered photon will still be able to resonate with another anomalous scatterer within the crystal, without breaking any conservation of energy theory. Since anomalous scattering is an elastic effect, if one accepts the explanation model of "photon interfering with itself" and "mini-waves"  in the case without resonators, then this model could be equally valid even in the presence of more than one anomalous scatterer.



I would like to thank Colin Nave to make me realize that I was mixing up anomalous scattering with inelastic scattering. I am pretty sure I had it clear while doing my PhD many moons ago.



I hope I understood correctly the original question and that this explanation to the query might make some sense to other people as well, rather than just me :-).



Best,



and sorry again for bringing this back,



D




________________________________
From: CCP4 bulletin board [[log in to unmask]] on behalf of [log in to unmask] [[log in to unmask]]
Sent: 31 August 2015 14:12
To: ccp4bb
Subject: [ccp4bb] AW: [ccp4bb] AW: [ccp4bb] Diffraction as a Single-Photon Process; was RE: [ccp4bb] Twinning Question

Dear Jacob,

You are not the only one who does not believe in quantum mechanics. Albert Einstein was probably the most famous non-believer.

I agree with you that since we observe interference and diffraction patterns, there must occur interference somewhere. Although Niels Bohr claimed that you cannot say anything about a quantum system between two measurements, my strong feeling is that we see interference between the different superimposed quantum states. This is for me the truly spooky part of quantum mechanics: instead of a single foton, as long as we do not measure, there can be hundreds of  fotons haunting our crystal. However, the moment we switch on the light, we find only one. The position of this foton will have been influenced by all other spooky fotons.

I do not see how quantum mechanics would not conserve energy, but would be interested to learn.

HS


Von: Keller, Jacob [mailto:[log in to unmask]]
Gesendet: Montag, 31. August 2015 13:06
An: Schreuder, Herman R&D/DE; [log in to unmask]
Betreff: RE: [ccp4bb] AW: [ccp4bb] Diffraction as a Single-Photon Process; was RE: [ccp4bb] Twinning Question

>This means that when a foton at the same moment interacts with 100 scatterers (or resonators), there are 100 or more different states and in each state the foton interacts with a different scatterer. In each state, one foton interacts with only one scatterer. The moment the measurement is made, we find only one discrete foton, corresponding to one of these states. The distribution of the states, and therefore the possible outcomes, depend on the presence of all scatters/resonators within coherent range.

Then I don’t see how interference or diffraction patterns can occur without resorting to what others have said, which I don’t understand really: that interference is not really happening at all, but something else with spooky probability distributions which don’t need to subscribe to conservation of energy.

JPK


Von: CCP4 bulletin board [mailto:[log in to unmask]] Im Auftrag von Keller, Jacob
Gesendet: Donnerstag, 20. August 2015 20:42
An: [log in to unmask]<mailto:[log in to unmask]>
Betreff: Re: [ccp4bb] Diffraction as a Single-Photon Process; was RE: [ccp4bb] Twinning Question

What I don’t understand is how a single photon, which I thought by definition was an indivisible quantum of energy, can be split up arbitrarily amongst a number of scatterers into these “mini-waves.” Doesn’t that self-contradict QM’s concept of quanta?

One might say that somehow there are two energy-related characteristics to the photon:

1. the actual amount of total energy in the photon, and then

2. the “type” or “color” or “frequency” of the photon’s energy.

If you will allow me this dichotomy, then I can understand how it can be distributed to different atoms—small portions of energy of the same “color” are distributed to all of the resonators. One would also have to presuppose another thing, that the resonators themselves are able to accept packets of energy of size 1/n, as long as it’s of a certain color. The problem is, however, that allowing photons and resonators to do these things violates what I thought was the central tenet of QM, that there are indivisibles known as quanta.

Maybe, then, one can just drop the bit about there being quanta, or at least put a star by it?

JPK


From: [log in to unmask]<mailto:[log in to unmask]> [mailto:[log in to unmask]]
Sent: Thursday, August 20, 2015 2:03 PM
To: Keller, Jacob; [log in to unmask]<mailto:[log in to unmask]>
Subject: RE: [ccp4bb] Diffraction as a Single-Photon Process; was RE: [ccp4bb] Twinning Question

Valid questions.

The phenomenon of resonance needs some explanation here, in terms we can imagine:

Take first the normal case: let all the n resonating electrons gain 1/n in energy from the disappearing photon. These n resonating electrons emit partial waves or whatever you want to call them totaling n*1/n in energy, which recombines into the new photon. But what happens when the phases lead to extinction? Where does the energy go? Well, it just does not happen, it won’t get scattered in THAT direction. So in the probabilistic picture again, IF a photon does gets elastically scattered, then it WILL appear again. WHERE it might appear, is given by its probability distribution, aka the Fs.  No contradiction here, although I fully admit that the mini-wave picture results from the need to explain, in experience-accessible terms, a non-experienceable process. That is the QM conundrum, but not a contradiction.

Now the anomalous (n.b.: not inelastic!) case: In this case the net effect is a change in the fs and thus Fs, and again all it does is change the probability distribution accordingly and above picture holds. But wait – where does the X-ray fluorescence come from, and if the photon uses all its energy to kick a photoelectron out, how can it reappear? It does not. The unlucky photon that generated the photoelectron is DEAD, otherwise we violate energy conservation. That photoelectron then causes either fluorescence via outer to core transitions or can be directly measured in case it manages to escape, or make Auger electrons, whatever satisfies energy conservation. The lucky photons, passing close to absorption energy, experience only the change in scattering factor. If you look at the theoretical QM calculations for absorption spectra (Cromer Lieberman etc.), you see that the dispersion curves actually show a singularity at precisely the orbital excitation energy. That absorption curve is again simply a probability function for photon death at a given energy. In solids, this curve can be more complicated and have more detail, but still the same.

So, you cannot simultaneously measure diffraction and fluorescence of one and the same photon. The fluorescence scan does not come from the anomalously but elastically scattered photons. It comes from the absorbed dead ones. There is no difference between the normal and anomalous ‘miniwave’ picture other than a change in fs and Fs.

Radiation damage, btw, is just a process cascade caused by that photon death.

I abstain from digressing into inelastic/incoherent processes.

Best, BR

From: CCP4 bulletin board [mailto:[log in to unmask]] On Behalf Of Keller, Jacob
Sent: Thursday, August 20, 2015 9:48 AM
To: [log in to unmask]<mailto:[log in to unmask]>
Subject: [ccp4bb] Diffraction as a Single-Photon Process; was RE: [ccp4bb] Twinning Question

Do you have any explanation of how a single photon, which contains x amount of energy, can cause multiple electrons (at least 1000’s!) in anomalously-scattering atoms to resonate at that energy?

We don’t find that the presence of different numbers of resonant scatterers requires x-rays of different energy; so why, if the energy is being divided into different numbers of resonators, does the same energy of x-rays work?

I believe that BR’s book says that the photon disappears or annihilates briefly, then re-emerges. This must be true, then, across thousands of electrons at once, both normal and anomalous?

Jacob


From: Edwin Pozharski [mailto:[log in to unmask]]
Sent: Thursday, August 20, 2015 9:36 AM
To: Keller, Jacob
Subject: Re: [ccp4bb] Twinning Question

typo indeed.  The point, of course, stands - with older sources there are *no* photons inside the crystal for over 99% of the time.  (Notice that diffraction pattern is still present, Bragg's law satisfied, etc))  X-ray diffraction is, for all intents and purposes, a single photon experiment.  Even with the brightest and most coherent sources, when you could have multiple photons within a large crystal, these are still separated in space by a distance that is at least 100x the coherence length.  Thus, X-ray photons do not interact with each other (and even if they would, it's still does not make them a wave, just good ole photons that due to their high spatial density would have detectable probability to engage in multi-photon events).


On Wed, Aug 19, 2015 at 5:13 PM, Keller, Jacob <[log in to unmask]<mailto:[log in to unmask]>> wrote:
>Also, if your X-ray source is not exactly the brightest synchrotron, you are probably looking at ~10^9 photons/sec at best (I am estimating here that it would take at least 15-20 minutes of data collection using early 2000s "RAxisIV" in-house system to get diffraction image of intensity similar to 0.5s exposure at 12-2).  That is one photon every nanosecond. Let's continue to ignore the fact that most photons just fly through.  A photon zips through a 1mm crystal in about 3fs.  Think about this - at a moderate intensity home source (and I can go to sealed tubes), the process of crystal illumination by X-rays is more like single photons flying through with about 300x long pauses between events.  To scale this, imagine that a single photon spends a whole second inside a crystal, probing it's electron density.  You would then have to wait five minutes for the next photon to arrive.
I was re-thinking through this, and I think one of these numbers is wrong, viz, “A photon zips through a 1mm crystal in about 3fs.” The speed of light is 3x10^8 m/s, so this leads to ~3.3 ps for a 1 mm path, and not 3 fs, a difference of ~10^6. Maybe it was just a typo? Anyway, it may not make a huge difference, since this would still make for an average of ~1 photon in the crystal at a time, assuming a high flux of 10^12 photons per second. But of course there would be some Poisson statistics involved, and there would be several photons a significant part of the time.
Also, I wonder about relativistic effects: in the famous train-in-tunnel thought experiment, a large train can fit in a short tunnel if it’s going close to the speed of light. Is this applicable here, such that many photons are in some sense in the crystal at once? Or maybe this is a red herring.
But, to change topics a bit: part of the reason I am wondering about this is anomalous scattering. Since the resonance energy of an atom is a fixed amount, how can one photon provide that energy simultaneously to the requisite number (at least thousands, I would think) of resonant scatterers? Something’s very funny here.
Or, come to think of it, perhaps resonant scattering is no worse than normal scattering: if the energy is divided up between the all the normally-scattering electrons, you even have a problem with the one-photon picture, since the emerging radiation is still of the same energy. You want to have everything being scattered with a certain energy, but you also want all the scatterers to scatter. The concept of “energy” seems to get strange. Does one then need two terms, in which “energy” is just a characteristic of radiation, like a color, and then there is some other attribute like “probabilistic intensity,” which describes how much “photon” is there?
It is striking to me how much depth these everyday occurrences really have when one starts wondering about them.
Jacob



-- 
This e-mail and any attachments may contain confidential, copyright and or privileged material, and are for the use of the intended addressee only. If you are not the intended addressee or an authorised recipient of the addressee please notify us of receipt by returning the e-mail and do not use, copy, retain, distribute or disclose the information in or attached to the e-mail.
Any opinions expressed within this e-mail are those of the individual and not necessarily of Diamond Light Source Ltd. 
Diamond Light Source Ltd. cannot guarantee that this e-mail or any attachments are free from viruses and we cannot accept liability for any damage which you may sustain as a result of software viruses which may be transmitted in or with the message.
Diamond Light Source Limited (company no. 4375679). Registered in England and Wales with its registered office at Diamond House, Harwell Science and Innovation Campus, Didcot, Oxfordshire, OX11 0DE, United Kingdom