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I received a helpful email from Chris, a physicist knowledgeable about current thinking and terminology regarding the double slit experiment. In case you don't speak physics, I'll do a short summary here: the main issue (other than out of date terminology) was my claim that physicists in the 1920s realized that it was the recording of the measured data rather than the measurement itself that was critical to the outcome of the double slit experiment. Indeed that is not true. As Chris points out, physicists assumed (and still do) that it is the interaction with the measurement apparatus, rather than the fact that measured data is recorded. The math works out the same either way  and since both were always done together, the experiment to determine whether their assumption was true has not been done. The bottom line is that my discussion of the history of the doubleslit experiment in the YouTube videos is incorrect  it would seem that contemporary physicists would not agree that creating information in PMR is the key ingredient rather than an invisible theoretical physical interaction. In any case, which ever assumption is shown to be correct, there is no impact on fundamental MBT theory presented in the books or presented in the workshops (YouTube). It is primarily a matter of me getting my history right at best and not attributing to current physics understandings that they don't have. Certainly at the Macro level, probability and information are more fundamental to the content and substance of our reality than the details of the ruleset which simply represent constraints on the information allowed expression within the virtual PMR  it would seem to make sense that that would be true of the micro world as well  but that remains an unproven assumption.
Tom

Chris: Hi. In a set of Youtube videos, you bring up the dual slit
experiment: You say that if we "leave on" the apparatus which detects which slit a photon goes through, and merely don't look at it, don't let it record its observation, then you still see an interference pattern. You claim that this problem is called the "measurement problem."
Both of these statements are incorrect. In the dualslit experiment, you couple some environment to the photon going through the two slits 0 and 1. This coupling changes the system from  0 > +  1 > to the entangled state  00 > +  11 >. This causes the density matrix to trace out as 0.5  0 >< 0  + 0.5  1 >< 1   which is not the density matrix for  0 > +  1 >. Thus, it is the fact that you are entangling another system /at all/ which causes the disappearance of the interference pattern.
In other words, the change in patterns that you see has nothing to do with the measurement problem, and is governed by the normal unitary evolution of quantum mechanics  the part that's not an "interpretation". When you perform that CNOT gate to take it from  00> +  10 > into the state  00 > +  11 >, you automatically destroy
the original interference pattern, no matter whether you record the second qubit or not.
 Chris

On Tue, Aug 3, 2010 at 11:05 PM, Tom Campbell < twcjr@comcast.net> wrote:
Thanks Chris. Tell me what this means: "you couple some environment to the photon going through the two slits 0 and 1."
You seem to assume an interaction of some sort (coupling) and that the photon is going through both slits which is, of course, the answer to the problem which gives an interference pattern.
I'd like to understand if I have misspoken  please explain your notation  its been a while.
Tom

Tom 
Well, to take the notation first: it's your standard normal Dirac braket notation that you see everywhere in QM; I have been sloppy with normalizing factors for the sake of conciseness. I will fix that
here: let s = sqrt(1/2).
I mean, just to review, we define an inner product space where every wave function ÃË† is represented by a ket vector  ÃË† > which has a conjugate transpose < ÃË† . There are basis states for a system  for a qubit, those states are  0 > and  1 >. The general state in QM is allowed to be any linear combination of these states, although we generally make the restriction that < ÃË†  ÃË† > = 1 (normalization). An observable operator A is a mapping from kets to kets, or equivalently from bras to bras, defined where < ÃË†  A  ÃË† > is the average value of the observable A seen across an ensemble of systems prepared in state
 ÃË† >.
With that said: the environmental coupling does not "give an interference pattern," but rather it takes it away. The system starts out in the s  0 > + s  1 > state, a superposition of going through both slits equally. The key to its interference pattern is that its density matrix looks like this:
 ÃË† >< ÃË†  = 1/2 (  0 >< 0  +  0 >< 1  +  1 >< 0  +  1 >< 1  )
...where you can see "cross terms" in the density matrix. Those cross terms represent the distinctively /quantum/ superposition, different from if I flip a coin and point my laser at each of the two slits based on its outcome.
Now, you have an experimental environment  a detector of "which slit did it go through?  which we will perhaps say starts in a neutral state  n > but can indicate either "the photon went through 0"  0 > or "the photon went through 1"  1 >. Thus, at the start, just as the photon goes through the slits, you have a state:
 ÃË† > = s  0n > + s  1n >.
Now you turn on the coupling: a unitary evolution U which allows you to use your environment to observe the photon. Ideally, you design your apparatus very well, so that it takes  0n > to  00 > and  1n > to  11 >. Thus, your apparatus produces the state:
U  ÃË† > = s  00 > + s  11 >.
This is the "interaction of some sort"  this stateevolution operator U which entangles your measurement apparatus with the photoninflight. The first bit now has lost the crossterms which created the initial entanglement; it has the same density matrix as if you just flipped a fair coin and sent it through  0 > on heads or  1
> on tails.
I hope this gives you an understanding of what I mean when I say that your "which slit?" detection apparatus is an "environment" which is being coupled to the photon's "qubit state".
Friendly greets,
Chris

2010/8/4 Tom Campbell < twcjr@comcast.net>:
Thanks Chris,
Your tutorial sounded familiar  I have worked the last 40 years in industry  an applied physicist  and there isn't much use for ket vectors there. So it has been a long time. Use it or lose it.
One question on notation:  ÃË† >< ÃË†  and < ÃË†  ÃË† > both would appear to be a multiplication of a ket vector with its conjugate transpose  refresh my memory about the difference the order of terms makes. Just using cross product vector multiplication doesn't seem to explain what you have written, so I must not fully understand this notation:  ÃË† >< ÃË† 
I did read up on the double slit experiment before talking about it on film and thought I had it correct. It was explained in several places much as I explained it on the video  with multiple references using the terminology "probability wave colapsing to a physical particle" and "the measurement problem"
So what is the bottom line here? I take it from your notation that you are saying that what makes the photon go through just one slit or the other is the fact of making a detection. Not the
fact of recording the detection?
Are you sure it's the appratus? Well it appears I need to do some more reading. I have presented this to a dozen physicists and not one saw that issue. And several of those were current Academic physicist  well known even.
Thanks for pointing out the problem and I will let you know what I find out  I can't call back movies, but can stop making the same mistake in the future. And thanks for the brush up on ket vector notation. I appreciate you taking the time to be helpful.
Tom

Tom 
Well, the difference if you like is that  a > < b  is a matrix while < a  b > is a scalar. If I were to rightmultiply by a ket  c > then they would become two different vectors:
< a  b >  c > vs.  a > < b  c >
The first is parallel to  c > and the second is parallel to  a >.
The first is 0 if a and b are orthogonal; the second is 0 if b and c are orthogonal. The scalars can then of course commute with the vector, but you can't commute a vector with a vector to turn an outer product into an inner product: it would be an abuse of the notation.
You ask, "I take it from your notation that you are saying that what makes the photon go through just one slit or the other is the fact of making a measurement (a detection). Not the fact of recording the measurement (detection)?"
It's not even "making a measurement." The moment you prepare the two qubits into the 00 + 11 quantum state, entangling the two with each other, the first qubit is not in the quantum 0 + 1 superposition any more, and in fact cannot be described as a pure quantum state any more. The entanglement comes at the price of decoherence. This is, if you like, why quantum computation is so hard: any environmental effect which couples with a set of quantum bits harms their ability to remain in their carefully prepared state. You need very high quality factors to make sure that as few shenanigans as possible happen.
Now there *is* a measurement problem lurking about  it is the nonunitary dynamics which attend a wavefunction collapse at the point of a measurement. But this disappearance of the interference pattern has nothing to do with it  it is a part of the unitary dynamics which prevail normally. The disappearance of the interference pattern is a decoherence problem, rather than a measurement problem.
If you like, QM says that my television can be in the state:
sqrt(1/2)  living room > + sqrt(1/2)  bedroom >
The above decoherence argument, when applied to photons and air molecules bouncing off of my television, can entangle so many alternate degrees of freedom that this becomes classical:
50% living room, 50% bedroom.
...while the environment is entangled with it (i.e. there is a hole in the gas in one room or the other to "make room" for the television, and so forth). Any interference effects between my bedroom and living room disappear. But now, nonetheless, when we go to observe it, we don't see it "half in the living room" and "half in the bedroom," but the television is in either one or the other place. That is the measurement problem.
Feynman gives a good discussion in his Messenger Lectures[1] and his big physics textbook about the dualslit experiment, and when he talks about the measurement problem it is a question of how the electron "makes up its mind" the way that a bullet does, to appear at one detector or the other. This is distinct from the question of the unitary decoherence. If you wish, the measurement problem is that "light comes in corpuscles", treated in his first Auckland lecture[2], whereas the decoherence problem is treated in microcosm at the beginning of his third Auckland lecture[3]. The measurement problem is "how does a photon decide whether it's reflected or not?"
 Chris
[1] Bill Gates has made these lectures available online, but I use Linux and his technology does not support my OS. Still, I have the memory that he treated them in those lectures: but I cannot give you a link saying "here is precisely in the videos where it is discussed."
[2] http://vega.org.uk/video/programme/45 , about 35 minutes in.
[3] http://vega.org.uk/video/programme/47 , about 5 minutes in.

2010/8/4 Tom Campbell <twcjr@comcast.net>:
Thanks Chris, I thought the cross/dot product thing was the key but wasn't familiar enough with the notation to be sure. It all makes sense.
Saying that it is not the measurement, but decoherence that "causes" the wave function to collapse to a particle (no diffraction pattern) sounds like a semantic issue (how you use the words  at what level of description)  indeed, With that bigger semantic picture in mind:
When the apparatus produces the state:
U  ÃË† > = s  00 > + s  11 >.
Where U represents an "interaction of some sort"  You say that the stateevolution operator U entangles the measurement apparatus with the photoninflight.
Could you not just as easily say that that the stateevolution operator U entangles the recording of the measurement data with the photoninflight. Has such an experiment ever been done or is it simply intuitively obvious that there is no point to doing such an experiment because we understand exactly how the measurement apparatus entangles with the photon in flight?
Though it may sound ridiculous and counterintuitive to you, (QM is known to be counterintuitive), is it possible that the fact that the "measurement apparatus is the source of entanglement" is a reasonable but untested assumption? Simply a belief. The information theory approach to reality (reality is information) might challenge that assumption since it is the information and not the measurement that is key. Whether U represents the measurement or the data produced by the measurement makes no difference to the math analysis. It seems to me that whether we interpret U as representing the 1) the measurement device interaction, or 2) the data/information produced by the measurement devise interaction, is entirely arbitrary since nothing changes in the equations except how we interpret them. So, Chris, Am I just howling at the moon here or do I have a valid point, no matter how farfetched and heretical to accepted opinion it might sound?
Tom

Tom,
You say that I am "Saying that it is not the measurement, but decoherence that "causes" the wave function to collapse to a particle (no diffraction pattern)." I cannot emphasize enough that what I'm saying is that "no diffraction pattern" is not equal to "collapse to a particle." The measurement problem  the nonunitary dynamics of wave function collapse  is totally different from the presence or absence of a diffraction pattern, which is ultimately part of the SchrÃƒÂ¶dinger equation.
[Tom: I understand your usage and how you are separating the two concepts, and I agree they are quite different phenomena. The language I use to describe the phenomena is, I confess, not very precise in terms of current physicsspeak (in which it appears that I am inappropriately mixing the metaphors of collapse and decoherence) but I am not confused about the difference between the two concepts  I am just confusing you about what I mean with my terminology]
"Could you not just as easily say that that the stateevolution operator U entangles the recording of the measurement data with the photoninflight. Has such an experiment ever been done or is it simply intuitively obvious that there is no point to doing such an experiment because we understand exactly how the measurement apparatus entangles with the photon in flight?"
Well, [I suppose that could be the case, however] there would be a massive problem in doing [the experiment] with photons, namely that it is very hard to count them without absorbing them. You could point a light at two slits while electrons passed through them, though, and then the photons scattering off the electron would potentially be visible at a detector. As for understanding exactly how the apparatus entangles  well, presumably, we /design/ it to entangle that way. We want to have an apparatus which, starting from a neutral state, gives us 1 if the particle went through 1, or 0 if the particle went through 0. If the apparatus doesn't entangle that way, then it is not prepared to give you a good measurement.
[Tom: Yes, We would design an apparatus which, starting from a neutral state, gives us 1 if the particle went through 1, or 0 if the particle went through 0, but that is exactly the case whether the entanglement comes from the measurement itself or from collecting the measured data.]
Anyway, yes, the light does indeed break the interference pattern of the electron  as far as I know, even when you switch off the photomultiplier detectors for those lights, though I have not done this experiment myself. If it did not do so, then that would be a significant violation of our current laws of quantum mechanics.
However, of course, the light comes as particles too, and thus there will be a class of results where the electron went through the slits but your counter remained in the neutral state  no photon got entangled with the passing electron. In this subset of cases, the system would be in the 0n + 1n state, which is separable, and you would again see an interference pattern. This will become more important as you slowly turn off the light source.
 Chris

On Wed, Aug 4, 2010 at 7:47 PM, Tom Campbell <twcjr@comcast.net> wrote:
Chris: " I cannot emphasize enough that what I'm saying is that "no diffraction pattern" is not equal to "collapse to a particle."
Tom: I must be wearing out your patience and willingness to be helpful  I apologize for that, but there is something I am missing and I am not sure what. One last try, then I am off to the links you provided.
OK, I must be dense. Here is My logic, where is it that I go wrong conceptually (as opposed to not using the proper terminology): 1) Waves traveling through two slits produce a diffraction pattern. 2) to make it simpler, particles with rest mass (e..g, an electron or neutron) whose position/trajectory is accurately known) must go though one slit or the other like any classical particle and thus do not produce a diffraction pattern. 3) a series of such particles of uncertain position approaching two slits one at a time will eventually produce a diffraction pattern. 4) Insert a measurement device into the experiment (causing decoherence) to detect the particle at a slit. The probability wave collapses to a physical particle (the particle is detected and its position/trajectory is known at a given slit), then like any classical particle, it passes on through the slit, producing no diffraction pattern. 5) From this I conclude that upon measurement, the probability waves of a series (in time) of individual particles collapse to classical physical particles which proceed to pass though the slit as any classical particles would, leaving a spot directly behind the slit rather than producing a diffraction pattern  in direct contrast to 3 above. Bottom line: classical particles cannot create diffraction patterns but probability waves can (if they remain waves and are not collapsed to physical particles by a measurement/decoherence).
Tom

Hi Tom,
"I must be wearing out your patience and willingness to be helpful"  eh, I've just had a bit of a bad day today; no worries.
As for (1)  (5):
"1) Waves traveling through two slits produce a diffraction pattern."
True in general, yes.
"2) to make it simpler, particles with rest mass (e..g, an electron or
neutron) whose position/trajectory is accurately known) must go though one slit or the other like any classical particle and thus do not produce a diffraction pattern."
This worries me for a couple reasons. The most important is that even going through one slit, there is a quantum effect which is not to be neglected[1]  and if the electron doesn't have enough momentum, both slits will sit within its de Broglie wavelength  in other words, you can't necessarily "target it" this way.
[Tom: I think what you are saying is that if there is enough uncertainty we move into statement number 3). Yes that is true but I wanted statement 2) to be only those particles that have no overlap: de Brogile wavelength << distance between slits. Any quantum effect in conceptual box 2) should be negligible.]
" 3) a series of such particles of uncertain position approaching two slits one at a time will eventually produce a diffraction pattern."
Well, yes. You measure their position along some axis  let us call it the yaxis. If an electron went through the first slit, then at the target it would have wave function Ãâ€ Ã‚Â¹(y), and it would have wave function Ãâ€ Ã‚Â²(y) going through the other. The wave function if it is allowed to go through both is just the quantum superposition:
ÃË† = sqrt(1/2) [Ãâ€ Ã‚Â¹(y) + Ãâ€ Ã‚Â²(y) ]
... and you measure:
ÃË†Ã‚Â² = 0.5  Ãâ€ Ã‚Â¹ + Ãâ€ Ã‚Â² Ã‚Â².
However, you only measure this distribution over y with a set of counters arranged along the yaxis, and those counters only fire one at a time. The particle "makes up its mind" which counter it is going to set off based on ÃË†Ã‚Â², by a process which is not unitary and not wellunderstood. Also, I hope you can see that there is only an effect when Ãâ€ Ã‚Â¹(y) and Ãâ€ Ã‚Â²(y) overlap. This is especially important if you read Feynman's version of QED, since he has a leastaction approach and thus assumes that the source and the detector are predetermined, and then sums up the different amplitudes for different paths that go from the source to the detector.
[Tom: I agree fully]
"4) Insert a measurement device into the experiment (causing
decoherence) to detect the particle at a slit. The probability wave collapses to a physical particle (the particle is detected and its position/trajectory is known at a given slit), then like any classical particle, it passes on through the slit, producing no diffraction pattern."
Here is where it gets touchy, and why I included the last formalism: I do not know what you precisely mean by "the probability wave collapses to a physical particle." What you observe due to the decoherence is something very different and wholly classical: when you measure
ÃË†(y)Ã‚Â², you ultimately measure:
ÃË†Ã‚Â² = 0.5  Ãâ€ Ã‚Â¹ Ã‚Â² + 0.5  Ãâ€ Ã‚Â² Ã‚Â².
...due to the decoherence. The difference between these two expressions is of course:
0.5 [ Ãâ€ Ã‚Â¹ (Ãâ€ Ã‚Â²)* + Ãâ€ Ã‚Â² (Ãâ€ Ã‚Â¹)* ]
Which is, if you like, < 1  2 > + < 2  1 >, and ultimately comes from the offdiagonal terms in the density matrix.
Of course, there is still the same old measurement problem: you have these wide distributions over y Ãâ€ Ã‚Â¹ and Ãâ€ Ã‚Â²  wide enough that they overlap for a significant portion  and somehow the particle "makes up its mind" which detector to enter. It's just that it now chooses the state by flipping a coin.
And, also: if you achieved this decoherence by entangling a qubit with the particle's slitchoice, to put them in the  11 > +  22 > state, then you may optionally measure this new qubit as 1 or 2. If you do, you will get a set of coincidence measurements (slit #, ydetector #).
The ones which went through slit 1 will have the distribution  Ãâ€ Ã‚Â¹ Ã‚Â², and the ones which went through slit 2 will have the distribution  Ãâ€ Ã‚Â²Ã‚Â². The measurement here is totally optional; the decoherence has
happened whether or not you make this measurement. It also doesn't matter what order you make these measurements in. You could leave these qubits lying around until *after* the yvalues are all aggregated and confirmed to look like a classical distribution.
"Bottom line: classical particles cannot create diffraction patterns but probability waves can (if they remain waves and are not collapsed to physical particles by a measurement/decoherence)"
It is absolutely crucial to understand that in quantum mechanics, there is no longer such a thing as a classical particle. Every particle has a de Broglie wavelength and is either a fermion or a boson. When the diffraction pattern appears or disappears, it is ultimately a quantum phenomenon amenable to the laws of quantum mechanics. Classical mechanics gets embedded in the quantum viewpoint by Ehrenfest's theorem. Even when you have  Ãâ€ Ã‚Â¹ Ã‚Â² +  Ãâ€ Ã‚Â² Ã‚Â², the same as you would classically expect, ultimately the prediction is probabilistic, and comes from quantum mechanics.
 Chris
[1] http://www.youtube.com/watch?v=KT7xJ0tjB4A

On thur, Aug 5, 2010 at 7:47 PM, Tom Campbell <twcjr@comcast.net> wrote:
OK, I got all that and I fully agree.
It is not so much that 1) through 4) are flat out wrong as it is that they don't tell the whole story, ignore some important details/possibilities/complications and use fuzzy poorly defined terms like "the probability wave collapses to a physical particle."
Given that my talk was at a nontechnical level to a general audience  some of that *might* be forgivable  From now on, I will try to make my presentations more accurate without sewing any more confusion than necessary among the audience.
As far as that information theory assumption goes  about the information being more fundamental to the nature of reality than the measurement of the information  I am now straight that such a position is NOT acceptable to current scientific belief. I say "Belief" because the experiment has never been done. The "reality is information" theorists are gaining respectability slowly  perhaps even enough someday to make it a thought worth considering  an experiment worth doing. Yes, upset the apple cart it would!  like QM upset Newtonian mechanics.
I don't mind being out on the fringe  that is what physicists are for. All paradigm breaking concepts must necessarily start out on the fringe. That being said, I certainly don't want to be spouting off nonsense and saying things that are wrong or misleading  that's counterproductive for everyone. I am serious about my science, and like most serious scientists, I hope one day the core of my work will be considered to be good science  I know, that might take a long, long time.
Chris, you have taught me a few things and set me straight in a way that I needed to be set straight. That has taken considerable time and energy on your part for which I am very grateful. If you read anything above that still sounds off track  please don't be shy about bringing it to my attention. I will try not to abuse your kindness in engaging me this way.
Tom
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